Problem 3
- Page ID
- 274057
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Before continuing, let's review the rules for including significant figures in calculations. When adding or subtracting, the result of the calculation is rounded to the last decimal place that is significant for all measurements. For example, the sum of 135.621, 0.33 and 21.2163 is 157.17 since the last decimal place that is significant for all three numbers (as shown below by the vertical line)
\[
\begin{array}{r|l}
135.62\hspace{-4pt} & \hspace{-4pt}1 \\[-2pt]
0.33\hspace{-4pt} & \\[-2pt]
21.21\hspace{-4pt} & \hspace{-4pt}63 \\
\hline
157.16\hspace{-4pt} & \hspace{-4pt}73 \\
\end{array}
\nonumber\]
is the hundredth's place. Note that rounding the answer to the correct number of significant figures occurs after completing the exact calculation.
When multiplying or dividing, the result of the calculation contains the same number of significant figures as that measurement having the smallest number of significant figures. Thus,
\[\dfrac{22.91\times0.152}{16.302}=0.21361\approx 0.214\nonumber\]
because 0.152, with three, has the fewest number of significant figures.
The reason for these rules is that the uncertainty in a result cannot be less than the uncertainty in the measurements used in calculating the result. One way to appreciate these rules is to calculate the largest possible results and the smallest possible result (the "Worst Case Scenarios") by taking into account the uncertainties in each measurements.
Consider the first problem at the top of the page. If we ignore significant figures, the exact answer to the problem is 157.1673. Suppose that the uncertainty in each value is ±1 in its last decimal place. For example, the measurement 135.621 could be as large as 135.622 or as small as 135.620. The largest possible sum of the three measurements, therefore, comes from adding together the largest possible measurements and the smallest possible sum comes from adding together the smallest possible measurements; thus
\[\begin{aligned}[t]
135&.622\\
0&.34\\
21&.2164\\
\hline
157&.1784
\end{aligned}
\qquad
\begin{aligned}[t]
135&.620\\
0&.32\\
21&.2162\\
\hline
157&.1562
\end{aligned}\nonumber\]
Comparing the two worst case results and the exact result, we see that rounding to the hundredth's place is the first instance where there is no agreement between the three calculations. The result of the exact calculation, therefore, is rounded to the hundredth's place, giving 157.17 as the appropriate result of the calculation.
For the second problem, the exact answer is 0.213613 (to six decimal places). To obtain the largest possible answer we divide the largest possible numerator by the smallest possible denominator, and to obtain the smallest possible answer we divide the smallest possible numerator by the largest possible denominator. If the uncertainty in each measurement is ±1 in its last decimal place, then the largest and smallest possible answers are
\[\dfrac{22.92\times0.153}{16.301}=0.215125\nonumber\]
\[\dfrac{22.90\times0.151}{16.303}=0.212102\nonumber\]
Comparing the two worst case results and the exact result, we see that rounding to the thousandth's place is the first instance where there is no agreement between the three calculations. The result of the exact calculation, therefore, is rounded to the thousandth's place, giving 0.214.
Try your hand at these two problems.
Task 1. The mass of water in a sample is found by weighing the sample before and after drying, yielding values of 0.4991 g and 0.3715 g, respectively. How many grams of water are in the sample and what is the percent water by mass? To how many significant figures should you report your answer?
Task 2. Assuming that the uncertainty in measuring mass is ±0.0001 g, what is the largest and the smallest possible mass of water in the sample? What is the largest and the smallest possible result for the percent water in the sample? To how many significant figures should you report your answer? Does this answer agree with the significant figure rules?
After you complete these tasks, read the module's summary.