Question

A welding fuel gas contains and hydrogen only. Burning a small sample of it in oxygen gives $3.38g$ 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

Answer 1

Let the molecular formula be $C_x{H}_y$

Step I - Estimation of Empirical formula

$C_x{H}_y+(2x+\frac{y}{2})O_2⟶xC{O}_2+\frac{y}{2}{H}_{2}O$

Moles of $C{O}_2=\frac{3.38}{44}=0.0768$

Moles of $H_{2}O=\frac{0.690}{18}=0.0384$

$⟹\frac{0.0768}{x}=\frac{0.0384}{\frac{y}{2}}$

$⟹x=y$

$1.$ Empirical Formula: $CH$

Step II - $2.$ Moles of gas at STP

$⟹\frac{10}{22.4}=0.4464$ $moles$

Let molecular weight be $W$

$\frac{11.6}{W}=0.4464$
$W=26$ $g/mol$

Step III - Molecular Formula

$EmpiricalformulamassofCH=13grams$

$n=\frac{molarmassofgas}{Empiricalformulamassofgas}$

$n=\frac{26}{13}=2$

$Thereforemolecularformulaofgas={\left(CH\right)}_2$

$Molecularformula=C_2{H}_2$