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Hydrated Crystals Lab. In the experiment of Hydrated Crystals the formula for the compounds had to be found including the water.

Extracts from this document...

Introduction

Substance before heating Mass (g) Evaporating dish 27.96 + 0.01g 3g of MgSO4 + evaporating dish 31.04 + 0.02g 3g of MgSO4 + hydrate 3.08 + 0.03g Evaporating dish 27.96 + 0.01g 3g of CuSO4 + evaporating dish 31.02 + 0.02g 3g of CuSO4 + hydrate 3.06 + 0.03g Substance after heating then cooling for the first time Mass(g) Evaporating dish 27.96 + 0.01g 3g of MgSO4 + evaporating dish 29.72 + 0.02g 3g of MgSO4 1.76 + 0.03g Evaporating dish 27.96 + 0.01g CuSO4 + evaporating dish 29.72 + 0.02g CuSO4 1.76 + 0.03g Substance after heating then cooling for the second time Mass (g) Evaporating dish 27.96 + 0.01g MgSO4 + evaporating dish 29.71 + 0.02g MgSO4 1.75 + 0.03g Evaporating dish 27.96 + 0.01g CuSO4 + evaporating dish 29.70 + 0.02g CuSO4 1.74 + 0.03g First time - second time after heating and cooling =Substance mass difference Mass(g) in difference MgSO4 0.01 + 0.04g CuSO4 0.02 + 0.04g Hydrated Crystals Lab table Standard Equation: Subtracting including uncertainties Subtraction of two mass numbers along with uncertainty. ...read more.

Middle

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 3) 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% 4) 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%. 5) 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% 6) ...read more.

Conclusion

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. ?? ?? ?? ?? ...read more.

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