Hydrated Crystals Lab. In the experiment of Hydrated Crystals the formula for the compounds had to be found including the water.
Hydrated Crystals Lab table
Standard Equation: Subtracting including uncertainties
Subtraction of two mass numbers along with uncertainty.
Question: Subtraction of mass of 3g of MgSO4 + evaporating dish + 0.02g - Evaporating dish ± 0.01g = ?
Subtraction of mass of :
3g of MgSO4 + evaporating dish ± 0.02g = 31.04 + 0.02g
Evaporating dish ± 0.01g = 27.96 + 0.01g
1) Subtract both of the masses together or insert in this formula: 3g of MgSO4 + evaporating dish ± 0.02g - evaporating dish ± 0.01g = ?
3g of MgSO4 + evaporating dish ± 0.02g = 31.04 + 0.02g
Evaporating dish ± 0.01g = 27.96 ...
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Standard Equation: Subtracting including uncertainties
Subtraction of two mass numbers along with uncertainty.
Question: Subtraction of mass of 3g of MgSO4 + evaporating dish + 0.02g - Evaporating dish ± 0.01g = ?
Subtraction of mass of :
3g of MgSO4 + evaporating dish ± 0.02g = 31.04 + 0.02g
Evaporating dish ± 0.01g = 27.96 + 0.01g
1) Subtract both of the masses together or insert in this formula: 3g of MgSO4 + evaporating dish ± 0.02g - evaporating dish ± 0.01g = ?
3g of MgSO4 + evaporating dish ± 0.02g = 31.04 + 0.02g
Evaporating dish ± 0.01g = 27.96 + 0.01g
31.04 + 0.02g - 27.96 + 0.01g = 3.08g
2) To find the uncertainty in any type of addition or subtraction, you just add both uncertainties of both numbers together.
3g of MgSO4 + evaporating dish = + 0.02g
Evaporating dish = + 0.01g
Add both uncertainties: ±0.02 + ±0.01 = ±0.03
3) Combine the mass of both object and substance with uncertainty so the answer is 3.08g ±0.03.
Analysis
- Lab table on above page.
- The number of moles of anhydrous salt of MgSO4 is 0.0142 ±0.04 while the number of moles of water is 0.0732 ±0.06. The number of moles of anhydrous salt of CuSO4 is 0.0110 ±0.04 and the number of moles of water is 0.0722 ±0.06.
Calculations:
MgSO4 + hydrate = 3.08 ±0.03g
MgSO4 after heated is = 1.76 ±0.03g
MgSO4 + hydrate - MgSO4 after heated = 3.08g - 1.76g = 1.32g ±0.06 of water
MgSO4 (anhydrous salt) = 1.76 ±0.03g/120.36 ±0.01g = 0.0142 ±0.04
Water = 1.32g ±0.06g/18.016 ±0.01g = 0.0732g ±0.06
CuSO4+ hydrate = 3.06 + 0.03g
CuSO4 after heated is = 1.76 + 0.03g
CuSO4 + hydrate - CuSO4 after heated = 3.06 + 0.03g - 1.76 + 0.03g = 1.3 + 0.06g
CuSO4 (anhydrous salt) = 1.76 + 0.03g/159.60 = 0.0110
Water = 0.0732g ±0.06/18.016 ±0.01g = 0.0722 ±0.06
- The percentage of water in the hydrated crystals of MgSO4 is 42.8%. The percentage of water in the hydrated crystals of CuSO4 is 42.5%.
Calculations:
1.32g of water
3.08g of hydrated crystal of MgSO4.
1.32/3.08 *100 = 42.8%
1.3g of water
3.06g of hydrated crystal of CuSO4
1.3/3.06 *100 = 42.5%
- The correct formula is CuSO4. 7 H2O and MgSO4. 5 H2O. The theoretical percentage of water is 42% in the MgSO4. 5 H2O and the percentage of water is 44%.
- The percentage of error is -.8% for MgSO4. 5 H2O and 1.5% for CuSO4. 7 H2O.
Calculations:
MgSO4. 5 H2O: original –experimental = change
42%-42.8% = -.8%
CuSO4. 7 H2O: original – experimental = change
44% - 42.5% = 1.5%
- The ratio of moles of water to the anhydrous compound of MgSO4. is 5 H2O to 1 MgSO4, so the formula for the hydrated MgSO4 is MgSO4. 5 H2O. The ratio of moles of water to the anhydrous compound of CuSO4 is 7 H2O to 1 CuSO4, so the formula for the hydrated CuSO4 is CuSO4. 7 H2O.
Calculations:
MgSO4 (anhydrous salt) = 1.76 ±0.03g/120.36 ±0.01g = 0.0142 ±0.04
Water = 1.32g ±0.06g/18.016 ±0.01g = 0.0732g ±0.06
0.0732/0.0142 = 5
0.0142/0.0142 = 1
5 moles of water and 1 mole of MgSO4
CuSO4 (anhydrous salt) = 1.76 + 0.03g/159.60 = 0.0110
Water = 0.0732g ±0.06/18.016 ±0.01g = 0.0722 ±0.06
0.0722/0.0110 = 7
0.0110/0.0110 = 1
7 moles of water and 1 mole of CuSO4.
- My results would have been different if the desiccator was not used as there might be vapor in the outside and this could have effected the measuring of the mass and our results. But in the desiccator, there wouldn’t be vapor going and it would be a anhydrous compound.
Conclusion
In the experiment of “Hydrated Crystals” the formula for the compounds had to be found including the water. By the hypothesis that the formula would be 5 H2O to 2 MgSO4 and 7 H2O to 1 CuSO4. As this was not the case, the correct formula was CuSO4. 7 H2O and MgSO4. 5 H2O. The experiment has also gotten the formula, it was done by a procedure of evaporating the water in the compound and finding the individual masses of each. Then trying to find the simplest formula which was CuSO4. 7 H2O and MgSO4. 5 H2O.
To better prove the data, it would be better to heat all the water from the substances of MgSO4. 5 H2O and CuSO4. 7 H2O. The change of data from the original and the theoretical was not in a range of zero. It had been in a range of 2, so this might have affected how the results were made. Also the exact measuring of MgSO4 and CuSO4 could have proven the need of zero range. The exact measuring was off by a range of one as well. This must have affected the change in percentage.
The lack of accuracy when experimenting could have affected the large difference in percentage from theoretical and experimental. Even though we had gotten the correct formula for the data, the range of difference makes a critical significant of the variation of the data even with the uncertainty. To better prove the data, it would be best if a more technological advanced balance scale was used so the uncertainty was not affected as much and the better measurements of each substance.