Chapter 4, Problem 42E

How many hydrogen atoms are in each of the following?

$\begin{array}{l}\text{a}{\text{. 5 H}}_{\text{2}}\text{O molecules}\hfill \\ \text{b}{\text{. 58 CH}}_{\text{4}}\text{ molecules}\hfill \\ \text{c}\text{. 1}\text{.4 }\times {\text{ 10}}^{\text{22}}{\text{C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{ molecules}\hfill \\ \text{d}{\text{. 14 dozen NH}}_{\text{3}}\text{ molecules}\hfill \end{array}$

Interpretation Introduction

Interpretation:

The number of hydrogen atoms is to be determined in the given molecules.

Concept Introduction:

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

A conversion factor is considered a numerical ratio in order to change one unit into another.

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

Answer to Problem 42E

Solution:

The number of hydrogen atoms present in the molecules is given as follows:

$\text{10}\text{H atoms}$

$232\text{H atoms}$

$3.1\times {10}^{23}\text{H atoms}$

$504\text{H atoms}$

Explanation of Solution

a) $5{\text{H}}_2\text{O}\text{molecules}$

The equivalence is obtained between $5{\text{H}}_2\text{O}\text{molecules}$ and atoms of hydrogen with the help of the chemical formula as follow:

$2\text{atoms H}\equiv 1{\text{molecule H}}_2\text{O}$

In order to obtain atoms of hydrogen, the conversion factor proceeds as follows:

$5\cancel{{\text{H}}_2\text{O}\text{molecules}}\times \frac{2\text{H atoms}}{1\cancel{{\text{H}}_2\text{O}\text{molecule}}}=10\text{H atoms}$

Therefore, the number of hydrogen atoms is $10$.

b) $58{\text{CH}}_4\text{molecules}$

The equivalence is obtained between $58{\text{CH}}_4\text{molecules}$ and atoms of hydrogen with the help of the chemical formula as follow:

$4\text{atoms H}\equiv 1{\text{molecule CH}}_4$

In order to obtain atoms of hydrogen, the conversion factor proceeds as follows:

$\left(58\cancel{{\text{CH}}_4\text{molecules}}\right)\times \left(\frac{\text{4}\text{H atoms}}{1\cancel{{\text{CH}}_4\text{molecule}}}\right)=232\text{H atoms}$

Therefore, the number of hydrogen atoms is $232$.

c) $1.4\times {10}^{22}{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\text{molecules}$

The equivalence is obtained between $1.4\times {10}^{22}{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\text{molecules}$ and atoms of hydrogen with the help of the chemical formula as follow:

$22\text{atoms H}\equiv 1{\text{molecule C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$

In order to obtain atoms of hydrogen, the conversion factor proceeds as follows:

$1.4\times {10}^{22}\cancel{{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\text{molecules}}\times \frac{22\text{H atoms}}{1\cancel{{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\text{molecule}}}=3.08\times {10}^{23}\text{H atoms}$

Therefore, the number of hydrogen atoms is $3.1\times {10}^{23}$.

d) $14{\text{dozen NH}}_3\text{molecules}$

The equivalence is obtained between $14{\text{dozen NH}}_3\text{molecules}$ and atoms of hydrogen with the help of the chemical formula as follow:

$3\text{atoms H}\equiv 1{\text{molecule NH}}_3$

$12\text{molecules}=1\text{dozen}$

In order to obtain atoms of hydrogen, the conversion factor proceeds as follows:

$14\text{dozen }\cancel{{\text{NH}}_3\text{molecules}}\times \frac{3\text{H atoms}}{1\cancel{{\text{NH}}_{\text{3}}\text{molecule}}}\times \frac{12\text{molecules}}{1\text{dozen}}=504\text{H atoms}$

Therefore, the number of hydrogen atoms is $504$.