A measurement provides information about both its magnitude and its uncertainty. Consider, for example, the three photos in Figure 2.1.1
, taken at intervals of approximately 1 sec after placing a sample on the balance. Assuming the balance is properly calibrated, we are certain that the sample’s mass is more than 0.5729 g and less than 0.5731 g. We are uncertain, however, about the sample’s mass in the last decimal place since the final two decimal places fluctuate between 29, 30, and 31. The best we can do is to report the sample’s mass as 0.5730 g ± 0.0001 g, indicating both its magnitude and its absolute uncertainty.
A measurement’s significant figures convey information about a measurement’s magnitude and uncertainty. The number of significant figures in a measurement is the number of digits known exactly plus one digit whose value is uncertain. The mass shown in Figure 2.1.1
, for example, has four significant figures, three which we know exactly and one, the last, which is uncertain.
Suppose we weigh a second sample, using the same balance, and obtain a mass of 0.0990 g. Does this measurement have 3, 4, or 5 significant figures? The zero in the last decimal place is the one uncertain digit and is significant. The other two zeros, however, simply indicates the decimal point’s location. Writing the measurement in scientific notation, \(9.90 \times 10^{-2}\), clarifies that there are three significant figures in 0.0990.
In the measurement 0.0990 g, the zero in green is a significant digit and the zeros in red are not significant digits.
How many significant figures are in each of the following measurements? Convert each measurement to its equivalent scientific notation or decimal form.
- 0.0120 mol HCl
- 605.3 mg CaCO3
- \(1.043 \times 10^{-4}\) mol Ag+
- \(9.3 \times 10^4\) mg NaOH
Solution
(a) Three significant figures; \(1.20 \times 10^{-2}\) mol HCl.
(b) Four significant figures; \(6.053 \times 10^2\) mg CaCO3.
(c) Four significant figures; 0.000 104 3 mol Ag+.
(d) Two significant figures; 93 000 mg NaOH.
There are two special cases when determining the number of significant figures in a measurement. For a measurement given as a logarithm, such as pH, the number of significant figures is equal to the number of digits to the right of the decimal point. Digits to the left of the decimal point are not significant figures since they indicate only the power of 10. A pH of 2.45, therefore, contains two significant figures.
The log of \(2.8 \times 10^2\) is 2.45. The log of 2.8 is 0.45 and the log of 102 is 2. The 2 in 2.45, therefore, only indicates the power of 10 and is not a significant digit.
An exact number, such as a stoichiometric coefficient, has an infinite number of significant figures. A mole of CaCl2, for example, contains exactly two moles of chloride ions and one mole of calcium ions. Another example of an exact number is the relationship between some units. There are, for example, exactly 1000 mL in 1 L. Both the 1 and the 1000 have an infinite number of significant figures.
Using the correct number of significant figures is important because it tells other scientists about the uncertainty of your measurements. Suppose you weigh a sample on a balance that measures mass to the nearest ±0.1 mg. Reporting the sample’s mass as 1.762 g instead of 1.7623 g is incorrect because it does not convey properly the measurement’s uncertainty. Reporting the sample’s mass as 1.76231 g also is incorrect because it falsely suggests an uncertainty of ±0.01 mg.
Significant figures are also important because they guide us when reporting the result of an analysis. When we calculate a result, the answer cannot be more certain than the least certain measurement in the analysis. Rounding an answer to the correct number of significant figures is important.
For addition and subtraction, we round the answer to the last decimal place in common for each measurement in the calculation. The exact sum of 135.621, 97.33, and 21.2163 is 254.1673. Since the last decimal place common to all three numbers is the hundredth’s place
\[\begin{align*}
&135.6{\color{Red} 2}1\\
&\phantom{1}97.3{\color{Red} 3}\\
&\underline{\phantom{1}21.2{\color{Red} 1}63}\\
&254.1673
\end{align*}\]
we round the result to 254.17.
The last common decimal place shared by 135.621, 97.33, and 21.2163 is shown in red.
When working with scientific notation, first convert each measurement to a common exponent before determining the number of significant figures. For example, the sum of \(6.17 \times 10^7\), \(4.3 \times 10^5\), and \(3.23 \times 10^4\) is \(6.22 \times 10^7\).
\[\begin{align*}
&6.1{\color{Red} 7} \phantom{323} \times 10^7\\
&0.0{\color{Red} 4}3 \phantom{23} \times 10^7\\
&\underline{0.0{\color{Red} 0}323 \times 10^7}\\
&6.21623 \times 10^7
\end{align*}\]
The last common decimal place shared by \(6.17 \times 10^7\), \(4.3 \times 10^5\) and \(3.23 \times 10^4\) is shown in red.
