I took the mass of the weighting boat alone which was 1.055g,
I then took the mass of the weighting boat with lithium metal 1.148g
This gives the mass of the lithium to be 0.093g
I then went on to put the Lithium into the distilled water in the conical flask and place the bung immediately to a void any lose of Hydrogen gas.
I then took the volume of the hydrogen produced which was 150.000cm3.
The balanced equation below shows the reaction between Lithium and distilled water.
2 Li(s) + 2 H2O(l) ________ 2LiOH(aq) + H2(g) .
From the volume of Hydrogen gas produced I can work out the number of moles of Hydrogen, and from this I can deduce the number of moles of Lithium in the reaction. This goes as;
1mole of gas occupies 24 000 cm3
Therefore 150cm3 = 150 ÷ 24000 = 0.00625moles of H2
From the equation 2li_______ H2
Therefore the ration is 2:1
This means the number of moles of lithium are twice the number of moles of hydrogen is in the reaction.
Therefore 0.006 × 2 = 0.012 moles
From here I will be able to find the atomic mass of lithium using the formula
Atomic mass = mass ÷ number of moles
I have the mass of lithium as 0.093g
And calculated the number of moles to be 0.012 moles
Therefore 0.093 ÷ 0.012 = 7.75 (AMU)
atomic mass of lithium is 7.75
Determining the atomic mass of lithium from method two.
When phenolphthalein indicator is added to the lithium hydroxide solution, it changes the colour into pink. In the titration there is a point where the colour will change from pink to still-grey. This is the end point and it is at this point that I took the burette's reading. My first titration (rough) was to give me a rough idea of where this point is so that I will be more careful with the rest when it nears the same volume by running the titration drop by drop. This will improve the accuracy of my results. I also repeated the titration of three times which improves the reliability of my results and read the bottom of the meniscus when taking readings and took it to the nearest 0.05cm3 for better precision and accuracy
RESULTS
Mean titration = (28.50+28.45+28.50) ÷ 3 = 28.48cm3
Calculating no of moles of HCl
No. of moles = concentration x volume
Volume of HCl = 28.48cm3 = 0.02848dm3
Concentration = 0.100dm3
- 0.100 × 0.02848 = 0.002848moles
From here I will find the number of moles of LiOH
The balanced chemical equation bellow shows the reaction between Lithium Hydroxide and Hydrochloric acid.
LiOH(aq) + HCl(aq) _______Cl(aq) + H2O(l).
The ratio of LiOH : HCL is 1 : 1 therefore the number of moles of
LiOH = 0.002848moles
The total volume of LiOH was 100cm3
Therefore 0.002848 × 4 = 0.011392 moles
Going back to the first equation
2 Li(s) + 2 H2O(l) ________ 2LiOH(aq) + H2(g) .
The ration of Li : LiOH is 2 : 2 = 1 : 1
Therefore the number of moles of Li = 0.011392 moles
From this I can deduce the atomic mass of Li with the following calculations
Mass of Li = 0.093g
Number of moles = 0.011392moles
Atomic mass = mass ÷ no. of moles
Atomic mass = 0.093 ÷ 0.011392 = 8.16 (2.d.p). This is the Ar of lithium
% error calculations (answers correct to 2.d.p)
Pipette: ± 0.06
Therefore % error = 0.06÷25×100 = 0.24%
Buirette: ± 0.01
Therefore % error = 0.1÷28.48×100 = 0.35%
- 0.24²+ 0.35² = 0.58 + 0.12 = 0.7 = 0.84%
Evaluation
- The experiment went fairly well because my results were not very far from each other. This small difference could have been brought about by a number of factors. This include:
-
When placing the Li in the conical flask, there is a risk that some of the H2 gas would escape before the bung could be placed. This would affect my final results. One way method to avoid this is to put the Li in a dry conical flask, place the bung then, Adding the 100cm³ of distilled water using a 'ladle'
- The Li oxidises in the air very rapidly so this will affect the mass. This is difficult to avoid and only possible in an inert atmosphere which is not possible to create with the apparatus available. The only thing I could have done was to ensure every thing was set up before taking out the Li and quicken the drying and measuring of it
- Another problem that could have affected the reliability of my final result was the equipments. Most of the equipments I used were class B. this could have affected accuracy of the measurement reading I could avoid this by using class A equipments so that the reading is as accurate as possible. Collecting the gas in a gas syringe could also give more accurate readings.
- There could be errors when reading the level of the burette and visually checking for the point at which the solution turned from pink to colourless. This fault is almost impossible to eradicate.
- It was required that only 5 drops of indicator were added, this was done carefully but the sizes of each drop may differ from the other, although very minimal this could affect the final result. Adding 20 drops to the 100cm³ before taking out anything would have ensured even distribution of the indicator
- Another problem was weighting the lithium. As it was stored under oil for it’s reactive nature
