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?
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.
If, during the procedure for forming magnesium oxide, some shiny metal remains in the crucible that is not heated any further, it would lead to an error in the calculation of the final product. The error occurs because the remaining shiny metal, which is likely unreacted magnesium, would be included in the total weight of the product.
When calculating the percentage error, the expected weight of the magnesium oxide is compared to the actual measured weight. If unreacted magnesium is included in the measured weight, it will cause the actual weight to be higher than the expected weight. This discrepancy means that the calculated percentage error will be significantly higher than it should be.
To avoid this error, it is essential to ensure thorough heating and oxidation of the magnesium. The shiny metal remaining in the crucible would need to be reheated until it forms magnesium oxide completely. By doing so, the final product would accurately reflect the correct weight of magnesium oxide, resulting in a more accurate calculation of the percentage error.