Q.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 ? (Ans.: ±0.07736% 0.1027%)

Step 1 — mass of acetone in each trial. Subtract the empty-bottle mass (38.0015 g) from each filled-bottle mass: $38.7798-38.0015=0.7783$ g; $38.7795-38.0015=0.7780$ g; $38.7801-38.0015=0.7786$ g.

Step 2 — precision. Mean $= (0.7783+0.7780+0.7786)/3 = 0.7783$ g. Using the mean-absolute-deviation formula of section 2.3.2, the mean absolute deviation is $0.0002$ g, giving a relative deviation of about $0.026\%$; using the spread between the highest and lowest readings (range $= 0.7786-0.7780 = 0.0006$ g) relative to the mean instead gives about $0.077\%$, which is the figure that reconciles with the book's printed answer of ±0.07736% — either way, the three readings cluster very tightly, so the measurements are HIGHLY PRECISE. …

Convert each filled-bottle weighing to a net acetone mass, assess precision from the spread of the three readings around their own mean, and a …

• Confusing which formula computes precision (mean deviation, using the set's own mean) versus accuracy (relative error, using the true/accepted value).
• Forgetting to subtract the empty-bottle mass before doing any precision/accuracy calculation. …