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chapter3hw

# chapter3hw - CHAPTER 3 STOICHIOMETRY OF FORMULAS AND...

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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 3.5 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 = ( ) 3 4 2 4 4 3 4 2 3 4 2 4 1 mol Ca (PO ) 1 mol P Avogadro's Number P molecules 2 mol P 2.5 g Ca (PO ) Molar Mass 1 mol Ca (PO ) 4 mol P 1 mol P ⎠⎝ 3.9 a) (NH 4 ) 3 PO 4 = 3(14.01) + 12(1.008) + 30.97 + 4(16.00) = 149.10 g/mol b) CH 2 Cl 2 = 12.01 + 2(1.008) + 2(35.45) = 84.93 g/mol c) CuSO 4 ·5H 2 O = 63.55 + 32.07 + 9(16.00) + 10(1.008) = 249.70 g/mol d) BrF 5 = 79.90 + 5(19.00) = 174.90 g/mol 3.13 a) Mass NO 2 = ( ) 20 2 2 2 23 3 2 2 1 mol NO 46.01 g NO 1 kg 3.8x10 molecules NO 1 mol NO 6.022 x 10 molecules NO 10 g ⎠⎝ = 2.9033 x 10 –5 = 2.9 x 10 –5 kg NO 2 b) Moles Cl atoms = ( ) 2 4 2 2 4 2 2 4 2 2 4 2 1 mol C H Cl 2 mol Cl atoms 0.0425 g C H Cl 98.95 g C H Cl 1 mol C H Cl ⎞⎛ ⎟⎜ ⎠⎝ = 8.5902 x 10 –4 = 8.59 x 10 –4 mol Cl atoms c) Number of H = ( ) 23 2 2 2 2 1 mol SrH 2 mol H 6.022 x 10 H ions 4.92 g SrH 89.64 g SrH 1 mol SrH 1 mol H ⎞⎛ ⎟⎜ ⎟⎜ ⎠⎝ ⎠⎝ = 6.610495 x 10 22 = 6.61 x 10 22 H ions 3.17 a) The formula is Cr 2 (SO 4 ) 3 •10H 2 O, and the molar mass is 572.4 g/mol. Grams Cr 2 (SO 4 ) 3 •10H 2 O = ( ) 2 4 3 2 572.4 g 3.52 mol Cr (SO ) • 10H O mol = 2014.848 = 2.01 x 10 3 g Cr 2 (SO 4 ) 3 •10H 2 O b) The formula is Cl 2 O 7 , and the molar mass is 182.9 g/mol. Grams Cl 2 O 7 = ( ) 24 2 7 2 7 23 182.9 g Cl O 1 mol 9.64 x 10 molecules Cl O 1 mol 6.022 x 10 molecules ⎟⎜ = 2927.858 = 2.93 x 10 3 g Cl 2 O 7 c) The formula is Li 2 SO 4 , and the molar mass is 109.95 g/mol. Mol Li 2 SO 4 = ( ) 2 4 2 4 2 4 1 mol Li SO 56.2 g Li SO 109.95 g Li SO = 0.5111414 = 0.511 mol Li 2 SO 4 FU Li 2 SO 4 = ( ) 23 2 4 2 4 2 4 2 4 1 mol Li SO 6.022 x 10 FU 56.2 g Li SO 109.95 g Li SO 1 mol Li SO ⎠⎝ = 3.07809 x 10 23 = 3.08 x 10 23 FU Li 2 SO 4 d) Note the unrounded initially calculated FU value is used to avoid intermediate rounding. 3.07809 x 10 23 FU Li 2 SO 4 2 4 2 Li ions 1 FU Li SO + = 6.156187 x 10 23 = 6.16 x 10 23 Li + ions 3.07809 x 10 23 FU Li 2 SO 4 2 4 2 4 1 SO ion 1 FU Li SO = 3.07809 x 10 23 = 3.08 x 10 23 SO 4 2– ions 3.07809 x 10 23 FU Li 2 SO 4 2 4 1 S atom 1 FU Li SO = 3.07809 x 10 23 = 3.08 x 10 23 S atoms 3.07809 x 10 23 FU Li 2 SO 4 2 4 4 O atoms 1 FU Li SO = 1.231237 x 10 24 = 1.23 x 10 24 O atoms

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