• question_answer 42) A welding fuel gas contains carbon and hydrogen only burning a sample of it in oxygen gives 3.38 g carbon dioxide, 0.69 g of water and no other products. A volume of 10 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.

Molar mass of the gas = Mass of 22.4 L gas at STP $=\frac{11.6}{10}\times 22.4=26$ Let the molecular formula of the gas is ${{C}_{x}}{{H}_{y}}$. It undergoes combustion as, ${{C}_{x}}{{H}_{y}}+\left( x+\frac{y}{4} \right){{O}_{2}}\to xC{{O}_{2}}(g)+\frac{y}{2}{{H}_{2}}O$ Number of moles of carbon = Number of moles of $C{{O}_{2}}$ $=\frac{3.38}{44}=0.0768$ Number of moles of hydrogen atom = Number of moles of ${{H}_{2}}O\times 2$ $=2\times \frac{0.69}{18}=0.076$ $C:H=0.076:0.076$ $=1:1$ $\therefore$ Empirical formula = CH Molecular formula = ${{(CH)}_{n}}$ $n=\frac{\text{Molar}\,\text{mass}}{\text{Empirical}\,\text{formula}\,\text{mass}}\text{=}\frac{26}{13}=2$ Molecular formula of the gas $={{C}_{2}}{{H}_{2}}$