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Posted by on Monday, October 4, 2010 at 9:24pm.

when a student removed the crucible cover afer the first strong heating in creating magnesium oxide, some of the white was knocked from the cover to the floor and was lost. did this loss cause the calculated number of moles of oxygen in the compound to be too high or too low?

  • chemistry - , Monday, October 4, 2010 at 9:56pm

    Too low but you can figure that yourself.
    Let's say we start with 5 g Mg and prepared MgO properly. We will get the following numbers (I calculated what they should be).
    mass Mg = 5.00 g
    mass MgO = 8.29 g
    mass Oxygen = 8.29- 5.00 = 3.29
    So:
    moles Mg = 5.00/24.3 = 0.206
    moles oxygen = 3.29/16 = 0.206
    and the ratio is 1:1 for the formula of MgO.

    Now that we have made up what SHOULD happen, we can calculate what happens if we lose some of the white powder.
    mass Mg = 5.00 g
    mass oxide = 6.65 g (instead of 8.29)
    mass oxygen = 6.65 - 5.00 = 1.65 g
    Now we determine moles
    moles Mg = 5.00/24.3 = 0.206
    moles oxygen = 1.65/16 = 0.103
    So the formula would be Mg2O.
    Thus losing some of the white powder would make moles oxygen too low.

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