A 25.0g sample of an unknown substance was burned in excess oxygen to produce 47.8 grams of carbon dioxide and 29.3 grams of water. What is the empirical formula?

Question

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Answer 1

It doesn’t say if the unknown contains C, H and O but one will assume it does.

moles C: 47.8 g CO2 x 1 mol CO2/44 g x 1 mol C/mol CO2 = 1.086

moles H: 29.3 g H2O x 1 mole/18g x 2 mol H/mol H2O = 3.26

To find moles O, we need grams of O. We find this as follows:

1.086 mol C x 12 g/mole = 13 g C

3.26 moles H x 1 g/mole = 3.26 g H

grams O = 25 g - 13 g - 3.3 g = 8.7 g O

moles O = 8.7 g x 1 mol/16g = 0.54 moles O

Summary:

1.1 moles C

3.3 moles H

0.54 moles O

Dividing all by 0.54.....

C = 1.1/0.54 = 2

H = 3.3/0.54 = 6

O = 0.54/0.54 = 1

Empirical formula is C₂H₆O