Empirical and Molecular Formulae?

Question

An organic compound X which contains carbon, hydrogen and oxygen only has an Mr of 85. When 0.43g of X are burned in excess oxygen, 1.10g of carbon dioxide and 0.45g of water are formed. Find the empirical and molecular formulae of compound?

Answer 1

Moles CO2 = 1.10 g / 44.0098 g/mol =0.0250

= moles C in the compound

Mass C = 0.0250 mol x 12.011 g/mol =0.300 g

Moles water = 0.45 g / 18.02 g/mol =0.0250

Moles H = 2 x 0.0250 = 0.0500

Mass H = 0.0500 mol x 1.008 g/mol =0.0504 g

Mass O = 0.43 - ( 0.300 + 0.0504) =0.0796 g

Moles O = 0.0796 / 15.9994 g/mol =0.00498

C 0.0250 H 0.0500 O 0.00498

we divide by the smallest number :

0.0250 / 0.00498 = 5 = C

0.0500 / 0.00498 = 10 => H

0.00498 / 0.00498 = 1 => O

C5H10 O is the empirical formula ( molar mass = 86 g/mol)

this is also the molecular formula

Answer 2

Let's calculate moles of C, H, and O in the products.

1.10 g CO2 x (1 mole CO2 / 44.0 g CO2) x (1 mole C / 1 mole CO2) = 0.0250 moles C

0.45 g H2O x (1 mole H2O / 18.0 g H2O) x (2 moles H / 1 mole H2O) = 0.0250 moles H

Oxygen comes from both CO2 and H2O:

1.10 g CO2 x (1 mole CO2 / 44.0 g CO2) x (2 moles O / 1 mole CO2) = 0.0500 moles O from CO2

0.45 g H2O x (1 mole H2O / 18.0 g H2O) x (1 mole O / 1 mole H2O) = 0.0250 moles O from H2O

Total moles O = 0.0500 + 0.0250 = 0.0750 moles

But O in the products comes from both compound X AND the O2 in which it was combusted. How much O came from the O2?

The total mass of products = 1.10 g CO2 + 0.45 g H2O = 1.55 g total. By the Law of Conservation of Mass, this must equal the mass of products. Since there was 0.43 g of X, then there must have been 1.55 - 0.43 = 1.12 g O2.

1.12 g O2 x (1 mole O2 / 32.0 g O2) x (2 moles O / 1 mole O2) = 0.0700 moles O from O2

So the O2 from X is total moles O - 0.0700 = 0.0750 - 0.0700 = 0.0050 moles O

The C:H:O ratio in X is 0.0250:0.0250:0.0050. Dividing by the smallest (0.0050) gives a ratio of 5:5:1. The empirical formula is C5H5O. This empirical formula has a formula weight of

(5 x 12) + (5 x 1) + (1 x 16) = 81, which is close to the experimental MW of 85. So C5H5O is also the molecular formula.