Question

0.262g of a substance gave, on combustion, 0.361g of CO, and 0.147g of H.O. What is the empirical formulaof the substance(EAMCET - 1996)| 1) CHÀO2) C,H,O3) C,H,O 4) CHOmini​

Answer 1

The Empirical Formula is CH2O

1) First calculate % of C, H AND O by using following formula

% C = [12/44][mass of carbon dioxide/mass of organic compound] * 100 = (12/44) * ( 0.361/0.262) *100 = 37.578 %.

% of H =[2/18][mass of water/mass of organic compound]*100 = (2/18)*(0.147/0.262) *100 = 6.234 %

% of O = 100 - (37.578 + 6.234 ) = 56.188 %

2) Calculate the moles of each substance from the given data :

moles of C = 37.578 /12 = 3.13

moles of H = 6.234/1  = 6.23

moles of O = 56.188/16 = 3.51

3) Divide all three by the lowest number of three

C = 3.13/3.13 = 1

H = 6.23/3.13 = 2

O = 3.51/3.13 = 1

4) The Empirical Formula is CH2O