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.

Answer:

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 \[={{C}_{2}}{{H}_{2}}\]