# 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?

##### 1 Answer

#### Answer:

#### 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**.