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  • Level: GCSE
  • Subject: Science
  • Essay length: 1237 words

The aim of this practical experiment is to calculate the formula of hydrated Iron(II) Sulphate crystals, using two methods.

Extracts from this essay...

Introduction

Chemistry practical Date: 26th September 2003 Name: Frazar Ngosa Aim: the aim of this practical experiment is to calculate the formula of hydrated Iron(II) Sulphate crystals, using two methods. Abstract Introduction and Theory Hydrated Iron(II) Sulphate crystals contains water of crystallization. Water of crystallization is the fixed amount of water that an hydrated salt contains. The formula of an hydrated salt is generally written as M.xH2O, where the M is an hydrous salt formula, and x is the number of moles of water per mole of hydrated salt. In this experimental investigation, we are going to determine the formula of hydrated Iron(II) Sulphate, which is given to us in the instruction as FeSO4.xH2O. Our main objective is to determine the value of x. Determining the value of x will involve the use of two different methods then comparing the results. One of the methods involves heating the hydrated to decompose it into an hydrous salt of Fe+2 and water, the calculating the value of x, employing the method called 'formula of calculating the empirical method'. This involves assuming the FeSO4 as an element, lets call it Y and H2O as another element and lets call it Z.

Middle

We had to use an electronic beam balance to measure it and the experiment was done away from the taps to prevent water plinking on to the crucible, which could influence our results. Diagram of apparatus Procedure The procedure is given in the instruction. It involves heating the salts, noting down the change in mass every 2 minutes, however, you have to wait after heating before measuring the mass, until the crucible is cool. Control of Variables Mass; using an electronic beam balance to measure the mass reduced the uncertainty in the measurement. Also the experiment is best done I a dry environment, keeping water and other liquids away to prevent the salt absorbing the water before weighing or else there could have been a more errors in the measurement. Data collections Mass of the crucible = 17.09g Crucible + FeSO4.xH2O = 18.49 FeSO4.xH2O = 18.49 -17.09 = 1.40g Time /s Mass of the crucible /g Mass of FeSO4 /g Mass of H2O /g 0 18.49 0.00 0.00 2 18.12 1.03 37 4 17.95 0.86 0.54 6 17.95 0.86 0.54 8 17.95 0.86 0.54 Method 2 Method II involved making the iron (II)

Conclusion

We had to measure the same mass three times and we were getting the same 3grams. The second method to find the formula of hydrated Iron(II) Sulphate salt involves dissolving the salt in sulphuric acid and water making up to 250cm3 solution. a method called titration is then employed, titrating it with potassium hydroxide. Full instructions are given in the question paper. Calculations: From method 1 Iron (II) sulphate decompose on heating giving anhydrous Iron(II) Sulphate salt and water. This can be represented by the equation FeSO4.xH2O(s) ? FeSO4(s) + xH2O (g) Mass of anhydrous salt that remained after decomposition is 18.49 -17.95 = 0.86g Number of moles of anhydrous FeSO2 salt = 0.85g/Mr(FeSO4) =o.85/152 = 0.00559 moles. Mass of water produced = 18.49g -17.95 = 0.55g Number of moles of water produced = 0.55./Mr(H2O) = 0.54/18 = 0.0306 moles Let H2O be X and FeSO4 be y There fore, the ratio of number of moles of Y to X is 0.00559 moles : 0.0306 moles 0.00559 moles 0.00559 moles 1 : 5.5 1 : 6 Hence x is equal to six and the formula becomes FeSO4.6H2O Calculation from method 2 Needs thorough editing before submitting, - Organise work,

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