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Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3 CHAPTER Mass Relationships in Chemical Reactions INTRODUCTION IN THIS CHAPTER WE WILL CONSIDER THE MASSES OF ATOMS AND MOL- 3.1 ATOMIC MASS ECULES AND WHAT HAPPENS TO THEM WHEN CHEMICAL CHANGES OCCUR. OUR 3.2 MOLAR MASS OF AN ELEMENT AND AVOGADRO’S NUMBER GUIDE FOR THIS DISCUSSION WILL BE THE LAW OF CONSER- 3.3 MOLECULAR MASS VATION OF MASS. 3.4 THE MASS SPECTROMETER 3.5 PERCENT COMPOSITION OF COMPOUNDS 3.6 EXPERIMENTAL DETERMINATION OF EMPIRICAL FORMULAS 3.7 CHEMICAL REACTIONS AND CHEMICAL EQUATIONS 3.8 AMOUNTS OF REACTANTS AND PRODUCTS 3.9 LIMITING REAGENTS 3.10 REACTION YIELD 69 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 70 MASS RELATIONSHIPS IN CHEMICAL REACTIONS 3.1 Section 3.4 describes a method for determining atomic mass. One atomic mass unit is also called one dalton. ATOMIC MASS In this chapter we will use what we have learned about chemical structure and formulas in studying the mass relationships of atoms and molecules. These relationships in turn will help us to explain the composition of compounds and the ways in which composition changes. The mass of an atom depends on the number of electrons, protons, and neutrons it contains. Knowledge of an atom’s mass is important in laboratory work. But atoms are extremely small particles — even the smallest speck of dust that our unaided eyes can detect contains as many as 1 1016 atoms! Clearly we cannot weigh a single atom, but it is possible to determine the mass of one atom relative to another experimentally. The first step is to assign a value to the mass of one atom of a given element so that it can be used as a standard. By international agreement, atomic mass (sometimes called atomic weight) is the mass of the atom in atomic mass units (amu). One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom. Carbon-12 is the carbon isotope that has six protons and six neutrons. Setting the atomic mass of carbon-12 at 12 amu provides the standard for measuring the atomic mass of the other elements. For example, experiments have shown that, on average, a hydrogen atom is only 8.400 percent as massive as the carbon-12 atom. Thus, if the mass of one carbon12 atom is exactly 12 amu, the atomic mass of hydrogen must be 0.0084 12.00 amu or 1.008 amu. Similar calculations show that the atomic mass of oxygen is 16.00 amu and that of iron is 55.85 amu. Thus, although we do not know just how much an average iron atom’s mass is, we know that it is approximately fifty-six times as massive as a hydrogen atom. AVERAGE ATOMIC MASS When you look up the atomic mass of carbon in a table such as the one on the inside front cover of this book, you will find that its value is not 12.00 amu but 12.01 amu. The reason for the difference is that most naturally occurring elements (including carbon) have more than one isotope. This means that when we measure the atomic mass of an element, we must generally settle for the average mass of the naturally occurring mixture of isotopes. For example, the natural abundances of carbon-12 and carbon-13 are 98.90 percent and 1.10 percent, respectively. The atomic mass of carbon13 has been determined to be 13.00335 amu. Thus the average atomic mass of carbon can be calculated as follows: average atomic mass of natural carbon (0.9890)(12.00000 amu) (0.0110)(13.00335 amu) 12.0 amu A more accurate determination gives the atomic mass of carbon as 12.01 amu. Note that in calculations involving percentages, we need to convert percentages to fractions. For example, 98.90 percent becomes 98.90/100, or 0.9890. Because there are many more carbon-12 atoms than carbon-13 atoms in naturally occurring carbon, the average atomic mass is much closer to 12 amu than to 13 amu. It is important to understand that when we say that the atomic mass of carbon is 12.01 amu, we are referring to the average value. If carbon atoms could be examined individually, we would find either an atom of atomic mass 12.00000 amu or one of Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.2 MOLAR MASS OF AN ELEMENT AND AVOGADRO’S NUMBER 71 13.00335 amu, but never one of 12.01 amu. The following example shows how to calculate the average atomic mass of an element. EXAMPLE 3.1 Copper, a metal known since ancient times, is used in electrical cables and pennies, among other things. The atomic masses of its two stable isotopes, 63Cu (69.09%) 29 and 65Cu (30.91%), are 62.93 amu and 64.9278 amu, respectively. Calculate the av29 erage atomic mass of copper. The percentages in parentheses denote the relative abundances. The first step is to convert the percentages to fractions. Thus 69.09 percent becomes 0.6909 and 30.91 percent becomes 0.3091. Next we calculate the average atomic mass as follows: Answer Copper. (0.6909)(62.93 amu) (0.3091)(64.9278 amu) 63.55 amu PRACTICE EXERCISE Similar problems: 3.5, 3.6. 11 The atomic masses of the two stable isotopes of boron, 10B (19.78%) and 05B 05 (80.22%), are 10.0129 amu and 11.0093 amu, respectively. Calculate the average atomic mass of boron. The atomic masses of many elements have been accurately determined to five or six significant figures. However, for our purposes we will normally use atomic masses accurate only to four significant figures (see the table of atomic masses inside the front cover). 3.2 The adjective formed from the noun “mole” is “molar.” MOLAR MASS OF AN ELEMENT AND AVOGADRO’S NUMBER Atomic mass units provide a relative scale for the masses of the elements. But since atoms have such small masses, no usable scale can be devised to weigh them in calibrated units of atomic mass units. In any real situation, we deal with macroscopic samples containing enormous numbers of atoms. Therefore it is convenient to have a special unit to describe a very large number of atoms. The idea of a unit to denote a particular number of objects is not new. For example, the pair (2 items), the dozen (12 items), and the gross (144 items) are all familiar units. Chemists measure atoms and molecules in moles. In the SI system the mole (mol) is the amount of a substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 grams (or 0.012 kilogram) of the carbon-12 isotope. The actual number of atoms in 12 g of carbon-12 is determined experimentally. The currently accepted value is 1 mole 6.022045 1023 particles This number is called Avogadro’s number, in honor of the Italian scientist Amedeo Avogadro.† Generally, we round Avogadro’s number to 6.022 1023. Thus, just as one † Lorenzo Romano Amedeo Carlo Avogadro di Quaregua e di Cerreto (1776–1856). Italian mathematical physicist. He practiced law for many years before he became interested in science. His most famous work, now known as Avogadro’s law (see Chapter 5), was largely ignored during his lifetime, although it became the basis for determining atomic masses in the late nineteenth century. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 72 MASS RELATIONSHIPS IN CHEMICAL REACTIONS FIGURE 3.1 One mole each of several common elements: copper (as pennies), iron (as nails), carbon (black charcoal powder), sulfur (yellow powder), and mercury (shiny liquid metal). In calculations, the units of molar mass are g/mol or kg/mol. However, it is also acceptable to say that the molar mass of Na is 22.99 g rather than 22.99 g/mol. dozen oranges contains twelve oranges, 1 mole of hydrogen atoms contains 6.022 1023 H atoms. (The Chemistry in Three Dimensions essay on p. 106 describes a method for determining Avogadro’s number.) Figure 3.1 shows 1 mole each of several common elements. We have seen that 1 mole of carbon-12 atoms has a mass of exactly 12 grams and contains 6.022 1023 atoms. This mass of carbon-12 is its molar mass, defined as the mass (in grams or kilograms) of 1 mole of units (such as atoms or molecules) of a substance. Note that the molar mass of carbon-12 (in grams) is numerically equal to its atomic mass in amu. Likewise, the atomic mass of sodium (Na) is 22.99 amu and its molar mass is 22.99 grams; the atomic mass of phosphorus is 30.97 amu and its molar mass is 30.97 grams; and so on. If we know the atomic mass of an element, we also know its molar mass. Using atomic mass and molar mass, we can calculate the mass in grams of a single carbon-12 atom. From our discussion we know that 1 mole of carbon-12 atoms weighs exactly 12 grams. This allows us to write the equality 12.00 g carbon-12 1 mol carbon-12 atoms Therefore, we can write the unit factor as 12.00 g carbon-12 1 mol carbon-12 atoms 1 (Note that we use the unit “mol” to represent “mole” in calculations.) Similarly, since there are 6.022 1023 atoms in 1 mole of carbon-12 atoms, we have 1 mol carbon-12 atoms 6.022 1023 carbon-12 atoms and the unit factor is 1 mol carbon-12 atoms 6.022 1023 carbon-12 atoms 1 We can now calculate the mass (in grams) of 1 carbon-12 atom as follows: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.2 FIGURE 3.2 The relationships between mass of an element and number of moles of an element, and between moles of an element and number of atoms of an element. MOLAR MASS OF AN ELEMENT AND AVOGADRO’S NUMBER Molar mass (g/mol) Mass (g) of element Avogadro's number 6.022 × 1023 Moles of element Number of atoms of element 12.00 g carbon-12 1 mol carbons-12 atoms 1.993 10 23 g carbon-12 1 mol carbon-12 atoms 6.022 1023 carbon-12 atoms 1 carbon-12 atom 73 We can use this result to determine the relationship between atomic mass units and grams. Since the mass of every carbon-12 atom is exactly 12 amu, the number of grams equivalent to 1 amu is gram amu 1.993 10 23 g 1 carbon-12 atom 1 carbon-12 atom 12 amu 24 g/amu 1.661 10 Thus 1 amu 1.661 10 24 g and 1g 6.022 1023 amu This example shows that Avogadro’s number can be used to convert from the atomic mass units to mass in grams and vice versa. The notions of Avogadro’s number and molar mass enable us to carry out conversions between mass and moles of atoms and between the number of atoms and mass and to calculate the mass of a single atom. We will employ the following unit factors in the calculations: 1 mol X molar mass of X The atomic masses of the elements are given on the inside front cover of the book. 1 6.022 1 mol X 1023 X atoms 1 where X represents the symbol of an element. Figure 3.2 summarizes the relationships between the mass of an element and the number of moles of an element and between moles of an element and the number of atoms of an element. Using the proper unit factors we can convert one quantity to another, as Examples 3.2–3.4 show. EXAMPLE 3.2 Helium (He) is a valuable gas used in industry, low-temperature research, deep-sea diving tanks, and balloons. How many moles of He are in 6.46 g of He? First we find that the molar mass of He is 4.003 g. This can be expressed by the equality 1 mole of He 4.003 g of He. To convert the amount of He in grams to He in moles, we write Answer 6.46 g He Comment Similar Problem: 3.15. Forward 1.61 mol He Thus there are 1.61 moles of He atoms in 6.46 g of He. Helium balloons. Back 1 mol He 4.003 g He Main Menu Since 6.46 g is greater than the molar mass of He, the answer is rea- sonable. TOC Study Guide TOC Textbook Website MHHE Website 74 MASS RELATIONSHIPS IN CHEMICAL REACTIONS PRACTICE EXERCISE How many moles of magnesium (Mg) are there in 87.3 g of Mg? EXAMPLE 3.3 Zinc (Zn) is a silvery metal that is used in making brass (with copper) and in plating iron to prevent corrosion. How many grams of Zn are in 0.356 mole of Zn? Since the molar mass of Zn is 65.39 g, the mass of Zn in grams is given Answer by 0.356 mol Zn 65.39 g Zn 1 mol Zn 23.3g Zn Thus there are 23.3 g of Zn in 0.356 mole of Zn. Zinc. Comment Similar Problem: 3.16. The mass in 0.356 mole of Zn should be less than the molar mass of Zn. PRACTICE EXERCISE Calculate the number of grams of lead (Pb) in 12.4 moles of lead. EXAMPLE 3.4 Sulfur (S) is a nonmetallic element. Its presence in coal gives rise to the acid rain phenomenon. How many atoms are in 16.3 g of S? Solving this problem requires two steps. First, we need to find the number of moles of S in 16.3 g of S (as in Example 3.2). Next, we need to calculate the number of S atoms from the known number of moles of S. We can combine the two steps as follows: Answer Sulfur. 1 mol S 32.07 g S 16.3 g S Similar Problem: 3.20. 6.022 1023 S atoms 1 mol S 3.06 1023 S atoms PRACTICE EXERCISE Calculate the number of atoms in 0.551 g of potassium (K). 