Chemistry » Essential Ideas in Chemistry » Measurement Uncertainty, Accuracy and Precision

# Significant Figures in Calculations

## Uncertainty in Measurements Continued

### Rules for Rounding Numbers

A second important principle of uncertainty is that results calculated from a measurement are at least as uncertain as the measurement itself. We must take the uncertainty in our measurements into account to avoid misrepresenting the uncertainty in calculated results. One way to do this is to report the result of a calculation with the correct number of significant figures, which is determined by the following three rules for rounding numbers:

1. When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (the least precise value in terms of addition and subtraction).
2. 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).
3. If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, we “round down” and leave the retained digit unchanged; if it is more than 5, we “round up” and increase the retained digit by 1; if the dropped digit is 5, we round up or down, whichever yields an even value for the retained digit. (The last part of this rule may strike you as a bit odd, but it’s based on reliable statistics and is aimed at avoiding any bias when dropping the digit “5,” since it is equally close to both possible values of the retained digit.)

Don’t worry if you didn’t understand the rules, we will look at examples shortly. The following examples illustrate the application of this rule in rounding a few different numbers to three significant figures:

• 0.028675 rounds “up” to 0.0287 (the dropped digit, 7, is greater than 5)
• 18.3384 rounds “down” to 18.3 (the dropped digit, 3, is less than 5)
• 6.8752 rounds “up” to 6.88 (the dropped digit is 5, and the retained digit is even)
• 92.85 rounds “down” to 92.8 (the dropped digit is 5, and the retained digit is even)

Let’s work through these rules with a few more examples.

### Rounding Numbers

Round the following to the indicated number of significant figures:

(a) 31.57 (to two significant figures)
(b) 8.1649 (to three significant figures)
(c) 0.051065 (to four significant figures)
(d) 0.90275 (to four significant figures)

#### Solution

(a) 31.57 rounds “up” to 32 (the dropped digit is 5, and the retained digit is even)
(b) 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5)
(c) 0.051065 rounds “down” to 0.05106 (the dropped digit is 5, and the retained digit is even)
(d) 0.90275 rounds “up” to 0.9028 (the dropped digit is 5, and the retained digit is even)

### Addition and Subtraction with Significant Figures

Rule: When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (i.e., the least precise value in terms of addition and subtraction).

(a) Add 1.0023 g and 4.383 g.
(b) Subtract 421.23 g from 486 g.

#### Solution (a)

$$\begin{array}{cccccccc} & 1 & . & 0 & 0 & 2 & 3 & \text{g} \\ + & 4 & . & 3 & 8 & 3 & & \text{g} \\ \hline & 5 & . & 3 & 8 & 5 & 3 & \text{g} \\ \end{array}$$

Answer is 5.385 g (round to the thousandths place; three decimal places)

#### Solution (b)

$$\begin{array}{cccccccc} & 4 & 8 & 6 & & & & \text{g} \\ − & 4 & 2 & 1 & . & 2 & 3 & \text{g} \\ \hline & & 6 & 4 & . & 7 & 7 & \text{g} \\ \end{array}$$

Answer is 65 g (round to the ones place; no decimal places)

Image credit: OpenStax, Chemistry

### 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)

$$0.628 \text{cm} × 6.6 \text{cm}$$$$= 4.11708 \text{cm}^2$$
result is 4.1 cm2 (round to two significant figures)
$$\text{four significant figures} ×$$$$\text{two significant figures}$$$$\rightarrow \text{two significant figures (answer)}$$

#### Solution (b)

$$\cfrac{421.23 \text{g}}{486 \text{mL}} = 0.86728… \text{g/mL}$$
result is 0.867 g/mL (round to three significant figures)
$$\cfrac{\text{five significant figures}}{\text{three significant figures}}$$$$\rightarrow \text{three significant figures (answer)}$$

### Note:

You might be wondering why we are doing mathematics in chemistry. Well, it turns out that mathematics helps us a lot in science. Also, in the midst of all these technicalities, it is important to keep in mind the reason why we use significant figures and rounding rules—which is 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

$$\begin{array}{lll} V & = & l × w × d \\ & = & 13.44 \text{ dm} × 5.920 \text{ dm} × 2.54 \text{ dm} \\ & = & 202.09459… \text{ dm}^3 \text{(value from calculator)} \\ & = & 202 \text{ dm}^3, \text{or }202 \text{ L } \text{(answer rounded to three significant figures)} \\ \end{array}$$

### 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.

Image credit: OpenStax, Chemistry

(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:

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

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

The density is the mass-to-volume ratio:

$$\text{density }$$$$= \cfrac{\text{mass}}{\text{volume}}$$$$= \cfrac{69.658 \text{g}}{8.9 \text{cm}^3}$$$$= 7.8 \text{g/cm}^3$$

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

From lesson 21, the density of iron is 7.9 g/cm3, very close to that of rebar, which lends some support to the fact that rebar is mostly iron.