If a sample of LiF is found to contain 7.73 x 10^24 Li+ ions how many grams of LiF formula units are present?

I tried to work this out and each time i got the incorrect answer. I know the answer is 332.959930255729 as Wiley gave me this however it didn't explain how. Any help would be very appreciated!!!

Mar 15, 2018

Explanation:

We know that Lithium fluoride is an ionic compound that contains negative fluoride ions and positive lithium ions in a 1:1 ratio.

1 mole of any substance contains $6.022 \times {10}^{23}$ molecules, and the molar mass for $L i F$ is 25.939 $g m o {l}^{-} 1$. The question is how many moles of $L i F$ does your amount correspond to?

$\frac{7.73 \times {10}^{24}}{6.022 \times {10}^{23}} = 12.836 m o l$

As lithium ions exist in a 1:1 ratio, this amount of moles of Lithium ions also correspond to the number of moles of the substance- $L i F$.

Hence to find the mass in grams we multiply the number of moles with the molar mass to find the mass (According to $n = \frac{m}{M}$)

$n = 12.836$
$M = 25.939$
m=?

$12.836 \times 25.939 = 332.9599$ grams

Which is consistent with what Wiley gave you :)