1.34. 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 S.T.P.) 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.
1.34. 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 S.T.P.) 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.
1.34. (i) 1 mole (44 g) of CO2 will have 12 g carbon.
So, 3.38 g of CO2 will have carbon = 12g/44g * 3.38
= 0.9217 g
18 g of water will have 2 g of hydrogen.
So, 0.690 g of water contain hydrogen = 2g/18g * 0.6902g
= 0.0767 g
Since carbon and hydrogen are the only constituents of the compound, the total
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Chemistry Ncert Solutions Class 11th 2023
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