The aim of this practical was to find the RMM of a group 2 carbonate, and to accurately guess the element from group 2 used.
Safety glasses Burette
Weighing bottle Burette funnel
Spatula Burette stand
3 x 250cm3 beakers White tile
Distilled water 3 x 250cm3 conical flasks
Stirring rod Hydrochloric acid
Volumetric flask Sodium Hydroxide
Filter funnel Methyl red indicator
Safety Precautions: –
Safety glasses were worn to prevent any substances entering the eye. Laboratory coats were also worn to prevent our clothes and skin, and hair was tied back so it did not interfere with our sight and experiment.
The weighing bottle and lid were weighed, and recorded. The required amount in the range of 1.55-1.65g of the unknown group 2 carbonate was added to the weighing bottle, and the new weight recorded. 50.0cm3 of standard hydrochloric acid were pipetted into a 250.0cm3 volumetric flask, and a funnel added at the top of the flask. The substance in the weighing bottle was transferred into the funnel and the weighing bottle reweighed. When the solution stopped reacting, water was added to the solution to make it up to 250.0cm3 mark and then it was shaken to ensure the solution was fully mixed. A series of 25.0cm3 of the carbonate solution were titrated against standard sodium hydroxide solution using methyl red as an indicator.
Molar Mass of compound A
Mass of empty weighing bottle and lid
Mass of bottle and contents before transfer, M1
Mass of bottle and contents after transfer, M2
Mass of compound A dissolved (M1-M2)
Actual concentration of hydrochloric acid used = 0.992moldm-3
Actual concentration of sodium hydroxide solution used = 0.0935moldm-3
The methyl red indicator changes from red to yellow, so the results are recorded when the solution instantaneously turns yellow.
To find the relative molecular mass (RMM) of a group II carbonate, we use this equation: –
RMM = mass in grams
In order to find the number of moles of the unknown substance (compound A), it is reacted with excess acid, which when left after the effervescence has finished, can be titrated against an alkali of known concentration.
Therefore to calculate the RMM of compound A, we do a series of calculations, which follow: –
No. of moles of NaOH that reacted with the excess acid
= 0.0935 x 0.0301 dm3
= 0.00281435M in 25.0cm3
NaOH and HCL react on a 1:1 mole ratio, so this gives the number of moles of acid in 25.0cm3
NaOH : HCL
1 : 1
0.00281435 : 0.00281435
To gain the number of moles in 250.0cm3, the answer of the previous step was multiplied by 10.
0.00281435M x 10 = 0.0281435 cm3 in 250.0cm3
The total number of moles of acid used initially before it was reacted with the unknown compound is now calculated: –
0.992M x 0.050 dm3
The number of moles of excess acid is subtracted from the total number of moles to find the number of moles that reacted with the unknown substance.
0.0496M – 0.0281435M = 0.0214565M
From the equation following, the number of moles of the unknown substance is calculated: –
XCO3 + 2HCL –> xCL2 + CO2 + H2O
1 : 2 –> products
0.01072825 : 0.0214565 –> products
The RMM can therefore be calculated now: –
mass in grams = 1.605
= 149.61 (to 2 dp)
Therefore, xCO3 = 149.61
CO3 = 60
x = 149.61 – 60
This means that the unknown substance was Strontium, as the RMM suggests, because the RAM of strontium is 88, which is the nearest to the calculated RMM of 89.61.
I can thus conclude that I have fulfilled my aim to find the unknown group 2 element in its compound, as I found that, from my calculations, that the element is most likely to be strontium as its relative atomic mass is 88, which is the nearest to my calculated figure. The percentage error for the results is: –
1.61 x 100 = 1.83% (to 2 dp)
This is a low percentage error, which could be attributed to the following errors in inaccuracy of the results.
The weighing balances were only accurate to 0.0005g, therefore the two readings that were taken amounts to an error of 0.001g. This in a percentage is: –
0.001 x 100 = 0.016%
The volumetric flask was only accurate to 1 drop, which is approximately 0.1cm3 in 250.0cm3. To work out the percentage error: –
0.1 x 100 = 0.04%
The pipette is only accurate to one drop, which is approximately 0.1cm3. This is in a total of 25.0cm3 so therefore
0.1 x 100 = 0.4%
The burette is accurate to 0.05cm3. This reading is taken three times, so the error becomes 0.1cm3 in a total of the mean titre.
0.1 x 100 = 0.33%
I think that there was some inaccuracy in my results because I achieved an answer of 89.16 when the actual figure is 88, so this could be due to any of the above errors, which have been calculated. To improve my results, I would take more repeats of the titration, and gain a more accurate average. I would also try to use a more accurate weighing balance, plus to make sure that the apparatus is clean. Otherwise, the experiment was fairly accurate as it fulfilled my aim.