3.3 MOLECULAR MASS If we know the atomic masses of the component atoms, we can calculate the mass of a molecule. The molecular mass (sometimes called molecular weight) is the sum of the atomic masses (in amu) in the molecule. For example, the molecular mass of H2O is 2(atomic mass of H) or 2(1.008 amu) atomic mass of O 16.00 amu 18.02 amu In general, we need to multiply the atomic mass of each element by the number of atoms of that element present in the molecule and sum over all the elements. Example 3.5 illustrates this approach. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.3 MOLECULAR MASS 75 EXAMPLE 3.5 Calculate the molecular masses of the following compounds: (a) sulfur dioxide (SO2), which is mainly responsible for acid rain; (b) ascorbic acid, or vitamin C (C6H8O6). (a) From the atomic masses of S and O we get Answer molecular mass of SO2 32.07 amu 2(16.00 amu) 64.07 amu (b) From the atomic masses of C, H, and O we get molecular mass of C6H8O6 6(12.01 amu) 8(1.008 amu) 6(16.00 amu) 176.12 amu Similar Problems: 3.23, 3.24. PRACTICE EXERCISE What is the molecular mass of methanol (CH4O)? From the molecular mass we can determine the molar mass of a molecule or compound. The molar mass of a compound (in grams) is numerically equal to its molecular mass (in amu). For example, the molecular mass of water is 18.02 amu, so its molar mass is 18.02 g. Note that 1 mole of water weighs 18.02 g and contains 6.022 1023 H2O molecules, just as 1 mole of elemental carbon contains 6.022 1023 carbon atoms. As the following two examples show, a knowledge of the molar mass enables us to calculate the numbers of moles and individual atoms in a given quantity of a compound. EXAMPLE 3.6 Methane (CH4) is the principal component of natural gas. How many moles of CH4 are present in 6.07 g of CH4? Answer First we calculate the molar mass of CH4: molar mass of CH4 Methane gas burning on a cooking range. 12.01 g 4(1.008 g) 16.04 g We then follow the procedure in Example 3.2: 6.07 g CH4 Similar Problem: 3.26. 1 mol CH4 16.04 g CH4 0.378 mol CH4 PRACTICE EXERCISE Calculate the number of moles of chloroform (CHCl3) in 198 g of chloroform. EXAMPLE 3.7 How many hydrogen atoms are present in 25.6 g of urea [(NH2)2CO], which is used as a fertilizer, in animal feed, and in the manufacture of polymers? The molar mass of urea is 60.06 g. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 76 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Answer First we follow the procedure in Example 3.4 to calculate the number of molecules present in 25.6 g of urea. Next we note that there are 4 hydrogen atoms in every urea molecule. Combining both steps we calculate the total number of hydrogen atoms present 25.6 g (NH2)2CO 1 mol (NH2)2CO 60.06 g (NH2)2CO 6.022 1023 molecules (NH2)2CO 1 mol (NH2)2CO 4 H atoms 1 molecules (NH2)2CO Urea. Similar Problems: 3.27, 3.28. 1.03 1024 H atoms We could calculate the number of nitrogen, carbon, and oxygen atoms by the same procedure. However, there is a shortcut. Note that the ratio of nitrogen to hydrogen atoms in urea is 2/4, or 1/2, and that of oxygen (and carbon) to hydrogen atoms is 1/4. Therefore the number of nitrogen atoms in 25.6 g of urea is (1/2)(1.03 1024), or 5.15 1023 atoms. The number of oxygen atoms (and carbon atoms) is (1/4)(1.03 1024), or 2.58 1023 atoms. PRACTICE EXERCISE How many H atoms are in 72.5 g of isopropanol (rubbing alcohol), C3H8O? 3.4 THE MASS SPECTROMETER The most direct and most accurate method for determining atomic and molecular masses is mass spectrometry, which is depicted in Figure 3.3. In a mass spectrometer, a gaseous sample is bombarded by a stream of high-energy electrons. Collisions between the electrons and the gaseous atoms (or molecules) produce positive ions by dislodging an electron from each atom or molecule. These positive ions (of mass m and charge e) are accelerated by two oppositely charged plates as they pass through the plates. The emerging ions are deflected into a circular path by a magnet. The radius of the path depends on the charge-to-mass ratio (that is, e/m). Ions of smaller e/m ratio trace a wider curve than those having a larger e/m ratio, so that ions with equal charges but different masses are separated from one another. The mass of each ion (and hence its parent atom or molecule) is determined from the magnitude of its deflection. Eventually the ions arrive at the detector, which registers a current for each type of ion. The amount of current generated is directly proportional to the number of ions, so it enables us to determine the relative abundance of isotopes. FIGURE 3.3 Schematic diagram of one type of mass spectrometer. Detecting screen Accelerating plates Electron beam Sample gas Filament Back Forward Main Menu TOC Ion beam Study Guide TOC Magnet Textbook Website MHHE Website 3.5 FIGURE 3.4 The mass spectrum of the three isotopes of neon. PERCENT COMPOSITION OF COMPOUNDS 77 Intensity of peaks 20 10 Ne(90.92%) 21 10 Ne(0.26%) 19 20 22 10 Ne(8.82%) 21 22 Atomic mass (amu) 23 The first mass spectrometer, developed in the 1920s by the English physicist F. W. Aston,† was crude by today’s standards. Nevertheless, it provided indisputable evidence of the existence of isotopes — neon-20 (atomic mass 19.9924 amu and natural abundance 90.92 percent) and neon-22 (atomic mass 21.9914 amu and natural abundance 8.82 percent). When more sophisticated and sensitive mass spectrometers became available, scientists were surprised to discover that neon has a third stable isotope with an atomic mass of 20.9940 amu and natural abundance 0.257 percent (Figure 3.4). This example illustrates how very important experimental accuracy is to a quantitative science like chemistry. Early experiments failed to detect neon-21 because its natural abundance is just 0.257 percent. In other words, only 26 in 10,000 Ne atoms are neon-21. The masses of molecules can be determined in a similar manner by the mass spectrometer. The Chemistry in Action essay on page 78 describes an interesting application of mass spectrometer. 3.5 PERCENT COMPOSITION OF COMPOUNDS As we have seen, the formula of a compound tells us the numbers of atoms of each element in a unit of the compound. However, suppose we needed to verify the purity of a compound for use in a laboratory experiment. From the formula we could calculate what percent of the total mass of the compound is contributed by each element. Then, by comparing the result to the percent composition obtained experimentally for our sample, we could determine the purity of the sample. The percent composition by mass is the percent by mass of each element in a compound. Percent composition is obtained by dividing the mass of each element in 1 mole of the compound by the molar mass of the compound and multiplying by 100 percent. Mathematically, the percent composition of an element in a compound is expressed as † Francis William Aston (1877 – 1945). English chemist and physicist. He was awarded the Nobel Prize in Chemistry in 1922 for developing the mass spectrometer. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 78 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Gold Fingerprinting by Mass Spectrometry Every year millions of dollars’ worth of gold is stolen. In most cases the gold is melted down and shipped abroad. This way the gold retains its value while losing all means of identification. However, a technique developed by scientists in Australia will soon enable law enforcement agents to identify the source of gold samples, even if it has been melted and reset, and apprehend the thieves. Gold is a highly unreactive metal that exists in nature in the uncombined form. Indeed, its low reactivity is one of the properties that makes gold suitable for use in jewelry. During the mineralization of gold, that is, the formation of gold nuggets from microscopic gold particles, various elements such as cadmium, lead, tellurium, and zinc, are incorporated into the nuggets. The amounts and types of the impurity or trace elements in gold vary according to the location where it was mined. To analyze a gold sample, scientists first heat a tiny spot (about 0.01 cm in diameter and in depth) of the sample with a high-power laser. The vaporized gold and its trace elements are swept by a stream of argon gas into a mass spectrometer. Comparison of the mass spectrum with a library of mass spectra of gold from known origins will identify the source of the gold, much as fingerprints are used to identify individuals. This technique can be used on large objects like bullion and nuggets as well as on small articles of jewelry. It can also be used to detect art forgery because the mass spectra of antique gold objects differ from the spectra of modern gold. (b) (a) Gold bar with marking for identification. (b) Gold melts at 1065°C. Liquid gold can be readily molded into other shapes. (a) percent composition of an element n molar mass of element molar mass of compound 100% (3.1) where n is the number of moles of the element in one mole of the compound. For example, in 1 mole of hydrogen peroxide (H2O2) there are 2 moles of H atoms and 2 moles of O atoms. The molar masses of H2O2, H, and O are 34.02 g, 1.008 g, and Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.5 PERCENT COMPOSITION OF COMPOUNDS 1200 1000 800 600 400 Cd Pb 200 0 40 60 80 100 120 140 160 Atomic mass (amu) 180 200 220 1400 1200 1000 Relative intensity Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Mass spectra of gold from two different sources. The amount of trace elements Cd and Pb differ greatly in these samples. The same holds for many other trace elements. 1400 Relative intensity 79 800 Pb 600 400 Cd 200 0 40 60 80 100 120 140 160 Atomic mass (amu) 180 200 220 16.00 g, respectively. Therefore, the percent composition of H2O2 is calculated as follows: %H %O Back Forward Main Menu TOC 2 1.008 g 34.02 g 100% 5.926% 2 16.00 g 34.02 g 100% 94.06% Study Guide TOC Textbook Website MHHE Website 80 MASS RELATIONSHIPS IN CHEMICAL REACTIONS The sum of the percentages is 5.926 percent 94.06 percent 99.99 percent. The small discrepancy from 100 percent is due to the way we rounded off the molar masses of the elements. If we had used the empirical formula HO for the calculation, we would have written %H 1.008 g 17.01 g 100% 5.926% %O 16.00 g 17.01 g 100% 94.06% Since both the molecular formula and the empirical formula tell us the composition of the compound, it is not surprising that they give us the same percent composition by mass. EXAMPLE 3.8 Phosphoric acid (H3PO4) is a colorless, syrupy liquid used in detergents, fertilizers, and toothpastes, and in carbonated beverages for a “tangy” flavor. Calculate the percent composition by mass of H, P, and O in this compound. The molar mass of H3PO4 is 97.99 g/mol. Therefore, the percent by mass of each of the elements in H3PO4 is Answer %H %P 30.97 g 97.99 g %O Similar Problem: 3.40. 3(1.008 g) 97.99 g 4(16.00 g) 97.99 g 100% 100% 3.086% 31.61% 100% 65.31% The sum of the percentages is (3.086% 31.61% 65.31%) crepancy from 100 percent is due to the way we rounded off. 100.01%. The dis- PRACTICE EXERCISE Calculate the percent composition by mass of each of the elements in sulfuric acid (H2SO4). The procedure used in the example can be reversed if necessary. Given the percent composition by mass of a compound, we can determine the empirical formula of the compound. Since we are dealing with percentages and the sum of all the percentages is 100 percent, it is convenient to assume that we started with 100 g of a compound, as Example 3.9 shows. EXAMPLE 3.9 Ascorbic acid (vitamin C) cures scurvy and may help prevent the common cold. It is composed of 40.92 percent carbon (C), 4.58 percent hydrogen (H), and 54.50 percent oxygen (O) by mass. Determine its empirical formula. We start by assuming there are 100 g of ascorbic acid. Therefore, in this sample there will be 40.92 g of C, 4.58 g of H, and 54.50 g of O. Next, we need to calculate the number of moles of each element in the compound. Let nC, nH, and nO be the number of moles of elements present. Using the molar masses of these elements, we write Answer Vitamin C tablets. