Calculate the amount (g) of each of the following salts that need to be added to a 200 mL volumetric flask to produce a 2 M solution of S.

CaSO(subscript 4)

[NH(subscript 4)subscript 2]SO(subscript 4) NOTE: the subscript 2 is outside the (NHsubscript 4)

Fe(subscript 2)[(SOsubscript 4)}subscript 3 NOTE: O has a subscript of 4, but the quantity SOsubscript 4 has a subscript of 3 outside.

Please help. I have no idea where to begin.

I'll do the CaSO4 (btw, you help clutter the problem with the explanation of subscripts etc. It would be nice if html was convenient for subscripts but it isn't. Most of us here recognize that number appearing BEFORE the formula means it is a coefficient and numbers AFTER a symbol means a subscript. For example, 2CaSO4 means 2 mols calcium sulfate.)

You want 200 mL of 2M solution of S and the S is to come from CaSO4.

How many mols S do you want? That is M x L = 2 x 0.200 = 0.400 mols.
g S = mols x atomic mass = 0.400 x 32 = about 13 but you need to do it more accurately. (Note: you may have been taught to use S8 as elemental S; if so, use S8 at this point.)
Then we convert 13 g S (or whatever number you have) to g CaSO4. That is done by
13g S x (molar mass CaSO4/atomic mass S) = ?
I'll be glad to check you numbers for the others.

Dr. Bob222,

Could you please check the following?
Thank you.

CaSO4: 54.45664 g

(NH4)2SO4: 12.8264 x 132.1402/32.066 (molar mass(NH4)2SO4/atomic mass S)= 52.85608 g

Fe2(SO4)3: 12.8264 x 399.8801/32.066
(molar mass Fe2(SO4)3/atomic mass S)=
159.95204 g

Thank you again.

To calculate the amount of each salt needed to prepare a 2 M solution of S in a 200 mL volumetric flask, you will need to use the formula:

Amount (g) = Concentration (M) × Volume (L) × Molar mass (g/mol)

First, let's calculate the molar mass for each salt:

1. CaSO₄ (Calcium sulfate):
- Molar mass of Ca: 40.08 g/mol
- Molar mass of S: 32.07 g/mol
- Molar mass of 4O: 4 × 16.00 g/mol = 64.00 g/mol

Molar mass of CaSO₄: 40.08 g/mol + 32.07 g/mol + 64.00 g/mol = 136.15 g/mol

2. [NH₄]₂SO₄ (Ammonium sulfate):
- Molar mass of 2NH₄: 2 × (14.01 g/mol + 4 × 1.01 g/mol) = 2 × 18.05 g/mol = 36.10 g/mol
- Molar mass of S: 32.07 g/mol
- Molar mass of 4O: 4 × 16.00 g/mol = 64.00 g/mol

Molar mass of [NH₄]₂SO₄: 36.10 g/mol + 32.07 g/mol + 64.00 g/mol = 132.17 g/mol

3. Fe₂(SO₄)₃ (Iron(III) sulfate):
- Molar mass of Fe: 2 × 55.85 g/mol = 111.70 g/mol
- Molar mass of 3S: 3 × 32.07 g/mol = 96.21 g/mol
- Molar mass of 12O: 12 × 16.00 g/mol = 192.00 g/mol

Molar mass of Fe₂(SO₄)₃: 111.70 g/mol + 96.21 g/mol + 192.00 g/mol = 399.91 g/mol

Now we can calculate the amount of each salt needed:

1. CaSO₄:
Amount (g) = 2 M × 0.200 L × 136.15 g/mol = 54.46 g

2. [NH₄]₂SO₄:
Amount (g) = 2 M × 0.200 L × 132.17 g/mol = 52.87 g

3. Fe₂(SO₄)₃:
Amount (g) = 2 M × 0.200 L × 399.91 g/mol = 159.96 g

Therefore, you need to add 54.46 grams of CaSO₄, 52.87 grams of [NH₄]₂SO₄, and 159.96 grams of Fe₂(SO₄)₃ to the 200 mL volumetric flask to produce a 2 M solution of S.