Sample Paper 2018: Question 2

2 (a) What is the difference between the accuracy and precision of an analytical method? (4 marks)

Accuracy can be defined as the closeness of an obtained value to that of a known value.

Precision can be defined as the closeness of two or more values to each other.

(b) What are the advantages and disadvantages of gravimetric and volumetric analytical methods? (6 marks)

Gravimetric analysis

Advantages

Accurate and precise (given a modern weighing balance)
Absolute method (involved direct measurements without the need for calibration)
Relatively inexpensive
It allows for little room of instrumental error

Disadvantages

Time consuming
Analysis of a single element or compound at a time.
Methods are often convoluted and a slight misstep in a procedure can often mean disaster for the analysis

Volumetric Analysis

Advantages

It is generally cheap, requiring little in the way of equipment.
Level of skill required is minimal.
Rapid.
Results are immediately available.

Disadvantages

Destructive method (cannot obtain the analyte after the reaction).
Large volumes normally required.
Reactions have to be done in the liquid phase.
Limited accuracy.

(c) During mixing of an active ingredient in powder form with other ingredients, also in powder form, samples were taken from the top (labelled Top Sample) and bottom(labelled Bottom Sample) of the mixer and analysed several times for active ingredient
Six replicate analysis were carried out on each sample and the results were statistically analysed to determine whether or not the mix could be considered homogeneous at the 95% confidence level:

Top 8.54; 8.66; 8.84; 8.92; 8.59; and 8.51 mg per gm
Bottom 8.92; 9.00; 8.86; 8.75; 9.06 and 8.84 mg per gm

(i) State the null hypothesis (2 marks)

There is no significant difference between the concentration of the analyte in the top layer and the concentration of the analyte in the bottom layer. Any difference between these two samples is due to random error and not due to determinate error or the mix not being homogenous.

(ii) Can the mix be considered as homogeneous at the 95% confidence level? (8 marks)

T-test is used as it is used to determine whether there is a significant difference between two groups.

$spooled=\sqrt{\frac{\sum x_{1}^{2}-\frac{(\sum x_{1})^2}{n_{1}}+\sum x_{2}^{2}-\frac{(\sum x_{2})^2}{n_{2}}}{n_{1}+{n_{2}}-2}}$

The data needed for the t-test can be seen in the below table.

	x1	x2	x1*x1	x2*x2
	8.54	8.92	72.9316	79.5664
	8.66	9.00	74.9956	81.00
	8.84	8.86	78.1456	78.4996
	8.92	8.75	79.5664	76.5625
	8.59	9.06	737881	82.0836
	8.51	8.84	72.4201	78.1456
Total	52.06	53.43	451.8474	475.8577

$spooled=\sqrt{\frac{451.847-\frac{52.06^2}{6}+475.8577-\frac{(53.43)^2}{6}}{10}}=0.1426$

The difference in the two means is:

$\overline{x}_1-\overline{x}_2=8.677-8.905=-0.22833$

The degrees of freedom in this question is 12-2, or 10. The t-value associated with 10 degrees of freedom at 95% confidence level is: 2.228

$\overline{x}_1-\overline{x}_2=t.spooled.\sqrt{\frac{N_1+N_2}{N_1.N_2}}$

$t.spooled.\sqrt{\frac{N_1+N_2}{N_1.N_2}}=2.228*0.1426*\sqrt{\frac{6+6}{36}}=0.1834$

Since $\overline{x}_1-\overline{x}_2> t.spooled.\sqrt{\frac{N_1+N_2}{N_1.N_2}}$ then the null hypothesis becomes invalid, showing that the mix is not homogenous.

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