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.5 PERCENT COMPOSITION OF COMPOUNDS nC nH 4.58 g H nO 1 mol C 12.01 g C 40.92 g C 54.50 g O 1 mol H 1.008 g H 1 mol O 16.00 g O 81 3.407 mol C 4.54 mol H 3.406 mol O Thus, we arrive at the formula C3.407H4.54O3.406, which gives the identity and the ratios of atoms present. However, since chemical formulas are written with whole numbers, we cannot have 3.407 C atoms, 4.54 H atoms, and 3.406 O atoms. We can make some of the subscripts whole numbers by dividing all the subscripts by the smallest subscript (3.406): C: 3.407 3.406 1 H: 4.54 3.406 1.33 O: 3.406 3.406 1 This gives us CH1.33O as the formula for ascorbic acid. Next, we need to convert 1.33, the subscript for H, into an integer. This can be done by a trial-and-error procedure: 1.33 1 1.33 1.33 2 2.66 1.33 3 3.99 4 Because 1.33 3 gives us an integer (4), we can multiply all the subscripts by 3 and obtain C3H4O3 as the empirical formula for ascorbic acid. Similar Problem: 3.44. Comment The molecular formula of ascorbic acid is C6H8O6. PRACTICE EXERCISE Determine the empirical formula of a compound having the following percent composition by mass: K: 24.75%; Mn: 34.77%; O: 40.51%. Chemists often want to know the actual mass of an element in a certain mass of a compound. Since the percent composition by mass of the element in the substance can be readily calculated, such a problem can be solved in a rather direct way. ” ion ers tV ex e-T in le ab ail Av ot “N EXAMPLE 3.10 Chalcopyrite (CuFeS2) is a principal ore of copper. Calculate the number of kilograms of Cu in 3.71 103 kg of chalcopyrite. Answer The molar masses of Cu and CuFeS2 are 63.55 g and 183.5 g, respectively, so the percent composition by mass of Cu is Chalcopyrite. %Cu 63.55 g 183.5 g 100% 34.63% To calculate the mass of Cu in a 3.71 103-kg sample of CuFeS2, we need to convert the percentage to a fraction (that is, convert 34.63% to 34.63/100, or 0.3463) and write mass of Cu in CuFeS2 0.3463 3.71 103 kg 1.28 103 kg This calculation can be simplified by combining the above two steps as follows: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 82 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Similar Problem: 3.45. 3.71 103 kg CuFeS2 1.28 mass of Cu in CuFeS2 63.55 g Cu 183.5 g CuFeS2 103 kg Cu PRACTICE EXERCISE Calculate the number of grams of Al in 371 g of Al2O3. 3.6 EXPERIMENTAL DETERMINATION OF EMPIRICAL FORMULAS The fact that we can determine the empirical formula of a compound if we know the percent composition allows us to identify compounds experimentally. The procedure is as follows. First, chemical analysis tells us the number of grams of each element present in a given amount of a compound. Then we convert the quantities in grams to number of moles of each element. Finally, using the method given in Example 3.9, we find the empirical formula of the compound. As a specific example let us consider the compound ethanol. When ethanol is burned in an apparatus such as that shown in Figure 3.5, carbon dioxide (CO2) and water (H2O) are given off. Since neither carbon nor hydrogen was in the inlet gas, we can conclude that both carbon (C) and hydrogen (H) were present in ethanol and that oxygen (O) may also be present. (Molecular oxygen was added in the combustion process, but some of the oxygen may also have come from the original ethanol sample.) The masses of CO2 and of H2O produced can be determined by measuring the increase in mass of the CO2 and H2O absorbers, respectively. Suppose that in one experiment the combustion of 11.5 g of ethanol produced 22.0 g of CO2 and 13.5 g of H2O. We can calculate the mass of carbon and hydrogen in the original 11.5-g sample of ethanol as follows: mass of C 22.0 g CO2 1 mol CO2 44.01 g CO2 1 mol C 1 mol CO2 12.01 g C 1 mol C 1 mol H2O 18.02 g H2O 2 mol H 1 mol H2O 1.008 g H 1 mol H 6.00 g C mass of H 13.5 g H2O 1.51 g H Ethanol O2 Heat ;;;;; ;;;;; ;;;;; Unused O2 H2O CO2 absorber absorber FIGURE 3.5 Apparatus for determining the empirical formula of ethanol. The absorbers are substances that can retain water and carbon dioxide, respectively. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.6 EXPERIMENTAL DETERMINATION OF EMPIRICAL FORMULAS 83 Thus, 11.5 g of ethanol contains 6.00 g of carbon and 1.51 g of hydrogen. The remainder must be oxygen, whose mass is mass of O mass of sample 11.5 g (mass of C (6.00 g mass of H) 1.51 g) 4.0 g The number of moles of each element present in 11.5 g of ethanol is moles of C 6.00 g C 1 mol C 12.01 g C 0.500 mol C moles of H 1.51 g H 1 mol H 1.008 g H 1.50 mol H moles of O 4.0 g O 1 mol O 16.00 g O 0.25 mol O The formula of ethanol is therefore C0.50H1.5O0.25 (we round off the number of moles to two significant figures). Since the number of atoms must be an integer, we divide the subscripts by 0.25, the smallest subscript, and obtain for the empirical formula C2H6O. Now we can better understand the word “empirical,” which literally means “based only on observation and measurement.” The empirical formula of ethanol is determined from analysis of the compound in terms of its component elements. No knowledge of how the atoms are linked together in the compound is required. DETERMINATION OF MOLECULAR FORMULAS The formula calculated from percent composition by mass is always the empirical formula because the coefficients in the formula are always reduced to the smallest whole numbers. To calculate the actual, molecular formula we must know the approximate molar mass of the compound in addition to its empirical formula. Knowing that the molar mass of a compound must be an integral multiple of the molar mass of its empirical formula, we can use the molar mass to find the molecular formula, as Example 3.11 demonstrates. EXAMPLE 3.11 A sample of a compound of nitrogen (N) and oxygen (O) contains 1.52 g of N and 3.47 g of O. The molar mass of this compound is known to be between 90 g and 95 g. Determine the molecular formula and molar mass of the compound. We first determine the empirical formula of the compound as outlined in Example 3.9. Let nN and nO be the number of moles of nitrogen and oxygen. Then Answer nN Forward Main Menu TOC 1 mol N 14.01 g N 0.108 mol N nO Back 1.52 g N 3.47 g O 1 mol O 16.00 g O 0.217 mol O Study Guide TOC Textbook Website MHHE Website 84 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Thus the formula of the compound is N0.108O0.217. We divide the subscripts by the smaller subscript, 0.108. After rounding off, we obtain NO2 as the empirical formula. The molecular formula will be the same as the empirical formula or some integral multiple of it (for example, two, three, four, or more times the empirical formula). The molar mass of the empirical formula NO2 is empirical molar mass 14.01 g 2(16.00 g) 46.02 g Next we determine the number of (NO2) units present in the molecular formula. This number is found by taking the ratio molar mass empirical molar mass Similar Problem: 3.52 95 g 46.02 g 2.1 2 Therefore the molar mass of the compound is twice the molar mass of the empirical formula. Consequently, there are two NO2 units in each molecule of the compound, and the molecular formula is (NO2)2, or N2O4. The molar mass of the compound is 2(46.02 g), or 92.04 g, which is between 90 g and 95 g. PRACTICE EXERCISE A sample of a compound of boron (B) and hydrogen (H) contains 6.444 g of B and 1.803 g of H. The molar mass of the compound is about 30 g. What is its molecular formula? 3.7 CHEMICAL REACTIONS AND CHEMICAL EQUATIONS Having discussed the masses of atoms and molecules, we turn next to what happens to atoms and molecules in a chemical reaction, a process in which a substance (or substances) is changed into one or more new substances. In order to communicate with one another about chemical reactions, chemists have devised a standard way to represent them using chemical equations. A chemical equation uses chemical symbols to show what happens during a chemical reaction. In this section we will learn how to write chemical equations and balance them. WRITING CHEMICAL EQUATIONS Consider what happens when hydrogen gas (H2) burns in air (which contains oxygen, O2) to form water (H2O). This reaction can be represented by the chemical equation H2 O2 88n H2O (3.2) where the “plus” sign means “reacts with” and the arrow means “to yield.” Thus, this symbolic expression can be read: “Molecular hydrogen reacts with molecular oxygen to yield water.” The reaction is assumed to proceed from left to right as the arrow indicates. Equation (3.2) is not complete, however, because there are twice as many oxygen atoms on the left side of the arrow (two) as on the right side (one). To conform with the law of conservation of mass, there must be the same number of each type of atom on both sides of the arrow; that is, we must have as many atoms after the reaction ends as we did before it started. We can balance Equation (3.2) by placing the appropriate coefficient (2 in this case) in front of H2 and H2O: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.7 FIGURE 3.6 Three ways of representing the combustion of hydrogen. In accordance with the law of conservation of mass, the number of each type of atom must be the same on both sides of the equation. CHEMICAL REACTIONS AND CHEMICAL EQUATIONS + Two hydrogen molecules + One oxygen molecule Two water molecules 2H2 When the coefficient is 1, as in the case of O2, it is not shown. 85 + O2 2H2O 2H2 O2 88n 2H2O This balanced chemical equation shows that “two hydrogen molecules can combine or react with one oxygen molecule to form two water molecules” (Figure 3.6). Since the ratio of the number of molecules is equal to the ratio of the number of moles, the equation can also be read as “2 moles of hydrogen molecules react with 1 mole of oxygen molecules to produce 2 moles of water molecules.” We know the mass of a mole of each of these substances, so we can also interpret the equation as “4.04 g of H2 react with 32.00 g of O2 to give 36.04 g of H2O.” These three ways of reading the equation are summarized in Table 3.1. We refer to H2 and O2 in Equation (3.1) as reactants, which are the starting materials in a chemical reaction. Water is the product, which is the substance formed as a result of a chemical reaction. A chemical equation, then, is just the chemist’s shorthand description of a reaction. In a chemical equation the reactants are conventionally written on the left and the products on the right of the arrow: reactants 88n products To provide additional information, chemists often indicate the physical states of the reactants and products by using the letters g, l, and s to denote gas, liquid, and solid, respectively. For example, 2CO(g) The procedure for balancing chemical equations is shown on p. 86. O2(g) 88n 2CO2(g) 2HgO(s) 88n 2Hg(l) O2(g) To represent what happens when sodium chloride (NaCl) is added to water, we write HO 2 NaCl(s) 88n NaCl(aq) where aq denotes the aqueous (that is, water) environment. Writing H2O above the arrow symbolizes the physical process of dissolving a substance in water, although it is sometimes left out for simplicity. Knowing the states of the reactants and products is especially useful in the laboratory. For example, when potassium bromide (KBr) and silver nitrate (AgNO3) react TABLE 3.1 Interpretation of a Chemical Equation 36.04 g reactants Back Forward Main Menu TOC Study Guide TOC 2H2 O2 88n 2H2O Two molecules one molecule 88n two molecules 2 moles 1 mole 88n 2 moles 2(2.02 g) 4.04 g 32.00 g 88n 2(18.02 g) 36.04 g 36.04 g product Textbook Website MHHE Website 86 MASS RELATIONSHIPS IN CHEMICAL REACTIONS in an aqueous environment, a solid, silver bromide (AgBr), is formed. This reaction can be represented by the equation: KBr(aq) AgNO3(aq) 88n KNO3(aq) AgBr(s) If the physical states of reactants and products are not given, an uninformed person might try to bring about the reaction by mixing solid KBr with solid AgNO3. These solids would react very slowly or not at all. Imagining the process on the microscopic level, we can understand that for a product like silver bromide to form, the Ag and Br ions would have to come in contact with each other. However, these ions are locked in place in their solid compounds and have little mobility. (Here is an example of how we explain a phenomenon by thinking about what happens at the molecular level, as discussed in Section 1.2.) BALANCING CHEMICAL EQUATIONS Suppose we want to write an equation to describe a chemical reaction that we have just carried out in the laboratory. How should we go about doing it? Because we know the identities of the reactants, we can write their chemical formulas. The identities of products are more difficult to establish. For simple reactions it is often possible to guess the product(s). For more complicated reactions involving three or more products, chemists may need to perform further tests to establish the presence of specific compounds. Once we have identified all the reactants and products and have written the correct formulas for them, we assemble them in the conventional sequence — reactants on the left separated by an arrow from products on the right. The equation written at this point is likely to be unbalanced; that is, the number of each type of atom on one side of the arrow differs from the number on the other side. In general, we can balance a chemical equation by the following steps: Identify all reactants and