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.

Expert Reply
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 – 11.2 =12.8 gm

Thu March 23, 2017

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