Question
Sun November 27, 2011 By:

# 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.

Wed November 30, 2011

(1) Calculation of mass percentage of different elements:

Calculation of percentage of carbon in the compound:-

1 mole of CO2 contains 12 gm of carbon

Or,   44 gm of CO2 contains 12 gm of carbon

35.2 gm of CO2 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 H2O contains 2 gm of hydrogen

Or,  18 gm of CO2 contains 2 gm of hydrogen

14.4 gm of CO2 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 = CH2O

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

= C2H4O2

(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

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