Determination of the Relative Atomic Mass of Lithium

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Rob Ayres

Determination of the Relative Atomic Mass of Lithium

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To determine the relative atomic mass of lithium I cut a piece of lithium, weighing 0.82g and reacted it with 100cm3 of distilled water. This reaction gives off hydrogen gas, which I collected in a 250cm3 measuring cylinder. I collected 129cm3 of hydrogen gas. To calculate the number of moles of hydrogen that I collected, assuming that 1 mole of gas occupies 24000cm3 at room temperature and pressure, I must divide the amount of gas I collected by 24000.

129 ÷ 24000 = 0.005375 moles

I can use this information to calculate the number of moles of lithium that reacted. I will multiply the number of moles of hydrogen I collected by 2 because of the 2:1 ratio of the lithium used to hydrogen given off.

0.005375 × 2 = 0.01075 moles

Now, by dividing the amount of lithium I used in the reaction, which was 0.082g, by the number of moles of Lithium that reacted, 0.01075 moles, I can work out an estimate for the relative atomic mass of lithium.

0.082 ÷ 0.01075 = 7.628

This is close to the actual relative atomic mass of Lithium but not as accurate as I had planned to get.

After doing this I then titrated 25.0cm3 of the lithium hydroxide made by the reaction, with a 0.100 moldm-3 solution of hydrochloric acid. I did the titration 3 times and these are the results.

 

The results are all within 0.1cm3 of each other so I am pleased with the accuracy.

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The equation for the neutralisation reaction is as follows:

LiOH(aq) + HCl(aq) → LiCl(aq) + H2O(l)

I will now calculate the number of moles of hydrochloric acid used by multiplying the average amount of HCl used in dm3, by the concentration of it, which is 0.100 moldm-3 

Average amount of HCl used in dm3 = 0.02253 dm3

0.02253 × 0.100 = 0.002253 mol

So the number of moles of LiOH used is also 0.002253 moles because of the 1:1 ratio.

To calculate the number of moles of LiOH present in ...

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