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Unformatted text preview: Deflection of cathode rays by an applied electric field. Chapter 3 Atomic Mass, The Mole, Stoichiometry Isotopes Isotopes Isotopes are atoms of the same element with different numbers of neutrons. Most Most elements have atoms of more than one isotope. Isotopes Isotopes of an element are very similar and have nearly identical properties The The number of protons and electrons, which are the same in all isotopes of an element, has much more to do with the properties of an element. Isotopes Atoms of the same element (Z) but different mass number (A). Hydrogen: Hydrogen: Hydrogen: one electron, one proton Deuterium Deuterium one electron, one proton, one neutron Tritium: Tritium: one electron, one proton, two neutrons: looses one “beta particle”, or neutrons: looses one beta particle or electron. electron. Given 100 atoms, 50 will loose one beta particle within 12.3 years What’s special about Hydrognen isotopes? Different Different properties as the mass is different (1 amu vs 2 or 3 amu) amu) Deuterium Deuterium and tritium are important in fusion reactions reactions 2H + 3H 4He + neutron Sun’s Sun’s power “Future” “Future” nuclear power? Hydrogen bomb 1 Figures: (right) A scientist injecting a sample into a mass spectrometer. (below left) Schematic diagram of a mass spectrometer. (below right) Mass spectrograph of Neon Mass Number The mass The mass number (A) is the sum of the protons and the neutrons in the nucleus of an atom. mass mass number = # protons + # neutrons The The name for an isotope of an element is the the element name followed by the mass number. The The symbol for an isotope is its element symbol along with its mass number (A) and atomic number (Z). Mass Number (cont.) Consider Consider the isotope Phosphorus-31. Phosphorussymbol: 31 A = 31 (mass number = the sum of protons and (mass neutrons) 15 P Z = 15 (atomic number = protons ) (atomic In In order to determine how many neutrons are in the nucleus of an atom simply subtract: # neutrons neutrons = mass number – atomic number = 31 – 15 = 16 neutrons 16 Atomic Mass Atoms Atoms are very small. For instance, 18 mL of water contains 602,200,000,000,000,000,000,000 molecules of H2O. A single water molecule has a mass of 2.99 x 10-23 single grams. Because atoms are so small, it is difficult to Because atoms are so small, it is difficult to express express their mass in the standard units of grams. Instead, Instead, we typically express the mass of atom in terms of atomic mass units, or amu, or simply u. atomic The The atomic mass unit is defined as 1/12 of the mass of an atom of the carbon –12 isotope. Atomic Mass (cont.) 1 amu = 1/12 mass of C-12 amu CBoth Both the neutron and the proton have a mass of approximately 1 amu: amu: 1 amu ≈ 1 proton ≈ 1 neutron Electrons Electrons are very small: 1 electron ≈ 0.00055 amu Because Because the mass number is the sum of protons and neutrons, the mass number is an approximation of the atomic mass of an approximation isotope. 2 Average Average Atomic Mass Most Most elements are a mixture of two or more isotopes. The The percentage of an isotope in a naturally occurring sample of an element is called the isotopic isotopic abundance (or percentage percentage abundance) of that that isotope. The isotopic The isotopic mass is the mass of a single atom of an isotope. The average The average atomic mass is the weighted weighted average of the masses of isotopes of an element. Calculating Average Atomic Mass from Isotope Data Average Average Atomic mass = Σ atomic mass x abundance Consider the element Magnesium, composed of Magnesium three isotopes of the following percentage abundances: abundances: Isotope Mass MgMg-24 MgMg-25 MgMg-26 % Abundance Isotope (amu) amu) 79.0 % 10.0 % 11.0 % 23.985 24.986 25.982 Calculating Percent Abundances from Mass Data Consider the two isotopes of Boron: Isotope BoronBoron-10 BoronBoron-11 Isotope Mass (amu) amu) Counting Atoms Chemistry is a quantitative science— science—we need a “counting unit.” % Abundance Abundance 10.0129 11.0093 MOLE 1 mole is the amount of substance