[Reducto]

The last post got deleted, I assume, because i listed the question and did not post the answers I got. So hopefully you guys can help me check my answers this time.

The Question was:

a) How many formula units are contained in 154.3g of K2CrO4 (Potassium Chromate)

b) How many potassium Ions ?

c) How many CrO4 Ions?

d) how many atoms of all kinds?

Here are the answers I got:

a) 154.3g / 194.3g/mole= .7945 mol  x  6.022x10^23=4.785x10^23 formula units

                    4.785 x 10^23 formula units

b) 154.3g / 78.2 g/mol= 1.97 mol x 6.022x10^23= 1.186x10^24 x 2 Potassium atoms= 2.37x10^24 K Ions

                     2.37 x 10^24 K Ions.

c)  154.3g / 116 g/mo; = 1.33 mol x 6.022x10^23= 8.01x10^23 x 5 Atoms in CrO4= 4.005 x 10^24 CrO4 Ions

                     4.005 x 10^24 CrO4 Ions.

d)  4.785x10^23 x 7 Atoms= 3.3495 x 10^24 Atoms

                     3.3495 x 10^24 Atoms (Im unsure about this one)

I want to make sure my methods are correct. Please tell me if I have verbage and units correct as well. Thanks.

[Borek]

a is OK. This answer should be your starting point for all others questions.

b and c are awfully wrong, but d is OK.

You should do b and c the same way you did d.

Thnk about it this way: in a you calculated how many 'particles' fo potassium chromate you have. Every particle contains two K⁺ ions, thus answer to b is 2*a.