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Answer Key Chapter 6: Standard Review Worksheet 1

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Answer Key Chapter 6: Standard Review Worksheet 1 Answer Key Chapter 6: Standard Review Worksheet 1. 1 amu = 1.66 _ 10–24 g. For example, the average atomic mass of sodium is 22.99 amu, which represents the average mass of all the sodium atoms in the world (including all the various isotopes and their relative abundances). So that we will be able to use the mass of a sample of sodium to count the number of atoms of sodium present in the sample, we consider that every sodium atom in a sample has exactly the same mass (the average atomic mass). The average atomic mass of an element is typically not a whole number of amu’s because of the presence of the different isotopes of the element, each with its own relative abundance. Since the relative abundance of an element can be any number, when the weighted average atomic mass of the element is calculated, the average is unlikely to be a whole number. 2. On a microscopic basis, one mole of a substance represents Avogadro’s number (6.022 _ 1023) of individual units (atoms or molecules) of the substance. On a macroscopic basis, one mole of a substance represents the amount of substance present when the molar mass of the substance in grams is taken (for example, 12.01 g of carbon will be one mole of carbon). 3. The molar mass of a compound is the mass in grams of one mole of the compound (6.022 _ 1023 molecules of the compound) and is calculated by summing the average atomic masses of all the atoms present in a molecule of the compound. For example, a molecule of the compound H3PO4 contains three hydrogen atoms, one phosphorus atom, and four oxygen atoms. The molar mass is obtained by adding up the average atomic masses of these atoms: Molar mass H3PO4 = 3(1.008 g) + 1(30.97 g) + 4(16.00 g) = 97.99 g. 4. When a new compound is prepared (the formula is not known), the percent composition must be determined on an experimental basis. An elemental analysis must be done of a sample of the new compound to see what mass of each element is present in the sample. For example, if a 1.000-g sample of a hydrocarbon is analyzed, and it is found that the sample contains 0.7487 g of C, then the percent by mass of carbon present in the compound is 0.7487 g C 100% = 74.87% C 1.000-g samp Since we can use the formula of a known compound to calculate the percent composition by mass of the compound, it is not surprising that we can go in the opposite direction—from experimentally determined percent compositions for an unknown compound, we can calculate the formula of the compound. 5. a. K2CrO4 (molar mass = 194.2 g) Mol K CrO = 5.00 g _ 1 mol = 0.0257 mol K CrO 2 4 194.2 g 2 4 23 6.022 10 formula units 22 0.0257 mol _ = 1.55 _ 10 formula units K2CrO4 1 mol Atoms K = 1.55 _ 1022 formula units _ 2 K atoms = 3.10 _ 1022 atoms K 1 formula un 22 1 Cr atom 22 Atoms Cr = 1.55 _ 10 formula units = 1.55 _ 10 atoms Cr 1 formula un 22 4 O atoms 22 Atoms O = 1.55 _ 10 formula units = 6.20 _ 10 atoms O 1 formula un b. AuCl3 (molar mass = 303.4 g) Mol AuCl = 5.00 g _ 1 mol = 0.0165 mol AuCl 3 303.4 3 23 6.022 10 formula units 21 0.0165 mol AuCl3 _ = 9.94 _ 10 formula units AuCl3 1 mol Atoms Au = 9.94 _ 1021 formula units _ 1 Au atom = 9.94 _ 1021 atoms Au 1 formula un Atoms Cl = 9.94 _ 1021 formula units _ 3 Cl atoms = 2.98 _ 1022 atoms Cl 1 formula un c. SiH4 (molar mass = 32.12 g) Mol SiH = 5.00 g _ 1 mol = 0.156 mol SiH 4 32.12 4 23 6.022 10molecules 22 0.156 mol SiH4 _ = 9.39 _ 10 molecules SiH4 1 mol Atoms Si = 9.39 _ 1022 molecules _ 1 Si atom = 9.39 _ 1022 atoms Si 1 molecul Atoms H = 9.39 _ 1022 molecules _ 4 H atom = 3.76 _ 1023 atoms H 1 molecul d. Ca3(PO4)2 (molar mass = 310.18 g) Mol Ca (PO ) = 5.00 g _ 1 mol = 0.0161 mol Ca (PO ) 3 4 2 310.18 3 4 2 23 6.022 10 formula units 21 0.0161 mol Ca3(PO4)2 _ = 9.70 _ 10 formula units Ca3(PO4)2 1 mol Atoms Ca = 9.70 _ 1021 formula units _ 3 Ca atoms = 2.91 _ 1022 atoms Ca 1 formula un Atoms P = 9.70 _ 1021 formula units _ 2 P atoms = 1.94 _ 1022 atoms P 1 formula un Atoms O = 9.70 _ 1021 formula units _ 8 O atoms = 7.76 _ 1022 atoms O 1 formula un 6. a. K CrO : %K = 2(39.10 g K _ 100% = 40.27% K 2 4 194.2 g %Cr = 52.00 g C _ 100% = 26.78% Cr 194.2 g %O = 4(16.00 g O _ 100% = 32.96% O 194.2 g b. AuCl : %Au = 197.0 g A _ 100% = 64.93% Au 3 303.4 g %Cl = 3(35.45 g C _ 100% = 35.05% Cl 303.4 g c. SiH : %Si = 28.09 g S _ 100% = 87.45% Si 4 32.12 g %H = 4(1.008 g H _ 100% = 12.55% H 32.12 g d. Ca (PO ) : %Ca = 3(40.08 g Ca _ 100% = 38.76% Ca 3 4 2 310.18 g %P = 2(30.97 g P _ 100% = 19.97% P 310.18 g %O = 8(16.00 g O _ 100% = 41.27% O 310.18 g 7. The empirical formula of a compound represents the smallest ratio of the relative number of atoms of each type present in a molecule of the compound, whereas the molecular formula represents the actual number of atoms of each type present in a real molecule of the compound. Once the empirical formula of a compound has been determined, it is also necessary to determine the molar mass of the compound before the actual molecular formula can be calculated. 8. Assume 100.0 g. 70.0% Fe = 70.0 g Fe _ 1 mol = 1.25 mol Fe 55.85 30.0% O = 30.0 g O _ 1 mol = 1.88 mol O 16.00 g 1.88/1.25 = 1.50 = 3/2. Thus the empirical formula is Fe2O3. 9. Assume 100.0 g. 30.4% N = 30.4 g N _ 1 mol = 2.17 mol N 14.01 g 69.6% O = 69.6 g O _ 1 mol = 4.35 mol O 16.00 g 4.35/2.17 = 2. The empirical formula is NO2. The molar mass of NO2 is about 46 g/mol, which falls in the range of half the molar mass of the compound. The molecular formula must be N2O4..
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