If 8.5g of pure ammonium phosphate , (NH4)3PO4(s), is dissolved in distilled water to make 400mL of solution, what are the concentrations (in moles per litre) of ions in the solution.

How would I solve this type of stoichiometric problem. I know the steps
Step 1: make the balanced equation and state the givens and unknowns
Step 2: find the moles using c=n/v
Step 3: find the mole ratio
Step 4: find the mass

Please guide me! Thanks!!:) I appreciate your time!

Steps 1, 2, 3 look good. I don't know why you need to find the mass.

Why don't you follow the steps and see what happens?

To solve this stoichiometric problem, follow these steps:

Step 1: Write the balanced equation and state the givens and unknowns:
The balanced equation for the dissolution of ammonium phosphate ((NH4)3PO4) in water is:
(NH4)3PO4(s) → 3NH4+(aq) + PO4-3(aq)

The given information:
- Mass of ammonium phosphate: 8.5 g
- Volume of solution: 400 mL (or 0.4 L)

The unknown is the concentration (in moles per liter) of ions in the solution.

Step 2: Find the moles using c = n/V:
To find the number of moles of ammonium phosphate, divide the mass by its molar mass (M):
n(NH4)3PO4 = m / M(NH4)3PO4

The molar mass of ammonium phosphate is calculated by adding the atomic masses of each element in the formula:
M(NH4)3PO4 = (3 x M(N)) + M(P) + (4 x M(H)) + (4 x M(O))

Step 3: Find the mole ratio:
Since ammonium phosphate dissociates into 3 ammonium ions (NH4+) and 1 phosphate ion (PO4-3), the mole ratio is 3:1. This means that for every 1 mole of ammonium phosphate that dissolves, it produces 3 moles of ammonium ions and 1 mole of phosphate ions.

Step 4: Find the concentration:
The concentration (C) is calculated by dividing the number of moles (n) by the volume (V) in liters:
C = n / V

For the ammonium ions (NH4+), the concentration would be:
C(NH4+) = n(NH4+) / V

Substituting the values:
C(NH4+) = n(NH4+) / V = (n(NH4+) / n(NH4)3PO4) * C(NH4)3PO4

Given that n(NH4+) = 3 * n(NH4)3PO4 (from the mole ratio), you can simplify the equation to:
C(NH4+) = 3 * C(NH4)3PO4

Similarly, for the phosphate ions (PO4-3):
C(PO4-3) = n(PO4-3) / V = (n(PO4-3) / n(NH4)3PO4) * C(NH4)3PO4

Given that n(PO4-3) = 1 * n(NH4)3PO4 (from the mole ratio):
C(PO4-3) = 1 * C(NH4)3PO4 = C(NH4)3PO4

So, to find the concentrations of ions in the solution, you need to calculate the concentration of ammonium phosphate (C(NH4)3PO4) and multiply it by 3 to get the concentration of ammonium ions (C(NH4+)). The concentration of phosphate ions (C(PO4-3)) is equal to the concentration of ammonium phosphate (C(NH4)3PO4).

After performing the calculations, you will have the concentrations (in moles per liter) of the ions in the solution.