Question

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

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 = \(\frac{0.7783+0.7780+0.7786}{3}\)

= 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 = \(\frac{0+0.0003+0.0003}{3}\)

= 0.0002

∴ Mean absolute deviation = ±0.0002 g

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

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.

i. Relative deviation = 0.0257% 

ii. Relative error = 0.1027%

[Note : i. As per the method given in textbook, the calculated value of relative deviation is 0.0257%. 

ii. The negative sign in -0.1027% indicates that the experimental result is lower than the true value.]