In an experiment 3.10g of carbon, hydrogen and oxygen produced 4.4g of C02 and 2.7g of water on complete combustion. Determine the empirical formula of the compound

( C= 12, 0= 16, H =1)

3 answers

Given compound with C, H, O => 4.4-g CO₂ + 2.7-g H₂O
Elemental %Composition of …
%C in CO₂ = (12/44)100% = 27.3% => Wt C from CO₂ = 27.3% of 4.4-g CO₂ = 1.206-g Carbon
%O in CO₂ = (32/44)100% = 72.7% => Wt O from CO₂= 72.7% of 4.4-g CO₂ = 3.199-g Oxygen
%O in H₂O = (16/18)100% = 89.0% => Wt O from H₂O = 89.0% of 2.7-g H₂O = 2.403-g Oxygen
%H in H₂O = (2/18)100% = 11.0% => Wt H from H₂O = 11.0% of 2.7-g H₂O = 0.300-g Hydrogen
=> Σ (masses of elements in compounds formed) = 7.103-g (which corresponds with Σ of compound masses given 4.4-g + 2.7-g).
Converting to %-per 100-wt …
%C = (1.206/7.103)100% = 16.9%
%H = (0.300/7.103)100% = 4.2%
%O =[(3.199-g + 2.403-g)/7.103-g]100% = 78.9%
%C = 16.9% =>16.9-g per 100wt/12-g∙moleˉ¹ = 1.408-mole
%H = 4.2% => 4.2-g per 100wt/1-g∙moleˉ¹ = 4.200-mole
%0 = 78.9% => 78.9-g per 100wt/16-g∙moleˉ¹ = 4.931-mole
%Σ = 100% => Σ = 100-g sample
Empirical Ratio is then reduced mole ratios by dividing by the smaller of the mole values.
C:H:O => (1.408/1.408):(4.200/1.408):(4.931/1.403) => 1:3:3.5 x 2 => 2:6:7
Empirical formula => C₂H₆O₇
Note: After converting given data to grams per 100wt, calculation follows the progression …
% per 100wt => grams per 100wt => moles each element => reduce mole values by dividing by smallest mole value => Empirical Ratio => Empirical Formula

Good attempt

Good approach