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NCERT Class XI Chemistry Some Basic Concepts of Chemistry Solutions

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Question : 34 of 36
Marks: +1, -0
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.
Solution:  
Number of moles of CO2\mathrm{CO}_2 = 3.3844\frac{3.38}{44} = 0.0768 mole
No. of moles of C = 0.0768 mole
No. of moles of H2O\mathrm{H_2O} = 0.69018\frac{0.690}{18} = 0.0383 mole
No. of moles of H = 2 × 0.0383 = 0.0766 mole
(i) The ratio of moles of C to H is 0.0768 : 0.0766 or 1 : 1
Therefore, empirical formula = CH
(ii) 10.0 L of fuel gas at STP weighs 11.6×22.410\frac{11.6 \times 22.4}{10} = 25.98 g
Molar mass of gas = 25.98 ≈ 26 g mol−1\mathrm{g\,mol}^{-1}
(iii) n = Molar massEmperical formula mass\frac{\text{Molar mass}}{\text{Emperical formula mass}} = 2313\frac{23}{13} = 2
Molecular formula = (Empirical formula)n = (CH)2\mathrm{(CH)_2} = C2H2\mathrm{C_2H_2}
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