che106_lecture7

che106_lecture7 - Chemistry 106 Lecture 7 Topics ...

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Unformatted text preview: Chemistry 106 Lecture 7 Topics: Mass and Moles Lectures 6+7+8 = Chapter 3.1 – 3.7 Announcements Ø།  Reminder: Third Homework and Tutorial sets are due on the MasteringChemistry website on Friday, September 23rd. Ø།  The on- line homework is part of your overall course grade. It is NOT opRonal. Exam Reminder Ø། The First Exam will be on September 28th (during normal class <me and loca<on). Ø། Exam #1 will cover Chapters 1 through 3. Ø། All exams will be mulRple choice. Ø། A pracRce exam is posted on blackboard.syr.edu. Ø། If you have any special needs or issues (such as sports) you must inform me in wri<ng as soon as possible (Before September 21st). Stoichiometry: QuanRtaRve RelaRons in Chemical ReacRons •  Stoichiometry is the calculaRon of the quanRRes of reactants and products involved in a chemical reacRon. –  It is based on balanced chemical equaRons and on the relaRonship between mass and moles. –  Such calculaRons are fundamental to most quanRtaRve work in chemistry. Chemical ReacRons: EquaRons •  A chemical equa<on is the symbolic representaRon of a chemical reacRon in terms of chemical formulas. –  For example, the reacRon of sodium and chlorine to produce sodium chloride is wriZen: 2Na + Cl2 → 2NaCl •  The reactants are starRng substances in a chemical reacRon. The arrow means “yields”. •  The formulas on the right side of the arrow represent the products. Mass and Moles of a Substance •  In order to use these balanced equaRons, we need to relate molecules and mass. •  Chemistry requires a method for determining the numbers of molecules in a given mass of a substance. –  This allows the chemist to carry out “recipes” for compounds based on the relaRve numbers of atoms involved. –  These calculaRons are called stoichiometry. Molecular Weight •  The molecular weight of a substance is the sum of the atomic weights of all the atoms in a molecule of the substance. –  For, example, a molecule of H2O contains 2 hydrogen atoms (at 1.0 amu each) and 1 oxygen atom (16.0 amu), giving a molecular weight of 18.0 amu. Formula Weight •  The formula weight of a substance is the sum of the atomic weights of all the atoms in one formula unit of the compound, whether molecular or not. –  For example, one formula unit of NaCl contains 1 sodium atom (22.99 amu) and 1 chlorine atom (35.45 amu), giving a formula weight of 58.44 amu. CalculaRng the Formula Weight AW = atomic weight, FW = formula weight •  Calculate (with 4 significant digits) the formula weight of glucose (C6H12O6). •  FW = (6 x AW of C) + (12 x AW of H) + (6 x AW of O) AW of C = 12.0107 amu AW of H = 1.00794 amu AW of O = 15.9994 amu •  FW = 180.156 amu FW = 180.2 amu The Mole Concept •  A mole (symbol is “mol”) is defined as the quanRty of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 grams of carbon–12. •  The number of atoms in a 12- gram sample of carbon– 12 is called Avogadro’s number (to which we give the symbol NA). •  The value of Avogadro’s number is 6.02214179 x 1023 , but ocen rounded to 6.02 x 1023. The Mole Concept •  A mole of a substance contains Avogadro’s number (6.02 x 1023) of molecules. •  Same idea as using the terms “dozen” or “gross” –  dozen = 12 –  gross = 144 •  When using moles, it is ocen important to indicate the formula of the substance. –  A mole of oxygen atoms (O) has 6.02 x 1023 O atoms –  A mole of oxygen molecules (O2) has 1.20 x 1024 O atoms 1- octanol (C8H17OH) mercury (II) iodide (HgI2) One mole each of various substances. sulfur (S8) methanol (CH3OH) How many sulfur atoms are present in 1.0 mole of Al2(SO4)3? A.  1 sulfur atom B.  3 sulfur atoms C.  4 sulfur atoms D.  6.0 x 1023 sulfur atoms E.  1.8 x 1024 sulfur atoms How many sulfur atoms are present in 1.0 mole of Al2(SO4)3? A.  1 sulfur atom B.  3 sulfur atoms C.  4 sulfur atoms D.  6.0 x 1023 sulfur atoms E.  1.8 x 1024 sulfur atoms Molar Mass •  The molar mass of a substance is the mass of one mole of a substance. –  For all substances, molar mass, in grams per mole, is numerically equal to the formula weight in atomic mass units. –  That is, one mole of any element weighs its atomic mass in grams. 1 mole of C = 12.0107 grams 1 mole of Cl = 35.453 grams 1 mole of Ca = 40.078 grams Mole CalculaRons •  ConverRng the number of moles of a given substance into its mass, and vice versa, is fundamental to understanding the quanRtaRve nature of chemical equaRons. mass of X moles of X = atomic (or molecular) mass of X Mole CalculaRons •  Suppose we have 100.0 grams of iron (Fe). The atomic weight of iron is 55.8 g/mol. How many moles of iron does this represent? 