Chapter 3 Solutions

Chapter 3 Solutions - CHAPTER 3 STOICHIOMETRY OF FORMULAS...

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3-1 CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS 3.1 Cl 35.45 amu 35.45 g/mol Cl Mass Cl = (3 mol Cl) x (35.45 g Cl/l mol Cl) = 106.4 g Cl Al 26.98 amu 26.98 g/mol Al Mass Al = (2 mol Al) x (26.98 g Al/l mol Al) = 53.96 g Al 3.2 Plan: The formulas are based on the mole ratios of the constituents. Avogadro’s number allows the change from moles to atoms. Solution: a) Moles of C atoms = () 12 22 11 12 mol C 1 mol Sucrose 1molC H O    = 12 mol C b) C atoms = 23 12 22 11 12 22 11 12 mol C 6.022 x 10 C atoms 1molC = 7.226 x 10 24 C atoms 3.3 “1 mol of nitrogen” could be interpreted as a mole of nitrogen atoms or a mole of nitrogen molecules. Specify which to avoid confusion. The same problem is possible with other diatomic or polyatomic molecules, e.g., Cl 2 , Br 2 , S 8 , and P 4 . For these elements, as for nitrogen, it is not clear if atoms or molecules are being discussed. 3.4 The molecular mass is the sum of the atomic masses of the atoms or ions in a molecule. The molar mass is the mass of 1 mole of a chemical entity. Both will have the same numeric value for a given chemical substance but molecular mass will have the units of amu and molar mass will have the units of g/mol. 3.5 The mole maintains the same mass relationship between macroscopic samples as exist between individual chemical entities. It relates the number of chemical entities (atoms, molecules, ions, electrons) to the mass. 3.6 P 4 molecules = 34 2 44 2 2 4 1molCa (PO ) 1 mol P Avogadro's Number P molecules 2mo lP 2.5 g Ca (PO ) Molar Mass 1 mol Ca (PO ) 4 mol P 1 mol P 3.7 Plan: It is possible to relate the relative atomic masses by counting the number of atoms. Solution: a) The element on the left (green) has the higher molar mass because only 5 green balls are necessary to counterbalance the mass of 6 yellow balls. Since the green ball is heavier, its atomic mass is larger, and therefore its molar mass is larger. b) The element on the left (red) has more atoms per gram. This figure requires more thought because the number of red and blue balls is unequal and their masses are unequal. If each pan contained 3 balls, then the red balls would be lighter. The presence of six red balls means that they are that much lighter. Because the red ball is lighter, more red atoms are required to make 1 gram. c) The element on the left (orange) has fewer atoms per gram. The orange balls are heavier, and it takes fewer orange balls to make 1 gram. d) Both the left and right elements have the same number of atoms per mole. The number of atoms per mole (6.022 x 10 23 ) is constant and so is the same for every element. 3.8 Plan: Locate each of the elements on the periodic table and record its atomic mass. The mass of the element times the number present in the formula gives the molar mass.
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3-2 Solution: a) The molar mass, M , is the sum of the atomic weights, expressed in g/mol: Sr = = 87.62 g Sr/mol Sr(OH) 2 O = (2 mol O) (16.00 g O/mol O) = 32.00 g O/mol Sr(OH) 2 H = (2 mol H) (1.008 g H/mol H) = 2.016 g H/mol Sr(OH) 2 = 121.64 g/mol of Sr(OH) 2 b) M = (2 mol N) (14.01 g N/mol N) + (1 mol O) (16.00 g O/mol O) = 44.02 g/mol of N 2 O c) M = (1 mol Na) (22.99 g Na/mol Na) + (1 mol Cl) (35.45 g Cl/mol Cl) + (3 mol O) (16.00 g O/mol O) = 106.44 g/mol of NaClO
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Chapter 3 Solutions - CHAPTER 3 STOICHIOMETRY OF FORMULAS...

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