mass of magnesium

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Fatima Omar                Determination of RAM of Mg

DETERMINATION OF RAM OF MAGNESIUM

Method1

Mass of Mg = 0.12g

Volume of Hydrogen = 128cm3

Treatment of results

                        Mg + 2HCl  MgCl2 +H2

  • n= v÷V       128 ÷24000 = 5.33×10-³ moles

  • mole ratio is 1:1, therefore moles of mg that reacted= 5.33×10-³ moles

  • RAM of Mg (n= m÷M  M= m÷n)    0.12 ÷ 5.33×10-³ = 22.51

Method2

                                

Mass of magnesium from method 1 = 0.12

Weight of boat = 44.17

Weight of boat with solution = 52.84

Weight of boat with salt (MgCl2) = 44.74

  • Mass of MgCl2 formed = 44.74  44.17 = 0.57

  • Mass of Cl- ions = 0.57  0.12 = 0.45g

  • n= m ÷M  = 0.45 ÷ 35.5 = 0.013moles

  • n= m ÷M   = 0.57 ÷ 24.3 = 0.023moles

                                

EVALUATION

Method                 Result

  1. 22.5
  2. 18.9

Expected                 24.3

From looking at the end results in both methods, I can clearly see that method 1 was the more accurate and appropriate way to determine the mass of magnesium. This is because out of both results, method 1 had the closer answer to the expected than method 2. Method1 only had a difference of 1.8 from the actual RAM of magnesium, whereas Method2 had a difference of 5.4- which is quiet a large difference. We cannot depend on just looking at the result that we got from the experiments to settle the more significant method, so I will work out the %errors for both experiment methods to see by what percentage my results were from the expected result.

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% error of experiment = difference between result & expected               x100

                                        Expected result

Method1: (1.8 ÷ 24.3) x 100=7.4%

Method2: (5.4 ÷ 24.3) x 100= 22%

The percentage errors for both methods are quiet high, so this suggests that maybe both methods were not very significant. Method 1 is still better than method2 even though the experiment error was 7.4%, which is a high and insignificant percentage, but it is still more accurate than method 2 as the %error was 22%, which is too high and the result is too different from the expected.

Surely, because ...

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