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

1 Answer
Jun 6, 2017

Answer:

#"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"))#

# = "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 color(red)(cancel(color(black)("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.