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# Determination of the relative atomic mass of magnesium.

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Introduction

Determination of the relative atomic mass of magnesium Method 1 - RSULTS: Assume that 1 mole of gas occupies 24 000 cm3 at room temperature and pressure. Mg(s) + 2HCl (aq) --> MgCl2(aq) + H2 (g) Calculate the number of moles of hydrogen you collected. Mg(s) + 2HCl (aq) --> MgCl2(aq) + H2 (g) Ratio Magnesium: Hydrogen is 1:1 Number of moles = Amount of hydrogen produced / 24000 Hydrogen = 142 / 24000 = 0.00592 moles Deduce the number of moles of magnesium Number of moles = 1 x 0.00592 Magnesium =0.00592 moles Calculate the relative atomic mass of magnesium RAM = Mass / Number of moles = 0.12 / 0.00592 = 20.27 Method 2 - RESULTS Mass (g) Weight of clean evaporating basin 31.70 Weight of evaporating basin + conical flask contents 41.64 Weight of evaporating basin + solid magnesium chloride 32.22 Calculate the mass of magnesium chloride formed Weight of evaporating basin and - Weight of clean evaporating basin solid magnesium chloride = 32.22 - 31.70 = 0.52 g Calculate the number of moles of magnesium chloride formed Number of moles of Magnesium chloride = Mass/ Mr = 0.52 / 91.27 = 0.00570 moles Deduce the ...read more.

Middle

Further Work If I was going to do further work I might use some different metals. I might also use different acid concentrations and different acids altogether. Also I might like to find the relative atomic mass of a metal using a more accurate method and achieve more accurate results. I would use the method of some kind of electrolysis. I feel that overall the results of my experiment were fairly accurate. I can test the accuracy by calculating the percentage of accuracy for each experiment. This is done by dividing the calculated result of the relative atomic mass by the actual atomic mass (24.1) of magnesium, then multiplying this by 100: Method 1: 20.27 / 24.1 x 100 = 84.1 % accurate Method 2: 20.27 / 24.1 x 100 = 84.1 % accurate This shows that my results have a high percentage of accuracy. There was however an anomalous result in the titration in Method 2. It was very different from the other results so, for accuracy I ignored the anomaly when calculating the average titre. This anomaly may have been caused by a number of factors. ...read more.

Conclusion

According to the percentage accuracies, Method 1 is more accurate than Method 2. There are a number of factors that could be responsible for the lower percentage accuracy of Method 2; error in measurement of the lithium or distilled water, misjudgement of water mark in the measuring tube, oxidation of the lithium. Also, I repeated Method 2 three times, allowing me to disreguard anomalies and take an average. But, because I had to use the solution from Method 1 in Method 2, I did not repeat it. Any error in measuring the lithium or distilled water or fault in the lithium during Method 1 is likely to affect the results of Method 2. This means that ideally the solution from Method 1 would have an accuracy as near to 100% as possible. This could be done by repeating Method 1, at least three times, each time keeping the solution. Calculate the atomic mass from each volume of gas produced, the solution with the highest accuracy can then be used for Method 2. The calculation of the atomic mass of lithium from Method 2 would then be expected to produce the highest accuracy and most reliable result, according to the calculated percentage accuracies. ...read more.

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