Molar mass, Percent Yield, theoretical yield.

Discussion in 'Chemistry' started by ljs88, Feb 22, 2009.

  1. ljs88 Registered Member

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    Never mind, I got it taken care of!
     
    Last edited: Feb 24, 2009
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  3. James R Just this guy, you know? Staff Member

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    This sounds like homework. What have you done so far?
     
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  5. Trippy ALEA IACTA EST Staff Member

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    I was just thinking/wondering the same thing.
     
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  7. ljs88 Registered Member

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    theoretical, molar mass, percentage yield

    Heres my work

    To figure out the molar mass of CaCl2•H2O (rounding to the hundredths place), I did

    Ca = 40.08
    Cl2 = (35.45) x 2 = 70.9
    2H2O = (18.02) x 2 = 36.04

    40.08 + 70.90 + 36.04 = 147.02 g/mol


    2.97g CaCl•2H2O x (1 mol CaCl•2H2O/147.02g CaCl•2H2O) x (1 mol CaCO3/1 mol CaCl•2H2O) x (100.09/1 mol CaCO3) = 2.04 g

    (The coefficients were all one, so I used 1 mol in all the neccessary conversion factors.)

    Now, the third question asked "What is the percent yield of calcium carbonate if your theoretical yield was 2.07 grams" even though I came out with 2.04 g as my theoretical yield.

    So, percentage yield, (1.46/2.07) X 100 = 70.5%.

    Was I accurate in all these equations? If not, could you guide me to where I need to make a modification? Thanks!
     

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