1.4 Uncertainty, Accuracy, and Precision

When measurements are made, there is uncertainty in those measurements. Uncertainty in measurements is dictated by the sensitivity of the instrument being used.

The first thing we need to understand is that there are two types of numbers in chemistry, exact numbers, and inexact numbers. Exact numbers are defined values. An example of an exact number would be a dozen, meaning exactly twelve. These are values that are counted: thirteen chairs in the room, fifteen people in the room, etc. Inexact numbers are numbers obtained from measurements. Whenever we make a measurement, the uncertainty from that measurement must be indicated. Significant figures or significant digits are the meaningful digits in a reported measured value. Understanding the uncertainty associated with the instrument. allows us to choose the proper instrument and measurements that we make in chemistry labs.

If measuring the length of a flashdrive there are a variety of width measuring instruments that can be chosen. Below, there are two different rulers being used.

The difference between the rulers is the measurement markings. The top ruler only has each centimeter marked whereas the bottom ruler has each centimeter marked but also has fractional markings. This indicates the bottom ruler would give a more accurate measurement of the length of this flashdrive.

If the top ruler is used to measure the width of the flashdrive, it is clear the width is greater than 2 but less than 3; therefore, in describing the width an estimation on the width of the flashdrive is required. In estimating, it would be reasonable to estimate that the length of the flashdrive is 2.5 centimeters.

If, however, the bottom ruler is used to measure the width of the flashdrive, we have more graduations, more measurement lines, to allow us to be more accurate. Again, we are past 2 and below 3, but it is clearer that the width is right before that halfway mark. This leads to a measured width of 2.48.

Notice for both measurements, the measurement was taken one place further than the lines that were provided on the measurement instrument. The top ruler had graduations in the ones places, leading to a an estimation of the tenths place. In the bottom Ruler, the graduations were to the tenths place, so an estimation to the hundredths place was made. As a rule we’re going to estimate between the markings for our uncertain digit. And the last number that we measure is going to be that uncertain value. Remember that estimation is going to be dependent on who’s making that measurement.

1.4.1 Significant Figures

We can determine the number of significant figures in measured numbers using specific guidelines. The following rules should be used in determining the number of significant figures in a reported value:

The first rule is that any non-zero digit is significant.
When there are zeros between non-zero digits, these zeros are significant.
Zeros to the left of the first non-zero digit are not significant.
Zeros to the right of the last nonzero digit are significant if a decimal is present.
Zeroes to the right of the last non-zero digit in a number that does not contain a decimal point may or may not be significant. When there is ambiguity, arising from rule 5, scientific notation is used to remove that ambiguity.

Remember when you are working with exact numbers, these can be as accurate as needed, therefore they have an infinite number of significant figures. metric-to-metric conversions are exact numbers; therefore, they have an infinite number of significant figures.

1.4.2 Significant Figures in Calculations

It is important to learn how to use this idea of significance in terms of calculations. The first types of calculations we’re going to discuss are addition and subtraction.

In addition or subtraction, the answer cannot have more digits to the right of the decimal point than any of the original numbers. Below is an example of adding 102.50 plus 0.230.

The first number has two digits after the decimal place or two digits to the right of the decimal place. The second number has three digits after the decimal place or to the right at the decimal place. Since the answer cannot have more digits to the right of the decimal point than any of the original numbers, the answer cannot have more than two digits to the right of the decimal place. To ensure our answer has two decimal places, the answer will be rounded at the hundredths place.

Below is an example subtraction calculation, where 143.29 minus 20.1.

Notice the first number has two digits to the right of the decimal point and our second number has one digit after the decimal point. Again, since the least number of decimal places in the original numbers is allowed in the answer, the answer should be rounded to have one digit. When the calculation is carried out, an answer of 123.19 is obtained, but the final answer can only have one decimal place, therefore the answer must be rounded to 123.2.

