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

(iii) molecular formula.

Given:

Mass of Carbon = 3.38g

Volume of welding gas = 10 L

Mass of 10 L of welding gas = 11.6g

Finding Percentage of Carbon:

44 parts of CO₂→12 parts of C

OR

44g of CO₂→12 g of C

Therefore, according to the question

3.38 g of CO₂→ X g of C

X = [12× 3.38] /44

=0.921g

18 g of water → 2g of hydrogen

So, 0.690 g of water → X g of hydrogen

X = [2× 0.690]/18

=0.0767g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is obtained as follows:

=0.9217 g + 0.0767 g = 0.9984 g

Percentage of carbon = [100×weight of carbon]/weight of compound

= [0.921× 100]/0.998

= 92.32%

Also percentage of hydrogen =[100×weight of hydrogen]/weight of compound

=[0.0766× 100]/0.998

=7.68%

(i) Finding Empirical Formula:

Empirical formula of the compound = CH

Therefore, the Empirical Formula of the compound is CH

(ii) Weight of 22.4 L of gas of STP = [11.6g×22.4L]/10.0L

= 25.985g

≈ 26 g

Therefore, the molecular mass of the gas is 26 g.

(iii) Finding Molecular formula:

Empirical formula mass = 12+1 = 13g

Also molecular mass =26 g [as obtained in step (ii)]

n = [Molecular Mass] / [Empirical Formula Mass]

=26/13

=2

Now molecular formula = n ×Empirical Formula = 2×CH = C₂H₂

Therefore, the molecular formula of the compound is C₂H₂.