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* The Rough was excluded during the calculation of the mean as the rough was a trial to indicate the general whereabouts of the end point, which is inaccurate of the exact end point.
Table 2. Observations collected during the experiment. Table contains observations which were recorded during each trial of the experiment.
* When water was added to the crystals of oxalic acid, it dissolved almost instantly with a light stirring of the beaker.
* When two drops phenolphthalein was added to the solution of sodium hydroxide, the solution turned pink.
* After adding about 20.8dm3 of oxalic acid into the sodium hydroxide solution, while swirling the conical flask, the solution turned completely clear – the oxalic acid was filled up to 21dm3 to have a rounded rough end point to work with.
* When two drops of phenolphthalein was added to the sodium hydroxide solution, the solution turned pink.
* After adding about 20dm3 of oxalic acid solution, while swirling the conical flask gently, each drop of oxalic acid solution began to turn the pink solution slightly clear.
* After about 20.6dm3 oxalic acid solution was added, the pink solution turned completely clear.
* Two drops of phenolphthalein was added to sodium hydroxide solution, producing a pink coloured solution.
* After about 20dm3 of oxalic acid solution was added, each drop began turning the pink solution slightly clear.
* After adding around 20.4dm3 of oxalic acid solution, the pink solution turned completely clear.
* Two drops of phenolphthalein was added to the sodium hydroxide solution, turning the solution in the conical flask pink.
* After adding about 20dm3 of oxalic acid solution, each drop began turning the pink solution slightly clear.
* When about 20.3dm3 of the acid was added, the pink solution turned completely clear.
1. Write an equation for the reaction.
(COOH)2 + 2NaOH –> 2NaCO3 + 2H2O
2. What amount (in moles) of NaOH is present in 20.0 cm3 of 0.097 mol dm-3 sodium hydroxide solution?
c = n/v
c = 0.097mol dm-3 ï¿½0.001
= 0.097mols ï¿½ (0.001/0.097 x 100%)
= 0.097mols dm-3 ï¿½1.03%
v = 20.0cm3 ï¿½0.04
= (20.0/1000) ï¿½ (0.04/20.0 x 100%)
= 0.020dm3 ï¿½0.2%
n = ?
n = c x v
n = 0.097 ï¿½1.03% x 0.020 ï¿½0.2%
n = 0.00194 ï¿½ (1.03 + 0.2)%
? n = 0.00194 mols ï¿½1.23%
3. What amount of oxalic acid was present in the average volume required to react exactly with the sodium hydroxide solution?
Using the molar ratio of 1:2 for acid : base respectively:
0.00194 ï¿½1.23% / 2 = 0.00097 mols
? amount of oxalic acid present = 0.00097 mols ï¿½1.23%
4. What amount of oxalic acid was present in your 250 cm3 volumetric flask?
Concentration of acid in 250mL is the same as the concentration of acid in 1L.
Therefore the concentration of acid in 1 dm-3 of acid solution is the same as concentration of acid in 250mL:
c = n/v
c = 0.0119ï¿½2.03% / 0.25 ï¿½(0.15/250 x 100%)
c = 0.0476 ï¿½ (2.03+0.06)%
? c = 0.05 mols dm-3 ï¿½2.09% (concentration of acid)
