# 1.5 Measurement uncertainty, accuracy, and precision  (Page 4/11)

 Page 4 / 11

## Multiplication and division with significant figures

Rule: When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).

(a) Multiply 0.6238 cm by 6.6 cm.

(b) Divide 421.23 g by 486 mL.

## Solution

(a) $\begin{array}{l}\begin{array}{l}\text{0.6238 cm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}6.6\phantom{\rule{0.2em}{0ex}}\text{cm}=4.11708\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{result is}\phantom{\rule{0.2em}{0ex}}4.1\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{2}\phantom{\rule{0.2em}{0ex}}\left(\text{round to two significant figures}\right)\hfill \\ \text{four significant figures}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{two significant figures}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{two significant figures answer}\hfill \end{array}\hfill \end{array}$

(b) $\begin{array}{l}\frac{\text{421.23 g}}{\text{486 mL}}\phantom{\rule{0.2em}{0ex}}=\text{0.86728... g/mL}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{result is 0.867 g/mL}\phantom{\rule{0.2em}{0ex}}\left(\text{round to three significant figures}\right)\\ \frac{\text{five significant figures}}{\text{three significant figures}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{three significant figures answer}\end{array}$

(a) Multiply 2.334 cm and 0.320 cm.

(b) Divide 55.8752 m by 56.53 s.

(a) 0.747 cm 2 (b) 0.9884 m/s

In the midst of all these technicalities, it is important to keep in mind the reason why we use significant figures and rounding rules—to correctly represent the certainty of the values we report and to ensure that a calculated result is not represented as being more certain than the least certain value used in the calculation.

## Calculation with significant figures

One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters.

## Solution

$\begin{array}{lll}V\hfill & =\hfill & l\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}w\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}d\hfill \\ & =\hfill & \text{13.44 dm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{5.920 dm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{2.54 dm}\hfill \\ & =\hfill & \text{202.09459}...\phantom{\rule{0.2em}{0ex}}{\text{dm}}^{3}\left(\text{value from calculator}\right)\hfill \\ & =\hfill & {\text{202 dm}}^{3}\text{, or 202 L}\phantom{\rule{0.2em}{0ex}}\left(\text{answer rounded to three significant figures}\right)\hfill \end{array}$

What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm 3 ?

1.034 g/mL

## Experimental determination of density using water displacement

A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown.

(a) Use these values to determine the density of this piece of rebar.

(b) Rebar is mostly iron. Does your result in (a) support this statement? How?

## Solution

The volume of the piece of rebar is equal to the volume of the water displaced:

$\text{volume}=\text{22.4 mL}-\text{13.5 mL}=\text{8.9 mL}={\text{8.9 cm}}^{3}$

(rounded to the nearest 0.1 mL, per the rule for addition and subtraction)

The density is the mass-to-volume ratio:

$\text{density}=\phantom{\rule{0.2em}{0ex}}\frac{\text{mass}}{\text{volume}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{69.658 g}}{{\text{8.9 cm}}^{3}}={\text{7.8 g/cm}}^{3}$

(rounded to two significant figures, per the rule for multiplication and division)

From [link] , the density of iron is 7.9 g/cm 3 , very close to that of rebar, which lends some support to the fact that rebar is mostly iron.

An irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown.

(a) Use these values to determine the density of this material.

(b) Do you have any reasonable guesses as to the identity of this material? Explain your reasoning.

(a) 19 g/cm 3 ; (b) It is likely gold; the right appearance for gold and very close to the density given for gold in [link] .

## Accuracy and precision

Scientists typically make repeated measurements of a quantity to ensure the quality of their findings and to know both the precision    and the accuracy    of their results. Measurements are said to be precise if they yield very similar results when repeated in the same manner. A measurement is considered accurate if it yields a result that is very close to the true or accepted value. Precise values agree with each other; accurate values agree with a true value. These characterizations can be extended to other contexts, such as the results of an archery competition ( [link] ).

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