Explain how #4"Ca"_3("PO"_4)_2# contains #52# total atoms?

Ca=?
P=?
O=?

(I'm getting 22 total and not quite understanding what I've done wrong)

1 Answer
Feb 19, 2018

Here's how I do it.

Explanation:

Assume you have one formula unit #"Ca"_3("PO"_4)_2#.

The subscripts represent the number of atoms or groups immediately before them.

Thus,

#"Ca"_3# means that you have three #"Ca"# atoms.

#("PO"_4)_2# means that you have two #"PO"_4# groups.

One #"PO"_4# group contains one #"P"# atom and four #"O"# atoms.

∴ Two #"PO"_4# group contain two #"P"# atoms and eight #"O"# atoms.

Let's count the number of atoms in one formula unit.

#"Ca" color(white)(m)= color(white)(l)"3 atoms"#
#"P"color(white)(mll) =color(white)(l) "2 atoms"#
#ul("O" color(white)(mll)= color(white)(l)"8 atoms")#
#"Total = 13 atoms"#

If 1 formula unit contains 13 atoms, then
#color(white)(ll)# 4 formula units contains #4 × 13 = 52# atoms.