on complete combustion on oxygen of 0.42grams of gaseous hydrocarbon x,it gave 1.32grams of carbon dioxide and 0.54grams of water determine the empirial formula

Question

On complete combustion in oxygen of 0.42grams of a gaseous hydrocarbon x,it gave 1.32grams of carbon dioxide and 0.54grams of water .Determine the empirial formula

Answer 1

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Moles of CO2 = 1.32 g / 44.01 g/mol = 0.03 mol

Moles of H2O = 0.54 g / 18.02 g/mol = 0.03 mol

Moles of C = 0.03 mol

Moles of H = 0.5 * 0.03 mol = 0.015 mol

Mass of C = 0.03 mol * 12.01 g/mol = 0.36 g

Mass of H = 0.015 mol * 1.01 g/mol = 0.015 g

To simplify the ratio, we divide both masses by the smaller mass (0.015 g):

Mass of C / 0.015 g = 0.36 g / 0.015 g = 24

Mass of H / 0.015 g = 0.015 g / 0.015 g = 1

The empirical formula of the hydrocarbon X is CH₄.

Answer 2

C2H2

Answer 3

Answer 

Answer 4

CH