Combustion of 28.30 g of a compound containing only carbon, hydrogen, and oxygen produces 32.75 gco2 and 13.41 gh2o. What is the empirical formula of the compound?

The empirical formula is C₂H₄O₃.

We must calculate the masses of C, H, and O from the masses given.

Mass of C = 32.75 g CO₂ * (12.01 g C/44.01 g CO₂) = 8.9372 g C

Mass of H = 13.4 g H₂O * (2.016 H/18.02 g H₂O) = 1.5067 g H

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

= (28.30 - 8.9372 - 1.5067) g = 17.856 g

Now, we must convert these masses to moles and find their ratios. If the ratios are not integers, we multiply them by a number to make them close to integers.

From here on, I like to summarize the calculations in a table.

Element m/g n/mol Ratio * 2 Integers

C 8.9372 0.74415 1 2 2

H 1.5067 1.4948 2.0086 4.0174 4

O 17.856 1.1160 1.4997 2.9994 3

The empirical formula is C₂H₄O₃.