Question

Solve the following problem:

A 1.000 mL sample of acetone, a common solvent used as a paint remover, was placed in a small bottle whose mass was known to be 38.0015 g.
The following values were obtained when the acetone - filled bottle was weighed: 38.7798 g, 38.7795 g and 38.7801 g. How would you characterise the precision and accuracy of these measurements if the actual mass of the acetone was 0.7791 g?

Answer 1

i. Precision:

Measurement	Mass of acetone observed (g)
1	38.7798 – 38.0015 = 0.7783
2	38.7795 – 38.0015 = 0.7780
3	38.7801 – 38.0015 = 0.7786

Mean = `(0.7783+0.7780+0.7786)/3` = 0.7783 g

Measurement	Mass of acetone observed (g)	Absolute deviation (g) = |Observed value – Mean|
1	0.7783	0
2	0.7780	0.0003
3	0.7786	0.0003

Mean absolute deviation = `(0+0.0003+0.0003)/2` = 0.0002

∴ Mean absolute deviation = ±0.0002 g

Relative deviation = `"Mean absolute deviation"/ "Mean"xx100%`

= `0.0002/0.7783xx100%`

= 0.0256%

ii. Accuracy:

Actual mass of acetone = 0.7791 g
Observed value (average) = 0.7783 g

a. Absolute error = Observed value – True value
= 0.7783 – 0.7791
= – 0.0008 g

b. Relative error = `"Absolute error"/"True value"xx 100%`

= `(-0.0008)/0.7791xx100%`

= –0.1026%

These observed values are close to each other and are also close to the actual mass. Therefore, the results are precise and as well accurate.

1. Relative deviation = 0.0256%
2. Relative error = 0.1026%