c = n/v
0.05ï¿½2.09% = n / 0.25 ï¿½0.06%
0.05ï¿½2.09% x 0.25ï¿½0.06% = n
? n = 0.0125 ï¿½ (2.09 + 0.06)%
? moles of oxalic acid in 250mL solution = 0.0125 mols ï¿½2.15%
5. What is the mass of one mole of oxalic acid?
n = m/MM
MM = m/n
MM = 1.5g ï¿½0.02 / 0.0125mols ï¿½2.15%
MM = 1.5 ï¿½(0.02/1.5 x100%) / 0.0125ï¿½2.15%
MM = 120 ï¿½(1.33+2.15)%
MM = 120 ï¿½3.48%
m = n x MM
m = 1 x 120ï¿½3.48%
? mass of one mole of oxalic acid = 120g ï¿½3.48%
6. How many molecules of water of crystallisation are present in one mole?
MM of (COOH)2 = 2(12.01 + 32 + 1) = 90.04
MM of H2O = (2.02 + 16) = 18.02
(COOH)2.xH2O = 120g ï¿½3.48%
? 90.04 + 18.02x = 120g ï¿½3.48%
? 18.02x = 120g ï¿½3.48% – 90.04
? 18.02x = 29.96ï¿½3.48% (/18.02)
? x = 1.6626ï¿½3.48%
? Molecules of water of crystallisation = 1.66ï¿½0.06
Conclusion and Evaluation
1. Estimate the degree of uncertainty in your readings using the balance, the volumetric flask, the pipette and the burette. How accurately can you quote your answer?
Balance = ï¿½0.02g
Volumetric flask = ï¿½0.15mL
Pipette = ï¿½0.04cm3
Burette = ï¿½0.15cm3
The equipments used were relatively accurate as the uncertainties compared to the collected values are small.
2. Compare your answer with the correct answer and work out the percentage error.
Percentage error = [(experimental value – actual value) / actual value] x 100%
= [(1.6626 ï¿½3.48% – 2) / 2] x 100%
= 16.87% ï¿½1.74%
3. Suggest any other reasons for possible error.
Other reasons for possible error include the uncertainty of equipments, human errors and mistakes, possibility of contaminated solutions, air bubbles and transference of solution between equipments.
This experiment was conducted to find the ratio of water in hydrate oxalic acid crystals. This included calculations of the theoretical and experimental values for the ratio of water in the hydrate oxalic acid crystals and the percentage errors of values compared.
The equation for the reaction between the sodium hydroxide solution (NaOH) and the oxalic acid solution ((COOH)2) was worked out, with the product being a salt (NaCO3) and water (H2O). As the oxalic acid solution is a dibasic acid solution, the reaction ratio with sodium hydroxide is 1:2. The amount of NaOH, in moles, was also calculated by multiplying the concentration of the sodium hydroxide solution and the volume of sodium hydroxide solution used. The equation c = n/v was used to calculate this. The result was 0.00194 moles of NaOH in the sodium hydroxide solution with an uncertainty of ï¿½1.23%. The amount of oxalic acid presented in the average volume required to react exactly with the sodium hydroxide solution was also calculated; by dividing the moles of NaOH (0.00194ï¿½1.23%) by 2, based on the 1:2 reaction ratio stated before. The result was 0.00097 moles with an uncertainty of ï¿½1.23%.
The amount of oxalic acid present in 250cm3 was also calculated. The concentration of acid in 250mL is the same as the concentration of acid in 1L, therefore, the concentration of acid in 1dm-3 of the oxalic acid solution is the same concentration as the acid in 250mL. With this understood, the equation c = n/v was used once again. The ‘concentration’ was first calculated by dividing the moles of NaOH by 0.25dm3.
The result (0.05mols dm-3 ï¿½2.09), was used in further calculations to determine the moles of oxalic acid by multiplying the previous calculated concentration (0.05mols dm-3 ï¿½2.09) with 0.25dm3, resulting in the moles of oxalic acid in the 250mL solution as 0.0125mols ï¿½2.15%. This value was used in the calculation of the mass of one mole of oxalic acid. The equation n = m/MM was used in this calculation. The MM (molar mass) was first calculated, dividing the mass of oxalic acid crystals by the moles of oxalic acid (1.5 / 0.0125). The mass (m = n x MM) was calculated with the result of this (1 x 120ï¿½3.48%), resulting in the mass of one mole of oxalic acid being 120ï¿½3.48%.
Finally, the molecules of water of crystallisation presented in one mole were calculated. This was done by using the mass of one mole of oxalic acid (120ï¿½3.48%), subtracting the MM of (COOH)2 (90.04) from it and dividing the result by MM of H2O (18.02). The final result of the number of molecules of water in crystallisation equalled 1.66 with an uncertainty of ï¿½3.48% or 0.06. With this final result, the percentage error was calculated, being 16.87%ï¿½1.74%.
All the uncertainties within the values used were calculated into percentage through dividing the uncertainty by the value and multiplying it by 100%. The uncertainties were converted into percentages before doing calculations were done.
