Question

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

Answer 1

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 `CuCO_3` we are considering, multiply this number of moles by the molar mass of `CuCO_3`, which is `63.5 + 12.0 + 48.0 = 123.5 g/(mol)`:

We get

`7.143 mol xx 123.5 g/(mol) = 882g`