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Mass percentage

  1. Jun 30, 2016 #1
    Hi! I got a question. I have been given the reaction of the equation and no other data. I have to calculate the mass percentage of H2CO3. The gas dissolves in water.

    1. The problem statement, all variables and given/known data

    CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)

    2. Relevant equations


    3. The attempt at a solution
    CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)
    1 mol 1 mol 1 mol 1 mol 1 mol
    m (CO2) = 1 mol * 44g/mol = 44 g
    m (H2O) = 1 mol * 18g/mol = 18 g
    m (solution) = 44 g + 18 g = 62 g
    % = (44 g / 66 g) * 100 = 70,96%
    or
    CaCO3 + H2SO4 = CaSO4 + H2O(l) + CO2(g)
    n (CO2) = 1 dm3 / 22,4 dm3/mol = 0,045 mol
    m (CO2) = 0,045 mol * 44 g/mol = 1,98 g
    n (H2O) = 0,045 mol (1:1)
    m (H2O) = 0,045 mol * 18 g/mol = 0,81 g
    m (solution) 1,98 g + 0,81 g = 2,79 g
    % = (1,98 g / 2,79 g) * 100 = 70,96%

    Does it even make sense?
     
    Last edited: Jun 30, 2016
  2. jcsd
  3. Jun 30, 2016 #2
    I would interpret "the mass percentage of the H2CO3" as: the ratio of the mass of the H2CO3 over the total mass of the products, expressed as a percentage.

    If that's what you want, then you will need to know the mass of one mol of CaSO4.
     
  4. Jun 30, 2016 #3
    What is the exact statement of your problem?
     
  5. Jun 30, 2016 #4
    Assume that it's 136 g. How do I calculate the ratio?
     
  6. Jun 30, 2016 #5
    The mass of the H2CO3 (62) over the total mass of the products (__ + __).

    But Chestermiller has a point. I am not convinced you have paraphrased the problem correctly and completely.
     
  7. Jun 30, 2016 #6
    It's possible that I forgot something. It was on the test.
     
  8. Jun 30, 2016 #7
    Your original computation was the percentage of mass of the CO2 in the solution to the solution (H2CO3).
    I interpreted what you reported as the percentage of mass of the solution (H2CO3) to the total mass of the product (CaSO4+H2CO3).
     
  9. Jun 30, 2016 #8
    Thanks! Now I got it. :)
     
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