# How many formula units are in #"0.335 g CaO"#?

##### 2 Answers

#### Answer:

#### Explanation:

Let's try to create a conversion factor that will take us *directly* from the number of grams of calcium oxide to the number of formula units present in the sample.

For starters, look up the **molar mass** of calcium oxide

#M_ "M CaO" = "56.0774 g mol"^(-1)#

This tells you that **mole** of calcium oxide has a mass of

#color(blue)("1 mole CaO")/"56.0774 g"#

Now, you know that in order to have **mole** of calcium oxide, you need to have **formula units** of calcium oxide **Avogadro's constant**.

This can be written as a conversion factor as well

#(6.022 * 10^(23)color(white)(.)"formula units")/color(blue)("1 mole CaO")#

As you can see, the two conversion factors have a common quantity. If you multiply these two conversion factors

#color(red)(cancel(color(blue)("1 mole CaO")))/"56.0774 g" * (6.022 * 10^(23)color(white)(.)"formula units")/color(red)(cancel(color(blue)("1 mole CaO")))#

you will get

#(6.022 * 10^(23)color(white)(.)"formula units")/"56.0774 g"#

You now have a conversion factor that takes you from *grams* to *formula units* or vice versa.

So, you know that your sample has a mass of

#0.335 color(red)(cancel(color(black)("g"))) * (6.022 * 10^(23)color(white)(.)"formula units")/(56.0774color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(3.60 * 10^(21)color(white)(.)"formula units")))#

The answer is rounded to three **sig figs**, the number of sig figs you have for the mass of calcium oxide.

#### Answer:

There are

#### Explanation:

There are

You need to determine the number of moles in

You need to ** #color(red)("determine the molar mass"#** of

** #color(red)("Determine the number of moles"#** by multiplying the given mass of

Multiply the mol