A sample of CuSO4*5H2O was heated to 110 degrees C, where it lost water and gave another hydrate of copper (II) ion that contains 32.50% Cu. A 98.77-mg sample of this new hydrate gave 116.66 mg or barium sulfate precipitate when treated with barium nitrate solution. What is the formula of the new hydrate?
The answer was given Copper Sulfate Dehydrate (CuSO4*2H2O) but how do I find this answer from this equation, it is extremely overwhelming and I have been reading my text for 3 hours now to find directions toward this solution with no help. Can someone please help???
This is as far as I have got in the past few hours:
CuSo4 x 5H2O --> CuSO4 x ? H2O
CuSO4 x ? H2O + Ba (NO2)2 --> BaSO4
116.66 mg BaSO4
98.77 mg CuSO4 x ? H2O
At some point once I can get the % of the CuSO4 x ? H2O I can turn it into an empirical formula problem and solve but I still cannot get there, any help?
(0.3250)(0.09877g) = 0.03210 g Cu - - - -> also moles of CuSO4
0.03210 g Cu / 63.55 g/mol = 5.05x10^-4 moles Cu
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1 mole BaSO4 = 233.4 g
0.11666g BaSO4 / 233.4 g/mole = 5.000x10^-4 moles BaSO4
(confirms mol Cu = moles SO4 as expected in CuSO4
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(0.0005 moles CuSO4)(159.6g/mol) = 0.0798g CuSO4
0.09877g hydrate - 0.07980 g CuSO4 = 0.01897 g H2O
0.01897 g H2O /18.015 g/mol H2O = 1.053x10^-3 mol H2O
1.053x10^-3 mol H2O / 5.05x10^-4 mol CuSO4 = 2.00 mol H2O / 1 mole CuSO4
What does the above ratio mean?
Notes on previous posting:
1) Correction: 5.05x10^-4 moles Cu - - - -> also moles of CuSO4
2) The data on BaSO4 confirms that the anhydrous salt is and remains CuSO4 since moles Cu = moles SO4
3) The last set of calculations gives the mole ratio of H2O/CuSO4 in the second hydrate. This ratio confirms its assumed formula.
To determine the formula of the new hydrate, we need to analyze the given information step-by-step. Let's break it down.
1. Start with the initial hydrate, CuSO4*5H2O, which was heated to 110 degrees Celsius. This heating caused the loss of water and formation of a new hydrate.
2. The new hydrate contains copper (II) ions, Cu, and has a composition of 32.50% Cu. This means that 32.50% of the total mass of the new hydrate is due to copper.
3. We are given that a 98.77-mg sample of the new hydrate produced 116.66 mg of barium sulfate precipitate when treated with barium nitrate solution.
4. Barium sulfate, BaSO4, is an insoluble compound that can be used to determine the presence of sulfate ions, SO4^2-. When barium nitrate, Ba(NO3)2, reacts with the new hydrate, it forms barium sulfate precipitate, BaSO4.
5. The mass of the barium sulfate precipitate is 116.66 mg.
Now, let's use this information to find the formula of the new hydrate.
Step 1: Calculate the mass of copper in the 98.77-mg sample.
Mass of Cu = 32.50% of 98.77 mg = 0.325 * 98.77 mg = 32.13 mg
Step 2: Calculate the number of moles of copper.
Molar mass of Cu = 63.55 g/mol
Moles of Cu = Mass of Cu / Molar mass of Cu = 32.13 mg / 63.55 g/mol = 0.506 mol
Step 3: Calculate the number of moles of sulfate ions (SO4^2-) from the barium sulfate precipitate.
Molar mass of BaSO4 = 233.39 g/mol
Moles of BaSO4 = Mass of BaSO4 / Molar mass of BaSO4 = 116.66 mg / 233.39 g/mol = 0.500 mol
Step 4: Determine the ratio of copper to sulfate ions.
From the balanced equation of the reaction between the new hydrate and barium nitrate, we know that 1 mole of copper reacts with 1 mole of sulfate ions to form 1 mole of barium sulfate.
So, the ratio of moles of copper to moles of sulfate ions is 1:1.
Step 5: From the ratio calculated above, we deduce that there is 0.500 mol of copper in the new hydrate.
Step 6: Use the moles of copper to calculate the number of moles of water lost during heating.
The initial hydrate, CuSO4*5H2O, contains 5 moles of water per mole of copper.
So, the new hydrate, CuSO4*xH2O, must contain x moles of water per mole of copper.
Therefore, 5 moles - x moles = 0.500 mol
Solving for x, we get x = 4.5 moles
Step 7: Now we have all the information to determine the formula of the new hydrate.
The formula of the new hydrate is CuSO4*4.5H2O.
However, since the formula of a hydrate should have whole numbers for the water molecules, we multiply the entire formula by 2 to get a whole number for water atoms:
2(CuSO4*4.5H2O) = 2CuSO4*9H2O
Therefore, the formula of the new hydrate is CuSO4*9H2O.
I apologize for the confusion earlier in the answer. The correct formula for the new hydrate is CuSO4*9H2O, not CuSO4*2H2O.