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CHEMISTRY. Mole Practice Test. % Composition. K 2 CO 3 mm = 2(39.1)+12.0 + 3(16.0) 138 au three sig figures %K = At. Mass K x Sub X 100% 1 pt mm = 39.1 x 2 x 100% = 56.7% 2 pt 138
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CHEMISTRY Mole Practice Test
% Composition • K2CO3 mm = 2(39.1)+12.0 + 3(16.0) • 138 au three sig figures • %K = At. Mass K x Sub X 100% 1 pt • mm • = 39.1 x 2 x 100% = 56.7% 2 pt • 138 • %C = 12.0 x 1 x 100% = 8.69% 3 pt • 138 • %O = 16.0 x 3 x 100% = 34.8% 3 pt • 138
Number 2 and 3 • Mass of K = % K x mass of sample • 100 • = 56.7% x 205.0g = 116 g • 100 • 3-a) Na2S b) GaBr3 c) CsNO2 • 2(23.0)+32.0 69.7+3(79.9) 133 + 14.0+2(16.0) • 78.0 au 309au 179 au
Problem 4 • 23.5 g Be x 1.00 mol Be = 2.61 mol Be • 9.01 g Be • 0.125 g Ar x 1.00 mol Ar = 0.00313 mol Ar • 39.9 g Ar • 12.6 L O2 x 1,00 mol O2 = 0.563 mol O2 • 22.4 L O2 • 1.25 x 1024molec Cl2 x 1.00 mol Cl2 = 2.08 mol Cl2 • 6.02 x 1023 molec Cl2
Problem 5-The Chart • 8.64 g NO3 x 1.00 mol = 0.139 mol x 6.02x1023 molec • 62.0 g 1.00 mol • = 8.40 x 1022molec x 4 atoms = 3.35 x 1023 atoms • 1.00 molec • 0.358 mol MgO x 40.3 g = 14.4 g, 0.358 mol x 6.02x1023 molec • 1.00 mol 1.00 mol • = 2.16 x 1023molec x 2 atoms = 4.32 x 1023 atoms • 1.00 molec • 55.9 g CCl4, 0.389 mol, 1.17 x 1024 atoms • 5.44 g C4H10 0.0936 mol, 5.64 x 1022 molec
Empirical Formula # 6 • Cu 33.88 g x 1.00 mol = 0.534 mol/0.534 = 1 • 63.5 g • N 14.94 g x 1.00 mol = 1.07 mol/0.534 = 2 • 14.0 g • O 51.18 g x 1.00 mol = 3.20 mol/0.534 = 6 • 16.0 g • CuN2O6 = Cu(NO3)2
Molecular Formula # 7 • Ca 2.461 g x 1.00 mol = 0.0614 mol/0.0614 = 1 • 40.1 g • Cl 4.353 g x 1.00 mol = 0.123 mol/0.0614 = 2 • 35.5 g • Empirical Formula = CaCl2 • Molecular Formula: • ( CaCl2 ) x = 110 amu • 40.1 = 2(35.5) x = 110 amu • 110 x = 110 • x = 1 CaCl2