Question

An automobile antifreeze consists of $38.7\%$ C; $9.7\%$ H and remaining oxygen by weight. When 0.93 g of it is vaporized at $200{}^\circ \text{C}$ and 1 atm pressure, 582 mL of vapor is formed. The molecular formula of the antifreeze is – ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}{{\text{O}}_{\text{2}}}$.
(A) True
(B) False

Answer 1

Find the empirical formula by using the composition data of elements and molar mass by the ideal gas equation. Then, use the following equations to get the molecular formula of antifreeze.

& \text{PV}=\text{nRT (Ideal gas equation)} \\\ & \text{Molecular formula}=\text{n}\times \text{(Empirical formula)} \\\ \end{aligned}$$ **Complete answer:** The Ideal Gas equation defines the relationship between pressure, volume, amount of gas, and temperature. It is used to study the behavior of gases and given as: $$\text{PV}=\text{nRT}$$ Where P – Pressure, V – Volume, n – number of moles of gas, R – Universal gas constant and T – temperature. The conditions given in the question are: $$\text{P}=1\text{ atm},\text{ V}=582\text{ mL},\text{ T}=200{}^\circ \text{C}+273=473K,\text{ m}=0.93\text{ g}$$ Putting the values in the ideal gas equation, we get: $$\begin{aligned} & 1\text{(atm)}\times \dfrac{582}{1000}\text{(L)}=\dfrac{0.93\text{ g}}{\text{molar mass}}\times 0.082\text{(L atm mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}})\times 473\text{(K)} \\\ & \Rightarrow \text{molar mass}=61.97\text{ g} \\\ \end{aligned}$$ The empirical formula is the simplest chemical formula of a compound in which the elements are present in their simplest whole-number ratios. It can be calculated by using the percentage composition value of each element in the compound. Given: $\%$ weight of C= $38.7\%$ $\%$ weight of H=$9.7\%$ $\%$ weight of O =$100\% - ( 38.7\%+9.7\%)=51.6\%$ Step (1) – Let us assume the total mass of the compound as 100 g. Then, convert the percentage composition of each element into grams. $$\begin{aligned} & \text{C}=38.7\text{ g} \\\ & \text{H}=9.7\text{ g} \\\ & \text{O}=51.6\text{ g} \\\ \end{aligned}$$ Step (2) – Convert the grams into the number of moles by dividing each by its molar mass. $$\begin{aligned} & {{\text{n}}_{\text{C}}}=\dfrac{38.7\text{ g}}{12\text{ g mo}{{\text{l}}^{-1}}}=3.225 \\\ & {{\text{n}}_{\text{H}}}=\dfrac{9.7\text{ g}}{1\text{ g mo}{{\text{l}}^{-1}}}=9.7 \\\ & {{\text{n}}_{\text{O}}}=\dfrac{\text{51}\text{.6 g}}{16\text{ g mo}{{\text{l}}^{-1}}}=3.225 \\\ \end{aligned}$$ Step (3) – Divide the number of moles by the smallest number and convert them to whole numbers. $$\begin{aligned} & \text{C}=\dfrac{3.225}{3.225}=1 \\\ & \text{H}=\dfrac{9.7}{3.225}=3 \\\ & \text{O}=\dfrac{3.225}{3.225}=1 \\\ \end{aligned}$$ The elements are present in the ratio - $\text{C}:\text{H}:\text{O}=1:3:1$. Thus, the empirical formula is $\text{C}{{\text{H}}_{\text{3}}}\text{O}$ and empirical formula mass of the compound is $\left( 12+3+16 \right)=31\text{ g}$. Now, $$\text{Molecular formula}=\text{n}\times \text{(Empirical formula)}$$ Where $$\begin{aligned} & \text{n}=\dfrac{\text{Molar mass}}{\text{Empirical formula mass}}=\dfrac{61.97\text{ g}}{31\text{ g}} \\\ & \Rightarrow \text{n}=1.99\approx 2 \\\ & \\\ & \therefore \text{Molecular formula}=2\times \left( \text{C}{{\text{H}}_{\text{3}}}\text{O} \right)={{\text{C}}_{2}}{{\text{H}}_{6}}{{\text{O}}_{2}} \\\ \end{aligned}$$ **Hence, the correct answer is (A) True.** **Note:** The empirical formulas of compounds denote the relative numbers of atoms present in the substance and not the actual composition. So, it is also possible that two or more different compounds have the same empirical formula.