You weigh a strip of Mg and the mass was 0.178g. You then heat the sample in a crucible until it completely burns. Once it cools the weight of the sample was 0.092g once the weight of the crucible has been subtracted. Determine the mass ratio of Mg to O, mole ration of Mg to O, and the empirical formula of magnesium oxide
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There's something seriously wrong with these numbers.
When magnesium burns in gains mass. So the final weight must be greater than 0.178 g, not less.
If you work the problem already knowing that the mole ratio of Mg to O is 1:1, you get:
(0.178 g Mg) / (24.30506 g Mg/mol) x (1/1) x (40.30449 g MgO/mol) = 0.295 g MgO
So "0.092g" is even more mysterious since it isn't just a typo.
When magnesium burns in gains mass. So the final weight must be greater than 0.178 g, not less.
If you work the problem already knowing that the mole ratio of Mg to O is 1:1, you get:
(0.178 g Mg) / (24.30506 g Mg/mol) x (1/1) x (40.30449 g MgO/mol) = 0.295 g MgO
So "0.092g" is even more mysterious since it isn't just a typo.