In our most recent mini lab, we set out to find the percentage of water in a hydrate. By boiling off the water, we were able to distinguish what weight was CuSO4 and what weight was H2O. We came to the conclusion that 36.8 percent of the hydrate is water. Our results are very reliable; we were careful while performing the experiment and while measuring and massing. That being said, we could have lost some mass that was considered to be water as a result of the substance sticking to the end of the stirring rod. We know that our results are reliable because we followed instructions meticulously and had little to no errors during experimentation. Also, we used significant digits accordingly, therefore improving precision.
Due to our calculations, we found that 2.2 grams of water would be ‘lost’ if you used 6.0g of hydrate. This result is based upon our earlier findings that established the percent composition of water of this sample to be 36.8%. If some of the hydrate/anhydrous solid were splattered out, our results would change, mainly our masses. If some of the hydrate splattered, the percent composition would be inaccurate because the lost mass would be attributed to water, meaning that it would seem more water had been present than actually was. If some of the anhydrous solid splatter out, then the same result would occur, because the lost mass would be seen as water weight.
(90g H2O/249.6g CuSO4 x 5H20) x 100= 36.05
The calculation above shows what we should have gotten as a percent composition for the hydrate in CuSo4 x 5H2O. We had an actual percent composition of 36.8% based off of our experiment.
(2.02g CuSO4 x 5H2O/1) x (90g H2O/ 249.6g CuSO4 x 5H2O)= .728
The equation above displays the calculation of the theoretical mass of hydrate in CuSO4. It was very close to our actual result of .74, meaning our percent error was very low. The equation for percent error can be seen below.
Percent Error (A-T/T) .74-.728/.728 x 100= .016%