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Embed code for: Chem Lab Report Unit 2
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Course: Chemistry 11 Name: XueTing(Shirley) Mao
Block: 4 Partner’s Names: Bela, Sarah D Unit 2: The Mole Date: Nov. 24th, 2016
FORMULA OF A HYDRATE LAB
To determine the percent of water in the given unknown hydrate.
Crucible and cover
Lab stand and ring clamp
5.00g of hydrate CuSO4 • XH2O
1. Used the electronic balance to obtain the mass of a clean dry crucible.
2. Used the electronic balance to obtain the mass of the crucible and the given sample of hydrate.
3. Placed the crucible above the bunsen burner. Heated it for 5 min.
4. Removed the heat to allow it cool completely for 5 min.
5. Recorded the mass of the crucible and contents.
6. Reheated the crucible for 2 min, then removed the heat to cool for 5 min.
7. Recorded the mass of the crucible and contents.
8. Repeated the last two steps one more time.
9. Added a few drops of water to the crucible. Noted any changes in the substance.
Before heating: Bright blue small crystals.
During first heating: White on the edges and blue in the middle.
During first cooling: White on the edges and light blue in the middle.
During second heating: More white on the edges and small amount of baby blue colour in the middle.
During second cooling: Same as second heating.
During third heating: The substance became fully white.
During third cooling: The substance is fully white.
After dropping the water: The parts where the water landed became bright blue and wet. Other places appeared the same.
Mass of empty crucible
Mass of crucible plus hydrate
Mass after first heating
Mass after second heating
Mass after third heating
PART I: Mass of Water in Hydrate
Mass of Hydrate: 25.69g-20.62g=5.07g±0.01g
Mass of Anhydrous Salt: 23.87g-20.62g=3.25g±0.01g
Mass of Water: 25.69g-23.87g=1.82g±0.01g
PART II: Ratio of Moles of H2O to moles of Anhydrous Salt:
(1) H2O=1.008g/mol x 2+15.999g/mol=18.02g/mol
1.82g x 1 mol =0.101mol H2O
(2) CuSO4=63.546g/mol+32.066g/mol+15.999g/mol x 4=159.61g/mol
3.25g x1 mol =0.0204mol CuSO4
(3) H2O = 1.82g = 35.9%
PART III: Simplest Ratio and Empirical Formula:
(1) 0.0204mol CuSO4•0.101mol H2O
(2) 0.0204 : 0.101= 204 : 1010 ≈ 1 : 4.95 ≈ 1 : 5
1. CuSO4 • 5H2O
It is a physical change because when you heat these crystals, they become white. However, when you add water it becomes blue again. Copper sulphate that has been dehydrated has the possibility to rehydrate. The water that came out of the copper sulphate when it was dehydrated can be added back to the material and the copper sulphate will be able to return to its original form
2. If the hydrate was not heated long enough to extract all of the water then the mass of water lost would be incorrect. If the mass of water lost is wrong, then the water lost in moles will most likely also be incorrect. Then, when getting the ratio of salt to water, since the water lost in moles is wrong, the ratio will be off and inaccurate. Given the incorrect ratio, the empirical formula will also be wrong.
3. SO4 • 5H2 O —> CuSO4 + 5H2O
If the hydrate was not heated for too long at too high of a temperature then the mass of hydrous salt would be incorrect. If the mass of hydrous salt is wrong, then the hydrous salt in moles will most likely also be incorrect. Then, when getting the ratio of salt to water, since the hydrous salt in moles is wrong, the ratio will be off and inaccurate. Given the incorrect ratio, the empirical formula will also be wrong.
4. Anhydrous is a substance that contains no water. In this lab CuSO4 is anhydrous.
(1) Sources of Error
• The anhydrous salt could have been exposed to air before measurement, and reabsorbed some moisture, which results in errant data.
•When heating the crystal because of using opaque crucible, parts of the crystal on the bottom might be burnt without noticing. This will cause the crystal to be over heated, which will cause inaccurate measurements and data.
(2) Percent Error
•Measured value: 4.95 Accepted value: 5.00
Percentage of error: (5.00 – 4.95) / 5.00=1.00%
(3) Relevant Theory
•The goal of this experiment is to determine the percentage of water in a hydrate, and to calculate the ratio of salt to water in a hydrated salt. To achieve this, a known mass of hydrate was heated to evaporating the water. It is known that the mass lost is equivalent to the mass of the water. By comparing the mass of the water to the mass of hydrate, we can calculate the percentage of water within the hydrate. We can also calculate the ratio of hydrous salt to water.
The calculated ratio between copper sulphate and water was found to be: 1:4.95. The accepted ratio between copper sulphate and water is: 1:5.00. The percent error within this measurement is 1.00 % .
The formula for copper sulphate hydrate is: CuSO4 • 5 H2O .
The calculated ratio between copper sulphate and water was found to be: 1:4.95. The accepted ratio between copper sulphate and water is: 1:5.00. The percent error within this measurement is 1.00 % .ore white on the edges and small amount of baby blue colour in the middle.
• The anhydrous salt could have been exposed to air before measurement, and reabsorbed some moisture, which results in errant da