A welding fuel gas contains carbon and hydrogen only. Burning a small space of it in oxygen gives 3.38 g carbon dioxide, 0·690g of water and no other products. A volume of 10·0L (measured at STP) of this welding gas is found to weigh 11·6g. Calculate: (i) empirical formula (ii) molar mass of the gas and (iii) molecular formula.

i.1mole(44g) of CO₂ contains 12g of carbon.

3.38 g of CO₂ will contain carbon  = (12g/44g) x3.38g =0.9217g

18g of water contains 2g of hydrogen

Therefore 0.690g of water will contain hydrogen=2g/18gx 0.690 =0.0767 g

Since carbon and hydrogen are the only constituents of the compounds, the total mass of the compounds is:

=0.9217+0.767

=0.9984g

Thus,

Percent of C in the compound = (0.9217/.9984) x100 =92.32%

Percent of H in the compound =(0.0767/0.9984) x100=7.68%

Moles of carbon in the compound =92.32/12.00 =7.69

Moles of hydrogen in the compound= 7.68/1=7.68

Since, ration of carbon to hydrogen in the compound=7.69:7.68=1:1

Hence, the empirical formula of the gas is CH.

ii) Given,

Weight of 10.0L of the gas (at S.T.P)=11.6 g

 Therefore, weight of 22.4L of gas at STP  =(11.6/10.0L) x22.4L

     =25.984g

Hence, the molar mass of the gas is 26.0 g.

iii) Empirical formula mass of CH =12+1 =13g

n= molar mass of gas/empirical formula mass of gas

n=26/13

n=2

Therefore, molecular formula of gas= (CH)_n­

=C₂H₂