Determination of the relative atomic mass of Lithium

   The problems, which could have caused my readings to be beyond the actual Ar of lithium (6.9), are:

• The specified amount of lithium supposed to be used was 0.10 grams. I think the fact that I received a piece of lithium that was of 0.11 grams, caused the measured relative atomic mass of lithium to increase because there was more lithium, hence there was a lot more hydrogen released. This therefore means that I would have a greater amount of hydrogen released when using 0.11 g of lithium, than if I was using 0.10 g of lithium.
• Also whilst doing my experiment I found that there was a bubble of air caught in the measuring cylinder which meant the volume of gas would increase.
• Some hydrogen could have escaped between the time the lithium touched the water and place the bung over the top of the conical flask.

Method Two

Results

   Below I have processed the evidence from the evidence in a table and I have found the titres:

   I found that because two titres were the same I didn’t need to take an average, so I found the titre to be 38.6.

Using these readings the aim was to do the following:

• Calculate the number of moles of HCl used in the titration.
• Deduce the number of moles of LiOH used in the titration/
• Calculate the number of moles of LiOH present in 100cm3 of the solution from Method One.
• Use this result and the original mass of lithium to calculate the relative atomic mass of lithium.

   This method is a volumetric analysis in other words a titration. The first solution is added from the burette until the end point, when all the second solution has reacted.      The volume of solution from the burette needed to reach the end point is the titre. This, the volume of solution in the flask and the known concentration of one solution are used to calculate the number of moles of lithium hydroxide.

Here I used the following formula:

3. Number of moles = Concentration X Volume  (n = CV)

So I used this formula in the equation below to complete the aims listed above:

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

                                              N = CV

                                            = 0.100 moldm3 X 0.0386 dm3

                                  = 0.00386 mols

                             1 : 1

0.00386mols                    0.00386 mols

• To work out the number of moles of HCl, I had to multiply the concentration by the volume of HCl used in the titration. The volume had to be given in dm3 so I divided the volume beforehand by 1000 to get 0.0386. Also the concentration of the HCl was already given.

In 25cm3 of LiOH there are 0.00386 mols present, and in Method One, I had 100cm3 of LiOH.

• To calculate the number of moles of LiOH present in 100cm3 of the solution in Method One, I saw that 25 multiplied by 4 gives me 100. This means that the ratio between the two is 1 : 4 so I have to multiply 25cm3 by 4,

25cm3 X 4 = 100cm3

Therefore I have to multiply the number of moles of LiOH by 4 as well,

0.00386mols X 4 = 0.01544mols

So the number of moles of LiOH present in 100cm3 of the solution from Method One is 0.01544 mols.

• To calculate the relative atomic mass of you have to divide the original mass of lithium (0.11 g) by the number of moles of LiOH present in 100cm3 of the solution (0.01544 mols).

0.11g ÷ 0.01544 mols = 7.124352332

      Therefore the Ar of lithium in method two is 7.12 gmol-1.

From here I can work out the average relative atomic mass of lithium by taking the average of the two relative atomic masses, which are 7.10 and 7.12.

((7.096774194 + 7.124352332) ÷ 2 = 7.110563263

                                                             = 7.11 (3sf)

You can see that the relative atomic mass of lithium is 7.11, but the actual relative atomic mass is 6.9, so why could there have been this difference in mass, the reasons why are listed below:

• The plan had told me to weigh about 0.10 grams of lithium on a scale, it should have given a range, because how can I tell how much about really is?
• The lithium had to be added to the distilled water as quickly as possible, without losing any hydrogen gas. It was all about your reaction time, if too slow then a lot more gas escapes than if you were fast.
• There was already air bubbles present in the measuring cylinder prior to the reaction, this meant that the end volume of hydrogen gas would be a percentage higher than it would be if there were no bubbles.
• The delivery tube may have had some substance inside (left over from a previous experiment), so the volume of gas entered cannot therefore, be to its maximum.
• The fact that the methods did not tell me to rinse the apparatus, i.e. the burette with HCl meant that there may have been other substances still in adhesion with the glass.  

EVALUATION

I am confident that my results are reliable because my final relative atomic mass came to be 7.11, which is 0.21 units away from the actual relative atomic mass of 6.9.