100.0 g Fe 1 mol Fe moles Fe = × 55.8 g Fe = 1.79 moles of Fe Mole CalculaRons •  Conversely, suppose we have 5.75 moles of magnesium (atomic wt. = 24.3 g/mol). What is its mass? 5.75 mol Mg 24.3 g Mg mass Mg = × 1 mol Mg = 140 grams of Mg Mole CalculaRons •  This same method applies to compounds. Suppose we have 100.0 grams of H2O (molecular weight = 18.0 g/mol). How many moles does this represent? 100.0 g H 2O 1 mol H 2O moles H 2O = × 18.0 g H 2O = 5.56 moles of H 2O Mole CalculaRons •  Conversely, suppose we have 3.25 moles of glucose, C6H12O6 (molecular wt. = 180.2 g/mol). What is its mass? 3.25 mol C6 H12O6 180.2 g C6 H12O6 mass C6 H12O6 = × 1 mol C6 H12O6 = 586 grams of C6 H12O6 Mass and Moles and Number of Molecules or Atoms •  The number of molecules or atoms in a sample is related to the moles of the substance: 1 mole HCl = 6.02 × 10 23 HCl molecules 1 mole Fe = 6.02 × 10 23 Fe atoms Suppose we have a 3.46-g sample of hydrogen chloride, HCl. How many molecules of HCl does this represent? 1 mole HCl 6.02 x 10 23 HCl molecules 3.46g HCl × 36.5g HCl × 1 mole HCl = 5.71×1022 HCl molecules The formula weight of Na3PO4 is: A.  B.  C.  D.  70 grams/mole 164 grams/mole 265 grams/mole 116 grams/mole From the Periodic Table Na = 22.989770 amu P = 30.973761 amu O = 15.9994 amu The formula weight of Na3PO4 is: A.  B.  C.  D.  70 grams/mole 164 grams/mole 265 grams/mole 116 grams/mole From the Periodic Table Na = 22.989770 amu P = 30.973761 amu O = 15.9994 amu Determining Chemical Formulas •  The percent composi<on of a compound is the mass percentage of each element in the compound. – We define the mass percentage of “X” as the parts of “X” per hundred parts of the total, by mass. That is, mass of X in whole mass % X = × 100% mass of the whole Mass Percentages from Formulas •  Let’s calculate the percent composiRon of butane, C4H10. First, we need the molecular mass of C4H10. 4 carbons @ 12.0 amu/atom = 48.0 amu 10 hydrogens @ 1.00 amu/atom = 10.0 amu 1 molecule of C4 H10 = 58.0 amu Now, we can calculate the percentages. 4 amu C % C = 58.8.0mu total × 100% = 82.8%C 0a 1 amu H % H = 58.0.0 mu total × 100% = 17.2%H 0a CombusRon Method for Determining the Percentages of Carbon and Hydrogen in a Compound If you have an unknown compound (unknown formula), elemental analysis is an experiment that can tell you the percentage composition. Sample + O2 → CO2 + H2O 100 g butane 17.2 g H 82.8 g C Determining Chemical Formulas •  Determining the formula of a compound from the percent composiRon. – The percent composiRon of a compound leads directly to its empirical formula. – An empirical formula (or simplest formula) for a compound is the formula of the substance wriZen with the smallest integer (whole number) subscripts. Determining Chemical Formulas •  Determining the empirical formula from the percent composiRon. –  Benzoic acid is a white, crystalline powder used as a food preservaRve. The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. What is its empirical formula? –  In other words, give the smallest whole- number raRo of the subscripts in the formula: Cx HyOz Determining the Empirical Formula from the Percent ComposiRon •  For the purposes of this calculaRon, we will assume we have 100.0 grams of benzoic acid. Remember: The compound contains 68.8% C, 5.0% H, and 26.2% O by mass. •  Then the mass of each element equals the numerical value of the percentage. •  Since x, y, and z in our formula represent mole- mole raRos, we must first convert these masses to moles. Cx HyOz Determining the Empirical Formula from the Percent ComposiRon •  Our 100.0 grams of benzoic acid would contain: C5.733H5.000O1.637 1 mol C 68.8 g C × = 5.733 mol C 12.0 g 1 mol H 5.0 g H × = 5.000 mol H 1.0 g 1 mol O 26.2 g O × = 1.637 mol O 16.0 g This isn’t quite a whole number ratio, but if we divide each number by the smallest of the three, the correct ratio will be found. Determining the Empirical Formula from the Percent ComposiRon •  Our 100.0 grams of benzoic acid would contain: 5.73 mol C ÷1.637 = 3.50 5.0 mol H ÷1.637 = 3.0 1.637 mol O ÷1.637 = 1.00 Now you can see that the smallest whole number raRo is: 7:6:2 The empirical formula is: C7H6O2 Determining the Molecular Formula from the Empirical Formula •  An empirical formula gives only the smallest whole- number raRo of atoms in a formula. •  The molecular formula should be a mulRple of the empirical formula (since both have the same percent composiRon). •  To determine the molecular formula, we must know the molecular weight of the compound. Molecular Formula vs. Empirical Formula Determining the Molecular Formula from the Empirical Formula •  For example, suppose the empirical formula of a compound is CH2O and its molecular weight is 60.0 g/mol. •  The molar weight of the empirical formula (the empirical weight) is only 30.0 g/mol. molecular weight Number of empirical formula units (n) = empirical formula weight •  This indicates that the molecular formula is actually the empirical formula doubled (n=2), or: C2H4O2 Next Lecture •  Chap 3 conRnues ...
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