# What is the mass (in grams) of 9.45 xx 10^24 molecules of methanol, "CH"_3"OH" ?

Jun 6, 2017

$\text{50.3 g}$

#### Explanation:

Start by converting the number of molecules of methanol to moles by using Avogadro's constant.

9.45 * 10^(24) color(red)(cancel(color(black)("molecules CH"_3"OH"))) * overbrace(("1 mole CH"_3"OH")/(6.022 * 10^(23)color(red)(cancel(color(black)("molecules CH"_3"OH")))))^(color(blue)("Avogadro's constant"))

$= \text{15.69 moles CH"_3"OH}$

Now, in order to go from moles to grams, you need to use the molar mass of the compound, i.e. the mass of exactly $1$ mole of methanol.

Use the molar mass as a conversion factor to get

$15.69 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CH"_3"OH"))) * "32.04 g"/(1color(red)(cancel(color(black)("mole CH"_3"OH")))) = color(darkgreen)(ul(color(black)("50.3 g}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the number of molecules present in the sample.