Chapter 4, Problem 55E

Interpretation Introduction

Interpretation:

The mass of ${\text{CO}}_2$ is to be determined for different amounts of methane.

Concept introduction:

The characteristic mass of a compound is the formula mass of the compound, which is obtained by adding the atomic masses of all atoms.

A conversion factor is a numerical ratio, which is used to change one unit into another.

The chemical formula represents the precise number of atoms in a molecule, but not their structural arrangement.

Answer to Problem 55E

Solution: The masses of ${\text{CO}}_2$ for different amounts of methane are given as follows:

$1.0\times {10}^2{\text{g CO}}_2$

$23{\text{g CO}}_2$

$30{\text{g CO}}_2$

$3.6\times {10}^3{\text{g CO}}_2$

Explanation of Solution

a) $2.3{\text{mol CH}}_4$

The equivalence is obtained between moles of ${\text{CH}}_4$ and moles ${\text{CO}}_2$ with the help of the chemical formula as follows:

$1{\text{mol CO}}_2\equiv 1{\text{mol CH}}_4$

The equivalence is obtained between grams of ${\text{CO}}_2$ and moles of ${\text{CO}}_2$ with the help of the chemical formula as follows:

$1{\text{mol CO}}_2\equiv 44{\text{g CO}}_2$

In order to obtain the mass of ${\text{CO}}_2$, the conversion factor proceeds as follows:

$2.3\cancel{{\text{mol CH}}_4}\times \left(\frac{1\cancel{{\text{mol CO}}_2}}{1\cancel{{\text{mol CH}}_4}}\right)\times \left(\frac{44{\text{g CO}}_2}{1\cancel{{\text{mol CO}}_2}}\right)=1.0\times {10}^2{\text{g CO}}_2$

Here,

$\text{molar}{\text{mass of CO}}_2=44\text{g/mol}$

Therefore, the amount of ${\text{CO}}_{\text{2}}$ is $1.0\times {10}^2\text{ g}$.

b) $0.52{\text{mol CH}}_4$

The equivalence is obtained between moles of ${\text{CH}}_4$ and moles of ${\text{CO}}_2$ with the help of the chemical formula as follows:

$1{\text{mol CO}}_2\equiv 1{\text{mol CH}}_4$

The equivalence is obtained between grams of ${\text{CO}}_2$ and moles of ${\text{CO}}_2$ with the help of the chemical formula as follows:

$1{\text{mol CO}}_2\equiv 44{\text{g CO}}_2$

In order to obtain the mass of ${\text{CO}}_2$, the conversion factor proceeds as follows:

$0.52\cancel{{\text{mol CH}}_4}\times \left(\frac{1\cancel{{\text{mol CO}}_2}}{1\cancel{{\text{mol CH}}_4}}\right)\times \left(\frac{44{\text{g CO}}_2}{1\cancel{{\text{mol CO}}_2}}\right)=23{\text{g CO}}_2$

$\text{molar}{\text{mass of CO}}_2=44\text{g/mol}$

Therefore, the amount of ${\text{CO}}_{\text{2}}$ is $23\text{g}\text{. }$

c) $11{\text{g CH}}_4$

The equivalence is obtained between grams of ${\text{CH}}_4$ and moles of ${\text{CH}}_4$ with the help of the chemical formula as follows:

$1{\text{mol CH}}_4\equiv 16{\text{g CH}}_4$

The equivalence is obtained between moles of ${\text{CH}}_4$ and moles of ${\text{CO}}_2$ with the help of the chemical formula as follows:

$1{\text{mol CO}}_2\equiv 1{\text{mol CO}}_2$

The equivalence is obtained between grams of ${\text{CO}}_2$ and moles of ${\text{CO}}_2$ with the help of the chemical formula as follows:

$1{\text{mol CO}}_2\equiv 44{\text{g CO}}_2$

In order to obtain the mass of ${\text{CO}}_2$, the conversion factor proceeds as follows:

$11\cancel{{\text{g CH}}_4}\times \left(\frac{1\cancel{{\text{mol CH}}_4}}{16\cancel{{\text{g CH}}_4}}\right)\times \left(\frac{1\cancel{{\text{mol CO}}_2}}{1\cancel{{\text{mol CH}}_4}}\right)\times \left(\frac{44{\text{g CO}}_2}{1\cancel{{\text{mol CO}}_2}}\right)=30{\text{g CO}}_2$

Here,

$\text{molar}{\text{mass of CH}}_4=16\text{g/mol}$

$\text{molar}{\text{mass of CO}}_2=44\text{g/mol}$

Therefore, the amount of ${\text{CO}}_{\text{2}}$ is $\text{30 g}$.

d) $1.3{\text{kg CH}}_4$

The equivalence is obtained between grams of ${\text{CH}}_4$ and moles of ${\text{CH}}_4$ with the help of the chemical formula as follows:

$1{\text{mol CH}}_4\equiv 16{\text{g CH}}_4$

The equivalence is obtained between moles of ${\text{CH}}_4$ and moles of ${\text{CO}}_2$ with the help of the chemical formula as follows:

$1{\text{mol CO}}_2\equiv 1{\text{mol CO}}_2$

The equivalence is obtained between grams of ${\text{CO}}_2$ and moles of ${\text{CO}}_2$ with the help of the chemical formula as follows:

$1{\text{mol CO}}_2\equiv 44{\text{g CO}}_2$

In order to obtain the mass of ${\text{CO}}_2$, the conversion factor proceeds as follows:

$\text{1}\text{.3}\cancel{\text{kg}}\cancel{{\text{CH}}_4}\times \left(\frac{1000\cancel{\text{g}}}{1\cancel{\text{kg}}}\right)\times \left(\frac{1\cancel{\text{mol}}\cancel{{\text{CH}}_4}}{16\cancel{\text{g}}\cancel{{\text{CH}}_4}}\right)\times \left(\frac{1\cancel{{\text{mol CO}}_2}}{1\text{mol }\cancel{{\text{CH}}_4}}\right)\times \left(\frac{44{\text{g CO}}_2}{1\cancel{{\text{mol CO}}_2}}\right)=3.6\times {10}^3{\text{g CO}}_2$

Here,

$\text{molar}{\text{mass of CH}}_4=16\text{g/mol}$

$\text{molar}{\text{mass of CO}}_2=44\text{g/mol}$

Therefore, the amount of ${\text{CO}}_{\text{2}}$ is $3.6\times {10}^3\text{ g}$.