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On combustion of 24g of a compound of carbon,oxygen and hydrogen gave 35.2g of CO2 and 14.4g of H2O. The molar mass of the compound was found to be 60.
(1) Calculation of mass percentage of different elements:
Calculation of percentage of carbon in the compound:
1 mole of CO_{2} contains 12 gm of carbon
Or, 44 gm of CO_{2} contains 12 gm of carbon
35.2 gm of CO_{2 }contains = (12 x 35.2)/44 gm of carbon
Percentage of carbon = (Weight of carbon x 100) / Weight of compound
= (12 x 35.2 x 100) /(44 x 24) = 40%
1 mole of H_{2}O contains 2 gm of hydrogen
Or, 18 gm of CO_{2} contains 2 gm of hydrogen
14.4 gm of CO_{2 }contains = (2 x 14.4)/18 gm of hydrogen
Percentage of hydrogen = (Weight of carbon x 100) / Weight of compound
= (2 x 14.4x 100) /(18 x 24) = 6.67%
Percentage of oxygen = [100 (40+6.67)] = 53.33%
Elements 
Percentage 
Atomic mass 
No. of moles 
Mole ratio 
C 
40 
12 
40/12=3.33 
3.33/3.33 =1 
H 
6.67 
1 
6.67/1=6.67 
6.67/3.33=2 
O 
53.33 
16 
53.33/16=3.33 
3.33/3.33=1 
Therefore, Empirical formula = CH_{2}O
Empirical formula mass = 12+2+16 =30
Given, molar mass of the compound =60
n = Molar mass / Empirical formula mass
= 60 /30 = 2
Molecular formula = n x Empirical formula
= C_{2}H_{4}O_{2}
(2)
The mass of carbon in 24g of the compound is = (12 x 35.2) /44 =9.6 gm
The mass of hydrogen in 24g of the compound is = (2 x 14.4) /18 = 1.6 gm
The mass of oxygen in 24g of the compound is = 24 Â– 11.2 =12.8 gm