Question

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?

Answer 1

`7.9 * 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

`color(blue)(ul(color(black)("volume" = l xx w xx h)))`

Here

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

In your case, you have

`"volume" = "3.0 mm" * "3.0 mm" * "0.50 mm"`

`"volume" = "4.5 mm"^3`

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

`color(blue)(ul(color(black)("1 cm" = 10color(white)(.)"mm")))`

Since you have

`"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 * 10^(-3) color(red)(cancel(color(black)("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.