ISM_Chapter_03

ISM_Chapter_03 - CHAPTER 3: STOICHIOMETRY: RATIOS OF...

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CHAPTER 3: STOICHIOMETRY: RATIOS OF COMBINATION CHAPTER 3 STOICHIOMETRY: RATIOS OF COMBINATION 3.3 Strategy : add the masses of all atoms in the formula. Remember that the absence of a subscript means that there is one atom of that element present. (a) CH 3 Cl 1(12.01 amu) + 3(1.01 amu) + 1(35.45 amu) = 50.49 amu (b) N 2 O 4 2(14.01 amu) + 4(16.00 amu) = 92.01 amu (c) SO 2 1(32.07 amu) + 2(16.00 amu) = 64.07 amu (d) C 6 H 12 6(12.01 amu) + 12(1.008 amu) = 84.16 amu (e) H 2 O 2 2(1.008 amu) + 2(16.00 amu) = 34.02 amu (f) C 12 H 22 O 11 12(12.01 amu) + 22(1.008 amu) + 11(16.00 amu) = 342.3 amu ( g ) NH 3 1(14.01 amu) + 3(1.008 amu) = 17.03 amu 3.4 Using the appropriate atomic masses, (a) C 6 H 6 O 6(12.01 amu) + 6(1.008 amu) + 1(16.00 amu) = 94.11 amu (b) H 2 SO 4 2(1.008 amu) + 1(32.07 amu) + 4(16.00 amu) = 98.07 amu (c) C 6 H 6 6(12.01 amu) + 6(1.008 amu) = 78.11 amu (d) C 6 H 12 O 6 6(12.01 amu) + 12(1.008 amu) + 6(16.00 amu) = 180.2 amu (e) BCl 3 1(10.81 amu) + 3(35.45 amu) = 117.2 amu (f) N 2 O 5 2(14.01 amu) + 5(16.00 amu) = 108.0 amu ( g ) H 3 PO 4 3(1.008 amu) + 1(30.97 amu) + 4(16.00 amu) = 97.99 amu 3.5 Using the appropriate atomic masses, (a) CH 4 1(12.01 amu) + 4(1.008 amu) = 16.04 amu (b) NO 2 1(14.01 amu) + 2(16.00 amu) = 46.01 amu (c) SO 3 1(32.07 amu) + 3(16.00 amu) = 80.07 amu (d) C 6 H 6 6(12.01 amu) + 6(1.008 amu) = 78.11 amu (e) NaI 1(22.99 amu) + 1(126.9 amu) = 149.9 amu (f) K 2 SO 4 2(39.10 amu) + 1(32.07 amu) + 4(16.00 amu) = 174.27 amu (g) Ca 3 (PO 4 ) 2 3(40.08 amu) + 2(30.97 amu) + 8(16.00 amu) = 310.18 amu 3.6 Using the appropriate atomic masses, (a) Li 2 CO 3 1(6.941 amu) + 1(12.01 amu) + 3(16.00 amu) = 73.89 amu (b) C 2 H 6 2(12.01 amu) + 6(1.008 amu) = 30.07 amu (c) NF 2 1(14.01 amu) + 2(19.00 amu) = 52.01 amu (d) Al 2 O 3 2(26.98 amu) + 3(16.00 amu) = 101.96 amu (e) Fe(NO 3 ) 3 1(55.85 amu) + 3(14.01 amu) + 9(16.00 amu) = 241.9 amu (f) PCl 5 1(30.97 amu) + 5(35.45 amu) = 208.2 amu (g) Mg 3 N 2 3(24.31 amu) + 2(14.01 amu) = 100.95 amu 3.9 Strategy: Recall the procedure for calculating a percentage. Assume that we have 1 mole of SnO 2 . The percent by mass of each element (Sn and O) is given by the mass of that element in 1 mole of SnO 2 40
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CHAPTER 3: STOICHIOMETRY: RATIOS OF COMBINATION divided by the molar mass of SnO 2 , then multiplied by 100 to convert from a fractional number to a percentage. Solution: Molar mass of SnO 2 = (118.7 g) + 2(16.00 g) = 150.7 g 118.7 g/mol 100% 150.7 g/mol = = %Sn 78.77% (2)(16.00 g/mol) 100% 150.7 g/mol = = %O 21.23% Check: Do the percentages add to 100%? The sum of the percentages is (78.77% + 21.23% 29 = 100.00%. 3.10 The molar mass of CHCl 3 = 12.01 g/mol + 1.008 g/mol + 3(35.45 g/mol) = 119.4 g/mol. The percent by mass of each of the elements in CHCl 3 is calculated as follows: 12.01 g/mol %C 100% 119.4 g/mol = = 10.06% 1.008 g/mol %H 100% 119.4 g/mol = = 0.8442% 3(35.45) g/mol %Cl 100% 119.4 g/mol = = 89.07% 3.11 Compound Molar mass (g) N% by mass (a) (NH 2 ) 2 CO 60.06 2(14.01 g)
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This note was uploaded on 09/16/2009 for the course CHEM 1001 taught by Professor Yi during the Fall '09 term at Marquette.

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ISM_Chapter_03 - CHAPTER 3: STOICHIOMETRY: RATIOS OF...

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