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 = () 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 ⎛⎞ ⎜⎟ ⎝⎠ 3.9 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 22 2 23 3 2 2 1 mol NO 46.01 g NO 1kg 3.8x10 molecules NO 1molNO 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 = 24 2 1molC 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 1molSrH 2 mol H 6.022 x 10 H ions 4.92 g SrH 89.64 g SrH 1 mol SrH 1molH −− = 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. G r a m s C r 2 (SO 4 ) 3 •10H 2 O = 24 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. G r a m s C l 2 O 7 = 24 27 23 182.9 g Cl O 1mol 9.64 x 10 molecules Cl O 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. M o l L i 2 SO 4 = 1molLi SO 56.2 g Li SO 109.95 g Li SO = 0.5111414 = 0.511 mol Li 2 SO 4 F U L i 2 SO 4 = 23 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 2L i ions 1FULi 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 1SO ion = 3.07809 x 10 23 = 3.08 x 10 23 SO 4 2– ions 3.07809 x 10 23 FU Li 2 SO 4 1Satom = 3.07809 x 10 23 = 3.08 x 10 23 S atoms 3.07809 x 10 23 FU Li 2 SO 4 4Oa toms = 1.231237 x 10 24 = 1.23 x 10 24 O atoms
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3-2 3.18 Plan: Determine the formula and the molar mass of each compound. The formula gives the number of atoms of each type of element present. Masses come from the periodic table. Solution: a) Ammonium bicarbonate is an ionic compound consisting of ammonium ions, NH 4 + and bicarbonate ions, HCO 3 . The formula of the compound is NH 4 HCO 3 . M of NH 4 HCO 3 = (14.01 g/mol) + (5 x 1.008 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol) = 79.06 g/mol NH 4 HCO 3 In 1 mole of ammonium bicarbonate, with a mass of 79.06 g, there are 5 H atoms with a mass of 5.040 g. 43 (5 mol H) (1.008 g/mol H) x 100% 79.06 g/mol NH HCO = 6.374905 = 6.375% H b) Sodium dihydrogen phosphate heptahydrate is a salt that consists of sodium ions, Na + , dihydrogen phosphate ions, H 2 PO 4 , and seven waters of hydration. The formula is NaH 2 PO 4 7H 2 O. Note that the waters of hydration are included in the molar mass.
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This note was uploaded on 03/18/2010 for the course CHEM 210 taught by Professor Mcomber during the Spring '10 term at Skyline College.

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chapter3hw - CHAPTER 3 STOICHIOMETRY OF FORMULAS AND...

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