Chapter 10.3, Problem 42PP

a.

Interpretation Introduction

Interpretation:

The number of carbon atoms in 45.6 g sample of ethanol needs to be determined.

Concept introduction:

In 1 mol of any substance, there are $6.022\times {10}^{23}$ units of that substance. These units can be atoms and molecules. The number of units in 1 mol of any substance is known as Avogadro’s number. It is denoted by symbol N_A.

Answer to Problem 42PP

$\text{1}\text{.2}\times {10}^{24}\text{ atoms}$

Explanation of Solution

Given information:

The mass of ethanol is 45.6 g and molecular formula of ethanol is CH₃CH₂OH.

Calculation: From the given formula, molar mass of ethanol can be calculated from atomic mass of carbon, hydrogen and oxygen atoms. The atomic mass of C, H and O is 12 u, 1 u and 16 u respectively. Also, there are 2 C atoms, 6 H atoms and 1 O atom.

Therefore,

${\text{M}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\text{=2}\times \text{12+6+16=46 g/mol}$

The mass of ethanol is given 45.6 mol. The number of moles can be calculated as follows:

$n=\frac{m}{M}=\frac{\text{45}\text{.6 g}}{\text{46 g/mol}}\text{=0}\text{.99 mol}$

From the molecular formula 1 mol of ethanol contains 2 mol of C atoms. Thus, the number of moles of C atoms will be:

$n=2\times 0.99=1.98\text{ mol}$

Now, in 1 mol of any substance there are $6.022\times {10}^{23}$ units thus, the number of carbon atoms can be calculated as follows:

$\begin{array}{l}N=n\times N_A\\ =1.\text{98 mol}\times 6.022\times {10}^{23}\text{ atoms/mol}\\ \text{=1}\text{.2}\times {10}^{24}\text{ atoms}\end{array}$

Therefore, number of carbon atoms is $\text{1}\text{.2}\times {10}^{24}\text{ atoms}$.

b.

Interpretation Introduction

Interpretation:

The number of hydrogen atoms present in the sample needs to be determined.

Concept introduction:

In 1 mol of any substance, there are $6.022\times {10}^{23}$ units of that substance. These units can be atoms and molecules. The number of units in 1 mol of any substance is known as Avogadro’s number. It is denoted by symbol N_A.

Answer to Problem 42PP

The number of hydrogen atoms present in the sample is $\text{3}\text{.58}\times {10}^{24}\text{ atoms}$.

Explanation of Solution

From the molecular formula 1 mol of ethanol contains 6 mol of H atoms. Thus, the number of moles of H atoms will be:

$n=6\times 0.99=5.94\text{ mol}$

Now, in 1 mol of any substance there are $6.022\times {10}^{23}$ units thus, the number of hydrogen atoms can be calculated as follows:

$\begin{array}{l}N=n\times N_A\\ =5.94\text{mol}\times 6.022\times {10}^{23}\text{ atoms/mol}\\ \text{=3}\text{.58}\times {10}^{24}\text{ atoms}\end{array}$

Therefore, number of hydrogen atoms is $\text{3}\text{.58}\times {10}^{24}\text{ atoms}$.

c.

Interpretation Introduction

Interpretation:

The number of oxygen atoms present in the sample needs to be determined.

Concept introduction:

In 1 mol of any substance, there are $6.022\times {10}^{23}$ units of that substance. These units can be atoms and molecules. The number of units in 1 mol of any substance is known as Avogadro’s number. It is denoted by symbol N_A.

Answer to Problem 42PP

The number of oxygen atoms present in the sample is $\text{5}\text{.96}\times {10}^{23}\text{ atoms}$.

Explanation of Solution

From the molecular formula 1 mol of ethanol contains 1 mol of O atoms. Thus, the number of moles of O atoms will be:

$n=1\times 0.99=0.99\text{mol}$

Now, in 1 mol of any substance there are $6.022\times {10}^{23}$ units thus, the number of oxygen atoms can be calculated as follows:

$\begin{array}{l}N=n\times N_A\\ =0.99\text{mol}\times 6.022\times {10}^{23}\text{ atoms/mol}\\ \text{=5}\text{.96}\times {10}^{23}\text{ atoms}\end{array}$

Therefore, number of oxygen atoms is $\text{5}\text{.96}\times {10}^{23}\text{ atoms}$.