products and write their correct formulas on the left side and right side of the equation, respectively. • Begin balancing the equation by trying different coefficients to make the number of atoms of each element the same on both sides of the equation. We can change the coefficients (the numbers preceding the formulas) but not the subscripts (the numbers within formulas). Changing the subscripts would change the identity of the substance. For example, 2NO2 means “two molecules of nitrogen dioxide,” but if we double the subscripts, we have N2O4, which is the formula of dinitrogen tetroxide, a completely different compound. • First, look for elements that appear only once on each side of the equation with the same number of atoms on each side: the formulas containing these elements must have the same coefficient. Therefore, there is no need to adjust the coefficients of these elements at this point. Next, look for elements that appear only once on each side of the equation but in unequal numbers of atoms. Balance these elements. Finally, balance elements that appear in two or more formulas on the same side of the equation. • Check your balanced equation to be sure that you have the same total number of each type of atoms on both sides of the equation arrow. • Let’s consider a specific example. In the laboratory, small amounts of oxygen gas can be prepared by heating potassium chlorate (KClO3). The products are oxygen gas (O2) and potassium chloride (KCl). From this information, we write Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.7 CHEMICAL REACTIONS AND CHEMICAL EQUATIONS KClO3 88n KCl 87 O2 (For simplicity, we omit the physical states of reactants and products.) All three elements (K, Cl, and O) appear only once on each side of the equation, but only for K and Cl do we have equal numbers of atoms on both sides. Thus KClO3 and KCl must have the same coefficient. The next step is to make the number of O atoms the same on both sides of the equation. Since there are three O atoms on the left and two O atoms on the right of the equation, we can balance the O atoms by placing a 2 in front of KClO3 and a 3 in front of O2. 2KClO3 88n KCl Heating potassium chlorate produces oxygen, which supports the combustion of wood splint. 3O2 Finally, we balance the K and Cl atoms by placing a 2 in front of KCl: 2KClO3 88n 2KCl 3O2 (3.3) As a final check, we can draw up a balance sheet for the reactants and products where the number in parentheses indicates the number of atoms of each element: REACTANTS PRODUCTS K (2) Cl (2) O (6) K (2) Cl (2) O (6) Note that this equation could also be balanced with coefficients which are multiples of 2 (for KClO3), 2 (for KCl), and 3 (for O2); for example, 4KClO3 88n 4KCl 6O2 However, it is common practice to use the simplest possible set of whole-number coefficients to balance the equation. Equation (3.3) conforms to this convention. Now let us consider the combustion (that is, burning) of the natural gas component ethane (C2H6) in oxygen or air, which yields carbon dioxide (CO2) and water. The unbalanced equation is C2H6 O2 88n CO2 H2O We see that the number of atoms is not the same on both sides of the equation for any of the elements (C, H, and O). In addition, C and H appear only once on each side of the equation; O appears in two compounds on the right side (CO2 and H2O). To balance the C atoms, we place a 2 in front of CO2: C2H6 O2 88n 2CO2 H2O To balance the H atoms, we place a 3 in front of H2O: C2H6 O2 88n 2CO2 3H2O At this stage, the C and H atoms are balanced, but the O atoms are not because there are seven O atoms on the right-hand side and only two O atoms on the left-hand side of the equation. This inequality of O atoms can be eliminated by writing 7 in front of 2 the O2 on the left-hand side: C2H6 7 2 O2 88n 2CO2 3H2O 7 2 The “logic” for using as a coefficient is that there were seven oxygen atoms on the right-hand side of the equation, but only a pair of oxygen atoms (O2) on the left. To balance them we ask how many pairs of oxygen atoms are needed to equal seven oxy- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 88 MASS RELATIONSHIPS IN CHEMICAL REACTIONS gen atoms. Just as 3.5 pairs of shoes equal seven shoes, 7 O2 molecules equal seven O 2 atoms. As the following tally shows, the equation is now balanced: REACTANTS PRODUCTS C (2) H (6) O (7) C (2) H (6) O (7) However, we normally prefer to express the coefficients as whole numbers rather than as fractions. Therefore, we multiply the entire equation by 2 to convert 7 to 7: 2 2C2H6 7O2 88n 4CO2 6H2O The final tally is REACTANTS PRODUCTS C (4) H (12) O (14) C (4) H (12) O (14) Note that the coefficients used in balancing the last equation are the smallest possible set of whole numbers. In Example 3.12 we will continue to practice our equation-balancing skills. EXAMPLE 3.12 When aluminum metal is exposed to air, a protective layer of aluminum oxide (Al2O3) forms on its surface. This layer prevents further reaction between aluminum and oxygen, and it is the reason that aluminum beverage cans do not corrode. (In the case of iron, the rust, or iron(III) oxide, that forms is too porous to protect the iron metal underneath, so rusting continues.) Write a balanced equation for the formation of Al2O3. Answer This beverage can is made of aluminum, which resists corrosion. The unbalanced equation is Al O2 88n Al2O3 We see that both Al and O appear only once on each side of the equation but in unequal numbers. To balance the Al atoms, we place a 2 in front of Al: 2Al O2 88n Al2O3 There are two O atoms on the left-hand side and three O atoms on the right-hand side of the equation. This inequality of O atoms can be eliminated by writing 3 in 2 front of the O2 on the left-hand side of the equation. 2Al 3 2 O2 88n Al2O3 As in the case of ethane described earlier, we multiply the entire equation by 2 to convert 3 to 3 2 4Al 3O2 88n 2Al2O3 The final tally is Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.8 AMOUNTS OF REACTANTS AND PRODUCTS REACTANTS PRODUCTS Al (4) O (6) Similar Problems: 3.59, 3.60. 89 Al (4) O (6) PRACTICE EXERCISE Balance the equation representing the reaction between iron(III) oxide, Fe2O3, and carbon monoxide (CO) to yield iron (Fe) and carbon dioxide (CO2). 3.8 AMOUNTS OF REACTANTS AND PRODUCTS A basic question raised in the chemical laboratory is, “How much product will be formed from specific amounts of starting materials (reactants)?” Or in some cases we might ask the reverse question: “How much starting material must be used to obtain a specific amount of product?” To interpret a reaction quantitatively, we need to apply our knowledge of molar masses and the mole concept. Stoichiometry is the quantitative study of reactants and products in a chemical reaction. Whether the units given for reactants (or products) are moles, grams, liters (for gases), or some other units, we use moles to calculate the amount of product formed in a reaction. This approach is called the mole method, which means simply that the stoichiometric coefficients in a chemical equation can be interpreted as the number of moles of each substance. For example, the combustion of carbon monoxide in air produces carbon dioxide: 2CO(g) O2(g) 88n 2CO2(g) For stoichiometric calculations we would read this equation as “2 moles of carbon monoxide gas combine with 1 mole of oxygen gas to form 2 moles of carbon dioxide gas.” The mole method consists of the following steps: 1. 2. 3. 4. 5. Write correct formulas for all reactants and products, and balance the resulting equation. Convert the quantities of some or all given or known substances (usually reactants) into moles. Use the coefficients in the balanced equation to calculate the number of moles of the sought or unknown quantities (usually products) in the problem. Using the calculated numbers of moles and the molar masses, convert the unknown quantities to whatever units are required (typically grams). Check that your answer is reasonable in physical terms. Step 1 is a prerequisite to any stoichiometric calculation. We must know the identities of the reactants and products, and their mass relationships must not violate the law of conservation of mass (that is, we must have a balanced equation). Step 2 is the critical process of converting grams (or other units) of substances to number of moles. This conversion allows us to analyze the actual reaction in terms of moles only. To complete step 3, we need the balanced equation furnished by step 1. The key point here is that the coefficients in a balanced equation tell us the ratio in which moles of one substance react with or form moles of another substance. Step 4 is similar to Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 90 MASS RELATIONSHIPS IN CHEMICAL REACTIONS FIGURE 3.7 Three types of stoichiometric calculations based on the mole method. Moles of reactant Moles of product Mass of reactant Moles of reactant Moles of product Mass of reactant Moles of reactant Moles of product Mass of product step 2, except that it deals with the quantities sought in the problem. Step 5 is often underestimated but is very important: Chemistry is an experimental science, and your answer must make sense in terms of real species in the real world. If you have set up the problem incorrectly or made a computational error, it will often become obvious when your answer turns out to be much too large or much too small for the amounts of materials you started with. Figure 3.7 shows three common types of stoichiometric calculations. In stoichiometry we use the symbol , which means “stoichiometrically equivalent to” or simply “equivalent to.” In the balanced equation for the formation of carbon dioxide, 2 moles of CO react with 1 mole of O2, so 2 moles of CO are equivalent to 1 mole of O2: 2 mol CO 1 mol O2 In terms of the factor-label method, we can write the unit factor as 2 mol CO 1 mol O2 1 or 1 mol O2 2 mol CO 1 Similarly, since 2 moles of CO (or 1 mole of O2) produce 2 mols of CO2, we can say that 2 moles of CO (or 1 mole of O2) are equivalent to 2 moles of CO2: 2 mol CO 2 mol CO2 1 mol O2 2 mol CO2 The following examples illustrate the use of the five-step method in solving some typical stoichiometry problems. EXAMPLE 3.13 All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide. A typical reaction is that between lithium and water: 2Li(s) Lithium reacting with water to produce hydrogen gas. 2H2O(l) 88n 2LiOH(aq) H2(g) (a) How many moles of H2 will be formed by the complete reaction of 6.23 moles of Li with water? (b) How many grams of H2 will be formed by the complete reaction of 80.57 g of Li with water? Answer (a) Step 1: The balanced equation is given in the problem. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.8 AMOUNTS OF REACTANTS AND PRODUCTS 91 Step 2: No conversion is needed because the amount of the starting material, Li, is given in moles. Step 3: Since 2 moles of Li produce 1 mole of H2, or 2 mol Li 1 mol H2, we cal- culate moles of H2 produced as follows: moles of H2 produced 6.23 mol Li 1 mol H2 2 mol Li 3.12 mol H2 Step 4. This step is not required. Step 5: We began with 6.23 moles of Li and produced 3.12 moles of H2. Since 2 moles of Li produce 1 mole of H2, 3.12 moles is a reasonable quantity. Answer (b) Step 1: The reaction is the same as in (a). Step 2: The number of moles of Li is given by moles of Li 80.57 g Li 1 mol Li 6.941 g Li 11.61 mol Li Step 3: Since 2 moles of Li produce 1 mole of H2, or 2 mol Li 1 mol H2, we cal- culate the number of moles of H2 as follows: moles of H2 produced 11.61 mol Li 1 mol H2 2 mol Li 5.805 mol H2 Step 4: From the molar mass of H2 (2.016 g), we calculate the mass of H2 produced: mass of H2 produced 5.805 mol H2 2.016 g H2 1 mol H2 11.70 g H2 Step 5: Since the molar mass of H2 is smaller than that of Li and two moles of Li Similar Problems: 3.63, 3.64. are needed to produce one mole of H2, we expect the answer to be smaller than 80.57 g. PRACTICE EXERCISE The reaction between nitric oxide (NO) and oxygen to form nitrogen dioxide (NO2) is a key step in photochemical smog formation: 2NO(g) O2(g) 88n 2NO2(g) (a) How many moles of NO2 are formed by the complete reaction of 0.254 mole of O2? (b) How many grams of NO2 are formed by the complete reaction of 1.44 g of NO? After some practice, you may find it convenient to combine steps 2, 3, and 4 in a single equation, as the following example shows. EXAMPLE 3.14 The food we eat is degraded, or broken down, in our bodies to provide energy for growth and function. A general overall equation for this very complex process represents the degradation of glucose (C6H12O6) to carbon dioxide (CO2) and water (H2O): Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 92 MASS RELATIONSHIPS IN CHEMICAL REACTIONS C6H12O6 6O2 88n 6CO2 6H2O If 856 g of C6H12O6 is consumed by a person over a certain period, what is the mass of CO2 produced? Answer Step 1: The balanced equation is given. Steps 2, 3, and 4: From the balanced equation we see that 1 mol C6H12O6 6 mol CO2. The molar masses of C6H12O6 and CO2 are 180.2 g and 44.01 g, respectively. We combine all of these data into one equation: mass of CO2 produced 856 g C6H12O6 1 mol C6H12O6 180.2 g C6H12O6 6 mol CO2 1 mol C6H12O6 44.01 g CO2 1 mol CO2 1.25 103 g CO2 Step 5: Since one mole of C6H12O6 produces six moles of CO2 and the molar mass of C6H12O6 is four times that of CO2, we expect the mass of CO2 formed to be greater than 856 g. Therefore, the answer is reasonable. Similar Problem: 3.68. PRACTICE EXERCISE Methanol (CH3OH) burns in air according to the equation 2CH3OH 3O2 88n 2CO2 4H2O If 209 g of methanol are used up in a combustion process, what is the mass of H2O produced? 