What is the percent composition of lead (IV) iodate?

Question

If you can do it can you try to explain it too?

Answer 1

Lead (IV) Iodate has a formula of Pb(IO3)4... that means for each molecule of this there are:

1 Lead (mass 207.2)

4 Iodine (mass 126.9, each)

12 Oxygen (mass 16.0, each)

Now, the total mass of this molecule is:

207.2 + 4(126.9) + 12 (16.0) = 906.8

To find the percent composition of any of the elements, all you have to do is divide the element mass by the total... so

for Lead 207.2/906.8 = 0.228 or 22.8%

for Iodine 507.6/906.8 = 0.560 or 56%

for Oxygen 192/906.8 = 0.212 or 21.2%

Hope this helps...