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a student did not completely burn all the magnesium and some un-burned magnesium remained. How will this affect the empirical formula found? Solution Mg(s) + N 2 (g) + O 2 (g) ---> MgO(s) + Mg 3 N 2 (s) If you left some Mg 3 N 2 in the crucible as product you would calculate a ratio of MgO to contain too little oxygen. When we weigh the final contents we wont know that the contents have more magnesium, we assume itâ€™s pure MgO. Mg 3 N 2 does not contain any oxygen, so there would be too little oxygen, but there is magnesium present so there is more magnesium than oxygen. Since we did not completely burn all the magnesium and some un-burned magnesium remained, hence we will find that Mg:O ratio will be too high. Hope this helped you! Thank You So Much! Please Rate this answer as you wish.(\"Thumbs Up\") .

a student did not completely burn all the magnesium and some un-burned magnesium remained.
How will this affect the empirical formula found?
Solution
Mg(s) + N 2 (g) + O 2 (g) ---> MgO(s) + Mg 3 N 2 (s)
If you left some Mg 3 N 2 in the crucible as product you would calculate a ratio of MgO to
contain too little oxygen. When we weigh the final contents we wont know that the contents have
more magnesium, we assume itâ€™s pure MgO. Mg 3 N 2 does not contain any oxygen, so there
would be too little oxygen, but there is magnesium present so there is more magnesium than
oxygen. Since we did not completely burn all the magnesium and some un-burned magnesium
remained, hence we will find that Mg:O ratio will be too high.
Hope this helped you!
Thank You So Much! Please Rate this answer as you wish.("Thumbs Up")

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