# How many moles of carbon are present in a 3.0 mm by 3.00 mm by 0.5 mm diamond that's cut in the shape of a rectangular prism?

Sep 25, 2017

$7.9 \cdot {10}^{20}$

#### Explanation:

The first thing that you need to do here is to figure out the volume of the sample by using the fact that you're dealing with a rectangular prism, which has

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{volume} = l \times w \times h}}}$

Here

• $h$ is the height of the prism
• $l$ is its length
• $w$ is its width

$\text{volume" = "3.0 mm" * "3.0 mm" * "0.50 mm}$

${\text{volume" = "4.5 mm}}^{3}$

Next, convert the volume of the prism to cubic centimeters by using the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 cm" = 10color(white)(.)"mm}}}}$

Since you have

$\text{4.5 mm"^3 = 4.5 * "1 mm" * "1 mm" * "1 mm}$

you will end up with

4.5 * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) * 1color(red)(cancel(color(black)("mm"))) * "1 cm"/(10color(red)(cancel(color(black)("mm")))) = 4.5 * 10^(-3)color(white)(.)"cm"^3

Next, use the density of diamond to calculate the mass of the sample

4.5 * 10^(-3) color(red)(cancel(color(black)("cm"^3))) * "3.52 g"/(1color(red)(cancel(color(black)("cm"^3)))) = 1.584 * 10^(-2)color(white)(.)"g"

Next, use the molar mass of carbon to find the number of moles of carbon present in the sample

1.584 * 10^(-2) color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 1.319 * 10^(-3)color(white)(.)"moles C"

Finally, to find the number of atoms of carbon present in the sample, use Avogadro's constant

$1.319 \cdot {10}^{- 3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles C"))) * (6.022 * 10^(23)color(white)(.)"atoms C")/(1color(red)(cancel(color(black)("mole C")))) = color(darkgreen)(ul(color(black)(7.9 * 10^(20)color(white)(.)"atoms C}}}}$

The answer is rounded to two sig figs.