How many formula units make up #"25.0 g"# of magnesium chloride?

1 Answer
Apr 8, 2018

Answer:

#1.58 * 10^(23)#

Explanation:

You know that magnesium chloride has a molar mass of #"95.211 g mol"^(-1)#, which means that #1# mole of magnesium chloride has a mass of #"95.211 g"#.

You also know that in order for your sample to contain #1# mole of magnesium chloride, it must contain #6.022 * 10^(23)# formula units of magnesium chloride #-># this is given by Avogadro's constant.

So if #1# mole of magnesium chloride has a mass of #"95.211 g"# and contains #6.022 * 10^(23)# formula units of magnesium chloride, you can say that

#color(white)(overbrace(underbrace(color(black)("95.211 g"))_ (color(red)("given by the molar mass")))^(color(blue)("= 1 mole MgCl"_ 2)) color(black)(=) " "overbrace(underbrace(color(black)(6.022 * 10^(23) quad "formula units MgCl"_ 2))_ (color(red)("given by Avogadro's constant")))^(color(blue)("= 1 mole MgCl"_2))#

This means that your sample will contain

#25.0 color(red)(cancel(color(black)("g"))) * (6.022 * 10^(23) quad "form. units MgCl"_2)/(95.211color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(1.58 * 10^(23) quad "form. units MgCl"_2)))#

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