Question

An unknown compound contains only C, H, and O. Combustion of 4.90 g of this compound produced 11.5 g of CO2 and 3.15 g of H2O. what is the empirical formula?

Answer 1

no of moles of CO2 = weight of CO2 / molar mass of CO2 = 11.5 g / 44 g/mol = 0.2614 moles

from each mole of CO2 will have one mole of C

so 0.2614 moles will have 0.2614 mole of Carbon

weight of the C = moles x molar mass = 0.2614 x 12.0107 = 3.14 grams

now percentage of mass = weight of C / total mass of C = 3.14/4.90 x 100 = 64%

similarly no of moles of H2O = weight of H2O / molar mass of H2O = 3.15 g / 18.0153 g/mole = 0.1748 moles

one mole of H2O will have 2 moles of H

so 0.1748 moles will have 2x 0.1748 = 0.3497 moles H

weight of H = moles x molar mass of H = 0.3497 x 1.0079 = 0.3524 grams

percentage of H = 0.3524 / 4.90 x 100 = 7.2 %

if you consider total % = 100 which is the combined % of C, H and O

64 + 7.2 + % of Oxyge = 100

% of Oxygen = 100 - 71.2 = 28.8 %

now C, H, O % = 64 % : 7.2% : 28.8 %

divide every thing with molar mass

64/ 12.0107 : 7.2 / 1.0079 : 28.8 / 16

5.33 : 7.14 : 1.86

divide every thing with leas no which is 1.86

5.33 / 1.86 : 7.14 / 1.86 : 1.86/1.86

2.86 : 3.84 : 1

3: 4:1

C3H4O is the empirical formula