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 product

Question

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.

Answer 1

CxHy + O₂ ==> CO₂ + H₂O

moles of C:  3.38 g CO2 x 1 mole CO2/44 g x 1 mole C/mole CO2 = 0.0768 moles C

moles of H:  0.690 g H2O x 1 mole H2O/18 g x 2 moles H/mole H2O = 0.0767 moles H

Since the mole ratio of C:H is 1:1, the empirical formula is CH

 

 

10.0 L x 1 mole/22.4 L = 0.446 moles

11.6 g/0.446 moles = 26.0 g/mole = molar mass of the gas

 

Molar mass of empirical formula = 12 + 1 = 13

26.0 g/13 g = 2

Molecular formula = C2H2 (acetylene)