Question

an automobile antifreeze consists of 38.7% C, 9.7% H and remaining oxygen by weight. When 0.93g of it are vaporized at STP 336 ml of vapour are formed. Find the molecular formula of the antifreeze.​

Answer 1

Given:

% weight of carbon = 38.7 %

% weight of hydrogen = 9.7 %

% weight of oxygen = 100 - (38.7+9.7) = 51.6 %

Weight of the compound = 0.93 g

Volume of vapour formed = 336 ml = 336 × $10^{-3}$ L

To find: Molecular formula of antifreeze

Solution:

• Number of moles of carbon ($n_{C}$) = $\frac{weight of carbon}{molar mass of carbon}$ = $\frac{38.7}{12}$ = 3.22 mol

• Number of moles of Hydrogen ($n_{H}$) = $\frac{weight of hydrogen}{molar mass of hydrogen}$ = $\frac{9.7}{1}$ = 9.7 mol

• Number of moles of Oxygen ($n_{O}$) = $\frac{weight of oxygen}{molar mass of oxygen}$ = $\frac{51.6}{16}$ = 3.22 mol

• Therefore, in the compound carbon, hydrogen and oxygen are in the ratio of $n_{C}$ : $n_{H}$ : $n_{O}$

                 = 3.22 : 9.7 : 3.22

                 =1 : 3 : 1

• Empirical formula of the compound is CH3O

• Molecular mass of Empirical formula = 31 g

• Using Ideal Gas equation:

         PV=nRT

         PV=$\frac{w}{M}$RT

      Where : P is Pressure at STP = 1 atm

                    T is Temperature at STP = 273.15 K

                    n is number of moles

                    V is Volume

                    R is Universal Gas Constant = 0.0821 L.atm.$K^{-1}$.$mol^{-1}$

                   w is weight of antifreeze

                   M is mass of antifreeze

      ⇒ 1 × 336 x $10^{-3}$ = $\frac{0.93}{molecular mass}$ × 0.0821 × 273.15

• Molecular mass of whole compound  ≈ 62 g

     ⇒Molecular mass of whole compound = 2 × Molecular mass of empirical    

                                                                                  formula

     ⇒Molecular formula = 2 × Empirical formula

                                        = 2 × CHO

• Molecular formula is C2H6O2