Question

Provide a reasonable estimate for the number of atoms in aadult human. Use the information given in Table 18.2.

Short answer

There are approximately$\text{6.45}\times {\text{10}}^{\text{27}}\text{atoms}$ in human body.

Step-by-step solution

Concept Introduction:

Chemical elements, often known as elements, are any compounds that, when subjected to ordinary chemical processes, cannot be broken down into simpler parts. Elements are the basic components of matter and serve as its building blocks.

Elements that are contributing:

First of all, take into account only elements contributing with one or more percent of body weight because elements with less than one percent of body weight do not influence the number of atoms. This will be a precondition.

$$\begin{gathered} \text{ω}\left(\text{O}\right)=0.65 \\ \text{ω}\left(\text{C}\right)=0.18 \\ \text{ω}\left(\text{H}\right)=0.10 \end{gathered}$$

$$\begin{gathered} \text{ω}\left(\text{N}\right)=0.03 \\ \text{ω}\left(\text{Ca}\right)=0.014 \\ \text{ω}\left(\text{P}\right)=0.01 \end{gathered}$$

$\text{m}=150\text{lb}=6800\text{0g}$

Mass of each element:

Determine the mass of each element by multiplying mass percentage with body weight.

Mass of Oxygen:

$\begin{array}{rcl}\text{m}\left(\text{O}\right)& =& \text{ω}\left(\text{O}\right)\times \text{m}\\ & =& \text{0.65}\times \text{68000g}\\ & =& \text{44200g}\end{array}$

Mass of Carbon:

$\begin{array}{rcl}\text{m}\left(\text{C}\right)& =& \text{ω}\left(\text{C}\right)\cdot \text{m}\\ & =& \text{0.18}\times \text{68000g}\\ & =& \text{12240g}\end{array}$

Mass of Hydrogen:

$\begin{array}{rcl}\text{m}\left(\text{H}\right)& =& \text{ω}\left(\text{H}\right)\cdot \text{m}\\ & =& \text{0.10}\times \text{68000g}\\ & =& \text{6800g}\end{array}$

Mass of Nitrogen:

$\begin{array}{rcl}\text{m}\left(\text{N}\right)& =& \text{ω}\left(\text{N}\right)\cdot \text{m}\\ & =& \text{0.03}\times \text{68000g}\\ & =& \text{2040g}\end{array}$

Mass of Calcium:

$\begin{array}{rcl}\text{m}\left(\text{Ca}\right)& =& \text{ω}\left(\text{Ca}\right)\cdot \text{m}\\ & =& \text{0.014}\times \text{68000g}\\ & =& \text{952g}\end{array}$

Mass of Phosphorus:

$\begin{array}{rcl}\text{m}\left(\text{P}\right)& =& \text{ω}\left(\text{P}\right)\cdot \text{m}\\ & =& \text{0.01}\times \text{68000g}\\ & =& \text{680g}\end{array}$

Number of atoms:

Now, form an equation to determine the number of atoms of each element in the human body,

$\begin{array}{rcl}n& =& \frac{N}{NA}\\ & =& \frac{m}{M}\end{array}$

$$\begin{gathered} \frac{N}{NA}=\frac{m}{M} \\ N=\frac{m\cdot NA}{M} \end{gathered}$$

The number of atoms of Oxygen is,

$\begin{array}{rcl}\text{N}\left(\text{O}\right)& =& \frac{\text{m}\left(\text{O}\right)\cdot \text{NA}}{\text{M}\left(\text{O}\right)}\\ & =& \frac{\text{44200g}\times \text{6.022}\times {\text{10}}^{\text{23}}{\text{mol}}^{-1}}{{\text{16.00gmol}}^{-\text{1}}}\\ & =& 1.66\times {10}^{27}\end{array}$

The number of atoms of Carbon is,

$\begin{array}{rcl}\text{N}\left(\text{C}\right)& =& \frac{\text{m(}\left(\text{C}\right)\cdot \text{NA}}{\text{M}\left(\text{C}\right)}\\ & =& \frac{\text{12240g}\times \text{6.022}\times {\text{10}}^{\text{23}}{\text{mol}}^{-1}}{{\text{12.01gmol}}^{-\text{1}}}\\ & =& \text{6.14}\times {\text{10}}^{\text{26}}\end{array}$

The number of atoms of Hydrogen is,

$\begin{array}{rcl}\text{N}\left(\text{H}\right)& =& \frac{\text{m}\left(\text{H}\right)\cdot \text{NA}}{\text{M}\left(\text{H}\right)}\\ & =& \frac{\text{6800g}\times \text{6.022}\times {\text{10}}^{\text{23}}{\text{mol}}^{-\text{1}}}{{\text{1.008gmol}}^{-\text{1}}}\\ & =& \text{4.06}\times {\text{10}}^{\text{27}}\end{array}$

The number of atoms of Nitrogen is,

$\begin{array}{rcl}\text{N}\left(\text{N}\right)& =& \frac{\text{m}\left(\text{N}\right)\cdot \text{NA}}{\text{M}\left(\text{N}\right)}\\ & =& \frac{\text{2040g}\times \text{6.022}\times {\text{10}}^{\text{23}}{\text{mol}}^{-\text{1}}}{{\text{14.01gmol}}^{-\text{1}}}\\ & =& \text{8.77}\times {\text{10}}^{\text{25}}\end{array}$

The number of atoms of Calcium is,

$\begin{array}{rcl}\text{N}\left(\text{Ca}\right)& =& \frac{\text{m}\left(\text{Ca}\right)\cdot \text{NA}}{\text{M}\left(\text{Ca}\right)}\\ & =& \frac{\text{952g}\times \text{6.022}\times {\text{10}}^{\text{23}}{\text{mol}}^{-\text{1}}}{{\text{40.01gmol}}^{-\text{1}}}\\ & =& \text{1.43}\times {\text{10}}^{\text{25}}\end{array}$

The number of atoms of Phosphorus­ is,

$\begin{array}{rcl}\text{N}\left(\text{P}\right)& =& \frac{\text{m}\left(\text{P}\right)\cdot \text{NA}}{\text{M}\left(\text{P}\right)}\\ & =& \frac{\text{680g}\times \text{6.022}\times {\text{10}}^{\text{23}}{\text{mol}}^{-\text{1}}}{{\text{30.97gmol}}^{-\text{1}}}\\ & =& \text{1.32}\times {\text{10}}^{\text{25}}\end{array}$

Calculate the sum:

The last step is to sum up these results.

$\begin{array}{rcl}\text{N}& =& \sum _{\text{n=1}}^{\text{6}}{\text{N}}_{\text{i}}\\ & =& (1.66\times {10}^{27})+(6.14\times {10}^{26})+(4.06\times {10}^{27})+(8.77\times {10}^{25})+(1.43\times {10}^{25})+(1.32\times {10}^{25})\\ & =& 6.45\times {10}^{27}\end{array}$

Hence, the value for total number of atoms is $\text{6.45}\times {\text{10}}^{\text{27}}$.