Chemistry, asked by Aakash55555, 8 months ago


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.

Answers

Answered by shubhangithakare47
3

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

2H2

Read this carefully.....Hope my answer will help u.

Answered by Anonymous
4

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

.

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