There were several limitations found during this experiment which would have been the cause of an inaccurate final result – the variation between the theoretical value of 2 and the experimental value of 1.66ï¿½0.06. These limitations include the uncertainties of the equipment, human errors and mistakes, possibility of contaminated solutions, air bubbles found in solutions and the transference of solution between equipments. The significance of these errors and the improvements are listed in the table below.
Uncertainty of equipments
Uncertainties were obtained during the experiment through the use of equipments. The inaccurate recording of data would have affected the results in the end. Eg. The readings on the electronic scale weren’t fully stable and consistent at one value, it varied – increasing and decreasing slightly.
In the case of the electronic scale, it was suggested that the variation in weight value was due to the air-conditioning blowing down on it. The air-conditioner could be turned off to reduce the uncertainties and variations. Students could also prevent themselves from breathing heavily down onto the scale while measuring, this could have contributed to the variation as well.
Human errors and mistakes
Mistakes could affect the outcome and accuracy of results. Eg. While adding water into the volumetric flask to dissolve the oxalic acid crystals, a student accidentally added a little too much, over the 250cm3 that was intended. This would have over diluted the oxalic acid.
This could have been avoided through patience while adding the water into the volumetric flask. This could also have been done slowly and carefully, probably using a pipette towards the end – at the 250cm3 mark.
Reading the measurement of the pipette, burette and volumetric flask could have been inaccurate, causing results to be inaccurate and in turn affecting the calculations.
Inaccuracy could have been due to improper way of reading the measurement. Students should have bent down slightly with the measurement at eye level. Student`s heads should not be tilted but level with the equipment as well. The measurement should also be read at the parallax, not above or below it.
Possibility of contaminated solutions
The experiment was conducted with the use of several different solutions. These solutions could have accidentally been mixed in with each other, causing the solution to be contaminated. The results obtained due to this could have been inaccurate, affecting the final results.
Being careful with the solutions used would help avoid this. Using clean equipments for each different solution, making sure the solutions that are not meant to be placed together do not get mixed with each other.
After the use of some of the equipments, they were to be washed and reused for another solution. The equipments were not dried before being used again; this could have diluted the solution and contaminated it.
Drying the equipment each time after it is washed/cleaned would prevent access water and dilution of solutions.
When transferring the dissolved oxalic acid into the burette, there were some tiny air bubbles. This would have because the measurement to be inaccurate as the air bubbles would have caused the reading of the solution to be greater than it actually is. Over time when the bubbles removed itself from the solution which would have caused a decrease in solution. The inaccurate measurement reading would have affected the final results of the experiment.
The burette could have been left alone for awhile to allow the bubbles to float to the top. Whatever the amount the bubbles made up for, fill the burette up again with the use of a pipette.
Transference of solution between equipments
When transferring solutions between two or more differing equipments would have caused an increase of decrease in measurement. An inaccurate amount of solution would have affected the results or recorded data.
When using a funnel during the transference of solution into the burette, it caused some increase in the desired amount. This is due to the tip of the funnel having some drops of solution left in it and when the funnel was removed, the movement could have caused those drops to drop into the burette, causing an increase in amount of solution. This could be prevented by either slowly removing the funnel or using the funnel up to a certain mark, remove it and fill up to the desired amount with the use of a small pipette – drop by drop.
When using the 20mL pipette to transfer the sodium hydroxide into the conical flask, the tip of the nozzle had a drop that was tempted to drop out of the pipette. Avoid touching the sides and move the pipette over the flask slowly would reduce the chances of it dripping out and altering the volume of solution.
A small amount of the solution was also stuck at the tip of the nozzle after being emptied. By touching the tip to the side of the flask would help that tiny bit of solution flow out. If possible have the nozzle flat on the side of the flask, this would allow it to flow out smoothly.
To avoid or reduce these significant errors, these solutions and improvements should be taken into consideration during future repetition of this experiment.
The experiment overall was invalid. Though the values in data collection may have been to a 0.10cm3 difference, the final calculation of the percentage error, 16.87%ï¿½1.74%, was much greater than the desired 1% causing the experiment to be invalid.