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Investigating the no. of moles of water of crystallization, in one mole of hydrated ethanedioic acid.

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Introduction

Investigating the no. of moles of water of crystallization, in one mole of hydrated ethanedioic acid Aim: The aim of this experiment is to find the number of moles of crystallized water, in one mole of ethanedioic acid, H C O � H O. Plan: Apparatus list The following items were required, to carry out the experiment: Equipment: Spatula Burette Glass rod Filter funnel Clamp Electronic weight Tong Pipette filler Safety goggles Pipette Bottle of distilled water Substances: 250 cm� volumetric flask Distilled water Thermometer Potassium manganate (VII) 50 ml measuring cylinder Sulphuric acid Beaker Ethanedioic acid 3 x conical flasks Small plastic bottle White tile Bunsen burner Gauze Tripod Heatproof mat Safety matches This is how the experiment was set out: Experimental procedure 1. Weigh out 1.5g of ethanedioic acid using a spatula, into a small plastic bottle. 2. You now have to make up 250 cm� of a solution made from the ethanedioic acid, and the 1.5 g of acid weighed out. ...read more.

Middle

You do this, till the temperature reaches between 65-70�C. Measure the temperature using a thermometer. 7. When it reaches the specified temperature, quickly remove the conical flask onto the white tile situated under the burette containing the potassium manganate (VII), and titrate this solution. 8. Titrate the solution till it reaches a pale pink colour. 9. You now repeat the steps 3-8 at least three times, to get a fair result. Safety During the experiment, some safety regulations were carried out. Safety goggles were worn throughout the experiment, hands were washed after contact with chemicals and long hair was tied back. If ethanedioic acid gets on your hands, it can be dangerous if your hands then get in contact with your eyes, or mouth. Also the Bunsen burner was kept on a yellow flame, when it wasn't in use. Fair Test To make it fair, I had to make sure, that exactly 1.5 g of ethanedioic acid was used weighed at the beginning of the experiment. ...read more.

Conclusion

Using the formula above, means that the concentration of hydrated ethanedioic acid would be: X 0.025 = 5 0.02 x 0.0232 2 So, x 0.025 = 2.5 0.00464 Then, x 0.025 = 2.5 x 0.02 x 0.0232 = 0.00116 = 0.00116 0.025 = 0.0464 dm� of C O Now we know the concentration, we can find the amount: n = cV = 0.0464 x 0.25 = 0.0116 mols The overall Mr of H C O � H O is: Mr = m n Mr = 1.5 1.5 0.0116 = 129.31 129.31 = Mr of H C O + H O 129.31 = 90 + 18 So the number of moles of H O is: = (129.31-90) 18 = 2.2 The number of moles of water of crystallization present in H C O � H O is 2, as I've rounded it down, because its under 0.4 whereas if it was 0.5 I would have rounded it up to the next whole number. Overall equation: H C O � 2H O Deyar Yasin ...read more.

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