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Question 34

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate

 

(i) empirical formula,

 

(ii) molar mass of the gas, and

 

(iii) molecular formula.

Answer

(i) percentage of C can be calculated as follows:

CO2 = C

i.e 44 parts of CO2= 12 parts of C

OR

44g of CO2 = 12 g of C

 

Therefore according to question

3.38 g of CO2 contains C = 12/44 * 3.38 = 0.921 g

18 g of water contains hydrogen = 2g

 

Therefore 0.690 g of water contains hydrogen = 2/18 * 0.690 = 0.0767g

0.690 g of water will contain hydrogen

= 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

= 0.9217 g + 0.0767 g = 0.9984 g

Now percentage of carbon = weight of carbon/weight of compound * 100

=0.921 /0.998 * 100= 92.32 %

Also percentage of hydrogen = weight of hydrogen/weight of compound *100

=0.0766/0.998 * 100 = 7.68 %

 

(ii) Given,

Weight of 10.0L of the gas (at S.T.P) = 11.6 g

Weight of 22.4 L of gas at STP

= 25.984 g

≈ 26 g

Hence, the molar mass of the gas is 26 g.

 

(iii) empirical formula

 

Element Percentage Atomic mass Atomic ratio Simplest ratio Simplest whole no ratio
C 92.32 12 92.32/12 = 7.69 7.69/7.65 = 1.00 1
H 7.65 1 7.65/1 = 7.65 7.65/7.65 = 1 1

 

Empirical formula of the compound  = CH

Now molecular formula  calculation

Empirical formula mass = 12 + 1 = 13 amu

Also molecular mass  = 26 g (calculated in previous step)

Therefore n = molecular mass/empirical formula mass = 26/13 = 2

Now molecular formula = n x empirical formula = 2 x CH = C2H2

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