I would appreciate very much If you guys could show me how to approach this problem.

A solution is prepared by dissolving 10.8g of ammonium sulfate in enough water to make 100.0mL of stock solution.
A 10.00mL sample of this stock solution is added to 50.00mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.
the way I approached this problem was by getting the molarity of ammonium sulfate(of corse after changing the grams to moles and the mL to L.)then I multiply the concentration by .01L since that is the amount of the solution that will be used. Then I calculated the number of moles again with the new liter and molarity, to obtain .01 moles of ammonium sulfate, I did the mole to mole ratio and got .01 moles of ammonium. Then I took the numberof moles and divide bi the water being add which is .05L which gave me .2 of the concentration of ammonium, which is wrong since the answer is .272M for ammonium and .136M for sulfate. Please help me.

calculate the molarity of the first solution. Now, .01L*XXM= new moles of ammonium sulfate.

Concentrated is now the new moles ammoniumslufate/.06

Overall, you have

Concentration= 2*(10.8/molmass)/.1 *.01/.06
Of course, the 2 is because there are two ammonium molecules per molecule of ammonium sulfate.

I do not understand where you got the .06 from?

when one adds 10 ml to 50 ml, the new volume is .06L.

To calculate the concentration of ammonium ions (NH4+) and sulfate ions (SO42-) in the final solution, follow these steps:

Step 1: Find the molarity of the stock solution of ammonium sulfate (NH4)2SO4:
The molarity (M) can be calculated using the formula:

Molarity (M) = Moles (mol) / Volume of solution (L)

Given that 10.8g of ammonium sulfate is dissolved in enough water to make 100.0mL of stock solution, we first need to convert grams to moles.

The molar mass of ammonium sulfate (NH4)2SO4 can be calculated as:
(NH4)2SO4 = (2 * (1.01 + 4.008)) + (32.06 + (4 * 1.01))

Molar mass = 132.14g/mol

Now, we can convert grams of ammonium sulfate to moles:

Moles (mol) = Mass (g) / Molar mass (g/mol)
Moles (mol) = 10.8g / 132.14g/mol

Calculate the moles of ammonium sulfate.

Step 2: Calculate the molarity of ammonium sulfate:
Molarity (M) = Moles (mol) / Volume (L)
Volume (L) = 100.0mL = 0.100L

Molarity (M) = Moles (mol) / Volume (L)
Molarity (M) = Moles of ammonium sulfate / Volume of solution

Step 3: Calculate the moles of ammonium in the 10.00mL sample:
Moles of ammonium = Molarity of ammonium sulfate × Volume of solution used (L)

As you correctly mentioned, the volume used is 0.01L.

Step 4: Calculate the concentration of ammonium ions in the final solution:
Concentration of ammonium (NH4+) = Moles of ammonium / Total volume (L)
The total volume is the sum of the sample volume (0.01L) and the water volume added (0.05L).

Concentration of ammonium (NH4+) = Moles of ammonium / Total volume

Step 5: Repeat steps 3 and 4 for the sulfate ions (SO4^2-).

Remember to consider the mole-to-mole ratio between ammonium sulfate and the respective ions (1:1 ratio) when calculating their concentrations.

Following these steps should give you the correct concentration values for both ammonium and sulfate ions in the final solution.