Chapter 12, Problem 12.107QP

The carbohydrate digitoxose contains 48.64% carbon and 8.16% hydrogen. The addition of 18.0 g of this compound to 100 g of water gives a solution that has a freezing point of −2.2°C.

1. a What is the molecular formula of the compound?
2. b What is the molar mass of this compound to the nearest tenth of a gram?

Interpretation Introduction

Interpretation:

The carbohydrate digitoxose is made up of $48.64\%$ Carbon and $8.16\%$ Hydrogen. $18\text{ g}$ of this substance is added to $100\text{ g}$ of water and the solution has freezing point of $-2.2°C.$

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 12.107QP

Molecular formula of the compound is determined as $C_6{H}_{12}{O}_4.$

Explanation of Solution

Given that freezing point of the solution is $-2.2°C.$ rewriting the depression in freezing point equation, determine the molality of the solution.

$\begin{array}{l}m=\frac{\Delta T_f}{K_f}\\ =\frac{\left[0°C-(-2.2°C)\right]}{1.858°C/m}\\ =\text{ 1}\text{.184 m}\end{array}$

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

$\begin{array}{l}moles=\frac{1.184\text{ mol}}{1\text{ kg of solvent}}\times 0.100\text{ kg}\\ =\text{ 0}\text{.1184 mol}\end{array}$

Molar mass of the compound is calculated as,

$\begin{array}{l}molar\text{ mass}\text{= }\frac{18.0\text{ g}}{0.1184\text{ mol}}\\ =\text{ 152}\text{.0 g/mol}\end{array}$

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

$\begin{array}{l}moles\text{ of C}\text{= 48}\text{.64 g C }\times \frac{1\text{ mol}}{12.01\text{ g C}}\\ =\text{ 4}\text{.0500 mol}\\ \\ moles\text{ of H}\text{= 8}\text{.16 g H }\times \frac{1\text{ mol}}{1.008\text{ g H}}\\ =\text{ 8}\text{.095 mol}\\ \\ moles\text{ of O}\text{= 43}\text{.20 g O }\times \frac{1\text{ mol}}{16.00\text{ g O}}\\ =\text{ 2}\text{.7000 mol}\end{array}$

From the above calculation we could deduce that mole ratio of the elements Carbon, Hydrogen and Oxygen is $1.5:3:1$.  Multiply this by two to ease the calculation.  Thus the mole ratio of the elements is $3:6:2$.  Hence the empirical formula of the compound could be $C_3{H}_6{O}_2.$  Unit formula mass of this compound is approximately equal to $74\text{ amu}\text{.}$ Previously we have calculated the molar mass of the compound as $152\text{ g/mol}\text{.}$ this is approximately equal to twice the value of $74\text{ amu}\text{.}$ hence the molecular formula of the compound could be ${(C_3{H}_6{O}_2)}_2$ which is $C_6{H}_{12}{O}_4.$

Interpretation Introduction

Interpretation:

The carbohydrate digitoxose is made up of $48.64\%$ Carbon and $8.16\%$ Hydrogen. $18\text{ g}$ of this substance is added to $100\text{ g}$ of water and the solution has freezing point of $-2.2°C.$

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 12.107QP

Molar mass of the compound nearest to tenth of a gram is calculated to be  $148.2g/mol.$

Explanation of Solution

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

$\begin{array}{l}6\text{ C }(12.01)+12\text{ H }(1.008)+4\text{ O }(16.00)=\text{ 148}\text{.156 g/mol}\\ \text{= 148}\text{.2 g/mol}\end{array}$