The multiplication and division rule differs from the addition/subtraction rule.
In multiplication and division, states that the final answer must contain the same number of significant figures as the smallest number of significant figures in the original numbers. Below is an example of multiplication, where 1.4 is multiplied by 8.011.

\[(1.4)(8.011) = 11.2154\]

To know how many significant figures should be in the final answer, the number of significant figures for each of the original digits must be identified. There are 2 significant figures in 1.4 and 4 significant figures in 8.011. Since the final answer must contain the same number significant figures as the smallest number of significant figures in the original digits, the final answer should contain 2 significant figures. The final answer will then need to be rounded to 11. Note that scientific notation can be used to obtain the correct number of significant figures. If the final answer with the correct number of significant figures is put into scientific notation, it would be 1.1×10¹.

An example of division is given below, where 11.57 is divided by 305.88.

\[\dfrac{11.57}{305.88} = 0.0378252\]

Again, to know how many significant figures should be in the final answer, the number of significant figures for each of the original digits must be identified. The numerator has 4 significant figures, and the denominator has 5 significant figures. Since the final answer must contain the same number significant figures as the smallest number of significant figures in the original digits, the final answer should contain 4 significant figures. The final answer will then need to be rounded to 0.03782.

Remember that exact numbers are a count, not a measurement, and therefore are considered to have an infinite number of significant figures. Exact number are not considered when identifying the number of significant figures in a calculation.

Practice – Significant Figures

Three pennies each have a mass of 2.52 g. What is the total mass?

Solution

The three pennies are exact since this value comes from counting. The mass of each penny is a measurement and therefore an inexact value.
Now that the number types are identified, the calculation can be performed.

\[\begin{align*} \left ( 3~\mathrm{pennies} \right ) \left ( \dfrac{2.5~\mathrm{g}}{1~\mathrm{penny}} \right ) = 7.5~\mathrm{g} \end{align*}\]

The answer retrieved is 7.5 g, but the final answer must reflect the correct number of significant figures. Since multiplication and division are used in this problem, the final answer must contain the same number of significant figures as the smallest number of significant figures in the original digits. The 3 pennies is an exact number and therefore will not be used to determine the final answer’s number of significant figures. The 2.5 grams per penny, however, is a measurement and has 2 significant figures. The least number of significant figures in the original numbers is 2, therefore, the final answer must be rounded to 2 significant figures.

Practice – Significant Figures

How many significant numbers should be in the final answer of the following:

\[\dfrac{4.5 + 8.23}{4.5661}\]
\[\dfrac{15.68-8.2}{5.33}\]
\[\dfrac{8.34+3.5}{16.97-8.8}\]

Solution

Practice

Determine the number of significant figures in the following measurements:

123 cm
0.0857 kg
50.0 mL
25.03 g
1.106 × 10^–7 L
0.1600 m

Solution

Practice

Determine the number of significant figures for each of the following numbers.

Number	Significant Figures
2.12
4.500
0.002541
0.00100
500
500.
5.0 × 10²
1.05 g
dozen eggs
0.90 × 10⁴⁵ L

Solution

Number	Significant Figures	Comment on Zeros
2.12	3
4.500	4	Not placeholders. Significant
0.002541	4	Placeholders. Not significant
0.00100	3	Only the last two are significant
500	1, 2, 3?	Ambiguous (go with 1 when info not given)
500.	3	Decimal point shows significance
5.0 × 10²	2	No ambiguity
1.05 g	3	Between non-zero digits, significant
dozen eggs	infinite	Exact count
0.90 × 10⁴⁵ L	2	Not placeholders. Significant

Practice

Three pennies each have a mass of 2.5 g. What is the total mass?

Solution

Identify exact and inexact quantities.

Three pennies (exact; counting)
2.5 g (inexact; measurement)

\[\left ( 3~\mathrm{pennies} \right ) \left ( \dfrac{2.5~\mathrm{g}}{\mathrm{penny}} \right ) = 7.5~\mathrm{g}\]

Practice

Calculate the following to the appropriate number of significant figures.

