a welding fuel gas contains carbon and hydrogen only .burning a small sample of it in oxygen gives,3.38g of carbon dioxide,0.690g og water and no other products.A volume of 10.0L (measured at STP)of this welding gas is found to weight 11.6g.calculate (1)empirical formula,(2)molar mass of gas,(3)molecular formula. (C=12u,H=1u)
Answers
Explanation:
welding of yogas content small simple oxygen gives 3.38 carbon dioxide and 0.64 water and other product of value of type of welding is the weight and control calculator formula and Vastu consultant formula C2 12 and H2 and 11 and 12 the volume of the 69 of water and carbon dioxide of the burning of simple simple
Answer:
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