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

1 Answer
Dec 22, 2016

Answer:

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#