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Home » 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. Find: (i) Empirical formula (ii) Molar mass of the gas, and (iii) Molecular formula
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. Find: (i) Empirical formula (ii) Molar mass of the gas, and (iii) Molecular formula
CBSE | Chemistry | Class 11 | NCERT | Some Basic Concepts of Chemistry
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. Find: (i) Empirical formula (ii) Molar mass of the gas, and (iii) Molecular formula

(i) Empirical formula

1 mole of CO_{ 2 } contains 12 g of carbon

Therefore, 3.38 g of CO_{ 2 } will contain carbon

= \frac{ 12 \; g }{ 44 \; g } \; \times 3.38 \; g

= 0.9217 g

 

18 g of water contains 2 g of hydrogen

Therefore, 0.690 g of water will contain hydrogen

= \frac{ 2 \; g }{ 18 \; g } \; \times 0.690

= 0.0767 g

 

As hydrogen and carbon are the only elements of the compound. Now, the total mass is:

= 0.9217 g + 0.0767 g

= 0.9984 g

 

Therefore, % of C in the compound

= \frac{ 0.9217 \; g }{ 0.9984 \; g } \; \times 100

= 92.32 %

% of H in the compound

= \frac{ 0.0767 \; g }{ 0.9984 \; g } \; \times 100

= 7.68 %

 

Moles of C in the compound,

= \frac{ 92.32 }{ 12.00 }

= 7.69

 

Moles of H in the compound,

= 7.68

 

Therefore, the ratio of carbon to hydrogen is,

7.69: 7.68

1: 1

 

Therefore, the empirical formula is CH.

 

(ii) Molar mass of the gas, and

Weight of 10 L of gas at STP = 11.6 g

Therefore, weight of 22.4 L of gas at STP

= \frac{ 11.6 \; g }{ 10 \; L } \; \times \; 22.4 \; L

= 25.984 g

\approx26 g

 

(iii) Molecular formula

Empirical formula mass:

CH = 12 + 1

= 13 g

n = \frac{ Molar \; mass \; of \; gas}{Empirical \; formula \; mass \; of \; gas}

= \frac{ 26 \; g }{ 13 \; g}

= 2

Therefore, molecular formula is (CH)_{ n } that is C_{ 2 }H_{ 2 }

 

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