What error in calculation would result if, in the procedure for forming the magnesium oxide, some shiny metal remained in the crucible which was not heated any further?
Well, wouldn't that throw the weights off which in turn would throw the whole thing off, so that when you went to calculate %error the amt. would be very high?
1 answer
The only problem with your reasoning is that you don't say which element is high. Of course the whole thing is off. The percent Mg will be high, the percent oxygen will be low, the formula will not be MgO but some other set of numbers. Note that the problem asks for error, not percent error.