Question

How would the following errors affect the empirical formula of the compound for a hypothetical student performing s experiment? Note: Writing the answer will be incorrect is not sufficient for these questions. This student is not E) you nor is their data assumed to be the same as yours. student ran out of time and did not do the second heating. Explain thoroughly how this error will affect the water in the hydrate? Will the moles of water calculated be artificially a) The calculation for the number of moles of high or low? (Please think about the purpose of the second heating) b) The student recorded the mass of the cup + sample incorrectly and started with 2.20 g of hydrated compound but used 2.00 g in the calculations. Explain thoroughly how this error will affect the calculation for the number of moles of water in the hydrate? Will the moles of water calculated be artificially high or low?

Answer 1

a) The heating is used in this procedure so as to eliminate to the most the molecules of water that the compound may have attached to it. If the second heating did not take place, you can safely assume that there's still water attached to the compound itself. You may continue with your calculations and report a higher molecular mass for the compound than it actually possesses, since there is still water considered in those calculations. Also, the moles of water that are artificially calculated will be lower than the actual amount of water moles, since there are unaccounted moles still attached to the compound and that will be accounted as the latter's own mass.

b) The artificially calculated moles of water will be again lower. This is because as the student considered a lower initial mass, the difference between said value (2g) and the final value (the heated and cooled now-dehydrated compound) will also be lower than it should be. Hence, the student will think less water was consumed by heating the compound, and those 0.2 g will be unaccounted as water moles. However, if the rest of the procedure takes place normally, the calculated values for the compound will not be affected as such, since the final mass will have to correspond to the actual mass of the compound.