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) its empirical formula, (ii) molar mass of the gas and (iii) its molecular formula.

Answer 1

(i) Mass of carbon in 3.38 g of CO2

= 3.38 g/44 x 12 = 0.922 g

Mass of hydrogen in 0.690 g of H2O

= 0.690 g/18 x 2 = 0.077 g

Total mass of the sample burnt = 0.922 g + 0.077 g = 0.999 g

Percentage of carbon in the fuel = {0.922}/{0.999 g} x 100 = 92.29%

Percentage of hydrogen in the fuel = {0.077 g}/{0.999 g} x 100 = 7.71%

| Element | MassPercent | Atomicmass | Relative numberof atoms | Simpleatomic ratio |
| Carbon(C) | 92.29 | 12.0 | 92.29/12.0 = 7.69 | 7.69/7.69 = 1 |
| Hydrogen(H) | 7.71 | 1.0 | 7.71/1.0 = 7.71 | 7.71/7.69 = 1 |

Therefore, empirical formula of the compound = CH

(ii) Volume of the gaseous fuel = 11.6 g

Molar mass of the fuel = {11.6 g}/{10.0 L} x 22.4 L/ml = 26.0 g mol-1

(iii) Empirical formula mass of the fuel = (12 + 1) g mol-1 = 13 g mol-1

Molar mass of the fuel = 26.0 g mol-1

n = {26.0 g mol-1}/{13 g mol-1} = 2

Molecular formula of the fuel = 2 x Empirical formula

= 2 x CH = C2H2