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Making iron II sulphate

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Introduction

MAKING IRON II SULPAHATE Iron sulphate is a compound formed by a reaction between copper sulphate and iron The reaction between CuSO4 and Fe is called redox reaction. This is a type of reaction where one of the reactants is reduced to form a product and one of the reactants is oxidised to form a product REDOX REACTION The redox is used for the simultaneous processes of reduction and oxidation and involves the loss or gain of electrons and the loss or gain of oxygen and hydrogen. In oxidation reaction, atoms of an element in the reaction lose electrons and increase their oxidation (Oxygen is added) While in reduction reaction atoms of an element gain electrons and its oxidation reduced (loses oxygen) In my experiment I will be using copper sulphate and iron. When copper sulphate is added to iron the equation of the reaction will be CuSO4 + Fe Cu + FeSO4 Oxidation state of the reaction CuSO4 + Fe Cu + FeSO4 2+ 0 0 2+ From the above equation we can find that the oxidation state of copper was reduced from 2+ to 0 Cu2�+ + 2e Cu This is called reduction On the other hand we can also find that the oxidation state of iron has increased from 0 to +2 Fe Fe + 2e This is called oxidation IONIC BONDING Forces that hold ionic compounds together form ionic bond. ...read more.

Middle

56+32+16(4) +7(2+16) FeSO4 7H2O 56+32+16(4) +7(2+16) 56+32+64+126 278g Calculating theoretical yield The theoretical yield is the maximum amount of product that can be made. In other words it is a 100% of the yield From the equations above (under the RMM subheading) we found that the relative molecular mass of; Copper sulphate 5H2O to be 249g and Iron II sulphate 7H2O to be 278g These means that 249g of CuSO4 makes 278g of FeSO4 In my experiment I used 5g of CuSO4. I know that 249g of CuSO4 makes 278g of FeSO4 So to find how many grams this makes I will divide 278 by 249 and multiply it by 5 278�279�5 5.58 this is know my 100 Why the weight is less * Transferring the substances from one container to another. This will affect the weight in that it is practically not possible to transfer a 100% of a given substance. * The more greater the number of operations the more the chance of losing some at every stage * Though measures were taken to avoid, when heating the solution there were times when it splashed out hence losing some substances Improving the yield I would think that taking the following into consideration would have improved my actual yield: * Giving more time for the reaction to end * Transferring the substances as few times as possible and Make sure that I transfer the as much of the substance as possible (this is to make sure that there are as few of the substance lost in the transferring as possible. ...read more.

Conclusion

I tried to work out the ratios to find this. Bellow is the calculation I did: RMM for CuSO4 5H2O 249g (see 'calculating RMM') RAM for Fe 56 1g of CuSO4 5H2O = 56�249 In my experiment I used 5g therefore 5g of CuSO4 5H2O = 56�249�5 = 1.12 This shows that the amount of iron is even in excess which is contrary to what I thought. Then I thought that there was an equilibrium and decided to add more irons to push it over. I first increased the amount of the iron filings to 4.8g but this did not work either and the problem was the same. I then tried to half the amount of copper sulphate (make it 2.5g) and 4.8g of iron filings to try again to push the equilibrium over. This also didn't turn the colour of the solution. The same problem came up after every trial. I now strongly feel that the problem lied on the reaction of the iron. It is possible that the copper coated the iron filings and stopped the most of it from the reaction. SUGESTIONS ON IMROVING THE REACTION * One of the things that may have help will be to get keep on increasing the amount of iron and try to push the equilibrium over. * Another thing that can be done is grinding the iron to very small particles so that it reacts well and quick 1 Zakaria Dahir Chemistry coursework ...read more.

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