Ryan ThomasChemistry HonorsPeriod 15/6/07

Determining the Formula of a Hydrate

Purpose

To determine the formula of a hydrate

Materials

Goggles Glass Stirring RodLab Apron Hot PlateEvaporating Dish (clean and dry) BalanceCuSO4 crystals Tongs

Safety

1. Handle the evaporating dish with tongs at all times.2. Heat the crystals slowly to avoid splattering and popping.3. Reduce the heat if the edges of the crystals begin to turn brown.4. Do not place a hot dish on the balance.5. Check with the instructor how to dispose of the anhydrous salt.6. Copper sulfate hydrate is toxic by ingestion and inhalation, and is a skin and body

tissue irritant. If any comes into contact with any part of your body, flush with water for 15 minutes and notify the instructor.

Procedure

1. Heat the evaporating dish for 3 minutes to remove any water that the porcelain may have absorbed. Allow the dish to cool for 2 minutes.

2. Find the mass of the evaporating dish to the nearest 0.01g using the balance. Record this mass.

3. Add the CuSO4 crystals to the dish, and mass it again. Record this mass.4. Heat the dish with the crystals in it by placing it on the hot plate at medium heat.

While heating, use the glass stirring rod to stir the crystals gently. Stop heating once the crystals have turned gray.

5. Allow dish to cool until it is safe to mass it, then do so, without removing any crystals. Record this mass.

6. Dispose properly of the remaining crystals, and properly clean and replace all materials.

7. Subtract the mass of the empty evaporating dish from the mass of that dish containing the hydrate crystals. This is the experimental mass of the hydrate.

8. Subtract the mass of the empty evaporating dish from the mass of that dish containing the anhydrous crystals. This is the experimental mass of the CuSO4 from the hydrate.

9. Subtract the mass of the CuSO4 from the mass of hydrate. This is the experimental mass of the H2O from the hydrate.

10. Calculate the moles of CuSO4 present in the hydrate by multiplying the experimental mass of the CuSO4 from the hydrate by the conversion factor “1 mol CuSO4 / 159.6g CuSO4”, since 159.6g/mol is the approximate molar mass of CuSO4.

11. Calculate the moles of H2O present in the hydrate by multiplying the experimental mass of the H2O from the hydrate by the conversion factor “1 mol H2O / 18.0g H2O”, since 18.0g/mol is the approximate molar mass of H2O.

12. Find the ratio of moles CuSO4 in the hydrate to moles H2O in the hydrate; simplify the ratio by dividing both sides by the number of moles of CuSO4, so that the ratio is in the form “1 mol CuSO4:x mol H2O”.

13. Round the number “x” from the previous step to the nearest whole number. This is the approximate number of moles H2O per mole CuSO4 in the hydrate. Therefore, the formula for the hydrate should be:

CuSO4 . (the number just found)H2O.

Data & Observations

Mass of evaporating dish: 44.01gMass of evaporating dish and hydrate: 48.28gMass of evaporating dish and anhydrous crystals: 46.80g

Some crystals cling to the stirring rod, and are removed from the mixture unavoidably It is hard to tell whether or not the crystals have truly completed their change, as they are not very gray Other than that, the equipment is clean and the experiment has run smoothly

Calculations (As outlined in procedure)

46.80 - 44.01 = 2.79Experimental mass of CuSO4: 2.79g

48.28g - 44.01 = 4.27Experimental mass of hydrate: 4.27g

4.27 – 2.79 = 1.48Experimental mass of H2O = 1.48g

4.27g CuSO4 * 1 mol CuSO4/ 159.6g CuSO4 = 0.0268 mol CuSO4

1.48g H2O * 1 mol H2O/18.0g H2O = 0.0822 mol H2O

0.0268 mol CuSO4: 0.0822 mol H2O = 1 mol CuSO4: 3.06 mol H2O

EXPERIMENTAL FORMULA OF HYDRATE:CuSO4 . 3H2O

Conclusion

The purpose of this lab was to determine the formula of a hydrate through experimentation and calculation. To accomplish this, the mass of the hydrate was found, the hydrate was heated to remove the water, and the mass of the remaining CuSO4 crystals was found. Then, the formula of the CuSO4 was calculated based on the experimental measurements. The formula was calculated to be CuSO4 . 3H2O, as it was calculated that there were roughly 3 moles H2O present in hydrate per mol CuSO4. However, as with all labs, the figures were not exact, due to inevitable error. The three sources that were identified as being responsible for this error were loss of crystals, possible incomplete evaporation, and rounding of figures. The first source results in error by causing too low a mass reading for the CuSO4 crystals. This explains why the mol ratio is not exact; too low a mass reading for the CuSO4 crystals means that more water than was present is calculated to have escaped, thus resulting in too many moles of H2O calculated as being present in the hydrate. The second source would result in the opposite; too high a reading for CuSO4 crystals, as they would still contain water, and would result in a calculation of too few moles H2O; however, since this would not be as prominent as the error caused by the first source, and the effect is not seen in the calculations, it can be assumed that if this source caused error, it merely detracted from the error caused by loss of crystals. Finally, the rounding of exact measurements may have slightly altered the readings from the actual amounts, but this is not a significant source of error, and since almost all the figures are rounded, any effect should balance out, and seems to. Overall, the data seems accurate enough, and I am extremely confident in the results.

Questions for Further Thought

1. Hydrates that are efflorescent will lose their water molecules if their vapor pressure is greater than that of their surroundings, as their high concentrations of loosely held water molecules are inclined to spread out if there is a lower concentration of water molecules in the surrounding environment.

2. Hydrates are formed at temperatures higher than the freezing point of water when they are forced to react with the anhydrous compounds they combine with.

3. 16.3% waterCaCl2 has a mass of (40.1 + 35.5 + 35.5)g/mol = 110.1g/molH2O has a mass of (1.0 + 1.0 + 16.0)g/mol = 18.0g/mol18.0/110.1 = .163 = 16.3%; 16.3% water