Lab 5B: Composition of a Copper Sulfate Hydrate Lab

Post a photo of your hydrate before heating and after heating. Include calculations for #1-#4 in your lab manual. For #5, report your data, including the empirical formula you obtained for the hydrate. If your percent error is high, include a prediction for whether or not your predicted coefficient on water is higher or lower than the actual value.

DATA: 

Mass of Evaporating Dish: 45.06 g

Mass of Dish + Hydrate: 45.83 g

First Heating Mass of Dish + Anhydrous Salt: 45.57 g 

Second Heating Mass of Dish + Anhydrous Salt: 45.59 g

1. Calculate mass of hydrate used.

45.83 g - 45.06 g = 0.77 g CuSO4

2. Calculate mass of water lost.

45.83 g - 45.58 g = 0.25 g H2O

3. Calculate percentage of water in hydrate.

0.25 g / 0.77 g = 32.4675% = 32.5%

4. Find the percent error and provide a possible explanation for your error.

Percent error = (( | Experimental Value - Accepted Value | ) / Accepted Value ) * 100

(( | 0.325 - 0.360 | ) / 0.360 ) * 100 = 9.72% error

Possible explanations for this percent error include how the scale fluctuated and the possible loss of anhydrous salt while stirring. Because of the fluctuating scale while finding the mass of the dish, then of the dish and hydrate, and then after the first and second heatings, my lab partner and I might not have gotten the most accurate readings, as we had to estimate the masses as closely as possible. While stirring the anhydrous salt during the heating, we also accidentally flicked a small piece out of the evaporating dish when we tried to break a large blue crystal down.

5. Moles of water evaporated: 0.014 mol H2O

Moles of CuSO4 that remain in evaporating dish: 0.0033 mol CuSO4

Ratio: 4.24 H2O : 1 CuSO4

Empirical formula: 1CuSO4 * 4H2O

Although our percent error wasn't too high and three could possibly be the correct coefficient, if it is wrong, I predict that our coefficient would be too low. Since getting closer to 36% would mean more water would have to be lost than what happened in our lab, the moles of H2O evaporated would have to increase, meaning the ratio would have to increase, which would then lead to the coefficient increasing.