Problem: Combustion of 29.52 g of a compound containing only carbon, hydrogen, and oxygen produces 34.17 g CO2 and 13.99 g H2O. What is the empirical formula of the compound?

FREE Expert Solution

Step 1. Determine the %C, %H, and %O.

mass C=34.17 g CO2(1 mol CO244.01 g CO2)(1 mol C1 mol CO2)(12.01 g C1 mol C)=9.32 g C

%C=9.32 g29.52 g×100=31.57%

mass H=13.99 g H2O (1 mol H2O18.016 g H2O)(2 mol H1 mol H2O)(1.008 g H1 mol H)=1.57 g H

%H=1.57 g29.52 g×100=5.32%

%O = 100 - (31.57 + 5.32) = 63.11%

Step 2. Determine mol C, H, and O.

Assume 100 g sample, therefore we have 31.57 g C, 5.32g H, and 63.11 g O.

mol C=31.57 g C(1 mol C12.01 g C)=2.63 mol C

mol H=5.32 g H(1 mol H1.008 g H)=5.28 mol H

mol O=63.11 g O(1 mol O16.00 g O)=3.94 mol O

Step 3. Determine the empirical formula.

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Problem Details

Combustion of 29.52 g of a compound containing only carbon, hydrogen, and oxygen produces 34.17 g CO2 and 13.99 g H2O. What is the empirical formula of the compound?

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