that contains as many particles (atoms, molecules) as there are in 12 12.0 g of 12C. Avogadro and the MOLE The Mole In In Chemistry, the mole is a convenient number to count a large quantity of particles. We We can talk about a mole of anything, but we usually use it to talk about atoms, molecules, atoms, ions, and formula units – Matter at the particle level. A mole of anything is 6.022 x 1023 of those items. mole particles 1 mole = 6.022 x 1023 particles mole 6.022 6.022 x 1023 is also called Avogadro’s number. What is a MOLE? A cute, burrowing animal? cute, An An overgrowth of epidermal tissue? A spy? sp A popular TV show? popular VWxNlg7N7go (look under krtek) krtek) It It is all these things and more… 3 Mole / Dozen Analogy Like Like the mole, a dozen of something is a convenient way to talk about the number of items we tend to buy in those quantities: 1 dozen donuts = 12 donuts 3 dozen eggs = 36 eggs The The mole and the dozen make it easier to talk about large quantities. The mole and counting particles We We can use Avogadro’s number to convert between particles and moles: moles: The The conversion factors are: 6.022 ×10 23 particles 1 mol or 1 mol 6.022 ×10 23 particles Mole – Particle Conversions 1) Convert 6.78 x 1024 atoms of Argon to moles. Atomic / Molar Masses We We express the masses of atoms and molecules in atomic mass units (amu). (amu). However, However, we rarely work with small numbers of atoms or molecules. Avogadro’s number relates the atomic mass unit and Avogadro number relates the atomic mass unit and the gram the gram: 6.022 x 1023 amu = 1.000 g amu 2) Convert 0.881 moles of Boron to atoms. Therefore: Therefore: 1 amu = 1 g/mol CarbonCarbon-12 has an atomic mass of 12 amu or a molar molar mass of 12 g/mol. 6.022 6.022 x 1023 Carbon-12 atoms will have a mass of 12 Carbong. Proceeding clockwise from the top samples containing one mole each of copper, aluminum, iron, sulfur, iodine, and (in the center) mercury. Mole – Mass Conversions 1) What is the mass of 3.11 mol of nickel atoms? 2) How many moles is 335 mg of aluminum? 4 Mass – Particle Conversions 1) How many atoms are in 1.000 grams of xenon? 2) What is the mass of 3.5 x 1022 atoms of gold? ld? 3) What is the mass of a single sodium-23 sodiumatom in grams? The isotopic mass of NaNa23 is 22.99 amu. amu. 1. Additional Additional Examples How many atoms of cobalt are in a cylinder of cobalt with a diameter of 2.00 cm that is 5.00 cm long? What is the density of metal if What is the density of a metal if 3.18 moles moles of of the metal atoms occupy 37.5 cm3? 2. Calculating Calculating Molar Masses for Compounds Simply Simply add together the atomic masses for all of the atoms in a molecule or formula unit of a compound. Calculating Percent Composition % Element = in compound Empirical and Molecular Formulas Empirical Formula - the smallest whole number ratio of atoms in a compound. Molecular Formula - the actual number of atoms in molecule (multiple of the atoms in a molecule (multiple of the empirical empirical formula). Example: Octane Molecular Molecular formula = C8H18 Empirical Empirical formula = C4H9 atomic mass of element x # atoms molar mass of the compound x 100% Calculate Calculate the percent composition of calcium calcium phosphate for all elements in the compound. Sample Problem: Empirical & Molecular Formulas What is the molecular formula of the sugar ribose if it is 40.0% C, 6.7% H, and 53.3% O and it has a molecular mass of 150.1 g/mol? First, determine the empirical formula: Step 1. Assume 100.0 g of the compound. 