= 0.020dm3 ±0.2%
n = ?
n = c x v
n = 0.097 ±1.03% x 0.020 ±0.2%
n = 0.00194 ± (1.03 + 0.2)%
∴ n = 0.00194 mols ±1.23%
- 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 ±1.23% / 2 = 0.00097 mols
∴ amount of oxalic acid present = 0.00097 mols ±1.23%
-
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 = 0.0119±2.03% / 0.25 ±(0.15/250 x 100%)
c = 0.0476 ± (2.03+0.06)%
∴ c = 0.05 mols dm-3 ±2.09% (concentration of acid)
c = n/v
0.05±2.09% = n / 0.25 ±0.06%
0.05±2.09% x 0.25±0.06% = n
∴ n = 0.0125 ± (2.09 + 0.06)%
∴ moles of oxalic acid in 250mL solution = 0.0125 mols ±2.15%
- What is the mass of one mole of oxalic acid?
n = m/MM
MM = m/n
MM = 1.5g ±0.02 / 0.0125mols ±2.15%
MM = 1.5 ±(0.02/1.5 x100%) / 0.0125±2.15%
MM = 120 ±(1.33+2.15)%
MM = 120 ±3.48%
m = n x MM
m = 1 x 120±3.48%
∴ mass of one mole of oxalic acid = 120g ±3.48%
- 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 ±3.48%
∴ 90.04 + 18.02x = 120g ±3.48%
∴ 18.02x = 120g ±3.48% – 90.04
∴ 18.02x = 29.96±3.48% (/18.02)
∴ x = 1.6626±3.48%
∴ Molecules of water of crystallisation = 1.66±0.06
Conclusion and Evaluation
Discussion
- 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 = ±0.02g
Volumetric flask = ±0.15mL
Pipette = ±0.04cm3
Burette = ±0.15cm3
The equipments used were relatively accurate as the uncertainties compared to the collected values are small.
- 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 ±3.48% – 2) / 2] x 100%
= 16.87% ±1.74%
- 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 ±1.23%. 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 (0.00194±1.23%) by 2, based on the 1:2 reaction ratio stated before. The result was 0.00097 moles with an uncertainty of ±1.23%.
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 ±2.09), was used in further calculations to determine the moles of oxalic acid by multiplying the previous calculated concentration (0.05mols dm-3 ±2.09) with 0.25dm3, resulting in the moles of oxalic acid in the 250mL solution as 0.0125mols ±2.15%. 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 120±3.48%), resulting in the mass of one mole of oxalic acid being 120±3.48%.
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 (120±3.48%), 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 ±3.48% or 0.06. With this final result, the percentage error was calculated, being 16.87%±1.74%.
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 1.66±0.06. 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.
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%±1.74%, was much greater than the desired 1% causing the experiment to be invalid.