For multiplication and division, we round the answer to the same number of significant figures as the measurement with the fewest number of significant figures. For example, when we divide the product of 22.91 and 0.152 by 16.302, we report the answer as 0.214 (three significant figures) because 0.152 has the fewest number of significant figures.
\[\frac {22.91 \times 0.{\color{Red} 152}} {16.302} = 0.2136 = 0.214\nonumber\]
There is no need to convert measurements in scientific notation to a common exponent when multiplying or dividing.
It is important to recognize that the rules presented here for working with significant figures are generalizations. What actually is conserved is uncertainty, not the number of significant figures. For example, the following calculation
101/99 = 1.02
is correct even though it violates the general rules outlined earlier. Since the relative uncertainty in each measurement is approximately 1% (101 ± 1 and 99 ± 1), the relative uncertainty in the final answer also is approximately 1%. Reporting the answer as 1.0 (two significant figures), as required by the general rules, implies a relative uncertainty of 10%, which is too large. The correct answer, with three significant figures, yields the expected relative uncertainty. Chapter 4 presents a more thorough treatment of uncertainty and its importance in reporting the result of an analysis.
Finally, to avoid “round-off” errors, it is a good idea to retain at least one extra significant figure throughout any calculation. Better yet, invest in a good scientific calculator that allows you to perform lengthy calculations without the need to record intermediate values. When your calculation is complete, round the answer to the correct number of significant figures using the following simple rules.
- Retain the least significant figure if it and the digits that follow are less than halfway to the next higher digit. For example, rounding 12.442 to the nearest tenth gives 12.4 since 0.442 is less than half way between 0.400 and 0.500.
- Increase the least significant figure by 1 if it and the digits that follow are more than halfway to the next higher digit. For example, rounding 12.476 to the nearest tenth gives 12.5 since 0.476 is more than halfway between 0.400 and 0.500.
- If the least significant figure and the digits that follow are exactly halfway to the next higher digit, then round the least significant figure to the nearest even number. For example, rounding 12.450 to the nearest tenth gives 12.4, while rounding 12.550 to the nearest tenth gives 12.6. Rounding in this manner ensures that we round up as often as we round down.
For a problem that involves both addition and/or subtraction, and multiplication and/or division, be sure to account for significant figures at each step of the calculation. With this in mind, report the result of this calculation to the correct number of significant figures.
\[\frac {0.250 \times (9.93 \times 10^{-3}) - 0.100 \times (1.927 \times 10^{-2})} {9.93 \times 10^{-3} + 1.927 \times 10^{-2}} = \nonumber\]
- Answer
-
The correct answer to this exercise is \(1.9 \times 10^{-2}\). To see why this is correct, let’s work through the problem in a series of steps. Here is the original problem
\[\frac {0.250 \times (9.93 \times 10^{-3}) - 0.100 \times (1.927 \times 10^{-2})} {9.93 \times 10^{-3} + 1.927 \times 10^{-2}} = \nonumber\]
Following the correct order of operations we first complete the two multiplications in the numerator. In each case the answer has three significant figures, although we retain an extra digit, highlight in red, to avoid round-off errors.
\[\frac {2.48{\color{Red} 2} \times 10^{-3} - 1.92{\color{Red} 7} \times 10^{-3}} {9.93 \times 10^{-3} + 1.927 \times 10^{-2}} = \nonumber\]
Completing the subtraction in the numerator leaves us with two significant figures since the last significant digit for each value is in the hundredths place.
\[\frac {0.55{\color{Red} 5} \times 10^{-3}} {9.93 \times 10^{-3} + 1.927 \times 10^{-2}} = \nonumber\]
The two values in the denominator have different exponents. Because we are adding together these values, we first rewrite them using a common exponent.
\[\frac {0.55{\color{Red} 5} \times 10^{-3}} {0.993 \times 10^{-2} + 1.927 \times 10^{-2}} = \nonumber\]
The sum in the denominator has four significant figures since each of the addends has three decimal places.
\[\frac {0.55{\color{Red} 5} \times 10^{-3}} {2.92{\color{Red} 0} \times 10^{-2}} = \nonumber\]
Finally, we complete the division, which leaves us with a result having two significant figures.
\[\frac {0.55{\color{Red} 5} \times 10^{-3}} {2.92{\color{Red} 0} \times 10^{-2}} = 1.9 \times 10^{-2} \nonumber\]