100 th If you have 100g of the material, you will have the following masses of C, H, & O: 40.0 g C 6.7 g H 53.3 g O Step 2. Convert grams to moles: 40.0 g C 1 mol C = 3.33 mol C 12.01 g C 6.7 g H 1 mol H mol 1.01 g H = 6.6 mol H mol 53.3 g O 1 mol O = 16.00 g O 3.33 mol O 5 Step 3. Find the smallest ratio of atoms. C 3.33 / 3.33 =1 H 6.6 / 3.33 = 1.98 ≈ 2 O 3.33 / 3.33 =1 Therefore the empirical formula of Therefore, the empirical formula of ribose ribose is Next, determine the molecular formula. Step 4. Calculate the empirical formula Calculate mass. C= 12.01 x 1 12 = 12.01 g/mol g/mol H= 1.01 x 2 = 2.02 g/mol O= 16.0 x 1 = 16.00 g/mol 30.03 g/mol CH2O Example: Empirical Formulas Step 5. Calculate the number of empirical formula units within the molecular formula. Multiply the subscripts of the empirical formula by this number to empirical formula by this number to get get the molecular formula. molecular 150.1 g/mol 30.03 g/mol = 4.998 ≈ 5 An oxide of osmium is a pale yellow solid. If 2.89 g of the compound contains 2.16g of osmium, what is its formula? C1x5H2x5O1x5 = C5H10O5 Example: Empirical & Molecular Formulas A 30.5-g sample of acrylic acid, used in 30.5the manufacture of acrylic plastics, is found to contain 15.25 g C, 1.71 g H, and 13.54 O. 13.54 g O. In In a separate experiment, the acrylic acid is found to have a molar mass of approximately 72 g/mol. What are the empirical and molecular formulas of acrylic acid? What is the mass percent of iron in iron(II) oxalate, FeC2O4? a. 14.29% b. 61.18% c. 32.07% d. 38.82% e. 81.17% 6 A compound contains 43.64% P with the remainder being oxygen. What is the empirical formula of the compound? a. a. PO5 b. P2O5 c. PO3 d. P2O3 b. c. d. e. e. P4O10 A molecule is found to contain 47.35% C, 10.60% H, and 42.05% O. What is the empirical formula for this molecule? a. a. C2H6O b. C2H6O2 c. C3H8O2 c. d. C3H6O3 e. C4H6O A 2.000 g sample of CoCl2·xH2O is dried in an oven. When the anhydrous salt is removed from the oven, its mass is 1.565 g. What is the value of x? a. a. 0.5 b. 1 c. 2 d. 4 e. 6 Combustion train for the determination of the chemical composition of organic compounds. CnHm + (n+ m ) O2 = n CO2(g) +m H2O(g) 2 2 The combustion of 0.1703 mole of an unknown hydrocarbon produces 12.28 g H2O and 29.98 g CO2. What is the molar mass of the hydrocarbon? a. a. 176.0 g/mol b. 248.2 g/mol c. 30.07 g/mol d. 72.11 g/mol 30 72 e. e. 56.11 g/mol Soft Soft drink bottles are made of polyethylene terephthalate (PET), a polymer composed of carbon, hydrogen, and oxygen. If 1.250 g PET is burned in oxygen it produces 0.4328 g H2O and 2.643 g CO2. What is the empirical formula for PET? PET? a. a. C8H10O b. C10H8O5 c. C5H7O c. d. CH7O5 e. CHO e. 7 Calculating Calculating Waters of Hydration Terephthalic Terephthalic acid, used in the production of polyester fibers and films, is composed of carbon, hydrogen, and oxygen. When 0.6943 g of terephthalic acid was subjected to combustion acid was subjected to combustion analysis analysis it produced 1.471 g CO2 and 0.226 g H2O. If its molar mass is between 158 and 167 g/mol, what is its molecular formula? A 10.00 – g sample sample of CoCl2 • x H2O 10.00 is heated. The mass of the resulting solid was found to be 5.46 g. How How many waters of hydration are in the formula unit? What What is the formula of the hydrate? Counting objects of fixed relative mass. Chapter 3 Stoichiometry 12 red marbles @ 7g each = 84g 12 yellow marbles @4g each=48g 55.85g Fe = 6.022 x 1023 atoms Fe 32.07g S = 6.022 x 1023 atoms S One mole of common substances. Oxygen, Calcium Carbonate, copper, water Information contained in the chemical formula of Glucose C6H12O6 MM= 180.16 g/mole Carbon (C) Atoms/molecule of compound Moles of atoms/ mole of compound Atoms/mole of compound Mass/molecule of compound Mass/mole of compound Hydrogen (H) Oxygen (O) 6 atoms 6 moles of atoms 6(6.022 x 1023) atoms 6(12.01 amu) =72.06 amu 72.06 g 12 atoms 12 moles of atoms 12(6.022 x 1023) atoms 12(1.008 amu) =12.10 amu 12.10 g 6 atoms 6 moles of atoms 6(6.022 x 1023) atoms 6(16.00 amu) =96.00 amu 96.00 g 8 Interconverting Moles, Mass, and Number of Chemical Entities Mass (g) = no. of moles x no. of grams 1 mol 1 mol no. of grams of grams No. of entities = no. of moles x 6.022x1023 entities 1 mol 1 mol 6.022x1023 entities g Quick talk on balancing Mainly Mainly covered in the next chapter No. of moles = mass (g) x M No. of moles = no. of entities x Example Translate the statement Balance the atoms Adjust the coefficients Check the atom balance Specify the state of matter 20.0 20.0 grams of Sulfur combine with an excess of fluorine to make SF6. What’s the mass of fluorine used and SF6 formed? Given Given S(s) + 3 F2(g) SF