Determination of the molecular formula by the percentage of elements and the molar mass

Exercise 2

   

A substance has a molar mass of $M=183.18\frac{g}{mol}$. It contains $45.90\% \; C$.   $26.20\% \; O$.    $2.75\% \; H$.   $17.50\% \; S$. and $7.65\% \; N$. Find its molecular formula.

One mole of the substance contains: $0.4590\cdot 183.18=\; 84.1g \; C$ = $\frac{84.1}{12.0}= 7\; mol C$ $0.2620\cdot 183.18=\; 48.0 g \; O$ = $\frac{48.0}{16.0} = 3\; mol O$ $0.0275\cdot 183.18=\; 5.0 g \; H$ = $\frac{5.0}{1.0} = 5\; mol H$ $0.175\cdot 183.18=\; 32.0 g \; S$ = $\frac{32.0}{32.0} = 1\; mol S$ $0.0765\cdot 183.18=14.0\; g \; N$ = $\frac{14.0}{14.0} = 1\; mol N$ Molecular formula: $C_7H_5O_3NS$