1.2.2: Significant Figures in Calculations
- Page ID
- 408949
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Use significant figures correctly in arithmetical operations.
Rounding
Before dealing with the specifics of the rules for determining the significant figures in a calculated result, we need to be able to round numbers correctly. To round a number, first decide how many significant figures the number should have. Once you know that, round to that many digits, starting from the left. If the number immediately to the right of the last significant digit is less than 5, it is dropped and the value of the last significant digit remains the same. If the number immediately to the right of the last significant digit is greater than or equal to 5, the last significant digit is increased by 1.
Consider the measurement \(207.518 \: \text{m}\). Right now, the measurement contains six significant figures. How would we successively round it to fewer and fewer significant figures? Follow the process as outlined in Table \(\PageIndex{1}\).
| Number of Significant Figures | Rounded Value | Reasoning |
|---|---|---|
| 6 | 207.518 | All digits are significant |
| 5 | 207.52 | 8 rounds the 1 up to 2 |
| 4 | 207.5 | 2 is dropped |
| 3 | 208 | 5 rounds the 7 up to 8 |
| 2 | 210 | 8 is replaced by a 0 and rounds the 0 up to 1 |
| 1 | 200 | 1 is replaced by a 0 |
Notice that the more rounding that is done, the less reliable the figure is. An approximate value may be sufficient for some purposes, but scientific work requires a much higher level of detail.
It is important to be aware of significant figures when you are mathematically manipulating numbers. For example, dividing 125 by 307 on a calculator gives 0.4071661238… to an infinite number of digits. But do the digits in this answer have any practical meaning, especially when you are starting with numbers that have only three significant figures each?
In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. An answer is no more precise than the least precise number used to get the answer.
Significant Figures in Calculations
How are significant figures handled in calculations? It depends on what type of calculation is being performed. In each case, you will complete the calculation as normal, then consider how the final answer should be rounded so as to only the known digits and estimated digit, while rounding off digits which are not known.
Multiplication and Division:
Count Significant Figures in each number used in the calculation.
Round your final answer to whichever number of significant figures is least.
Addition and Subtraction:
Determine the least precise number (do not count the total number of significant figures).
Round your final answer to the same place as the least precise number.
Multiplication and Division
Addition and Subtraction
Write the answer for each expression using scientific notation with the appropriate number of significant figures.
- 23.096 × 90.300
- 125 × 9.000
Solution
a
| Explanation | Answer |
|---|---|
| The calculator answer is 2,085.5688, but we need to round it to five significant figures. Because the first digit to be dropped (in the tenths place) is greater than 5, we round up to 2,085.6. | \(2.0856 \times 10^3\) |
b
| Explanation | Answer |
|---|---|
| The calculator gives 1,125 as the answer, but we limit it to three significant figures. | \(1.13 \times 10^3\) |
- 13.77 + 908.226
- 1,027 + 611 + 363.06
Solution
a
| Explanation | Answer |
|---|---|
| The calculator answer is 921.996, but because 13.77 has its farthest-right significant figure in the hundredths place, we need to round the final answer to the hundredths position. Because the first digit to be dropped (in the thousandths place) is greater than 5, we round up to 922.00 | \(922.00 = 9.2200 \times 10^2\) |
b
| Explanation | Answer |
|---|---|
| The calculator gives 2,001.06 as the answer, but because 611 and 1027 has its farthest-right significant figure in the ones place, the final answer must be limited to the ones position. | \(2,001.06 = 2.001 \times 10^3\) |
Write the answer for each expression using scientific notation with the appropriate number of significant figures.
- 217 ÷ 903
- 13.77 + 908.226 + 515
- 255.0 − 99
- 0.00666 × 321
- Answer a:
- \(0.240 = 2.40 \times 10^{-1}\)
- Answer b:
- \(1,437 = 1.437 \times 10^3\)
- Answer c:
- \(156 = 1.56 \times 10^2\)
- Answer d:
- \(2.14 = 2.14 \times 10^0\)
Remember that calculators do not understand significant figures. You are the one who must apply the rules of significant figures to a result from your calculator.
Calculations Involving Multiplication/Division and Addition/Subtraction
In practice, chemists generally work with a calculator and carry all digits forward through subsequent calculations. When working on paper, however, we often want to minimize the number of digits we have to write out. Because successive rounding can compound inaccuracies, intermediate rounding needs to be handled correctly. When working on paper, always round an intermediate result so as to retain at least one more digit than can be justified and carry this number into the next step in the calculation. The final answer is then rounded to the correct number of significant figures at the very end.
In the worked examples in this text, we will often show the results of intermediate steps in a calculation. In doing so, we will show the results to only the correct number of significant figures allowed for that step, in effect treating each step as a separate calculation. This procedure is intended to reinforce the rules for determining the number of significant figures, but in some cases it may give a final answer that differs in the last digit from that obtained using a calculator, where all digits are carried through to the last step.
- 2(1.008 g) + 15.99 g
- 137.3 s + 2(35.45 s)
- \( {118.7 g \over 2} - 35.5 g \)
Solution
a.
| Explanation | Answer |
|---|---|
|
2(1.008 g) + 15.99 g = Perform multiplication first.2 (1.008 g 4 sig figs) = 2.016 g 4 sig figs The number with the least number of significant figures is 1.008 g; the number 2 is an exact number and therefore has an infinite number of significant figures. Then, perform the addition.2.016 g thousandths place + 15.99 g hundredths place (least precise) = 18.006 g Round the final answer.Round the final answer to the hundredths place since 15.99 has its farthest right significant figure in the hundredths place (least precise). |
18.01 g (rounding up) |
b.
| Explanation | Answer |
|---|---|
|
137.3 s + 2(35.45 s) = Perform multiplication first.2(35.45 s 4 sig figs) = 70.90 s 4 sig figs The number with the least number of significant figures is 35.45; the number 2 is an exact number and therefore has an infinite number of significant figures. Then, perform the addition.137.3 s tenths place (least precise) + 70.90 s hundredths place = 208.20 s Round the final answer.Round the final answer to the tenths place based on 137.3 s. |
208.2 s |
c.
| Explanation | Answer |
|---|---|
|
\( {118.7 g \over 2} - 35.5 g \) = Perform division first.\( {118.7 g \over 2} \) 4 sig figs = 59.35 g 4 sig figs The number with the least number of significant figures is 118.7 g; the number 2 is an exact number and therefore has an infinite number of significant figures. Perform subtraction next.59.35 g hundredths place − 35.5 g tenths place (least precise) = 23.85 g Round the final answer.Round the final answer to the tenths place based on 35.5 g. |
23.9 g (rounding up) |
Complete the calculations and report your answers using the correct number of significant figures.
- 5(1.008s) - 10.66 s
- 99.0 cm+ 2(5.56 cm)
- Answer a
- -5.62 s
- Answer b
- 110.2 cm
Summary
- Rounding
- If the number to be dropped is greater than or equal to 5, increase the number to its left by 1 (e.g. 2.9699 rounded to three significant figures is 2.97).
- If the number to be dropped is less than 5, there is no change (e.g. 4.00443 rounded to four significant figures is 4.004).
- The rule in multiplication and division is that the final answer should have the same number of significant figures as there are in the number with the fewest significant figures.
- The rule in addition and subtraction is that the answer is given the same number of decimal places as the term with the fewest decimal places.