Combustion of 1.000 g of an organic compound known to contain only carbon, hydrogen, and oxygen produces 2.360 g of carbon dioxide and 0.640 g of water. What is the empirical formula of the compound?

The empirical formula is C3H4O

Explanation:

Step 1: Data given

Mass of the compound = 1.000 grams

The compound contains:

- Carbon

- hydrogen

- oxygen

The combustion of this compound gives:

2.360 grams of CO2

0.640 grams of H2O

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 2.360 grams / 44.01 g/mol

Moles CO2 = 0.05362 moles

In CO2 we have 1 mol

This means for 1 mol CO2 we have 1 mol C

For 0.05362 moles CO2 we have 0.05362 moles C

We have 0.05362 moles of C in the compound

Step 3: Calculate mass of C

Mass C = moles C * molar mass C

Mass C = 0.05362 moles * 12.0 g/mol

Mass C = 0.643 grams

Step 4: Calculate moles of H2O

Moles H2O = 0.640 grams / 18.02 g/mol

Moles H2O = 0.0355 moles H2O

For 1 mol H2O we have 2 moles of H

For 0.0355 moles H2O we have 2*0.0355 = 0.071 moles H

Step 5: Calculate mass of H

Mass H = moles H * molar mass H

Mass H = 0.071 moles * 1.01 g/mol

Mass H = 0.072 grams

Step 6: Calculate mass of O

Mass of O = Mass of compound - mass of C - mass of H

Mass of O = 1.000 g - 0.643 - 0.072 = 0.285 grams

Step 7: Calculate moles of O

Moles O = 0.285 grams / 16.0 g/mol

Moles O = 0.0178 moles

Step 8: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.05362 / 0.0178 = 3

H: 0.071 / 0.0178 = 4

O: 0.0178/0.0178 = 1

The empirical formula is C3H4O