# What is the mass of 1.532 moles of a compound with a molar mass of 44 g/mol?

Jul 30, 2016

$\text{67.41 g}$

#### Explanation:

The idea here is that the molar mass of a compound can be used as a conversion factor between grams and moles.

As you know, the molar mass tells you the mass of one mole of a given compound. In your case, the compound is said to have a molar mass equal to ${\text{44 g mol}}^{- 1}$.

This means that if you were to measure out exactly one mole of this compound, its mass would be equal to $\text{44 g}$.

Now, your sample contains $1.532$ moles of this unknown compound. To use the molar mass a conversion factor, place the unit that you have on the bottom of the fraction and the unit that you need on the top of the fraction

You will thus have

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{have moles" * "44 grams"/"1 mole" = "get grams} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

or, when you have grams and need moles

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{have grams" * "1 mole"/"44 grams" = "get moles} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to find

1.532 color(red)(cancel(color(black)("moles"))) * "44 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("67.41 g")color(white)(a/a)|)))

I'll leave the answer rounded to four sig figs because you can say that the molar mass of the compound is a constant, which implies that it has an "infinite" number of sig figs.