34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it
in oxygen gives 3.38 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
Calculate (1) empirical formula. (11) molar mass of the gas, and (iii) molecular
formula
Answers
ANSWER
Since, 3.38 g of carbon dioxide are obtained, the mass of C present in the sample of welding fuel gas is
44
12
×3.38=0.92 g.
Since, 0.690 g of water are obtained, the mass of hydrogen present in the welding fuel gas sample is
18
2
×0.690=0.077g.
The percentage of C is
0.92+0.077
0.92
×100=92.3 %.
The percentage of H is
0.92+0.077
0.077
×100=7.7%.
(i) The number of moles of carbon =
12
92.2
=7.7.
The number of moles of hydrogen =
1
7.7
=7.7.
The mole ratio C:H=
7.7
7.7
=1:1.
Hence, the empirical formula of the welding fuel gas is CH.
(ii) The weight of 10.0 L of gas at N.T.P is 11.6 g.
22.4 L pf gas at N.T. P will weigh
10.0
11.6
×22.4=26g/mol
This is the molar mass.
(iii) Empirical formula mass is 12+1=13.
Molecular mass is 26.
The ratio, n of the molecular mass to empirical formula mass is
13
26
=2.
The molecular formula is 2(CH)=C
2
H
2
.
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