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I just did a lab on the heating of magnesium which result in magnesium oxide.Well the question is:

If some of the magnesium oxide had escaped from the crusible, would you percentage composition calculation for magnesium be too high or low. Explain.

I know the magnesium percentage will be greater. Right?
But why?
Im having difficultly tring to explain.

2007-10-24 08:17:47 · 2 answers · asked by Anonymous in Science & Mathematics Chemistry

i still dont get what you are saying can you put it in simpler plz

2007-10-24 08:35:10 · update #1

2 answers

if some of the magnesium oxide escaped, the product mass will be smaller. when you find the mass of oxygen, you subtract the magnesium from the product mass. in other words, you assume ALL magnesium was in the crucible. (of course, it is not true, Mg and O escaped together. ) thus, it appears less oxygen react with magnesium. (for this reason, sometimes students get a very weird mass for oxygen.) therefore, the calculated value will give magnesium a higher percentage.

2007-10-24 08:23:28 · answer #1 · answered by Carborane 6 · 1 0

I don't quite see why the percentage is higher, but I will believe you on that. Try doing a "thought experiment" with say 24 grams (1 g-atom) of magnesium. If all was converted to MgO in the lab, you should have 40 grams of MgO. Now, say 10 percent got lost. Redo your calculation.

I would guess that if you DIDNT KNOW material was lost, you would assume you still had 24 grams of magnesium, and thus you would have determined that the rest (oxygen) comprised the difference.

2007-10-24 08:30:59 · answer #2 · answered by cattbarf 7 · 0 0

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