Chapter 3, Problem 138QRT

(a)

Interpretation Introduction

Interpretation:

0.1235-g of ${\text{H}}_{\text{2}}\text{X}$ (diacid) is neutralized with $\text{15}\text{.55 mL of 0}\text{.1087-M NaOH}$. Then, 0.3469-g sample is burned with oxygen and producing $\text{0}{\text{.6268 g CO}}_{\text{2}}$ and $\text{0}{\text{.2138 g H}}_{\text{2}}\text{O}$. The molar mass of ${\text{H}}_{\text{2}}\text{X}$ has to be calculated.

Concept introduction:

Number of moles can be calculated by using following formula,

  $\text{No}\text{.of Moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Explanation of Solution

Given,

0.1235-g of ${\text{H}}_{\text{2}}\text{X}$ (diacid) is neutralized with $\text{15}\text{.55 mL of 0}\text{.1087-M NaOH}$. Then, 0.3469-g sample is burned with oxygen and producing $\text{0}{\text{.6268 g CO}}_{\text{2}}$ and $\text{0}{\text{.2138 g H}}_{\text{2}}\text{O}$.

The given acid diacid, therefore two moles of base is neutralizing the mole of acid, so, the balanced chemical equation is given below,

    ${\text{2NaOH + H}}_{\text{2}}\text{X }\to {\text{2H}}_{\text{2}}{\text{O(l) + Na}}_{\text{2}}\text{X}$

Moles of ${\text{H}}_{\text{2}}\text{X }$ is calculated as follows,

  $\begin{array}{l}\text{Moles}\text{of}{\text{H}}_{\text{2}}\text{X }\text{=}\text{0}\text{.0155}\text{L}\text{×}\frac{\text{0}\text{.1087}\text{mol}\text{NaOH}}{\text{1}\text{L}\text{NaOH}}\text{×}\frac{\text{1 mol}{\text{H}}_{\text{2}}\text{X}}{\text{2}\text{mol}\text{NaOH}}\\ \text{Moles}\text{of}{\text{H}}_{\text{2}}\text{X }\text{=}\text{8}{\text{.451×10}}^{\text{-4}}\text{mol}\end{array}$

Molar mass of ${\text{H}}_{\text{2}}\text{X }$ is calculated as follows,

  $\begin{array}{l}\text{Molar}\text{mass = }\frac{\text{Mass}}{\text{No}\text{.of Moles}}\\ \text{Molar}\text{mass = }\frac{\text{0}\text{.1235}\text{g/mol}}{\text{8}{\text{.451×10}}^{\text{-4}}\text{mol}}\\ \text{Molar}\text{mass = 146}\text{.13}\text{g}\end{array}$

Therefore, the molar mass is $\text{ 146}\text{.13}\text{g}$

(b)

Interpretation Introduction

Interpretation:

The Empirical formula of diacid has to be calculated.

Concept introduction:

Formula for determine the molecular formula of the compound is given below,

  $\text{Empirical formula / compound}\text{=}\frac{\text{g / mol compound}}{\text{g / mol empirical formula}}$

Combustion reaction:

Combustion is an exothermic reaction between a fuel (the reductant) and an oxidant. Oxidant is usually atmospheric oxygen.

Generally organic molecule undergoes combustion reaction which produces carbon dioxide and water molecule.

  ${\text{Hydrocarbons + O}}_{\text{2}}\text{(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(g)}$

Explanation of Solution

0.1235-g of ${\text{H}}_{\text{2}}\text{X}$ (diacid) is neutralized with $\text{15}\text{.55 mL of 0}\text{.1087-M NaOH}$. Then, 0.3469-g sample is burned with oxygen and producing $\text{0}{\text{.6268 g CO}}_{\text{2}}$ and $\text{0}{\text{.2138 g H}}_{\text{2}}\text{O}$.

  ${\text{H}}_{\text{2}}{\text{X + O}}_{\text{2}}\text{(g) }\to {\text{ CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(g)}$

Molar mass of carbon dioxide = $\text{44}\text{.0095 g/mol}$, Molar mass of water = $\text{18}\text{.0152 g/mol}$. Molar mass of nitrogen = $14.00\text{ g/mol}$, Molar mass of oxygen = $15.9988\text{ g/mol}$.

