A welding fuel gas contain carbon and hydrogen only. burning a small sample of it in oxygen gives 3.38g carbon dioxide, 0.690 g of water and no other products .A volume of 10 litre ( measured at S T P ) of this welding gas is found to weigh 11.6 g. calculate (1) emperical formula (2) molar mass of gas, and (3) molecular formula.
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Explanation:
3.38 g of CO₂ is obtained.
∴ Mass of C =12/44*3.38g = 0.92 g
Similarly, Mass of H = 1/44*3.38= 0.077 g
∴ Percentage of C = 0.92/0.92+0.077 = 92.3 %
and Percentage of H = 100-93.3= 7.7 %
(i) Empirical Formula:-
No. of moles of C = 92.2/12 = 7.7
No. of moles of H = 7.7/1= 7.7
∴ C : H = 7.7 : 7.7 = 1
Hence, Empirical Formula = (CH)₁ = CH (Ans.)
(ii) The weight of 10 L of gas at STP = 11.6 g
∴ 22.4 L of gas at STP = 11.6/10*22.4 g mol⁻¹ = 26 g mol⁻¹ (Ans.)
(iii) Let us suppose the molecular formula of the compound (CH)
∴ (CH)n = 26
or, (12 + 1) n= 26
or, = n = 2
Hence, Molecular formula = C₂H₂
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