Determining the relative atomic mass of lithium

H2 produced                =        172cm3                 =        7.167×10-3moles

24000cm3(1 mole of gas)                24000cm3

2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) – molar ratio of Li to H2 is 2:1 therefore;

the amount of moles of lithium that reacted = 7.167×10-3moles × 2 = 0.0143moles

Relative atomic mass of lithium = mass        =        0.1000g        = 6.99g mol-1

                                         moles                0.0143mol

• Method 2 results:

test run                        average titre = 36.1 + 36.0 + 36.0

                                                        3

= 36.03cm3

On average, 25.00cm3 of LiOH(aq) required 36.03cm3 of 0.10mol dm-3 HCl. So;

        moles of HCl used in the titration = concentration × volume

                                               = 0.1000mol dm-3 × 36.03 dm3

                                                                    1000

                                               = 3.603×10-3 moles

As 1 mole of LiOH reacts with 1 mole of HCl, [LiOH(aq) + HCl(aq) → LiCl(aq) + 2H2O(l)] then 3.603×10-3 moles of LiOH were used in the titration. Only 25cm3 of the LiOH solution at a time during the titration, so the number of moles in the 100cm3 solution from method one = 4 × 3.603×10-3 = 0.0144moles.

Therefore, the relative atomic mass of lithium = moles = 0.1000g     = 6.94g mol-1.

                                                                0.0144dm3

• Evaluation:

Overall my results are quite accurate and in the case of method two correct, when rounded to three significant figures. In my calculations I used four significant figures in order to improve accuracy, and by using the average of three titrations (all within one decimal place of each other) helped the reliability of the results.

Procedural errors:

Firstly the lithium. Exposure to oxygen in the air forms a layer of lithium oxide that coats the sample and when put in the water will affect the amount of lithium hydroxide produced. To overcome this problem the lithium should be added to the distilled water straight after it has been weighed. The oil the lithium is stored in is useful to stop this process but it will contaminate the lithium hydroxide as well as causing errors in measurement. A polarised, un-reactive solution could be used instead. Secondly, in method one the tube that carries the hydrogen to the measuring cylinder is susceptible to contain trapped air that would give an inaccurate reading of gas produced. This makes method two more relatively reliable as the indicator will give a more sensitive, accurate reading whilst you must also consider gas that escapes in the time taken to place the plug on the flask in method 1.

Errors in measurement:

The oil on the lithium affects the weight and it must be mostly removed to get a correct lithium weight measurement, whilst at the same time prevent the lithium reacting with the oxygen in the air and causing a negligible contamination of the lithium hydroxide. Ideally a polarized, un-reactive solution should be used instead. Another measurement error is the resolution of the apparatus used; the scales measured to the nearest 10mgram, the burette measured to the nearest 1cm3 whilst the measuring cylinder to the nearest 2cm3. With a higher resolution method two will have more accurate results. The most significant measurement error is the amount of gas given off in method one. Aside from the procedure the measurement can be modified to reduce the error:

H2 gas produced = 172cm3 ± 1cm3 (as the resolution is to the nearest 2cm3).

                                Minimum possible value: 171cm3. Relative atomic mass of lithium: 171/24000 = 7.125×10-3moles. 7.125×10-3 × 2 = 0.01425moles.

0.1/0.01425 = 7.02g mol-1(3s.f).

                                Maximum possible value: 173cm3. Relative atomic mass of lithium: 173/24000 = 7.208×10-3moles. 7.208×10-3 × 2 = 0.0144moles.

                                                        0.1/0.0144 = 6.94g mol-1(3s.f).

ROB HENDERSON