Chapter 12, Problem 12.108QP

Analysis of a compound gave 39.50% C, 2.21% H, and 58.30% Cl. When 0.855 g of this solid was dissolved in 7.50 g of naphthalene, the solution had a freezing point of 78.0°C. The pure solvent freezes at 80.0°C; its molal freezing point constant is 6.8°C/m.

1. a What is the molecular formula of the compound?
2. b What is its molecular weight to the nearest 0.1 g?

Interpretation Introduction

Interpretation:

A compound contains $39.50\%$ Carbon , $2.21\%$ Hydrogen and $58.30\%$ Chlorine. $0.855\text{ g}$ of this substance is dissolved in $7.50\text{ g}$ of naphthalene and the solution has freezing point of $78°C.$ Molecular formula of the compound has to be determined.

Concept Introduction:

Depression in freezing point is the phenomenon of lowering of freezing point of a substance, which is considered as solvent, by adding a non-volatile solute.   It is expresses as –

$\Delta T_f=K_f\times m$

Where,

$\begin{array}{l}\Delta T_f\text{ = depression in freezing point}\\ {\text{K}}_f\text{ = cryoscopic constant}\\ \text{m = molality of the solute}\end{array}$

Number of moles of a substance is related to molar mass of the substance as,

$no.of\text{ moles}\text{= }\frac{mass}{molar\text{ mass}}$

Answer to Problem 12.108QP

Molecular formula of the compound is determined as $C_{12}{H}_{8}C{l}_6.$

Explanation of Solution

Given that freezing point of the solution is $78°C.$ freezing point of pure naphthalene is $80°C$. $K_f$ for naphthalene is $6.8°C/m.$ rewriting the depression in freezing point equation, determine the molality of the solution.

$\begin{array}{l}m=\frac{\Delta T_f}{K_f}\\ =\frac{\left[80°C-78°C)\right]}{6.8°C/m}\\ =\text{ 0}\text{.294 m}\end{array}$

Number of moles of the compound considering $1\text{ kg}$ of solvent,

$\begin{array}{l}moles=\frac{\text{0}\text{.294 mol}}{1\text{ kg of solvent}}\times 0.00750\text{ kg}\\ =\text{ 2}\text{.21}\times {\text{10}}^{-3}\text{ mol}\end{array}$

Molar mass of the compound is calculated as,

$\begin{array}{l}molar\text{ mass}\text{= }\frac{\text{0}\text{.855 g}}{\text{2}\text{.21}\times {\text{10}}^{-3}\text{ mol}}\\ =\text{ 387 g/mol}\end{array}$

Number of moles of elements in $100\text{ g}$ of the compound are calculated as follows -

$\begin{array}{l}moles\text{ of C}\text{= 39}\text{.50 g C }\times \frac{1\text{ mol}}{12.01\text{ g C}}\\ =\text{ 3}\text{.2889 mol}\\ \\ moles\text{ of H}\text{= 2}\text{.21 g H }\times \frac{1\text{ mol}}{1.008\text{ g H}}\\ =\text{ 2}\text{.192 mol}\\ \\ moles\text{ of Cl}\text{= 58}\text{.30 g Cl }\times \frac{1\text{ mol}}{16.00\text{ g Cl}}\\ =\text{ 1}\text{.6466 mol}\end{array}$

From the above calculation we could deduce that mole ratio of the elements Carbon, Hydrogen and Chlorine is $2:1.33:1$.  Multiply this by three to ease the calculation.  Thus the mole ratio of the elements is $6:4:3$.  Hence the empirical formula of the compound could be $C_6{H}_{4}C{l}_3.$  Unit formula mass of this compound is approximately equal to $\text{182}\text{.5 amu}\text{.}$ Previously we have calculated the molar mass of the compound as $\text{387 g/mol}\text{.}$ this is approximately equal to twice the value of $74\text{ amu}\text{.}$ hence the molecular formula of the compound could be ${(C_6{H}_{4}C{l}_3)}_2$ which is $C_{12}{H}_{8}C{l}_6.$

Interpretation Introduction

Interpretation:

A compound contains $39.50\%$ Carbon , $2.21\%$ Hydrogen and $58.30\%$ Chlorine. $0.855\text{ g}$ of this substance is dissolved in $7.50\text{ g}$ of naphthalene and the solution has freezing point of $78°C.$ Molar mass of the compound nearest to the tenth of a gram has to be determined.

Concept Introduction:

Depression in freezing point is the phenomenon of lowering of freezing point of a substance, which is considered as solvent, by adding a non-volatile solute.   It is expresses as –

$\Delta T_f=K_f\times m$

Where,

$\begin{array}{l}\Delta T_f\text{ = depression in freezing point}\\ {\text{K}}_f\text{ = cryoscopic constant}\\ \text{m = molality of the solute}\end{array}$

Number of moles of a substance is related to molar mass of the substance as,

$no.of\text{ moles}\text{= }\frac{mass}{molar\text{ mass}}$

Answer to Problem 12.108QP

Molar mass of the compound nearest to tenth of a gram is calculated to be  $364.9g/mol.$

Explanation of Solution

Molar mass of the compound nearest to tenth of a gram is calculated as,

$\begin{array}{l}12C(12.01)+8\text{ H }(1.008)+6\text{ Cl }(16.00)=\text{ 364}\text{.884 g/mol}\\ =\text{ 364}\text{.9 g/mol}\end{array}$