# Question #a4544

##### 1 Answer

#### Explanation:

The first thing to note here is that you're dealing with an **ionic compound**, which means that you're actually looking for the mass of *formula units* o sodium hydroxide,

Now, the *unified atomic mass unit*,

More importantly, it's worth remembering that one unified atomic mass unit is *approximately equal* to the mass of **one proton** or of **one neutron**, i.e. of one **nucleon**.

In simple terms, one unified atomic mass unit is equivalent to

#color(blue)(|bar(ul(color(white)(a/a)"1 u " = " 1 g mol"^(-1)color(white)(a/a)|)))#

Use the **molar mass** of sodium hydroxide to find how many unified atomic mass units you get **per formula unit**

#39.997color(red)(cancel(color(black)("g mol"^(-1)))) * "1 u"/(1color(red)(cancel(color(black)("g mol"^(-1))))) = "39.997 u"#

This means that **three formula units** of sodium hydroxide will have a mass of

#3color(red)(cancel(color(black)("formula units"))) * "39.997 u"/(1color(red)(cancel(color(black)("formula unit")))) = color(green)(|bar(ul(color(white)(a/a)"119.991 u"color(white)(a/a)|)))#

The calculation you did in your suggested solution gives you the mass of three formula units of sodium hydroxide expressed in *grams*, **not** in unified atomic mass units.

Since the **molar mass** of a compound tells you the mass of **one mole** of said compound, you can use **Avogadro's number** to determine the mass of a single formula unit of sodium hydroxide.

#color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"formula units"color(white)(a/a)|))) -># Avogadro's number

In this case, you would have

#39.997"g"/color(red)(cancel(color(black)("mol"))) * overbrace((1color(red)(cancel(color(black)("mol"))))/(6.022 * 10^(23)"f. units"))^color(purple)("Avogadro's number") = 6.642 * 10^(-23)"g/f. unit"#

This means that **three formula units** will have a mass of

#3color(red)(cancel(color(black)("f. units"))) * (6.642 * 10^(-23)"g")/(1color(red)(cancel(color(black)("f. unit")))) = 1.993 * 10^(-22)"g"#