Answer to Question #73145 in General Chemistry for dylan

Question

The number of oxygen atoms in 40.0 g of Ca3(PO4)2 (FW 310.2)

Answer 1

.m (Ca3(PO4)2)=40 g
Mr(Ca3(PO4)2)=310.2
.n atoms (O)=?
Find the amount of substance of Ca3(PO4)2
.n(Ca3(PO4)2)= m/Mr=40/310.2=0.13 mol
.n (O)= 8 n (Ca3(PO4)2)= 8 *0.13=1.03 mol
. n atoms (O) = n *Na=1.03 *6*10^23=6.18 *10^23