Practice question for upcoming exam!!! I did it a bunch of times and my answer keeps turning out wrong...

Question

A solution is prepared by dissolving 10.8 g ammonium sulfate

in enough water to make 100.0 mL of stock solution. A

10.00-mL sample of this stock solution is added to 50.00 mL

of water. Calculate the concentration of ammonium ions and

sulfate ions in the final solution.

Answer 1

So molarity is what tells us the concentration of the chemical substance.

Recall that Molarity = moles⁄ liters or:

M = mol/ L

First, let's find the moles of ammonium sulfate which is (NH₄)₂SO₄

ammonium comes from the polyatomic ion NH₄ ⁺

and sulfate comes from the polyatomic ion SO₄ ^2-

Using the criss cross method for the charges, the compound is (NH₄)₂SO₄

To find moles we must divide the grams of ammonium sulfate with it's molar mass which is about 132.144 g/mol.

10.8 grams of (NH₄)₂SO₄ × [1 mol of ((NH₄)₂SO₄ ⁄ 132.144 g (NH₄)₂SO₄] = 0.0817 moles of (NH₄)₂SO₄

100.0 mL of stock solution × [ 1 liter ⁄ 1000 mL] = 0.100 liters of stock solution

0.0817 moles / 0.100 L = 0.817 M (NH₄)₂SO₄

Since we know in 1L = 1000 mL (milli means a thousandth)

We can convert 100.0 milliliters to liters:

10.0 mL of added stock solution × [ 1 liter ⁄ 1000 mL] = 0.010 liters of added stock solution

50.0 mL of water × [ 1 liter ⁄ 1000 mL] = 0.050 liters of water

Taking the sum of the volume of stock solution including water we have:

0.010 liters of stock solution + 0.050 liters of water = 0.060 liters in total

To find the molarity, or the concentration, of the ions, we must set up a proportion:

(Molarity of Ammonium Sulfate Initially) × (Volume of added Stock Solution in Liters) = (Molarity of Ammonium Sulfate Diluted) × (Total Volume of stock solution and water)

M₁V₁ = M₂V₂

0.817 M (NH₄)₂SO₄ × 0.010 liters of added stock solution = Molarity of diluted (NH₄)₂SO₄ × 0.060 liters in total

Letting M₂ being the molarity of the ammonum sulfare compound

0.817 M × 0.010 L = M₂ × 0.060 L

M₂ = (0.817 M × 0.010 L ) / (0.060 L) = 0.136 M of (NH₄)₂SO₄

For the individual ions:

Since there's two moles of ammonium

0.136 M × 2 = 0.272 M of NH₄ ⁺

0.136 M of SO₄ ^2-

Answer 2

So molarity is what tells us the concentration of the chemical substance.

Recall that Molarity = moles⁄ liters or:

M = mol/ L

First, let's find the moles of ammonium sulfate which is (NH₄)₂SO₄

ammonium comes from the polyatomic ion NH₄ ⁺

and sulfate comes from the polyatomic ion SO₄ ^2-

Using the criss cross method for the charges, the compound is (NH₄)₂SO₄

To find moles we must divide the grams of ammonium sulfate with its molar mass which is about 132.144 g/mol.

10.8 grams of (NH₄)₂SO₄ × [1 mol of ((NH₄)₂SO₄ ⁄ 132.144 g (NH₄)₂SO₄] = 0.0817 moles of (NH₄)₂SO₄

100.0 mL of stock solution × [ 1 liter ⁄ 1000 mL] = 0.100 liters of stock solution

0.0817 moles / 0.100 L = 0.817 M (NH₄)₂SO₄

Since we know in 1L = 1000 mL (milli means a thousandth)

We can convert 100.0 milliliters to liters:

10.0 mL of added stock solution × [ 1 liter ⁄ 1000 mL] = 0.010 liters of added stock solution

50.0 mL of water × [ 1 liter ⁄ 1000 mL] = 0.050 liters of water

Taking the sum of the volume of stock solution including water we have:

0.010 liters of stock solution + 0.050 liters of water = 0.060 liters in total

To find the molarity, or the concentration, of the ions, we must set up a proportion:

(Molarity of Ammonium Sulfate Initially) × (Volume of added Stock Solution in Liters) = (Molarity of Ammonium Sulfate Diluted) × (Total Volume of stock solution and water)

M₁V₁ = M₂V₂

0.817 M (NH₄)₂SO₄ × 0.010 liters of added stock solution = Molarity of diluted (NH₄)₂SO₄ × 0.060 liters in total

Letting M₂ being the molarity of the ammonum sulfate compound

0.817 M × 0.010 L = M₂ × 0.060 L

M₂ = (0.817 M × 0.010 L ) / (0.060 L) = 0.136 M of (NH₄)₂SO₄

For the individual ions:

Since there's two moles of ammonium

0.136 M × 2 = 0.272 M of NH₄ ⁺

0.136 M of SO₄ ^2-