SF6(g) Balance the following AlCl AlCl3 + Na2SO4 NaHCO NaHCO3 + HCl Al + O2 → Al2O3 Al Al Al Al2(SO4)3 + NaCl CO CO2 + H2O + NaCl Balance the following equation: B2O3(s) + HF(l) → BF3 (g) + H2O(l) HF UO UO2 (s) + HF(l) → UF4(s) + H2O(l) HF C8H18O3 (l) + O2 (g) → H2O(g) + CO2 (g) CO 9 Mix balancing and stoichiometry Aluminum Aluminum will react with bromine to form aluminum bromide (used as an acid catalyst in organic synthesis). Al(s) + Br2(l) → Al2Br6(s) [unbalanced] Br [unbalanced] How many moles of Al are needed to form 2.43 mol of Al2Br6? Balancing and convert to moles Ammonia, Ammonia, an important source of fixed nitrogen that can be metabolized by plants, is produced using the Haber process in which nitrogen and hydrogen combine. ___N2(g) + ___H2(g) → ___NH3(g) How many grams of nitrogen are needed to produce 325 grams of ammonia? Figure 3.8 Summary of the mass-mole-number relationships in a chemical reaction. MASS(g) of compound B M (g/mol) of compound B AMOUNT(mol) of compound B Avogadro’s number (molecules/mol) MOLECULES (or formula units) of compound B Figure 3.10 An ice cream sundae analogy for limiting reactions. MASS(g) of compound A M (g/mol) of compound A AMOUNT(mol) of compound A molar ratio from ratio from balanced equation Avogadro’s number (molecules/mol) MOLECULES (or formula units) of compound A Consider the reaction of 10.0 g of zinc nitrate with excess with excess sodium phosphate in aqueous solution. What mass of zinc phosphate precipitate will be formed? Unbalanced: Unbalanced: Zn(NO Zn(NO3)2 (aq) + Na3PO4 (aq) 3Zn(NO3)2 (aq) +2 Na3PO4 (aq) (aq) The effect of side reactions on yield. A +B (reactants) C (main product) D (side products) Zn Zn3(PO4)2 (s)+ NaNO3 (aq) Start here: finish here Zn Zn3(PO4)2 (s)+6 NaNO3 Also consider that not all reactions go 100% forward 10 Limiting Reagent Questions: Limiting Limiting reagents: If you are given “too much” information or quantities of two or more “given”, you have to solve both ways and then use the smaller amount ways and then use the smaller amount to to get the answer. Solve Solve only far enough to get a common answer. Then take the smaller one forward Back Back tracking to get “excess” is then easier Aluminum oxide (used as an adsorbent or a catalyst for organic reactions) forms when aluminum reacts with oxygen. 4Al( 4Al(s) + 3O2(g) → 2Al2O3(s) A mixture of 82.49 g of aluminum ( = 26.98 g/mol) and 117.65 g of oxygen ( = 32.00 g/mol) is allowed to react. What mass of aluminum oxide ( = 101.96 g/mol) can be formed? Excess Reagent What What is the excess reagent for the reaction below given that you start with 10.0 g of Al and 19.0 grams of O2? How much excess? Reaction: Reaction: 4Al + 3O2 → 2Al2O3 Aluminum Aluminum reacts with oxygen to produce aluminum oxide which can be used as an adsorbent, desiccant or catalyst for organic reactions. 4Al( 4Al(s) + 3O2(g) → 2Al2O3(s) A mixture of 82.49 g of aluminum ( = 26.98 g/mol) and 117.65 g of oxygen ( = 32.00 g/mol) is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete. Percent Yield questions: Percent Percent yield = actual units/stoich units You You will be given a percent or an actual amount. You calculate the stoichiometric amount and plug it into the formula to get the other. Disulfur Disulfur dichloride can be made by reacting chlorine gas with molten sulfur. S8( ) + 4 Cl2(g) → 4 S2Cl2(g) Cl What What is the percent yield if 4.88 g S2Cl2 is isolated from the reaction of 10.0 g S8 and 6.00 g Cl2? a. a. 11.4% b. 16.9% c. 21.1% d. 23.2% e. 42.7% 11 Generation of NO The The reaction for the oxidation of NH3 is given as: 4 NH3 + 5 O2 → 4 NO + 6 H2O Under Under certain conditions the reaction will proceed at 29.8% yield of NO. How many grams of NH3 must be made to react with excess oxygen to yield 70.5 g of NO? Tetraphosphorus Tetraphosphorus hexaoxide ( = 219.9 g/mol) is formed by the reaction of phosphorus with oxygen gas. P4( P4(s) + 3O2(g) → P4O6(s) If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of P4O6, what is the percent yield for the reaction? Note: Note: Limiting reagent question, as well. 12 ...
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