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Aim: To determine the concentration of Ethanoic acid in a sample of vinegar and compare this to the literature value.

Independent Variable: N/A
Dependent Variable: Volume of diluted sample of vinegar used
Controlled Variables: Volume of NaOH in the conical flask, no. of drops of indicator added

Results
Table showing the volume of sample usage in each test
Test Initial Volume (cm3) Final Volume (cm3) Volume Used (cm3)
Trial 0 32.8 32.8
1 0 27.6 27.6
2 0 27.7 27.7
3 0 27.4 27.4
4 0 27.7 27.7
Avg. 27.6

Calculations & Analysis
CH3COOH + NaOH → CH3COO- Na+ + H2O
No. of moles of NaOH used in each test = 25/1000 x 0.1
= 0.0025 moles

Since the mole ratio of CH3COOH and NaOH are 1:1, therefore No. of moles of CH3COOH in the diluted sample = no. of moles of NaOH used, which is 0.0025 moles.

Using the equation No. of moles = Concentration x Volume, we can deduce that the diluted concentration of CH3COOH is 0.0025 = C x (27.6 / 1000)

Therefore C = 0.0025 / (27.6 / 1000)
C = 0.0906 moldm-3 (3 s.f)

Applying the formula C1V1 = C2V2
Then 250 x 0.0906 = 25 x C2
C2 = 0.906 moldm-3
So C2 is the original concentration in moldm-3 of CH3COOH in the Vinegar sample.

In 25cm3 of vinegar sample the no. of mole of CH3COOH = CV
= 0.906 x 0.025
= 0.0227 moles (3s.f)

Mass of CH3COOH = 0.0227 x 60.06
= 1.36g (3 s.f)

Concentration of Ethanoic acid by mass = (1.36 / 25) x 100%
= 5.44%
The reference concentration of Ethanoic acid is 5.34%, comparing to the concentration of Ethanoic acid by mass I’ve calculated, the difference is only 0.1% and so the difference is not significant.

The percentage error would be...