On average, 25.0 cm3 of LiOH(aq) required 14.80 cm3 of 0.100 mol dm-3. This value was obtained by using our three concordant results.
Treatment of results:
The chemical equation for this reaction is:
LiOH(aq) + HCl(aq) → LiCl(aq) + H2O(l)
To find the Relative Atomic Mass of Lithium, we must first calculate the number of moles of HCl used in the titration. This can be done by using the following equation:
Number of Moles = Concentration x Volume = 0.100 x 14.80 = 1.480 x 10-3 moles
1000 1000
We can now deduce the number of moles of lithium hydroxide (LiOH) using the mole ratio. The mole ratio of LiOH to HCl is 1:1. This means that the number of moles of LiOH is the same as the number of moles of HCl.
So, the number of moles of LiOH is 1.480 x 10-3.
We now need to calculate the number of moles of LiOH present in 100cm3 of the solution. We used 10.0 cm3 of the solution from Method 1. Since there are 10 portions of 10.0 cm3 in 100cm3, to find the number of moles of LiOH present in 100cm3, we need to multiply the number of moles of LiOH in 10.0 cm3 by 10:
Number of moles of LiOH in 100cm3 = 1.480 x 10-3 x 10
= 0.0148 moles
Using this result we can now calculate the Relative Atomic Mass of Lithium. For this we must use the following formula:
Relative Atomic Mass = Original Mass (g) = 0.099 = 6.6892
Number of moles 0.0148
So the Relative Atomic Mass of Lithium is 6.69 (to 3 significant figures).
Hazard Of Chemicals:
Lithium Metal:
Flammable: The metal burns in air. It reacts violently with water and acids, liberating a highly flammable gas (hydrogen*).
Corrosive: Can cause burns. It reacts vigorously with water to form Lithium Hydroxide which is corrosive.
Evaluation
I am confident that my results are reliable due to my final relative atomic mass values using both methods coming to be 6.67 and 6.69, which are both very close to the actual relative atomic mass of 6.94. I think that with improvement in procedure and method this could have been closer to the true value of 6.94. It all depends on the accuracy, limitations in a procedure and the uncertainties in an experiment, taking all the evidence into account, such as the results and readings along with the likely errors, I can state whether or not the readings were reliable. An analysis of the errors in experimental results indicates how reliable they are. It can also suggest which aspects of the experimental method could be altered to reduce the error in the final result. This is calculated by the percentage error as below:
Percentage error = Absolute error x 100
Value of quantity
The main sources of error in procedure and in measurements are random errors and systematic errors. These are:
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Random errors are associated with most measurements. They can never be eliminated entirely but reducing them increases the precision of the final result. Improving techniques, making multiple measurements or using instruments with a higher degree of precision can reduce them. An example of this is a reading error which is an error due to guesswork involved in taking a reading from a scale when the reading lies between the scale divisions. Such as reading on the measuring cylinder when between 26 cm3 and 27 cm3, best estimate of the next figure is half a division to give reading of 26.5 cm3.
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Systematic errors result in readings, which are all too high, or all too low. They can be reduced or totally eliminated by the use of better technique and instruments and this increases the accuracy of the final result. An example of a systematic error is a reading error which occurs when the eye is not placed directly perpendicular and opposite a scale when a reading is being taken. This relates to the meniscus which formed a concave shape. The meniscus is brought about because of adhesion. To avoid the reading errors, readings of liquid levels must be taken with the eye lined up with the bottom of the meniscus, when it is a concave shape.
The limitations due to our procedure include the following:
- Lack of purity of chemicals due to reaction with air and more specifically, Lithium metal reacting with air to form an Oxide surface.
- The oil that the lithium is stored in will act as a barrier against the reaction.
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Time delay between the reaction starting (putting the Lithium in) and recording the result (putting the bung in) meaning some of the product in the form of hydrogen (H2) is lost into the atmosphere.
The errors in measurement and equipment include the following:
- Small quantities have larger uncertainties due to some calculations amplifying the smaller errors in previous steps.
- Sources of error such as the reading error can give unreliable readings, if scales are misread then we would get the same problems which in turn will give us an error in the final relative atomic mass. This is where percentage errors come in.
The reading error is the most significant source of error as it is quite easy to misread a reading of volume or any other measurement, and sometimes it is guesswork when the reading is situated between two values. Here you would take the value exactly in between the two. It is this guesswork that results in bringing about the percentage errors and so reinforces the need to use equipment with the smallest scale divisions.
The above limitations to procedure can be improved by:
- To prevent the lithium having an oxide surface and not reacting correctly with the distilled water, the piece of lithium used should be cut from the centre of the larger piece so that none of its surface has reacted to form lithium oxide. To prevent the reformation of the oxides, the lithium must be handled with speed when in the presence of air.
- To prevent any oil residue on the lithium, it must be ensured that the lithium is thoroughly wiped with a paper towel.
- Insulating the conical flask used prior to the reaction to prevent heat loss.
- Repeat the experiment more times to attain a greater number of concordant results meaning anomalies are discounted and reliability of results is increased.
The above errors in measurement and equipment can be improved by:
- Following correct procedure when taking a reading off the scales by keeping the eye lined up with the bottom of the meniscus.
- Percentage errors can be reduced by scaling up quantities or use of equipment which have smaller divisions. For example a small measuring cylinder has smaller divisions and so would be more accurate than a larger measuring cylinder with larger divisions, consequently reducing percentage errors and increasing accuracy in the readings made.
These improvements will minimise the significant sources of error as it excludes the previous errors that might have been made. Smaller divisions in measuring apparatus mean that if errors are made then they will be to a smaller percentage error. Using equipment like syringes may make the procedure easier to use and more accurate depending on the scale divisions.
Calculating percentage errors is a vital step in identifying the error margins and gives a very precise account of the accuracy of our results.
Instances in which percentage errors come into account are highlighted and calculated below:
Mass of Lithium used was measured by a balance. The value that was recorded alongside the absolute value was 0.099 ± 0.001g which means using the formula mentioned above, we reach a percentage error as below to 3 s.f.:
Percentage error = 0.001 x 100 = 1.01 %
0.099
Volume of Hydrogen produced was measured using a 250 cm3 measuring cylinder. The value that was recorded alongside the absolute value was 178 ± 1 cm3 which means using the formula mentioned above, we reach a percentage error as below to 3 s.f.:
Percentage error = 1 x 100 = 0.562 %
178
Volume of HCl used was measured using a 50 cm3 burette. The value that was recorded alongside the absolute value was 14.80 ± 0.5 cm3 which means using the formula mentioned above, we reach a percentage error as below to 3 s.f.:
Percentage error = 0.5 x 100 = 3.38%
14.80
Volume of LiOH used was measured using a 10 cm3 burette. The value that was recorded alongside the absolute value was 10.0 ± 0.04 cm3 which means using the formula mentioned above, we reach a percentage error as below to 3 s.f.:
Percentage error = 0.04 x 100 = 0.4%
10
As can be seen, the greatest percentage error is clearly with the measurement of the volume of HCl used (3.38%). To lower this percentage error and consequently improve the accuracy of our results, a larger volume of HCl would need to be used to neutralise the LiOH so therefore a larger amount of LiOH would be need to used. Not only would this decrease the percentage error in the volume of HCl used but will also lower the percentage error in the volume of LiOH used seeing as a larger measurement of LiOH would need to be used. The mass of the lithium contributes the second largest percentage error (1.01%), this can also be lowered by using a larger amount of substance, in this case lithium, which would result in a lower percentage error and a more accurate result.