(The hydrate before heating) (The hydrate after heating)
1.Mass of hydrate used: Mass of hydrate and the Evaporating Dish-Mass of the Evaporating Dish=Mass of Hydrate.
2.13-1.21 = 0.92 g
2. Mass of water lost: (Mass of Hydrate + Dish) - (Mass of Dish + Anhydrous salt (final variable in this step).
2.13-1.96= 0.17 g
3. Percentage of Water in hydrate: 0.17/0.92 = 18.5%
4. Percent error. An acceptable percentage of water in the hydrate was 36.0%
| 18.5 - 36.0 | /36.0= 48.7% error.
Some possible reasons for the high discrepancy are as follows:
-We completed the lab to the specifications given in 2 attempts, so it's possible there was still some water remaining
-The scale on which we weighed the evaporating dish and the salt was fluctuating greatly (the average seemed to be 1.96 g for the second measurement, so we went with that), so it's possible that the salt had, in fact, lost more than .05g (we were supposed to stop once the fluctuations between measurements had gone under .05g, which we did).
5.
Moles of water evaporated: 0.17/x=18.01/1, 18.01x=0.17, x=9.44 * 10^-3 (18.01 is the molar mass of all of the components in water).
Moles of CuSO4 remaining: 0.75/x=159.607/1, 0.75 = 159.607x, x=4.70 * 10^3 (159.607 is in this equation as was 18.01 was in the equation directly above this one).
Ratio of Moles: H2O: 9.44 * 10^-3/4.70 * 10^-3 = 2, CuSO4: 4.70 * 10^-3/4.70 *10^-3 = 1.
The ratio of moles is 2:1
Empirical Formula: CuSO4 * 2H2O
Going back to the percent error, I have a feeling that because the error is so high, and my coefficient is relatively low, that the coefficient I gave is lower than the actual coefficient.
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