3.9 SF6 is used as a gaseous insulator in the electronic industry and as a thermal insulator in three-pane glass windows. LIMITING REAGENTS When a chemist carries out a reaction, the reactants are usually not present in exact stoichiometric amounts, that is, in the proportions indicated by the balanced equation. Consequently some reactant will be used up while others will be left over at the end of the reaction. The reactant used up first in a reaction is called the limiting reagent, since the maximum amount of product formed depends on how much of this reactant was originally present (Figure 3.8). When this reactant is used up, no more product can be formed. Excess reagents are the reactants present in quantities greater than necessary to react with the quantity of the limiting reagent. The concept of the limiting reagent is analogous to the relationship between men and women in a dance contest at a club. If there are fourteen men and only nine women, then only nine female/male pairs can compete. Five men will be left without partners. The number of women thus limits the number of men that can dance in the contest, and there is an excess of men. Sulfur hexafluoride (SF6) is a colorless, odorless, and extremely stable compound. It is formed by burning sulfur in an atmosphere of fluorine: S(l) 3F2(g) 88n SF6(g) This equation tells us that 1 mole of S reacts with 3 moles of F2 to produce 1 mole of SF6. Suppose that 4 moles of S are added to 20 moles of F2. Since 1 mol S 3 mol F2, the number of moles of F2 needed to react with 4 moles of S is Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.9 FIGURE 3.8 A limiting reagent is completely used up in a reaction. Before reaction has started LIMITING REAGENTS 93 After reaction is complete Limiting reagent Excess reagent 4 mol S 3 mol F2 1 mol S 12 mol F2 But there are 20 moles of F2 available, more than needed to completely react with S. Thus S must be the limiting reagent and F2 the excess reagent. The amount of SF6 produced depends only on how much S was originally present. Alternatively, we could determine the limiting reagent by calculating the number of moles of S needed to react with 20 moles of F2. In this case we write 20 mol F2 1 mol S 3 mol F2 6.7 mol S Since there are only 4 moles of S present, we arrive at the same conclusion that S is the limiting reagent and F2 is the excess reagent. In stoichiometric calculations involving limiting reagents, the first step is to decide which reactant is the limiting reagent. After the limiting reagent has been identified, the rest of the problem can be solved as outlined in Section 3.8. The following example illustrates this approach. We will not include step 5 in the calculations, but you should always examine the reasonableness of any chemical calculation. EXAMPLE 3.15 Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide: 2NH3(g) CO2(g) 88n (NH2)2CO(aq) H2O(l) In one process, 637.2 g of NH3 are allowed to react with 1142 g of CO2. (a) Which of the two reactants is the limiting reagent? (b) Calculate the mass of (NH2)2CO formed. (c) How much of the excess reagent (in grams) is left at the end of the reaction? (a) Since we cannot tell by inspection which of the two reactants is the limiting reagent, we have to proceed by first converting their masses into numbers of moles. The molar masses of NH3 and CO2 are 17.03 g and 44.01 g, respectively. Thus Answer moles of NH3 637.2 g NH3 1 mol NH3 17.03 g NH3 37.42 mol NH3 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 94 MASS RELATIONSHIPS IN CHEMICAL REACTIONS moles of CO2 1 mol CO2 44.01 g CO2 1142 g CO2 25.95 mol CO2 From the balanced equation we see that 2 mol NH3 1 mol CO2; therefore, the number of moles of NH3 needed to react with 25.95 moles of CO2 is given by 2 mol NH3 1 mol CO2 25.95 mol CO2 51.90 mol NH3 Since there are only 37.42 moles of NH3 present, not enough to react completely with the CO2, NH3 must be the limiting reagent and CO2 the excess reagent. (b) The amount of (NH2)2CO produced is determined by the amount of limiting reagent present. Thus we write mass of (NH2)2CO 37.42 mol NH3 1 mol (NH2)2CO 2 mol NH3 60.06 g (NH2)2CO 1 mol (NH2)2CO 1124 g (NH2)2CO (c) The number of moles of the excess reagent (CO2) left is 25.95 mol CO2 37.42 mol NH3 1 mol CO2 2 mol NH3 7.24 mol CO2 and mass of CO2 left over 7.24 mol CO2 44.01 g CO2 1 mol CO2 319 g CO2 Similar Problem: 3.80. PRACTICE EXERCISE The reaction between aluminum and iron(III) oxide can generate temperatures approaching 3000°C and is used in welding metals: 2Al Fe2O3 88n Al2O3 2Fe In one process 124 g of Al are reacted with 601 g of Fe2O3. (a) Calculate the mass (in grams) of Al2O3 formed. (b) How much of the excess reagent is left at the end of the reaction? Example 3.15 brings out an important point. In practice, chemists usually choose the more expensive chemical as the limiting reagent so that all or most of it will be consumed in the reaction. In the synthesis of urea, NH3 is invariably the limiting reagent because it is much more expensive than CO2. 3.10 REACTION YIELD The amount of limiting reagent present at the start of a reaction determines the theoretical yield of the reaction, that is, the amount of product that would result if all the limiting reagent reacted. The theoretical yield, then, is the maximum obtainable yield, predicted by the balanced equation. In practice, the actual yield, or the amount of product actually obtained from a reaction, is almost always less than the theoretical yield. There are many reasons for the difference between actual and theoretical yields. For instance, many reactions are reversible, and so they do not proceed 100 percent from Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 3.10 95 REACTION YIELD left to right. Even when a reaction is 100 percent complete, it may be difficult to recover all of the product from the reaction medium (say, from an aqueous solution). Some reactions are complex in the sense that the products formed may react further among themselves or with the reactants to form still other products. These additional reactions will reduce the yield of the first reaction. To determine how efficient a given reaction is, chemists often figure the percent yield, which describes the proportion of the actual yield to the theoretical yield. It is calculated as follows: actual yield theoretical yield %yield 100% (3.4) Percent yields may range from a fraction of 1 percent to 100 percent. Chemists strive to maximize the percent yield in a reaction. Factors that can affect the percent yield include temperature and pressure. We will study these effects later. In Example 3.16 we will calculate the yield of an industrial process. EXAMPLE 3.16 Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets, aircraft, jet engines, and bicycle frames. It is prepared by the reaction of titanium(IV) chloride with molten magnesium between 950 C and 1150 C: TiCl4(g) The frame of this bicycle is made of titanium. 2Mg(l) 88n Ti(s) 2MgCl2(l) In a certain industrial operation 3.54 107 g of TiCl4 are reacted with 1.13 107 g of Mg. (a) Calculate the theoretical yield of Ti in grams. (b) Calculate the percent yield if 7.91 106 g of Ti are actually obtained. Answer (a) First we calculate the number of moles of TiCl4 and Mg initially present: moles of TiCl4 3.54 107 g TiCl4 moles of Mg 1.13 107 g Mg 1 mol TiCl4 189.7 g TiCl4 1 mol Mg 24.31 g Mg 1.87 4.65 105 mol TiCl4 105 mol Mg Next, we must determine which of the two substances is the limiting reagent. 2 mol Mg; therefore, the From the balanced equation we see that 1 mol TiCl4 number of moles of Mg needed to react with 1.87 105 moles of TiCl4 is 1.87 105 mol TiCl4 2 mol Mg 1 mol TiCl4 3.74 105 mol Mg Since 4.65 105 moles of Mg are present, more than is needed to react with the amount of TiCl4 we have, Mg must be the excess reagent and TiCl4 the limiting reagent. The equation shows that 1 mol TiCl4 1 mol Ti; thus the theoretical mass of Ti formed is mass of Ti formed 3.54 107 g TiCl4 1 mol TiCl4 189.7 g TiCl4 1 mol Ti 1 mol TiCl4 47.88 g Ti 1 mol Ti Back Forward Main Menu TOC Study Guide TOC Textbook Website 8.93 106 g Ti MHHE Website 96 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chemical Fertilizers Feeding the world’s rapidly increasing population requires that farmers produce ever-larger and healthier crops. Every year they add hundreds of millions of tons of chemical fertilizers to the soil to increase crop quality and yield. In addition to carbon dioxide and water, plants need at least six elements for satisfactory growth. They are N, P, K, Ca, S, and Mg. The preparation and properties of several nitrogen- and phosphorus-containing fertilizers illustrate some of the principles introduced in this chapter. Nitrogen fertilizers contain nitrate (NO3 ) salts, ammonium (NH4 ) salts, and other compounds. Plants can absorb nitrogen in the form of nitrate directly, but ammonium salts and ammonia (NH3) must first be converted to nitrates by the action of soil bacteria. The principal raw material of nitrogen fertilizers is ammonia, prepared by the reaction between hydrogen and nitrogen: 3H2(g) N2(g) 88n 2NH3(g) (This reaction will be discussed in detail in Chapters 13 and 14.) In its liquid form, ammonia can be injected directly into the soil. Alternatively, ammonia can be converted to ammonium nitrate, NH4NO3, ammonium sulfate, (NH4)2SO4, or ammonium hydrogen phosphate, (NH4)2HPO4, in the following acid-base reactions: NH3(aq) HNO3(aq) 88n NH4NO3(aq) 2NH3(aq) H2SO4(aq) 88n (NH4)2SO4(aq) 2NH3(aq) H3PO4(aq) 88n (NH4)2HPO4(aq) Another method of preparing ammonium sulfate requires two steps: 2NH3(aq) CO2(aq) (NH4)2CO3(aq) H2O(l) 88n (NH4)2CO3(aq) (1) CaSO4(aq) 88n (NH4)2SO4(aq) CaCO3(s) (2) This approach is desirable because the starting materials — carbon dioxide and calcium sulfate — are less costly than sulfuric acid. To increase the yield, am- (b) To find the percent yield, we write %yield actual yield theoretical yield 7.91 8.93 106 g 106 g 100% 100% 88.6% Similar Problems: 3.83, 3.84. PRACTICE EXERCISE Industrially, vanadium metal, which is used in steel alloys, can be obtained by reacting vanadium(V) oxide with calcium at high temperatures: 5Ca V2O5 88n 5CaO 2V In one process 1.54 103 g of V2O5 react with 1.96 103 g of Ca. (a) Calculate the theoretical yield of V. (b) Calculate the percent yield if 803 g of V are obtained. Industrial processes usually involve huge quantities (thousands to millions of tons) of products. Thus even a slight improvement in the yield can significantly reduce the cost of production. A case in point is the manufacture of chemical fertilizers, discussed in the Chemistry in Action essay above. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website SUMMARY OF FACTS AND CONCEPTS Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chemistry in monia is made the limiting reagent in Reaction (1) and ammonium carbonate is made the limiting reagent in Reaction (2). The table below lists the percent composition by mass of nitrogen in some common fertilizers. The preparation of urea was discussed in Example 3.15. Percent Composition by Mass of Nitrogen in Five Common Fertilizers FERTILIZER % N BY MASS NH3 NH4NO3 (NH4)2SO4 (NH4)2HPO4 (NH2)2CO 82.4 35.0 21.2 21.2 46.7 Several factors influence the choice of one fertilizer over another: (1) cost of the raw materials needed to prepare the fertilizer; (2) ease of storage, transportation, and utilization; (3) percent composition by mass of the desired element; and (4) suitability of the compound, that is, whether the compound is soluble in water and whether it can be readily taken up by plants. Considering all these factors together, we find that NH4NO3 is the most important nitrogen-containing fertilizer compound in the world, even though ammonia has the highest percentage by mass of nitrogen. Phosphorus fertilizers are derived from phosphate rock, called fluorapatite, Ca5(PO4)3F. Fluorapatite is insoluble in water, so it must first be converted to water-soluble calcium dihydrogen phosphate [Ca(H2PO4)2]: SUMMARY OF KEY EQUATIONS SUMMARY OF FACTS AND CONCEPTS Forward Main Menu Liquid ammonia being applied to the soil before planting. 