Question

Upon combustion, a 0. 8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1. 6004 g co2 and 0. 6551 g h2o. Find the empirical formula of the compound.

Answer 1

Consider a generic combustion reaction,

$\mathrm{C}_x\mathrm{H}_y\mathrm{O}_z + \mathrm{O}_2 \implies \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2$

Convert the given masses of products to moles:

• (1.6005 g CO₂) (1/44.009 mol/g) ≈ 0.03637 mol CO₂

• (0.6551 g H₂O) (1/18.015 mol/g) ≈ 0.03636 mol H₂O

I'll just take both of these amounts to be 0.03636 mol.

Now, 1 mole of CO₂ contains 1 mole of C and 2 moles of O; similarly, 1 mole of H₂O contains 2 moles of H and 1 mole of O. So among the products, we have

• (0.03636 mol C) (12.011 g/mol) ≈ 0.43672 g C

• (0.07272 mol H) (1.008 g/mol) ≈ 0.0733 g H

so that the remaining mass in the sample consists of

• 0.8009 g - (0.43672 g C) - (0.0733 g H) ≈ 0.29088 g O

So, the sample contains

• 0.03636 mol C

• 0.07272 mol H

• (0.29088 g O) (1/15.999 mol/g) ≈ 0.01818 mol O

so that C, H, and O occur in a ratio of

0.03636 : 0.07272 : 0.01818 = 2 : 4 : 1

moles, which means the empirical formula for the compound is C₂H₄O.