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.
CaSO4
(NH4)2SO4
Fe2(SO4)3
Dr. BOB222,
Could you please check the following?
Thank you.
CaSO4: moles x atomic mass = .0400 x 32.066 amu = 12.8264 g
12.8264 gS x molar mass CaSO4/atomic mass S =
12.8264 g x 136.1416/32.066 =
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.
I believe you are missing something here.
You want mols S = 0.200L x 2M = 0.4 mol S.
0.4mol S x (1 mol CaSO4/1 mol S) = 0.4 mol CaSO4
0.4 mol CaSO4 x molar mass CaSO4 = g CaSO4.
I'm confused again. Can you show me how to get the CaSO4 so I can do the other two? What did I do wrong? I showed you my work. I thought I followed your instructions from Saturday. You said I needed a number near 13. I got the 12.8264. If you could please do the CaSO4, I could probably use that as an example to finish the other two. Please help me.
Thank you.
I believe I read your answer as 12 and not as 55.
Your answer for a is 54.45664 and I don't see anything wrong with that except you have carried it out to more places than you are allowed. You can have 3 significant figures; I would round that to 54.5 grams.
I will check the others and try to read the right line this time.
b is ok except you need to round it.
I believe #3 is wrong. Let me show you how I would do it. Either I showed you something wrong earlier or you misunderstood somewhere along the line. At any rate, here is how to do these.
You want mols S = M x L = 2 x 0.200 = 0.400 mol S.
0.400 mol S x (1 mol CaSO4/1 mol S) = 04 mol CaSO4.
0.4 mol CaSO4 x 136.14 = 54.5 g CaSO4.
0.4 mol S x (1 mol (NH4)2SO4/1 mol S) = 0.4 mol (NH4)2SO4.
g(NH4)2SO4 = mol x molar mass = 0.400 x 132.14 = 52.9g
0.4 mol S x (1 mol Fe2(SO4)3/3 mol S) = 0.4 x 1/3 = 0.1333 mol Fe2(SO4)3
g = mols x molar mass = 0.1333 x 399.88 = 53.3g
If you know where that earlier post is give me a link. I need to look and see if I goofed. If I did perhaps I can delete my response. We don't need to keep wrong answers on the board.
To calculate the amount (g) of each salt that needs to be added to a 200 mL volumetric flask to produce a 2 M solution of S, you can follow these steps:
1. Determine the moles of sulfur (S) required for a 2 M solution:
- The molarity (M) of a solution is defined as moles of solute per liter of solution.
- Since you have a 200 mL (0.2 L) volumetric flask, you need to calculate the moles of S required for 0.4 moles (2 M x 0.2 L) of the solution.
2. Calculate the grams of each salt needed to provide the required moles of S:
- Take into account the molar mass of S in each salt formula.
Using your calculations as an example:
For CaSO4:
- You calculated that 0.0400 moles of CaSO4 are required to make a 2 M solution.
- To convert to grams, multiply by the molar mass of CaSO4 over the atomic mass of S, which is 136.1416 g/mol / 32.066 g/mol.
- This gives you the amount of CaSO4 needed: 0.0400 moles x (136.1416 g/mol / 32.066 g/mol) = 54.45664 g.
For (NH4)2SO4:
- Using the same method, you calculated that 0.0400 moles of (NH4)2SO4 are required.
- Multiply by the molar mass of (NH4)2SO4 over the atomic mass of S, which is 132.1402 g/mol / 32.066 g/mol.
- This gives you the amount of (NH4)2SO4 needed: 0.0400 moles x (132.1402 g/mol / 32.066 g/mol) = 52.85608 g.
For Fe2(SO4)3:
- Again, using the same method, you calculated that 0.0400 moles of Fe2(SO4)3 are required.
- Multiply by the molar mass of Fe2(SO4)3 over the atomic mass of S, which is 399.8801 g/mol / 32.066 g/mol.
- This gives you the amount of Fe2(SO4)3 needed: 0.0400 moles x (399.8801 g/mol / 32.066 g/mol) = 159.95204 g.
So, to make a 2 M solution of S in a 200 mL volumetric flask, you would need to add:
CaSO4: 54.45664 g
(NH4)2SO4: 52.85608 g
Fe2(SO4)3: 159.95204 g