2.4: Calculations and Significant Figures
- Page ID
- 451495
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- When using measured physical quantities in calculations report the correct number of significant figures in the answer.
Combining Numbers
For addition or subtraction, the rule is to stack all the numbers with their decimal points aligned and then limit (round to) the answer’s significant figures to the rightmost column for which all the numbers have significant figures. Consider the following:
The arrow points to the rightmost column in which all the numbers have significant figures—in this case, the tenths place. Therefore, we will limit our final answer to the tenths place. Is our final answer therefore 1,459.0? No, because when we drop digits from the end of a number, we also have to round the number. Notice that the first dropped digit, in the hundredths place, is 8. This suggests that the answer is actually closer to 1,459.1 than it is to 1,459.0, so we need to round up to 1,459.1. The standard rules for rounding numbers are simple: If the first dropped digit is 5 or higher, round up. If the first dropped digit is lower than 5, do not round up.
For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count. An example is as follows:
The final answer, limited to four significant figures, is 4,094. The first digit dropped is 1, so we do not round up.
Scientific notation provides a way of communicating significant figures without ambiguity. You simply include all the significant figures in the leading number. For example, the number 4,000 has one significant figure and should be written as the number 4 × 103. The number 450 has two significant figures and would be written in scientific notation as 4.5 × 102, whereas 450.0 has four significant figures and would be written as 4.500 × 102. In scientific notation, all reported digits are significant.
Write the answer for each expression using scientific notation with the appropriate number of significant figures.
- 23.096 × 90.300
- 125 × 9.000
- 1,027 + 610.0 + 363.06
- Answer a
-
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 hundredths place) is greater than 5, we round up to 2,085.6, which in scientific notation is 2.0856 × 103.
- Answer b
-
The calculator gives 1,125 as the answer, but we limit it to three significant figures and convert into scientific notation: 1.13 × 103.
- Answer c
-
The calculator gives 2,000.06 as the answer, but because 1,027 has its farthest-right significant figure in the ones column, our answer must be limited to the ones position: 2,000 which in scientific notation is 2.000 × 103.
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 x 10-1
- Answer b
- 1437 = 1.437 x 103
- Answer c
- 156 = 1.56 x 102
- Answer d
- 2.14 = 2.14 x 100
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
When working through calculations that contain both multiplication/division and addition/subtraction, the calculation is solved by following the order of operations math rules. Significant figures are then calculated at each step using the appropriate rule for that operation. Even when the number of significant figures is limited due to these rules, all digits are carried forward to minimize rounding errors. 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.
Solve the following problem accounting for correct significant digits. The number 3 is a counted number and has infinite significant figures in this problem.
\[\frac{117.7\:g}{3} - 35.5\:g\]
Solution
According to rules for order of operations, the division problem should be solved before proceeding with subtraction.
\[\frac{117.7\:g}{3} = \underline{39.23}33333...g\]
Because 117.8 has 4 sig figs and the number 3 has infinite sig figs, the answer to this problem follows the multiplication/division rule and has the least number of sig figs. In this case that would be 4 sig figs. Since this is not the end of the calculation, we want to keep track of this by underlining the 4 significant digits but keeping the rest. Because writing all the 3's would be time consuming on paper, keep at least one extra digit in the calculation, 39.233.
Now our problem is 39.233 g - 35.5 g, and we need to use the addition/subtraction rule to solve this step. We know 39.233 g to the hundredths place (the last digit that is significant). We know 35.5 g to the tenths. According to the addition/subtraction rule we use the least precise (leftmost column) of these in the answer. Therefore we will round the answer to the tenths.
\[\underline{39.23}3\:g - 35.5\:g = \underline{3.7}33\:g\] when rounded to the tenths equals 3.7 g. This is the final answer.