(1.278 + 0.3480)/1.08881
7846 × 92437 × 235.649 × 3.3×10³
583.00/83
(57.6 × 3)/(34×3.00times;87.507)

Solution

1.493
5.6×10¹⁴
7.0
0.2

Practice

Convert 14.62 in³ to cm³ given that 1 in = 2.54 cm (exact).

Solution

\[\begin{align*} \left ( \dfrac{14.62~\mathrm{in}^3}{} \right ) \left ( \dfrac{\left ( 2.54~\mathrm{cm}\right )^3}{\left (\mathrm{in}\right )^3} \right ) = 239.58~\mathrm{cm}^3 \end{align*}\]

Practice

A chair lift at the Divide Ski Resort in Cold Springs, WY is 4,806 ft long and takes about 9 minutes to ride. Determine the following:

What is the average speed (in mph or mi h^–1)? There are 5,280 ft in a mile.
How many feet per second (ft ^–1) does the lift travel?

Solution

Determine mph.

\[\begin{align*} \left ( \dfrac{4806~\mathrm{ft}}{9~\mathrm{min}} \right ) \left ( \dfrac{60~\mathrm{min}}{1~\mathrm{h}} \right ) \left ( \dfrac{1~\mathrm{mi}}{5280~\mathrm{ft}} \right ) &= 6.07~\mathrm{mi~h^{-1}} \end{align*}\]

Determine ft s^–1.

\[\begin{align*} \left ( \dfrac{4806~\mathrm{ft}}{9~\mathrm{min}} \right ) \left ( \dfrac{1~\mathrm{min}}{60~\mathrm{s}} \right ) &= 9~\mathrm{ft~s^{-1}} \end{align*}\]

Practice

An average human heart beats 60 times per minute. If an average person lives to the exact age of 75 y, how many times does the average heartbeat in this average lifetime?

Solution

\[\begin{align*} \left ( \dfrac{75~\mathrm{y}}{1~\mathrm{lifetime}} \right ) \left ( \dfrac{365~\mathrm{d}}{1~\mathrm{y}} \right ) \left ( \dfrac{24~\mathrm{h}}{1~\mathrm{d}} \right ) \left ( \dfrac{60~\mathrm{min}}{1~\mathrm{h}} \right ) \left ( \dfrac{60~\mathrm{beats}}{1~\mathrm{min}} \right ) &= 2.3652\times 10^5~\mathrm{beats~per~lifetime} \end{align*}\]

Practice

A car travels 14 miles in 15 minutes.

How fast is the car going in miles per hour?
How fast is the car going in meters per second? Note that 1 mile is approximately 1.60933 km.

Solution

Determine mi h^–1.

\[\begin{align*} \left ( \dfrac{14~\mathrm{mi}}{15~\mathrm{min}} \right ) \left ( \dfrac{60~\mathrm{min}}{1~\mathrm{h}} \right ) &= 56~\mathrm{mi~h^{-1}} \end{align*}\]

Determine m s^–1.

\[\begin{align*} \left ( \dfrac{14~\mathrm{mi}}{15~\mathrm{min}} \right ) \left ( \dfrac{1.60933~\mathrm{km}}{1~\mathrm{mi}} \right ) \left ( \dfrac{10^3~\mathrm{m}}{1~\mathrm{km}} \right ) \left ( \dfrac{1^3~\mathrm{min}}{60~\mathrm{s}} \right ) &= 25~\mathrm{m~s^{-1}} \end{align*}\]

Practice – Significant Figures

Solve the following equation to the appropriate number of significant figures.

\[\dfrac{(143.032-23.3)}{5.1123} =\]

Solution

23.4

Practice – Significant Figures

Report the answer with the correct number of sig figs for the given calculation.

\[4.003 + 0.23 + 22.0 = \]

Solution

26.2

Practice – Dimensional Analysis

You decide to buy a swimming pool. You are told by a contractor that your swimming pool can not be greater than 70 tons total. You want to build a 20 ft long, 10 ft wide, and 6 ft high swimming pool. The concrete for this pool would have a mass of 40 tons. If the swimming pool is filled to the top with water, will the total weight be under the 70 ton maximum?