Moles of carbon is calculated as follows,

  $\begin{array}{l}\text{Moles}\text{of}\text{Carbon}\text{=}0.6268\text{g }{\text{CO}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}\text{C}}{44.0095\text{g}{\text{CO}}_{\text{2}}}\\ \text{Moles}\text{of}\text{Carbon}\text{=}0.0142\text{mol}\end{array}$

Moles of hydrogen is calculated as follows,

  $\begin{array}{l}\text{Moles}\text{of}\text{hydrogen}\text{=}\text{0}\text{.2138 g}{H}_{2}O\text{×}\frac{2\text{mol}\text{hydrogen}}{\text{18}\text{.0152}\text{g}\text{hydrogen}}\\ \text{Moles}\text{of}\text{hydrogen}\text{=}\text{0}\text{.02373}\text{Moles}\end{array}$

Grams of carbon is calculated as follows,

  $\begin{array}{l}\text{Grams}\text{of}\text{Carbon}\text{=}\text{0}\text{.0142}\text{mol}\text{C ×}\frac{\text{12}\text{.0107 g C}}{\text{1}\text{mol}\text{CO}}\\ \text{Grams}\text{of}\text{Carbon}\text{=0}\text{.171 g}\end{array}$

Grams of hydrogen is calculated as follows,

  $\begin{array}{l}\text{Grams}\text{of}\text{hydrogen}\text{=}\text{0}\text{.02373}\text{Moles}\text{g}H\text{×}\frac{1.0079g\text{hydrogen}}{\text{1}\text{mol}\text{hydrogen}}\\ \text{Grams}\text{of}\text{hydrogen}\text{=}\text{0}\text{.239 g}\end{array}$

Grams of oxygen is calculated as follows,

  $\begin{array}{l}\text{Mass Oxygen = 0}{\text{.3469 g C}}_{\text{x}}{\text{H}}_{\text{y}}{\text{O}}_{\text{z}}\text{ - grams of carbon - grams of hydrogen}\\ \text{Mass Oxygen = 0}{\text{.3469 g C}}_{\text{x}}{\text{H}}_{\text{y}}{\text{O}}_{\text{z}}\text{ - 0}\text{.1711 g of carbon - 0}\text{.02392 g of hydrogen}\\ \text{Mass Oxygen = 0}\text{.1519 g}\end{array}$

Moles of oxygen is calculated as follows,

$\begin{array}{l}\text{Moles}\text{of}\text{oxygen}\text{=}0.1519\text{g }{\text{CO}}_{\text{2}}\text{×}\frac{\text{1}\text{mol}O}{15.9994\text{g}\text{O}}\\ \text{Moles}\text{of}\text{oxygen}\text{=}9.49\times {10}^{-3}\text{mol}\end{array}$

Divide through by smallest number:

  $\begin{array}{l}\text{C = }\frac{0.0142\text{mol}}{9.49\times {10}^{-3}\text{Mol}}\text{ = 1}\text{.5}\\ \text{C = 1}\text{.5}\end{array}$

  $\begin{array}{l}\text{H = }\frac{\text{0}\text{.02373}}{9.49\times {10}^{-3}\text{Mol}}\text{ = 2}\text{.5}\\ \text{H = 2}\text{.5}\end{array}$

  $\begin{array}{l}\text{O = }\frac{9.49\times {10}^{-3}\text{Mol}}{9.49\times {10}^{-3}\text{Mol}}\text{ = 1}\\ \text{O = 1}\end{array}$

Therefore,

The Empirical formula for propionic acid is calculated as follows,

$\text{1}\text{.5C : 2}\text{.5H : O}$.

Multiply with two to get the whole numbers,

The empirical formula =${\text{C}}_3{\text{H}}_5{\text{O}}_2$

(b)

Interpretation Introduction

Interpretation:

The molecular formula of diacid has to be calculated.

Concept introduction:

Formula for determine the molecular formula of the compound is given below,

$\text{Empirical formula / compound}\text{=}\frac{\text{g / mol compound}}{\text{g / mol empirical formula}}$

Explanation of Solution

The empirical formula is ${\text{C}}_3{\text{H}}_5{\text{O}}_2$

The molar mass of the compound is calculated as follows,

  $\begin{array}{l}\text{Molar mass = 3(12}\text{.0107 g/mol C) + 5(1}\text{.0079 g/mol H) + 2(15}\text{.9994 g/mol O) }\\ \text{Molar mass = 73}\text{.0704 g/mol }\end{array}$

Formula for the compound is given below,

  $\begin{array}{l}\text{Empirical formula / compound}\text{=}\frac{\text{g / mol compound}}{\text{g / mol empirical formula}}\\ \text{1 empirical formula / compound}\text{=}\frac{\text{146}\text{.13 g / mol }}{\text{73}\text{.0704 g/mol emp}\text{. formula}}\\ \text{1 empirical formula / compound}\text{=2}\end{array}$

Therefore, to get the molecular formula, multiple with two.

Hence, the molecular formula of the compound is ${\text{C}}_6{\text{H}}_{10}{\text{O}}_4$.