Chapter 2, Problem 38QAP

A cylindrical piece of pure copper $\left(d=8.92\text{g}/{\text{cm}}^{\text{2}}\right)$ has diameter 1.15 cm and height 4.00 inches. How many atoms are in that cylinder? (Note: the volume of a right circular cylinder of radius r and height h is $V=\pi r^{2}h$ .)

Interpretation Introduction

Interpretation:

The number of atoms in given cylinder of copper needs to be determined.

Concept introduction:

Relation between mass, volume and density is as follows:

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

Formula to calculate volume of cylinder is given as follows:

$\begin{array}{l}\text{Volume}\text{of}\text{cylinder}\text{=}{\text{πr}}^{\text{2}}\text{h}\\ \text{where}\\ \text{r}\text{=}\text{radius}\text{of}\text{base}\text{of}\text{cylinder}\\ \text{h=}\text{height}\text{of}\text{cylinder}\end{array}$

Relation between moles, mass of substance and number of atoms is as follows:

$\begin{array}{l}\text{moles}\text{=}\frac{\text{mass}\text{(g)}}{\text{molar}\text{mass}\text{(g/mol)}}\text{=}\frac{\text{number}\text{of}\text{atoms}}{{\text{N}}_{\text{A}}}\\ \text{where,}\\ {\text{N}}_{\text{A}}\text{represents}\text{avagadro's}\text{number}\text{equal}\text{to}\text{6}{\text{.023×10}}^{\text{23}}\text{atoms/mol}\end{array}$

Answer to Problem 38QAP

Number of atoms in given cylinder of pure copper = $8.92{\text{×10}}^{\text{23}}\text{atoms}$

Explanation of Solution

Diameter of cylinder = 1.15 cm

Since,

$\begin{array}{l}\text{ radius}\text{=}\frac{\text{Diameter}}{\text{2}}\\ \text{Therefore,}\\ \text{radius}\text{of}\text{base}\text{of}\text{cylinder}\text{=}\frac{\text{1}\text{.15}\text{cm}}{\text{2}}\text{=}\text{0}\text{.575}\text{cm}\end{array}$

Height of cylinder = 4.00 inches

As 1 inches = 2.54 cm

Therefore,

$\text{4}\text{.0}\text{inches}\text{=}\frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}}\text{×4}\text{.0}\text{in}\text{=}\text{10}\text{.16}\text{cm}$

Formula to calculate volume of cylinder is given as:

$\begin{array}{l}\text{Volume}\text{of}\text{cylinder}\text{=}{\text{πr}}^{\text{2}}\text{h}\\ \text{where}\\ \text{r}\text{=}\text{radius}\text{of}\text{base}\text{of}\text{cylinder}\\ \text{h=}\text{height}\text{of}\text{cylinder}\end{array}$

Putting the values in above formula to get the volume of cylinder as:

$\begin{array}{l}\text{Volume}\text{of}\text{cylinder}\text{=}\text{(3}\text{.14)×(0}{\text{.575)}}^{\text{2}}\text{×(10}\text{.16)}\\ \text{Volume}\text{of}\text{cylinder}\text{=}\text{10}\text{.55}{\text{cm}}^{\text{3}}\end{array}$

According to question, density of copper = 8.92 g/cm³

Relation between mass, volume and density:

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

Putting the values in above formula to get the mass of copper as:

$\begin{array}{l}\text{8}\text{.92}{\text{g/cm}}^{\text{3}}\text{=}\frac{\text{Mass}}{\text{10}\text{.55}{\text{cm}}^{\text{3}}}\\ \text{Mass}\text{=}\text{8}\text{.92}\frac{\text{g}}{{\text{cm}}^{\text{3}}}\text{×10}\text{.55}{\text{cm}}^{\text{3}}\\ \Rightarrow \text{Mass}\text{=}\text{94}\text{.106}\text{g}\end{array}$

Since,

$\text{moles}\text{=}\frac{\text{mass}\text{(g)}}{\text{molar}\text{mass}\text{(g/mol)}}\text{=}\frac{\text{number}\text{of}\text{atoms}}{{\text{N}}_{\text{A}}}$

Therefore,

$\frac{\text{mass}\text{(g)}}{\text{molar}\text{mass}\text{(g/mol)}}\text{=}\frac{\text{number}\text{of}\text{atoms}}{{\text{N}}_{\text{A}}}$

Also,

Mass of copper = 94.11 g

Molar mass of copper = 63.55 g/mol

Putting the values to get the number of atoms as:

$\begin{array}{l}\frac{\text{mass}\text{(g)}}{\text{molar}\text{mass}\text{(g/mol)}}\text{=}\frac{\text{number}\text{of}\text{atoms}}{{\text{N}}_{\text{A}}}\\ \frac{\text{94}\text{.11}\text{(g)}}{\text{63}\text{.55}\text{(g/mol)}}\text{=}\frac{\text{number}\text{of}\text{atoms}}{\text{6}{\text{.023×10}}^{\text{23}}}\end{array}$

Thus,

$\begin{array}{l}\text{number}\text{of}\text{atoms}\text{=}\frac{\text{94}\text{.11×6}{\text{.023×10}}^{\text{23}}}{\text{63}\text{.55}}\\ \text{number}\text{of}\text{atoms}\text{=}\text{8}{\text{.92×10}}^{\text{23}}\text{atoms}\end{array}$

Conclusion

Thus, the number of atoms in given cylinder of copper= $8.92{\text{×10}}^{\text{23}}\text{atoms}$