Experiment To Find The Amount Of Water In A Hydrated Compound

1066 words - 5 pages

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Experiment to find the amount of water in a hydrated compound |
Calculating Reacting Quantities and Formulae |
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Contents

Aim 2
Introduction 2
Hypothesis 3
Apparatus 3
Method 3
Results 4
Calculation 4
Conclusion 5
Evaluation 5
Bibliography 6

Aim

The aim of this experiment is to find the number of molecules of water of crystallisation combined with one molecule of magnesium sulphate and to ultimately find the formula of the hydrated salt.
Introduction

Within crystals of certain salts, there are a fixed number of water molecules, combined chemically in a fixed proportion. When a salt crystallises from a solution, it forms hydrated salts which ...view middle of the document...

Hypothesis

The hypothesis is to prove that magnesium sulphate has seven molecules of water and is a water of crystallisation and its formula is MgSO4.7H2O
Apparatus

* Lab coat
* Goggles
* Matches
* Bunsen burner
* Heatproof mat
* Tripod
* Pipe clay triangle
* Crucible and lid
* Magnesium sulphate
* Electronic scales
* Tongs
Method

1. Firstly, a clean dry crucible and lid was weighed and results recorded
2. The crucible was half filled with hydrated magnesium sulphate
3. The crucible, magnesium sulphate and lid were then re-weighed and results recorded
4. The crucible and contents were placed on a pipe clay triangle and heated gently
5. The heat was increased over the course of a few minutes
6. After five minutes heating, the crucible was removed from the heat with tongs and left to cool
7. Once cooled, the crucible and its contents and lid were re-weighed and results were recorded.
Results

Mass of crucible and lid | 31.29g |
Mass of crucible, lid and hydrated salt | 36.08g |
Mass of crucible, lid and anhydrous salt | 34.43g |

Calculation

To find the mass(g) of water in magnesium sulphate, we subtracted the mass of the crucible, lid and anhydrous salt from the mass of the crucible, lid and hydrated salt:
Mass of water contained in the hydrated salt = 36.08 – 34.43 = 1.65g
To find the mass(g) of the anhydrous salt left after heating, we subtracted the mass of the crucible and lid from the mass of the crucible, lid and anhydrous sale:
Mass of anhydrous salt left after heating = 34.43 – 31.29 = 3.14g
To find the amount of water, we ascertained the relative molecular mass of water (H20 =18) and divided this amount by the mass of water:
RMM of H2O = 1+1+16 = 18
Amount of Water = 1.65/18 (H2O) = 0.09mol
We used the same calculation to find out the amount of anhydrous salt by finding out the relative molecular mass of magnesium sulphate (MgSO4 = 120) and dividing this by the mass of anhydrous salt:
RMM of MgSO4 = 24+32+16+16+16+16 = 120
Amount of anhydrous salt = 3.14/120 (MgSO4) = 0.03 mol
To calculate the amount of water combined with 1 mole of salt, we divided the amount...

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