2Ca5(PO4)3F(s) 7H2SO4(aq) 88n 3Ca(H2PO4)2(aq) 7CaSO4(aq) actual yield theoretical yield 2HF(g) For maximum yield, fluorapatite is made the limiting reagent in this reaction. The reactions we have discussed for the preparation of fertilizers all appear relatively simple, yet much effort has been expended to improve the yields by changing conditions such as temperature, pressure, and so on. Industrial chemists usually run promising reactions first in the laboratory and then test them in a pilot facility before putting them into mass production. • percent composition of an element • %yield Back 97 n molar mass of element molar mass of compound 100% 100% (3.1) (3.4) 1. Atomic masses are measured in atomic mass units (amu), a relative unit based on a value of exactly 12 for the C-12 isotope. The atomic mass given for the atoms of a particular element is usually the average of the naturally occurring isotope distribution of that element. The molecular mass of a molecule is the sum of the atomic masses of the atoms in the molecule. Both atomic mass and molecular mass can be accurately determined with a mass spectrometer. TOC Study Guide TOC Textbook Website MHHE Website 98 MASS RELATIONSHIPS IN CHEMICAL REACTIONS 2. A mole is Avogadro’s number (6.022 1023) of atoms, molecules, or other particles. The molar mass (in grams) of an element or a compound is numerically equal to its mass in atomic mass units (amu) and contains Avogadro’s number of atoms (in the case of elements), molecules, or simplest formula units (in the case of ionic compounds). 3. The percent composition by mass of a compound is the percent by mass of each element present. If we know the percent composition by mass of a compound, we can deduce the empirical formula of the compound and also the molecular formula of the compound if the approximate molar mass is known. 4. Chemical changes, called chemical reactions, are represented by chemical equations. Substances that undergo change — the reactants — are written on the left and the substances formed — the products — appear to the right of the arrow. Chemical equations must be balanced, in accordance with the law of conservation of mass. The number of atoms of each type of element in the reactants must equal the number in the products. 5. Stoichiometry is the quantitative study of products and reactants in chemical reactions. Stoichiometric calculations are best done by expressing both the known and unknown quantities in terms of moles and then converting to other units if necessary. A limiting reagent is the reactant that is present in the smallest stoichiometric amount. It limits the amount of product that can be formed. The amount of product obtained in a reaction (the actual yield) may be less than the maximum possible amount (the theoretical yield). The ratio of the two is expressed as the percent yield. KEY WORDS Actual yield, p. 94 Atomic mass, p. 70 Atomic mass unit (amu), p. 70 Avogadro’s number, p. 71 Chemical equation, p. 84 Mole method, p. 89 Molecular mass, p. 74 Percent composition by mass, p. 77 Percent yield, p. 95 Chemical reaction, p. 84 Excess reagent, p. 92 Limiting reagent, p. 92 Molar mass, p. 72 Mole (mol), p. 71 Product, p. 85 Reactant, p. 85 Stoichiometric amount, p. 92 Stoichiometry, p. 89 Theoretical yield, p. 94 QUESTIONS AND PROBLEMS ATOMIC MASS Review Questions 3.1 What is an atomic mass unit? Why is it necessary to introduce such a unit? 3.2 What is the mass (in amu) of a carbon-12 atom? Why is the atomic mass of carbon listed as 12.01 amu in the table on the inside front cover of this book? 3.3 Explain clearly what is meant by the statement “The atomic mass of gold is 197.0 amu.” 3.4 What information would you need to calculate the average atomic mass of an element? Problems 3.5 The atomic masses of 35Cl (75.53%) and 37Cl 17 17 (24.47%) are 34.968 amu and 36.956 amu, respectively. Calculate the average atomic mass of chlorine. The percentages in parentheses denote the relative abundances. 3.6 The atomic masses of 6Li and 7Li are 6.0151 amu and 3 3 Back Forward Main Menu TOC 7.0160 amu, respectively. Calculate the natural abundances of these two isotopes. The average atomic mass of Li is 6.941 amu. 3.7 What is the mass in grams of 13.2 amu? 3.8 How many amu are there in 8.4 g? MOLAR MASS AND AVOGADRO’S NUMBER Review Questions 3.9 Define the term “mole.” What is the unit for mole in calculations? What does the mole have in common with the pair, the dozen, and the gross? What does Avogadro’s number represent? 3.10 What is the molar mass of an atom? What are the commonly used units for molar mass? Problems 3.11 Earth’s population is about 6.0 billion. Suppose that every person on Earth participates in a process of counting identical particles at the rate of two parti- Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20 3.21 3.22 cles per second. How many years would it take to count 6.0 1023 particles? Assume that there are 365 days in a year. The thickness of a piece of paper is 0.0036 in. Suppose a certain book has an Avogadro’s number of pages; calculate the thickness of the book in lightyears. (Hint: See Problem 1.47 for the definition of light-year.) How many atoms are there in 5.10 moles of sulfur (S)? How many moles of cobalt atoms are there in 6.00 109 (6 billion) Co atoms? How many moles of calcium (Ca) atoms are in 77.4 g of Ca? How many grams of gold (Au) are there in 15.3 moles of Au? What is the mass in grams of a single atom of each of the following elements? (a) Hg, (b) Ne. What is the mass in grams of a single atom of each of the following elements? (a) As, (b) Ni. What is the mass in grams of 1.00 1012 lead (Pb) atoms? How many atoms are present in 3.14 g of copper (Cu)? Which of the following has more atoms: 1.10 g of hydrogen atoms or 14.7 g of chromium atoms? Which of the following has a greater mass: 2 atoms of lead or 5.1 10 23 mole of helium. MOLECULAR MASS Problems 3.23 Calculate the molecular mass (in amu) of each of the following substances: (a) CH4, (b) NO2, (c) SO3, (d) C6H6, (e) NaI, (f) K2SO4, (g) Ca3(PO4)2. 3.24 Calculate the molar mass of the following substances: (a) Li2CO3, (b) CS2, (c) CHCl3 (chloroform), (d) C6H8O6 (ascorbic acid, or vitamin C), (e) KNO3, (f) Mg3N2. 3.25 Calculate the molar mass of a compound if 0.372 mole of it has a mass of 152 g. 3.26 How many molecules of ethane (C2H6) are present in 0.334 g of C2H6? 3.27 Calculate the number of C, H, and O atoms in 1.50 g of glucose (C6H12O6), a sugar. 3.28 Urea [(NH2)2CO] is used for fertilizer and many other things. Calculate the number of N, C, O, and H atoms in 1.68 104 g of urea. 3.29 Pheromones are a special type of compound secreted by the females of many insect species to attract the males for mating. One pheromone has the molecular formula C19H38O. Normally, the amount of this pheromone secreted by a female insect is about Back Forward Main Menu TOC 99 1.0 10 12 g. How many molecules are there in this quantity? 3.30 The density of water is 1.00 g/mL at 4 C. How many water molecules are present in 2.56 mL of water at this temperature? MASS SPECTROMETRY Review Questions 3.31 Describe the operation of a mass spectrometer. 3.32 Describe how you would determine the isotopic abundance of an element from its mass spectrum. Problems 3.33 Carbon has two stable isotopes, 12C and 13C, and flu06 06 orine has only one stable isotope, 19F. How many 09 peaks would you observe in the mass spectrum of the positive ion of CF4 ? Assume that the ion does not break up into smaller fragments. 3.34 Hydrogen has two stable isotopes, 1H and 2H, and 1 1 sulfur has four stable isotopes, 32S, 33S, 34S, 36S. How 16 16 16 16 many peaks would you observe in the mass spectrum of the positive ion of hydrogen sulfide, H2S ? Assume no decomposition of the ion into smaller fragments. PERCENT COMPOSITION AND CHEMICAL FORMULAS Review Questions 3.35 Use ammonia (NH3) to explain what is meant by the percent composition by mass of a compound. 3.36 Describe how the knowledge of the percent composition by mass of an unknown compound can help us identify the compound. 3.37 What does the word “empirical” in empirical formula mean? 3.38 If we know the empirical formula of a compound, what additional information do we need in order to determine its molecular formula? Problems 3.39 Tin (Sn) exists in Earth’s crust as SnO2. Calculate the percent composition by mass of Sn and O in SnO2. 3.40 For many years chloroform (CHCl3) was used as an inhalation anesthetic in spite of the fact that it is also a toxic substance that may cause severe liver, kidney, and heart damage. Calculate the percent composition by mass of this compound. 3.41 Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its molecular formula is C9H10O. (a) Calculate the percent composition by mass of C, H, and O in cinnamic alcohol. (b) How many molecules of cinnamic alcohol are contained in a sample of mass 0.469 g? Study Guide TOC Textbook Website MHHE Website 100 MASS RELATIONSHIPS IN CHEMICAL REACTIONS 3.42 All of the substances listed below are fertilizers that contribute nitrogen to the soil. Which of these is the richest source of nitrogen on a mass percentage basis? (a) Urea, (NH2)2CO (b) Ammonium nitrate, NH4NO3 (c) Guanidine, HNC(NH2)2 (d) Ammonia, NH3 3.43 Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4%; H: 6.21%; S: 39.5%; O: 9.86%. Calculate its empirical formula. What is its molecular formula given that its molar mass is about 162 g? 3.44 Peroxyacylnitrate (PAN) is one of the components of smog. It is a compound of C, H, N, and O. Determine the percent composition of oxygen and the empirical formula from the following percent composition by mass: 19.8 percent C, 2.50 percent H, 11.6 percent N. 3.45 The formula for rust can be represented by Fe2O3. How many moles of Fe are present in 24.6 g of the compound? 3.46 How many grams of sulfur (S) are needed to react completely with 246 g of mercury (Hg) to form HgS? 3.47 Calculate the mass in grams of iodine (I2) that will react completely with 20.4 g of aluminum (Al) to form aluminum iodide (AlI3). 3.48 Tin(II) fluoride (SnF2) is often added to toothpaste as an ingredient to prevent tooth decay. What is the mass of F in grams in 24.6 g of the compound? 3.49 What are the empirical formulas of the compounds with the following compositions? (a) 2.1 percent H, 65.3 percent O, 32.6 percent S, (b) 20.2 percent Al, 79.8 percent Cl. 3.50 What are the empirical formulas of the compounds with the following compositions? (a) 40.1 percent C, 6.6 percent H, 53.3 percent O, (b) 18.4 percent C, 21.5 percent N, 60.1 percent K. 3.51 The molar mass of caffeine is 194.19 g. Is the molecular formula of caffeine C4H5N2O or C8H10N4O2? 3.52 Monosodium glutamate (MSG), a food-flavor enhancer, has been blamed for “Chinese restaurant syndrome,” the symptoms of which are headaches and chest pains. MSG has the following composition by mass: 35.51 percent C, 4.77 percent H, 37.85 percent O, 8.29 percent N, and 13.60 percent Na. What is its molecular formula if its molar mass is 169 g? CHEMICAL REACTIONS AND CHEMICAL EQUATIONS 3.55 Use the formation of water from hydrogen and oxy- Forward Main Menu TOC Problems 3.59 Balance the following equations using the method outlined in Section 3.7: (a) C O2 88n CO (b) CO O2 88n CO2 (c) H2 Br2 88n HBr (d) K H2O 88n KOH H2 (e) Mg O2 88n MgO (f) O3 88n O2 (g) H2O2 88n H2O O2 (h) N2 H2 88n NH3 (i) Zn AgCl 88n ZnCl2 Ag (j) S8 O2 88n SO2 (k) NaOH H2SO4 88n Na2SO4 H2O (l) Cl2 NaI 88n NaCl I2 (m) KOH H3PO4 88n K3PO4 H2O (n) CH4 Br2 88n CBr4 HBr 3.60 Balance the following equations using the method outlined in Section 3.7: (a) KClO3 88n KCl O2 (b) KNO3 88n KNO2 O2 (c) NH4NO3 88n N2O H2O (d) NH4NO2 88n N2 H2O (e) NaHCO3 88n Na2CO3 H2O CO2 (f) P4O10 H2O 88n H3PO4 (g) HCl CaCO3 88n CaCl2 H2O CO2 (h) Al H2SO4 88n Al2(SO4)3 H2 (i) CO2 KOH 88n K2CO3 H2O (j) CH4 O2 88n CO2 H2O (k) Be2C H2O 88n Be(OH)2 CH4 (l) Cu HNO3 88n Cu(NO3)2 NO H2O (m) S HNO3 88n H2SO4 NO2 H2O (n) NH3 CuO 88n Cu N2 H2O AMOUNTS OF REACTANTS AND PRODUCTS Review Questions 3.61 On what law is stoichiometry based? Why is it essential to use balanced equations in solving stoichiometric problems? 