Question

Complete combustion of 4.90 g of a hydrocarbon produced 15.7 g of CO2 and 5.61 g of H2O. What is the empirical formula for the hydrocarbon?

Answer 1

The carbon in the produced CO2 and the hydrogen in the produced H2O is from the hydrocabon.

Molar mass of  CO2 = 44 g/mol

Mass of carbon in 44g of = 12g

Therefore, mass of carbon in 15.7 g of  CO2= (12/44)x 15.7 g =4.282 g

Mass of carbon in given sample of hydrocarbon=4.282 g

No. of moles of carbon in given sample of hydrocarbon=4.282 g / 12g mol^-1=0.357 mol

[molar mass of carbon= 12g mol^-1]

Molar mass of  H2O= 18g/mol

Mass of hydrogen in 18g of  H2O= 2 g

Mass of hydrogen in 5.61g of  H2O= (2 /18) x 5.61 g

=0.6233 g

No. of moles of hydrogen in given sample of hydrocarbon= 0.6233 g/1g mol^-1= 0.6233 mol

   [molar mass of hydrogen= 1g mol^-1]

	Carbon	Hydrogen
No. of moles	0.357 mole	0.6233 moles
No. of atoms in each element in empirical formula	0.357/0.357 =1	0.6233/0.357 =1.74 $\approx$ 2

Therefore empirical formula of the compound will be CH₂.