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Homework Help: Chemistry Help - Magnesium Carbonate decomposition

  1. Sep 23, 2006 #1
    Hey everyone,

    I have a homework question that I would appreciate some clarification on.


    A 3.96g sample of magnesium carbonate decomposed to produce 1.89g of magnesium oxide. What mass of magnesium was in the sample of magnesium carbonate? Calculate the mass percent of of magnesium, by mass, in the magnesium oxide produced. How much CO2 (carbon dioxide) was released into the atmosphere?


    In an example in the textbook it states that there is 0.301g of magnesium in 0.500g of magnesium oxide. Using this information,

    since MgCO3 --------(arrow) MgO + CO2

    1.89g magnesium oxide X 0.301g magnesium/0.500 magnesium oxide
    = 1.14g of magnesium in 1.89g magnesium oxide

    therefore, there is 1.14g of magnesium in 3.96g of magnesium carbonate.

    and, 3.96g magnesium carbonate - 1.89g magnesium oxide = 2.07g Carbon dioxide released into atmosphere.

    Can anyone please verify if my solution is correct, please?
  2. jcsd
  3. Sep 24, 2006 #2


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    Staff Emeritus
    Science Advisor

    The solutions are correct. One used the ratio of Mg/MgO correctly. And certainly the difference in mass between MgCO3 and MgO must be the CO2.

    Another way to look at this is determine how much Mg is in the CO2 by mass using the atomic masses.

    In mole of MgCO3, one would have 1 mole of Mg, 1 mole of C, and 3 moles of O. Mass wise, one would have 24.3 g of Mg, 12 g of C and 48 g (3 x 16 g) of O, which gives a mass of one mole of MgCO3 as 84.3 grams. That gives a ratio of 24.3 g Mg/84.3 g MgCO3 = 0.288 g Mg/ g MgCO3.

    3.96 g * 0.288 = 1.14 g, which is the answer one obtained.

    As for CO2, one found the mass but difference between masses of MgCO3 and MgO, which is correct.

    However, one could also find the moles of CO2 released, by the virture the one mole of MgCO3 releases one mole of CO2.

    To determine the moles of MgCO3, one simply takes the mass 3.96 g (MgCO3) and divides by the mass of one mole (84.3 g/ g-mole of MgCO3). This would yield - 0.047 moles of MgCO3, which would yield 0.047 moles of CO2.
  4. Sep 25, 2006 #3
    I didn't even think of doing it like that. Thanks Astronuc!
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