3.62 Describe the steps involved in the mole method. Problems Review Questions Back gen to explain the following terms: chemical reaction, reactant, product. 3.56 What is the difference between a chemical reaction and a chemical equation? 3.57 Why must a chemical equation be balanced? What law is obeyed by a balanced chemical equation? 3.58 Write the symbols used to represent gas, liquid, solid, and the aqueous phase in chemical equations. 3.63 Consider the combustion of carbon monoxide (CO) in oxygen gas: Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 2CO(g) O2(g) 88n 2CO2(g) Starting with 3.60 moles of CO, calculate the number of moles of CO2 produced if there is enough oxygen gas to react with all of the CO. 3.64 Silicon tetrachloride (SiCl4) can be prepared by heating Si in chlorine gas: Si(s) 2Cl2(g) 88n SiCl4(l) In one reaction, 0.507 mole of SiCl4 is produced. How many moles of molecular chlorine were used in the reaction? 3.65 The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other sources is about 26 million tons. The equation for the reaction is S(s) O2(g) 88n SO2(g) How much sulfur, present in the original materials, would result in that quantity of SO2? 3.66 When baking soda (sodium bicarbonate or sodium hydrogen carbonate, NaHCO3) is heated, it releases carbon dioxide gas, which is responsible for the rising of cookies, donuts, and bread. (a) Write a balanced equation for the decomposition of the compound (one of the products is Na2CO3). (b) Calculate the mass of NaHCO3 required to produce 20.5 g of CO2. 3.67 When potassium cyanide (KCN) reacts with acids, a deadly poisonous gas, hydrogen cyanide (HCN), is given off. Here is the equation: KCN(aq) HCl(aq) 88n KCl(aq) HCN(g) If a sample of 0.140 g of KCN is treated with an excess of HCl, calculate the amount of HCN formed, in grams. 3.68 Fermentation is a complex chemical process of wine making in which glucose is converted into ethanol and carbon dioxide: C6H12O6 88n 2C2H5OH glucose 2CO2 ethanol Starting with 500.4 g of glucose, what is the maximum amount of ethanol in grams and in liters that can be obtained by this process? (Density of ethanol 0.789 g/mL.) 3.69 Each copper(II) sulfate unit is associated with five water molecules in crystalline copper(II) sulfate pentahydrate (CuSO4 5H2O). When this compound is heated in air above 100 C, it loses the water molecules and also its blue color: CuSO4 5H2O 88n CuSO4 5H2O If 9.60 g of CuSO4 are left after heating 15.01 g of Back Forward Main Menu TOC 101 the blue compound, calculate the number of moles of H2O originally present in the compound. 3.70 For many years the recovery of gold — that is, the separation of gold from other materials — involved the use of potassium cyanide: 4Au 8KCN O2 2H2O 88n 4KAu(CN)2 4KOH What is the minimum amount of KCN in moles needed to extract 29.0 g (about an ounce) of gold? 3.71 Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide. Calculate how many grams of quicklime can be produced from 1.0 kg of limestone. 3.72 Nitrous oxide (N2O) is also called “langhing gas.” It can be prepared by the thermal decomposition of ammonium nitrate (NH4NO3). The other product is H2O. (a) Write a balanced equation for this reaction. (b) How many grams of N2O are formed if 0.46 mole of NH4NO3 is used in the reaction? 3.73 The fertilizer ammonium sulfate [(NH4)2SO4] is prepared by the reaction between ammonia (NH3) and sulfuric acid: 2NH3(g) H2SO4(aq) 88n (NH4)2SO4(aq) How many kilograms of NH3 are needed to produce 1.00 105 kg of (NH4)2SO4? 3.74 A common laboratory preparation of oxygen gas is the thermal decomposition of potassium chlorate (KClO3). Assuming complete decomposition, calculate the number of grams of O2 gas that can be obtained from 46.0 g of KClO3. (The products are KCl and O2.) LIMITING REAGENTS Review Questions 3.75 Define limiting reagent and excess reagent. What is the significance of the limiting reagent in predicting the amount of the product obtained in a reaction? Can there be a limiting reagent if only one reactant is present? 3.76 Give an everyday example that illustrates the limiting reagent concept. Problems 3.77 Nitric oxide (NO) reacts instantly with oxygen gas to form nitrogen dioxide (NO2), a dark-brown gas: 2NO(g) O2(g) 88n 2NO2(g) In one experiment 0.886 mole of NO is mixed with 0.503 mole of O2. Calculate which of the two reactants is the limiting reagent. Calculate also the number of moles of NO2 produced. Study Guide TOC Textbook Website MHHE Website 102 MASS RELATIONSHIPS IN CHEMICAL REACTIONS 3.78 The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from the highaltitude jet plane, the SST. The reaction is O3 NO 88n O2 NO2 If 0.740 g of O3 reacts with 0.670 g of NO, how many grams of NO2 will be produced? Which compound is the limiting reagent? Calculate the number of moles of the excess reagent remaining at the end of the reaction. 3.79 Propane (C3H8) is a component of natural gas and is used in domestic cooking and heating. (a) Balance the following equation representing the combustion of propane in air: C3H8 O2 88n CO2 H2O (b) How many grams of carbon dioxide can be produced by burning 3.65 moles of propane? Assume that oxygen is the excess reagent in this reaction. 3.80 Consider the reaction MnO2 4HCl 88n MnCl2 Cl2 REACTION YIELD Review Questions 3.81 Why is the yield of a reaction determined only by the amount of the limiting reagent? 3.82 Why is the actual yield of a reaction almost always smaller than the theoretical yield? Problems 3.83 Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is prepared by the reaction H2SO4 88n CaSO4 2HF In one process 6.00 kg of CaF2 are treated with an excess of H2SO4 and yield 2.86 kg of HF. Calculate the percent yield of HF. 3.84 Nitroglycerin (C3H5N3O9) is a powerful explosive. Its decomposition may be represented by 4C3H5N3O9 88n 6N2 12CO2 10H2O O2 This reaction generates a large amount of heat and many gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that Back Forward Main Menu FeTiO3 H2SO4 88n TiO2 FeSO4 H2O Its opaque and nontoxic properties make it suitable as a pigment in plastics and paints. In one process 8.00 103 kg of FeTiO3 yielded 3.67 103 kg of TiO2. What is the percent yield of the reaction? 3.86 Ethylene (C2H4), an important industrial organic chemical, can be prepared by heating hexane (C6H14) at 800 C: C6H14 88n C2H4 other products If the yield of ethylene production is 42.5 percent, what mass of hexane must be reacted to produce 481 g of ethylene? 2H2O If 0.86 mole of MnO2 and 48.2 g of HCl react, which reagent will be used up first? How many grams of Cl2 will be produced? CaF2 produces the explosion. (a) What is the maximum amount of O2 in grams that can be obtained from 2.00 102 g of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of O2 generated is found to be 6.55 g. 3.85 Titanium(IV) oxide (TiO2) is a white substance produced by the action of sulfuric acid on the mineral ilmenite (FeTiO3): TOC ADDITIONAL PROBLEMS 3.87 Use the symbols m for the mass of a sample of an element in grams, N for the number of atoms of the element in the sample, n for the number of moles of the element, for the molar mass of the element in grams per mole, and NA for Avogadro’s number to derive the expressions that would allow us to calculate: (a) the number of atoms in the sample in terms of n and NA, (b) the molar mass of the element in terms of m and n, (c) the number of moles of the element in terms of m and , and (d) Avogadro’s number in terms of N and n. 3.88 A sample of a compound of Cl and O reacts with an excess of H2 to give 0.233 g of HCl and 0.403 g of H2O. Determine the empirical formula of the compound. 3.89 The atomic mass of element X is 33.42 amu. A 27.22g sample of X combines with 84.10 g of another element Y to form a compound XY. Calculate the atomic mass of Y. 3.90 How many moles of O are needed to combine with 0.212 mole of C to form (a) CO and (b) CO2? 3.91 A research chemist used a mass spectrometer to study the two isotopes of an element. Over time, she recorded a number of mass spectra of these isotopes. On analysis, she noticed that the ratio of the taller peak (the more abundant isotope) to the shorter peak (the less abundant isotope) gradually increased with time. Assuming that the mass spectrometer was func- Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 3.92 3.93 3.94 3.95 3.96 3.97 3.98 tioning normally, what do you think was causing this change? The aluminum sulfate hydrate [Al2(SO4)3 xH2O] contains 8.20 percent Al by mass. Calculate x, that is, the number of water molecules associated with each Al2(SO4)3 unit. Mustard gas (C4H8Cl2S) is a poisonous gas that was used in World War I and banned afterward. It causes general destruction of body tissues, resulting in the formation of large water blisters. There is no effective antidote. Calculate the percent composition by mass of the elements in mustard gas. The carat is the unit of mass used by jewelers. One carat is exactly 200 mg. How many carbon atoms are present in a 24-carat diamond? An iron bar weighed 664 g. After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust (Fe2O3). Calculate the final mass of the iron bar and rust. A certain metal oxide has the formula MO where M denotes the metal. A 39.46-g sample of the compound is strongly heated in an atmosphere of hydrogen to remove oxygen as water molecules. At the end, 31.70 g of the metal is left over. If O has an atomic mass of 16.00 amu, calculate the atomic mass of M and identify the element. An impure sample of zinc (Zn) is treated with an excess of sulfuric acid (H2SO4) to form zinc sulfate (ZnSO4) and molecular hydrogen (H2). (a) Write a balanced equation for the reaction. (b) If 0.0764 g of H2 is obtained from 3.86 g of the sample, calculate the percent purity of the sample. (c) What assumptions must you make in (b)? One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is Fe2O3 3CO 88n 2Fe Forward Main Menu 3.102 3.103 3.104 3.105 3.106 3.107 3CO2 Suppose that 1.64 103 kg of Fe are obtained from a 2.62 103-kg sample of Fe2O3. Assuming that the reaction goes to completion, what is the percent purity of Fe2O3 in the original sample? 3.99 Carbon dioxide (CO2) is the gas that is mainly responsible for global warming (the greenhouse effect). The burning of fossil fuels is a major cause of the increased concentration of CO2 in the atmosphere. Carbon dioxide is also the end product of metabolism (see Example 3.14). Using glucose as an example of food, calculate the annual human production of CO2 in grams, assuming that each person consumes 5.0 102 g of glucose per day. The world’s population is 6.0 billion, and there are 365 days in a year. 3.100 Carbohydrates are compounds containing carbon, hydrogen, and oxygen in which the hydrogen to oxy- Back 3.101 TOC 3.108 3.109 103 gen ratio is 2:1. A certain carbohydrate contains 40.0% carbon by mass. Calculate the empirical and molecular formulas of the compound if the approximate molar mass is 178 g. Which of the following has the greater mass: 0.72 g of O2 or 0.0011 mol of chlorophyll (C55H72MgN4O5)? Analysis of a metal chloride XCl3 shows that it contains 67.2% Cl by mass. Calculate the molar mass of X and identify the element. Hemoglobin (C2952H4664N812O832S8Fe4) is the oxygen carrier in blood. (a) Calculate its molar mass. (b) An average adult has about 5.0 liters of blood. Every milliliter of blood has approximately 5.0 109 erythrocytes, or red blood cells, and every red blood cell has about 2.8 108 hemoglobin molecules. Calculate the mass of hemoglobin molecules in grams in an average adult. Myoglobin stores oxygen for metabolic processes in muscle. Chemical analysis shows that it contains 0.34% Fe by mass. What is the molar mass of myoglobin? (There is one Fe atom per molecule.) Calculate the number of cations and anions in each of the following compounds: (a) 8.38 g of KBr, (b) 5.40 g of Na2SO4, (c) 7.45 g of Ca3(PO4)2. A sample containing NaCl, Na2SO4, and NaNO3 gives the following elemental analysis: Na: 32.08%; O: 36.01%; Cl: 19.51%. Calculate the mass percent of each compound in the sample. (a) For molecules having small molecular masses, mass spectrometry can be used to identify their formulas. To illustrate this point, identify the molecule which most likely accounts for the observation of a peak in a mass spectrum at: 16 amu, 17 amu, 18 amu, and 64 amu. (b) Note that there are (among others) two likely molecules which would give rise to a peak at 44 amu, namely, C3H8 and CO2. In such cases, a chemist might try to look for other peaks generated when some of the molecules break apart in the spectrometer. For example, if a chemist sees a peak at 44 amu and also one at 15 amu, which molecule is