Solution

Water has a density of about 1 g mL^–1 at room temperature. Recall that we do not round numbers until the very end of a multi-step calculation.

Determine the volume of the pool (length × width × height).

\[\begin{align*} V_{\mathrm{pool}} &= l_{\mathrm{pool}} \times w_{\mathrm{pool}} \times h_{\mathrm{pool}} \\[2ex] &= 20~\mathrm{ft} \times 10~\mathrm{ft} \times 6~\mathrm{ft} \\[2ex] &= 1200~\mathrm{ft}^3 \end{align*}\]

Convert cubic feet into milliliters. 1 ft = 0.3048 m and 1 mL = 1 cm³

\[\begin{align*} \left ( \dfrac{1200~\mathrm{ft}^3}{} \right ) \left ( \dfrac{(0.3048~\mathrm{m})^3}{(\mathrm{ft})^3} \right ) \left ( \dfrac{(100~\mathrm{cm})^3}{(1~\mathrm{m})^3} \right ) \left ( \dfrac{1~\mathrm{mL}}{\mathrm{cm}^3} \right ) = 3.39802159\times 10^{7}~\mathrm{mL} \end{align*}\]

Determine the mass (in g) of the water that would fill the pool.

\[\begin{align*} d &= \dfrac{m}{V} \longrightarrow \\[2ex] m &= dV \\[2ex] &= \left ( \dfrac{1.0~\mathrm{g}}{\mathrm{mL}} \right ) \left ( \dfrac{3.39802159\times 10^{7}~\mathrm{mL}}{} \right ) \\[2ex] &= 3.39802159\times 10^{7}~\mathrm{g} \end{align*}\]

Determine the mass of the water in tons. 1 lb = 0.4535924 kg and 1 ton = 2000 lb.

\[\begin{align*} \left ( \dfrac{3.39802159\times 10^{7}~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{lb}}{0.4535924~\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{ton}}{2000~\mathrm{lb}} \right ) = 37.46~\mathrm{tons} \end{align*}\]

Find the combined mass of the pool and the water.

\[\begin{align*} m_{\mathrm{tot}} &= m_{\mathrm{pool}} + m_{\mathrm{water}} \\[2ex] &= 40~\mathrm{tons} + 37.46~\mathrm{tons}\\[2ex] &= 77.46~\mathrm{tons} \end{align*}\]

No, the current pool design would result in a mass that was too large.

1.4.3 Rounding Rules

When rounding, if the first digit to be dropped is less than five round down, if the first digit to be dropped is greater than five round up. If the digit to be dropped is five, the number should be rounded up or down to yield a value that is even.

The last rounding rule to discuss is to not round after every step of a calculation but instead to wait until the end of the calculation to round. Rounding at each step in a calculation introduces error. Enter all steps of a calculation into the calculator and then round the final answer using the significant figure calculation rules discussed above.

1.4.4 Accuracy and Precision

Accuracy tells us how close a measurement is to a true value, but precision tells us how close multiple measurements are to one another. This gives three scenarios for measurements:

Good accuracy/Good precision
Poor Accuracy/Good precision
Poor Accuracy/Poor precision

In scenario one, all measurements are close together, precise, and close to the true value, accurate. In scenario two, all measurements are close together, precise, but they do not match the true value. In the third scenario, the measurements do not match each other nor do they the true value.

What does this look like in terms of data. The data obtained from three students taking three measurements of the mass of an aspirin tablet are plotted below.

The black data point represents the true mass of the tablet, 0.270 g. Student A’s measurements are shown with the pink dots, student B’s measurements are shown with the green dot, and student C’s measurements are shown with the purple dots.

For student A, all three measurements are close to one another, good precision, but they are not close to the true value, good accuracy. Student B’s measurements are scattered from one another, poor precision, and they are not close to the true mass, poor accuracy. Student C’s measurements are close to one another, good precision, and they are close to the true mass, good accuracy.