Question

Upon combustion, a compound containing only carbon and hydrogen produced 0.660 g CO2 and 0.270 g H2O.

Find the empirical formula of the compound.

Answer 1

Hi,

Following formulas are used to get the percentage of Carbon, Hydrogen and Oxygen. Remember, only in compound with C, H and O does this method work.

% of Carbon= (Mass of CO2 / Mass of organic compound) x (12/44) x 100
% of Hydrogen= (Mass of H2O/Mass of organic compound) x (2.016/18) x 100
% of Oxygen= 100 - (% of Carbon) - (% of Hydrogen)

Solution:-

Mass of Organic Compound= 0.290g
Mass of CO2= 0.660g
Mass of H2O= 0.270

% of Carbon= (0.660/0.290) x (12/44) x 100= 62.1%
% of Hydrogen= (0.270/0.290) x (2.016/18) x 100= 10.4%
% of Oxygen= 100 - 62.1 - 10.4= 27.5%

To find the No. of moles for each element, we divide their percentages with their respective Mr.

Moles of Carbon = 62.1/12=5.175
Moles of Hydrogen= 10.4/1= 10.4
Moles of Oxygen= 27.5/16= 1.719

To find their Atomic ratios we divide all moles by the smallest no. of moles, which is in this case Oxygen's 1.719. So,

Atomic Ratio:-
Atomic Ratio of Carbon= 5.175/1.719= 3
Atomic Ratio of Hydrogen= 10.4/1.719= 6
Atomic ratio of Oxygen= 1.719/1.719= 1

Empirical formula: C3H6O