producing the 44-amu peak? Why? (c) Using the following precise atomic masses: 1H (1.00797 amu), 12 C (12.00000 amu), and 16O (15.99491 amu), how precisely must the masses of C3H8 and CO2 be measured to distinguish between them? Calculate the percent composition by mass of all the elements in calcium phosphate [Ca3(PO4)2], a major component of bone. Lysine, an essential amino acid in the human body, contains C, H, O, and N. In one experiment, the complete combustion of 2.175 g of lysine gave 3.94 g CO2 and 1.89 g H2O. In a separate experiment, 1.873 Study Guide TOC Textbook Website MHHE Website 104 MASS RELATIONSHIPS IN CHEMICAL REACTIONS 3.110 3.111 3.112 3.113 3.114 3.115 3.116 3.117 3.118 3.119 Back g of lysine gave 0.436 g NH3. (a) Calculate the empirical formula of lysine. (b) The approximate molar mass of lysine is 150 g. What is the molecular formula of the compound? Does 1 g of hydrogen molecules contain as many H atoms as 1 g of hydrogen atoms? Avogadro’s number has sometimes been described as a conversion factor between amu and grams. Use the fluorine atom (19.00 amu) as an example to show the relation between the atomic mass unit and the gram. The natural abundances of the two stable isotopes of hydrogen (hydrogen and deuterium) are 1H: 99.985% 1 and 2H: 0.015%. Assume that water exists as either 1 H2O or D2O. Calculate the number of D2O molecules in exactly 400 mL of water. (Density 1.00 g/mL.) A compound containing only C, H, and Cl was examined in a mass spectrometer. The highest mass peak seen corresponds to an ion mass of 52 amu. The most abundant mass peak seen corresponds to an ion mass of 50 amu and is about three times as intense as the peak at 52 amu. Deduce a reasonable molecular formula for the compound and explain the positions and intensities of the mass peaks mentioned. (Hint: Chlorine is the only element that has isotopes in comparable abundances: 35Cl: 75.5%; 37Cl: 24.5%. 17 17 For H, use 1H; for C, use 12C.) 1 06 In the formation of carbon monoxide, CO, it is found that 2.445 g of carbon combine with 3.257 g of oxygen. What is the atomic mass of oxygen if the atomic mass of carbon is 12.01 amu? What mole ratio of molecular chlorine (Cl2) to molecular oxygen (O2) would result from the breakup of the compound Cl2O7 into its constituent elements? Which of the following substances contains the greatest mass of chlorine? (a) 5.0 g Cl2, (b) 60.0 g NaClO3, (c) 0.10 mol KCl, (d) 30.0 g MgCl2, (e) 0.50 mol Cl2. Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium chloride (KCl) and potassium sulfate (K2SO4). Potash production is often reported as the potassium oxide (K2O) equivalent or the amount of K2O that could be made from a given mineral. (a) If KCl costs $0.055 per kg, for what price (dollar per kg) must K2SO4 be sold in order to supply the same amount of potassium on a per dollar basis? (b) What mass (in kg) of K2O contains the same number of moles of K atoms as 1.00 kg of KCl? Platinum forms two different compounds with chlorine. One contains 26.7 percent Cl by mass, and the other contains 42.1 percent Cl by mass. Determine the empirical formulas of the two compounds. Heating 2.40 g of the oxide of metal X (molar mass Forward Main Menu TOC 3.120 3.121 3.122 3.123 3.124 3.125 3.126 3.127 3.128 of X 55.9 g/mol) in carbon monoxide (CO) yields the pure metal and carbon dioxide. The mass of the metal product is 1.68 g. From the data given, show that the simplest formula of the oxide is X2O3 and write a balanced equation for the reaction. A compound X contains 63.3 percent manganese (Mn) and 36.7 percent O by mass. When X is heated, oxygen gas is evolved and a new compound Y containing 72.0 percent Mn and 28.0 percent O is formed. (a) Determine the empirical formulas of X and Y. (b) Write a balanced equation for the conversion of X to Y. The formula of a hydrate of barium chloride is BaCl2 xH2O. If 1.936 g of the compound gives 1.864 g of anhydrous BaSO4 upon treatment with sulfuric acid, calculate the value of x. It is estimated that the day Mt. St. Helens erupted (May 18, 1980), about 4.0 105 tons of SO2 were released into the atmosphere. If all the SO2 were eventually converted to sulfuric acid, how many tons of H2SO4 were produced? A mixture of CuSO4 5H2O and MgSO4 7H2O is heated until all the water is lost. If 5.020 g of the mixture gives 2.988 g of the anhydrous salts, what is the percent by mass of CuSO4 5H2O in the mixture? When 0.273 g of Mg is heated strongly in a nitrogen (N2) atmosphere, a chemical reaction occurs. The product of the reaction weighs 0.378 g. Calculate the empirical formula of the compound containing Mg and N. Name the compound. A mixture of methane (CH4) and ethane (C2H6) of mass 13.43 g is completely burned in oxygen. If the total mass of CO2 and H2O produced is 64.84 g, calculate the fraction of CH4 in the mixture. Leaded gasoline contains an additive to prevent engine “knocking.” On analysis, the additive compound is found to contain carbon, hydrogen, and lead (Pb) (hence, “leaded gasoline”). When 51.36 g of this compound are burned in an apparatus such as that shown in Figure 3.5, 55.90 g of CO2 and 28.61 g of H2O are produced. Determine the empirical formula of the gasoline additive. Because of its detrimental effect on the environment, the lead compound described in Problem 3.126 has been replaced in recent years by methyl tert-butyl ether (a compound of C, H, and O) to enhance the performance of gasoline. When 12.1 g of the compound are burned in an apparatus like the one shown in Figure 3.5, 30.2 g of CO2 and 14.8 g of H2O are formed. What is the empirical formula of the compound? Suppose you are given a cube made of magnesium (Mg) metal of edge length 1.0 cm. (a) Calculate the Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 3.129 3.130 3.131 3.132 Back number of Mg atoms in the cube. (b) Atoms are spherical in shape. Therefore, the Mg atoms in the cube cannot fill all of the available space. If only 74 percent of the space inside the cube is taken up by Mg atoms, calculate the radius in picometers of a Mg atom. (The density of Mg is 1.74 g/cm3 and the volume of a sphere of radius r is 4 r3.) 3 A certain sample of coal contains 1.6 percent sulfur by mass. When the coal is burned, the sulfur is converted to sulfur dioxide. To prevent air pollution, this sulfur dioxide is treated with calcium oxide (CaO) to form calcium sulfite (CaSO3). Calculate the daily mass (in kilograms) of CaO needed by a power plant that uses 6.60 106 kg of coal per day. Air is a mixture of many gases. However, in calculating its “molar mass” we need consider only the three major components: nitrogen, oxygen, and argon. Given that one mole of air at sea level is made up of 78.08 percent nitrogen, 20.95 percent oxygen, and 0.97 percent argon, what is the molar mass of air? A die has an edge length of 1.5 cm. (a) What is the volume of one mole of such dice? (b) Assuming that the mole of dice could be packed in such a way that they were in contact with one another, forming stacking layers covering the entire surface of Earth, calculate the height in meters the layers would extend outward. [The radius (r) of Earth is 6371 km and the area of a sphere is 4 r2.] The following is a crude but effective method for estimating the order of magnitude of Avogadro’s number using stearic acid (C18H36O2). When stearic acid is added to water, its molecules collect at the surface Forward Main Menu TOC 105 and form a monolayer; that is, the layer is only one molecule thick. The cross-sectional area of each stearic acid molecule has been measured to be 0.21 nm2. In one experiment it is found that 1.4 10 4 g of stearic acid is needed to form a monolayer over water in a dish of diameter 20 cm. Based on these measurements, what is Avogadro’s number? (The area of a circle of radius r is r2.) 3.133 Octane (C8H18) is a component of gasoline. Complete combustion of octane yields CO2 and H2O. Incomplete combustion produces CO and H2O, which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 1.000 gallon of octane is burned in an engine. The total mass of CO, CO2, and H2O produced is 11.53 kg. Calculate the efficiency of the process; that is, calculate the fraction of octane converted to CO2. The density of octane is 2.650 kg/gallon. 3.134 Industrially, hydrogen gas can be prepared by reacting propane gas (C3H8) with steam at about 400°C. The products are carbon monoxide (CO) and hydrogen gas (H2). (a) Write a balanced equation for the reaction. (b) How many kilograms of H2 can be obtained from 2.84 103 kg of propane? Answers to Practice Exercises: 3.1 10.81 amu. 3.2 3.59 moles. 3.3 2.57 103 g. 3.4 8.49 1021 K atoms. 3.5 32.04 amu. 3.6 1.66 moles. 3.7 5.81 1024 H atoms. 3.8 H: 2.055%; S: 32.69%; O: 62.25%. 3.9 KMnO4 (potassium permanganate). 3.10 196 g. 3.11 B2H6. 3.12 Fe2O3 3CO 88n 2Fe 3CO2. 3.13 (a) 0.508 mole, (b) 2.21 g. 3.14 235 g. 3.15 (a) 234 g, (b) 234 g. 3.16 (a) 863 g, (b) 93.0%. Study Guide TOC Textbook Website MHHE Website C HEMISTRY IN T D HREE IMENSIONS Determining Avogadro’s Number from the Structure of a Solid Iron has a body-centered cubic unit cell. A unit cell is the basic structural unit of the larger, three-dimensional crystal lattice. Many solids have highly ordered structures. An example is the ionic compound sodium chloride. If you turn to Figure 2.11, you will see that the Na and Cl ions are arranged in a regular manner. This three-dimensional structure is called a crystal lattice. Similar crystal lattices also exist in metals such as aluminum (Al) and iron (Fe). The basic structural unit of a crystal lattice is the unit cell. A crystalline solid consists of many unit cells extending in all directions. If we know the unit cell dimensions (that is, the edge lengths) and the number of atoms (or ions in the case of an ionic compound) within each cell, we can calculate the number of atoms (or ions) in one mole of a substance, that is, Avogadro’s number. As an example let us consider the crystal structure of iron. The unit cell of iron has a body-centered cubic structure. There are a total of two Fe atoms within the cell (one from the eight corners and one from the center). Using a technique called X-ray diffraction, we find the length of the body-centered cubic unit cell of Fe to be 286.7 pm. Given that iron’s density is 7.874 g/cm3 and its molar mass is 55.85 g/mol, we can calculate Avogadro’s number as follows. First we find the volume of a unit cell. Then, from the molar mass of Fe and its density, we determine the volume of a mole of Fe atoms (its molar volume). The ratio of the molar volume to the volume of a unit cell gives us the total number of unit cells in a mole of Fe atoms. Finally, knowing the number of Fe atoms within a unit cell we can determine Avogadro’s number. The calculations are shown below. Step 1: Calculate the volume of the unit cell. The volume of a cube of length a is a3. Therefore, the volume of a body-centered cubic unit cell with an edge length of 286.7 pm is volume of unit cell 1 (286.7 pm)3 2.357 10 29 10 29 m3 2.357 10 23 3 m3 2.357 10 12 m 1 pm cm3 1 102 cm 1m 3 106 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website In the extended body-centered cubic unit cell, each corner atom is shared by eight unit cells, and so each of the eight corner atoms contributes only 1⁄8 atom to the total number of atoms in one unit cell. This atom plus the center atom gives a total of two atoms per unit cell. a X-ray diffraction tells us that the side, or edge length, of a unit cell of iron is 286.7 pm. Using this number, we can calculate the volume of the unit cell and go on to determine the number of unit cells in a mole of Fe atoms and then the number of atoms in a mole. Step 2: Calculate the volume of one mole of Fe atoms. The ratio of the molar mass of Fe to its density gives the molar volume of Fe: molar volume of Fe 55.85 g/mol 7.874 g/cm3 7.093 cm3/mol Step 3: Use the results from Steps 1 and 2 to determine the number of unit cells in one mole of Fe. 7.093 cm3/mol number of unit cells in one mole of Fe 3.009 1023 unit cells/mol Step 4: Find the number of Fe atoms per mole — Avogadro’s number — by multiplying the answer in step 3 (the number of unit cells per mole) by the number of Fe atoms per unit cell (2): 3.009 1023 unit cells 1 mol 6.018 Avogadro’s number 1023 Fe atoms/mol 2 Fe atoms 1 unit cell The small difference between the above number and 6.022 1023 is the result of rounding off and using rounded-off values for density and other constants. X-ray diffraction and crystal structure are discussed in more detail in Chapter 11. 107 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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