Complete combustion of 7.80 g of a hydrocarbon produced 25.1 g of CO2 and 8.55 g of H2O. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary

The empirical formula is C3H5

Explanation:

Step 1: Data given

Mass of the compound = 7.80 grams

Mass of CO2 = 25.1 grams

Molar mass of CO2 = 44.01 g/mol

Mass of H2O = 8.55 grams

Molar mass of H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 25.1 grams / 44.01 g/mol

Moles CO2 = 0.570 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.570 moles CO2 we have 0.570 moles C

Step 4: Calculate mass C

Mass C = 0.570 moles * 12.01 g/mol

Mass C = 6.846 grams

Step 5: Calculate moles H2O

Moles H2O = 8.55 grams / 18.02 g/mol

Moles H2O = 0.474 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.474 moles H2O we have 2*0.474 = 0.948 moles H

Step 7: Calculate mass H

Mass H = 0.948 moles * 1.01 g/mol

Mass H = 0.957 grams

Step 8: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.570 moles / 0.570 = 1

H: 0.948 moles / 0.570 = 1.66

This means for 1 mol C we have 1.66 moles H   OR for 3 moles C we have 5 moles H

The empirical formula is C3H5