A 43.0 g sample of hydrated cobalt (II) sulfite is heated to drive off the water. The anhydrous cobalt (II) sulfite has a mass of 26.1 g. Calculate the mass of water. Calculate the moles of water. Calculate the moles of cobalt (II) sulfite. Calculate the mole ratio of water to cobalt (II) sulfite and write the correct formula.
Please help. I have been stuck on this for the past hour.
Ok. Please. Check. Thank you both so much.
Mass of water:
43.0-26.1=16.9 g H2O
Moles of water:
(16.9 g H2O)(1 mol H2O/18.0 g H2O)= 0.939 mol H2O
Moles of cobalt (II) sulfite:
CoSO3= 58.9332+32.066+3(15.9994)=138.9974 g/mol
(26.1 g CuSO3)(1 mol CuSO4/138.9974 g CuSO3)= 0.188 mol CuSO3
Mole ratio of water to cobalt (II) sulfite and correct formula:
0.188 mol CuSO3/0.188 mol CuSO3= 1.000 mol CuSO3
0.939 mol H20/0.188= 4.99 (round up)= 5 H2O
CuSO3. 5H2O
a. masswater=difference wet-dry mass
b. moleswater=masswater/18
c. moles cobaltII sulfite=dry mass/molmassCobaltII sulfite
d. ratio water=moleswater/molessulfite
formula CoSO3.nH2O where n is found in step d.
CoSO3.xH2O
mass hydrated salt = 43.0
mass anhydrous salt = 26.1
mass H2O lost = 43.0-26.1 = 16.9g
mols H2O = 16.9/molar mass H2O = ?
mols CoSO3 = 26.1/molar mass CoSO3=?.
Find the ratio of the two with CoSO3 being 1.00. The easy way to do that is to divide the mols of BOTH by mols CoSO3. That way CoSO3 = 1.000000000, divide the H2O by the same number and round to a whole number. Then the formula for the hydrate will be CoSO3.xH2O where x is the whole number.
Post your work if you get stuck and I can help you through it. Actually this is pretty simple.
To solve this problem, we'll need to use the concept of stoichiometry and the molar masses of the compounds involved.
First, we need to calculate the mass of water by finding the difference between the mass of the hydrated compound and the anhydrous compound.
Mass of water = Mass of hydrated compound - Mass of anhydrous compound
Mass of water = 43.0 g - 26.1 g
Mass of water = 16.9 g
Next, let's calculate the moles of water by using the molar mass of water (H2O), which is approximately 18.015 g/mol.
Moles of water = Mass of water / Molar mass of water
Moles of water = 16.9 g / 18.015 g/mol
Moles of water ≈ 0.938 mol
Moving on to the moles of cobalt (II) sulfite, we need to use the molar mass of cobalt (II) sulfite (CoSO3), which is approximately 119.01 g/mol.
Moles of cobalt (II) sulfite = Mass of anhydrous compound / Molar mass of cobalt (II) sulfite
Moles of cobalt (II) sulfite = 26.1 g / 119.01 g/mol
Moles of cobalt (II) sulfite ≈ 0.219 mol
Finally, to determine the mole ratio of water to cobalt (II) sulfite, we need to divide the number of moles of water by the number of moles of cobalt (II) sulfite.
Mole ratio of water to cobalt (II) sulfite = Moles of water / Moles of cobalt (II) sulfite
Mole ratio of water to cobalt (II) sulfite ≈ 0.938 mol / 0.219 mol
Mole ratio of water to cobalt (II) sulfite ≈ 4.28 mol
Therefore, the mole ratio of water to cobalt (II) sulfite is approximately 4.28 mol.
The correct formula for hydrated cobalt (II) sulfite can be determined by analyzing the mole ratio. Since the mole ratio is approximately 4.28, we can write the formula as CoSO3·4.28H2O.
I hope this helps! Let me know if you have any further questions.