Determining the Relative Atomic Mass of Lithium

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Determining the Relative Atomic Mass of Lithium

An experiment has been carried out to determine the relative atomic
mass of Lithium by using two different types of methods

The first method that was carried out was to determine the volume of
Hydrogen produced. In this experiment a fixed amount of Lithium was
used, in my case it was 0.11g. At the end of this experiment, the
volume of Hydrogen gas I collected was 185cm³.

Then using the solution of lithium hydroxide made from experiment one,
I used it in the titrating experiment, to find out the total volume of
Hydrochloric acid used to titrate the lithium hydroxide.

RESULTS TABLE

Experiment

Initial

Volume

( cm³)

Final

Volume

( cm³)

Total volume

Of HCl used

( cm³)

Rough

0.2

30.3

30.1

1

6.3

35.8

29.5

2

2.7

32.0

29.3

Average

29.6

CONCLUSION

Method 1

[IMAGE]2Li (s) + 2H20(l) LiOH(aq) + H2(g)

Number of moles of Hydrogen.

Volume of hydrogen gas was 185 cm³. Weight of Lithium was 0.11g.

N = __V__ _185_ = 0.0077 MOLES

24000 24000

Number of moles of Lithium.

2Li : H2 2 : 1 ratio

0.0154 : 0.077 Lithium = 0.0154 moles

Relative atomic mass of Lithium.

Ar = MASS _0.

MLA Citation:
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11_ = 7.14

MOLES 0.0154

METHOD 2

[IMAGE]LiOH (aq) + HCl (aq) LiCl(aq) + H20(l)

Number of moles of Hydrochloric acid.

29.6 = 0.0296 dm³

1000

N = CV no. of moles = 0.100 x 0.0296

= 0.00296 moles of HCl

Number of moles of Lithium Hydroxide.

LiOH : HCl

0.00296 : 0.00296 LiOH = 0.00296 moles

Took 25cm³ of LiOH each time in the conical flask

LiOH = 0.00296 moles

So in 100cm³ it will be 0.00296 x 4

= 0.01184 moles

Relative atomic mass of Lithium

Ar = MASS 0.11_ = 9.29

MOLES 0.01184

EVALUATION

While doing the experiment I was trying to make sure my procedure was
accurate. When doing method 1 I made sure that the delivery tube was
right in the measuring cylinder so no air bubbles could get out of the
measuring cylinder and I made sure that the measuring cylinder had no
air bubbles in it, as this would affect the outcome of my results.

In method 2 I made sure I used a constant amount of phenolphthalein
indicator all the way through the experiment, and the same amount of
lithium hydroxide all the way through also.

While doing these experiments I may have made errors even though I was
trying to avoid them, these are human errors. While reading the
burette I may have made an error in measurements if I was not at eye
level with the burette. So I may have had an error of in or out by
0.1cm³. The sources of error which may have occurred in the procedure
is that I may have not used the same amount of phenolphthalein
indicator each time, one drop more or less could have gone in, or in
one drop more may have gone in, then in another drop affecting the
outcome of my results. Also when filling the pipette with lithium
hydroxide I may not have put exactly 25cm³ each time I may have missed
the mark by a bit which can alter the whole outcome of my results.

To overcome the errors I could make sure that every time I read from
the burette, I was at eyelevel with the burette so I could read it
accurately each time and also before taking down the reading to make
sure the funnel was out of the top of the burette as this would alter
the reading. Each time I am putting the indicator in, I would make
sure that I put in the same amount of drops each time as this would
make my results more reliable as I would be using a constant amount of
indicator each time. When putting the lithium hydroxide in the conical
flask, I would make sure it was exactly at 25cm³ in the pipette before
transferring it into the conical flask, then I would rinse out the
rest of the contents left in the conical flask with distilled water
just to make sure all of the lithium hydroxide was used. I would use
distilled water, as this would not interfere with the reaction.