A compound is composed of only C, H, and O. The combustion of a 0.519-g sample of the compound yields 1.24 g of CO2 and 0.255 g of H2O. What is the empirical formula of the compound

C4H4O

Explanation:

1.24 g x 1mol/(44)g = 0.03mol C;

Mass of C: 0.03mol x 12.g/1mol

= 0.36g of C

mol of H: 0.255g x 1mol /(18) x2= 0.028mol

Mass of H; = 0.028g × 1 of H

Mass of O: 0.519 g – (0.36g +0.028g)= 0.131g;

0.131g/16 =0.008 mol

Divide by the smallest number of moles;

0.028/0.008 mol = 3.5; 0.03/0.008 mol =3.75, 0.008 mol/0.008 mol =1

C4H4O

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