I think that the overall accuracy of my experiments was fair because the Ar was 7.11, but I think that with improvement this could have been closer to 6.9. 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 ÷ value of quantity) X 100%

The main sources of error in procedure and in measurements are:

• Random errors, they 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 the reading error:

The reading error, this is an error due to guesswork involved in taking a reading from a scale when the reading lies between the scale divisions.

Reading on the measuring cylinder is between 26 cm3 and 27 cm3, best estimate of the next figure is half a division to give reading of 26.5cm3.

• Systematic errors result in readings, which are all too high, or all too low. They can be reduced of eliminated entirely, and this increases the accuracy of the final result. They can be reduced or totally eliminated by the use of better technique and instruments.

An example of a systematic error is the parallax error:

The parallax error, this is the error, which occurs when the eye is not placed directly opposite a scale when a reading is being taken. This relates to the meniscus, which I saw to be concave shaped. The meniscus is brought about because of adhesion; this is an intermolecular force of attraction between molecules of different substance.

Capillary action is a result of adhesion.

Upward capillary action

Concave meniscus

HCl moves in the glass tube (burette).

HCl molecules attracted to glass

molecules because forces of adhesion stronger than attraction between HCl molecules.

To avoid the parallax 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 in a procedure, materials, and strategy include the following:

• Temperature change due to heat loss from the solution not being insulated.
• Loss of small mass of solid when making standard solution.
• Lack of purity of chemicals due to reaction with air. (Lithium oxide)

The inbuilt uncertainties in measuring equipment includes the following:

• Small quantities have larger uncertainties.
• Percentage errors can be reduced by scaling up quantities or use equipment which have smaller divisions, i.e. a small measuring cylinder has smaller divisions and so would be more accurate than a larger measuring cylinder with larger divisions, thus percentage errors decreases.
• Sources of error such as the parallax error can give unreliable readings, which in turn will give me an error in my final relative atomic mass.

E.g. If the volume of hydrogen gas produced in Method One was misread, say it was smaller than 0.186 dm3, then the number of moles of hydrogen would decrease, thus giving a smaller number of moles of lithium, this then gives a larger Ar.

• The reading error also applies to the above statement, as if scales are misread then I would get the same problems, 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 higher value, the lower value or the value in between. It is much more accurate to take the middle value, it is this guesswork which brings about the percentage error. That is why it is easier to use apparatus with smaller divisions.

The limitation in procedures/ materials/ strategies and inbuilt certainties in measuring equipment can be improved in a lab situation by:

• Using measuring equipment with smaller percentage errors such as small measuring cylinders, which would increase accuracy in the readings made.
• Use a syringe instead of a pipette filler and pipette, because it is hard to get the liquid to the mark on the neck of the 25cm3-pipette and keep it there.
• Use a larger trough/ tray because it is easier to fill the 250cm3-measuring cylinder with water without having air bubbles in it prior to starting the actual reaction, preferably its length should be bigger than the height of the measuring cylinder, so that it can submerge fully underwater.
• Insulate the conical flask used in prior to the reaction to prevent heat loss.
• Repeat the experiment once more. As I had 100cm3 of solution of lithium hydroxide, I could have repeated the experiment a fourth time because in each reading I extracted 25cm3, so at the end of the 3rd reading I still had 25cm3 left over.

These improvements will minimise the significant sources of error as it excludes the previous errors that might have been made, such as the air bubbles present in the measuring cylinder (solved by using a larger tray of water).

Smaller divisions in measuring apparatus means that if errors are made then they will be to a small percentage error. Using equipment like syringes can be easier to use and more accurate depending on the scale divisions.

The uncertainties in the evidence in terms of their effect on the measure of confidence, which can be placed on my final conclusion drawn, are:

• The percentage errors, below I have drawn a table which states the percentage errors along with the absolute values:

So therefore the percentage error can affect the validity of the conclusions drawn and also determine how misreading the scale (random errors (reading errors)) or reading a measurement through the wrong point (systematic errors (parallax error)) can make an error no matter how small or big the error is.

For equipment such as the pipette is hard to work out the percentage error, because all I knew was that the mark on the neck meant that the volume of liquid up to that point was 25cm3. As there are no other markings such as scales, I couldn’t work out the absolute value.

Secondary sources used:

• The Usbourne science dictionary,
• Chemistry 1
• Encarta