How many grams of copper (ll) carbonate are present in 4.3*10^24 formula units?

Dec 22, 2016

882 g

Explanation:

Start by dividing the number of formula units by Avogadro's number:

#4.3 xx 10^(24)/6.02xx10^(23) = 7.143 moles

Next, since it is $C u C {O}_{3}$ we are considering, multiply this number of moles by the molar mass of $C u C {O}_{3}$, which is $63.5 + 12.0 + 48.0 = 123.5 \frac{g}{m o l}$:

We get

$7.143 m o l \times 123.5 \frac{g}{m o l} = 882 g$