Combustion analysis was run on a 0.80391 gram sample of an unknown compound containing only C, H, and O. The results indicate that 2.007 grams of CO2 and 0.9856 grams of H2O were produced. What is the empirical formula of the compound

The empirical formula of the compound is C5H12O

Explanation:

Step 1: Data given

Mass of a sample = 0.80391 grams

Mass of CO2 = 2.007 grams

Molar mass CO2 = 44.01 g/mol

Mass of H2O = 0.9856 grams

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles CO2

Moles = mass / molar mass

Moles CO2 = 2.007 grams / 44.01 g/mol

Moles CO2 = 0.0456 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.0456 moles CO2 we have 0.0456 moles C

Step 4: Calculate mass C

Mass C = 0.0456 moles * 12.01 g/mol

Mass C = 0.548 grams

Step 5: Calculate moles H2O

Moles H2O = 0.9856 grams / 18.02 g/mol

Moles H2O = 0.0547 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.0547 moles H2O we have 2* 0.0547 = 0.1094 moles H

Step 7: Calculate mass H

Mass H = 0.1094 moles * 1.01 g/mol

Mass H = 0.1105 grams

Step 8: Calculate mass O

Mass O = 0.80391 - 0.548 - 0.1105 = 0.14541 grams

Step 9: Calculate moles O

Moles O = 0.14541 grams / 16.0 g/mol

Moles O = 0.00909 moles

Step 10 : Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.0456 moles / 0.00909 moles =  5

H: 0.1094 moles / 0.00909 moles = 12

O: 0.00909 moles / 0.00909 = 1

This means for 1 O atom we have 5 C atoms and 12 H atoms

The empirical formula of the compound is C5H12O