Need help on an important chem question regarding concentration. PLEASE EXPLAIN. I have the answer already?

Question

a solution is prepared by dissolving 10.8g ammonium sulfate in enough water to make 100.0mL of stock solution. A 10.00mL sample of this stock solultion is added to 50.00mL of water. calculate the concentration of ammonium ions and sulfate ions in the final solution.

Answer 1

You have the weight, but you need to convert that to moles, using the molecular weight.

(NH4)2SO4 is the formula, right?

Calculate the molecular weight

2N= 28 g/mol
8H= 8 g/mol
S=32 g/mol
4O= 64 g/mol

Total (without sig figs)= 132 g/mol
Then (follow the units, they make it easy)

convert your 10.8 g into moles (divide by the molecular weight)

10.8 g/ 132 g/mol= .08 mol

Then you used 10 ml/ 100 ml of that for the next solution, so that's .008 moles in the final solution.

So, in molarity you have
.008 moles of SO4 ion/.060 L=
and
.016 moles of NH4 ion/.060 L=

The math is left for you. But you have 2x the ammonium ion as sulfate ion due to the stoichometry of the molecule.

Answer 2

Ok...lets see here...
The first solution is 10.8 g per 100 ML. SO, that means the 10.00 ML sample is 1.08 g per 10.00 mL. Mix this into a 50. ML and you have 1.08 g per 60 mL.

That close enough?