Explain how each of the following experimental errors would affect the value for the percentage of magnesium.?

I think I know what experiment you are working on so I will go from there.

Lets say you weigh out l.2 grams of Mg ribbon and put it into a crucible and heat the crucible

The Mg reacts with oxygen from the air forming MgO. If you have a perfect lab partner who never makes a mistake. (no such person), you should get 2.0 grams of pure Mg0 as a product.

Then you find the percent of Mg this way. l.2 grams Mg over 2.0 grams product = .60 or 60% Mg

Now suppose you have the typical lab partner who is having a bad day and he/she does not heat the crucible long enough to drive off any moisture remaining in the crucible, so when you weigh it you find that the mass of your product is 2.5 grams instead of 2.0 grams of pure MgO

so then you divide l.2 gms Mg ribbon over 2.5 grams product and get only .48 or 48% Mg in the product, so the % Mg you think is in your product is lowered.

B. now suppose you weighed out l.2 grams Mg but your lab partner accidently broke off a small piece so you really only started with l.0 grams of Mg.

This time if you did everything perfectly you would have ended up with l.60 grams of Mg0.

So to get % Mg you divide what you thought was your original mass of l.2 gms Mg over l.60 gms product and get .75 or 75% Mg, which is too much.