Question

MgO Lab: How would your calculated value for the percent composition of magnesium oxide been affected if all the magnesium in the crucible had not reacted?

Calculation:
`% Mg = (m_(Mg))/m_(MgO)` x 100%

Will the %Mg be over-reported? Under-reported?

Answer 1

Would it not be under-reported?

Explanation:

Look at your quotient.......

`"%Metal" ="Mass of magnesium"/"Mass of magnesium oxide"xx100%`......

Incomplete reaction DETRACTS from the denominator, and reduces the measured mass of magnesium oxide...., i.e. we measure `"mass of magnesium oxide + mass of unreacted metal"`. Since the starting mass of metal is UNCHANGED, the quotient should reduce...and %yield should be under-reported.....

Since magnesium oxidizes fairly fiercely, this circumstance is unlikely to occur.....of course, given the reaction some mass loss of the magnesium oxide ash could occur.

Here is a short video which summarizes this experiment.