Question

Combustion Analysis of an unknown hydrocarbon resulted in the capture of 216.00 g of water vapor and 440.00 g of CO2. The total mass of the hydrocarbon before combustion was 144 g.

(1. How many moles of carbon dioxide are present after combustion?)

(2. What is the empirical formula for the unknown hydrocarbon?)

(3. How many water molecules does the water vapor trap capture?)

(4. The gram-formula mass for the unknown hydrocarbon was determined to be 72 g/mol. How many moles of the unknown hydrocarbon were present in the original sample?)

(5. infrared spectral analysis determines that the unknown hydrocarbon does NOT contain oxygen. How many moles of O2 are used in combusting the unknown sample?)

Answer 1

1: 10 mol CO2

2: C10H24

3: 12 mol H2O

4: 0.5 mol

5: 16 mol O2

Answer 2

The compound is C5H10.

From the information  available in the question;

Number of moles of carbon dioxide = 440.00 g/44 g/mol = 10 moles

Number of moles of water vapor = 216.00 g/18 g/mol = 12 moles

Mass of carbon = 10 moles × 12 g/mol = 120 g of carbon

Moles of hydrogen = 12 moles  × 2 g/mol = 24 g of hydrogen

Number of moles of carbon = 120 g/12 g/mol = 10 moles

Number of moles of hydrogen = 24 g/1 g/mol =24 moles

Divide through by the lowest number of moles =

C - 10/10   H - 24/10

C - 1          H - 2

The empirical formula is CH2

Molar mass of the compound = 72 g/mol

[12 + 2(1)]n = 72

14n = 72

n = 5

The compound is C5H10.

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