Post a New Question


posted by .

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.

  • Chemistry -

    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.

  • Chemistry -

    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.

  • Chemistry -

    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

  • Chemistry -

    I didn't check the math step by step but the formula is correct at CoSO3.5H2O.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question