# 2.1: Measurements in Analytical Chemistry


Analytical chemistry is a quantitative science. Whether determining the concentration of a species, evaluating an equilibrium constant, measuring a reaction rate, or drawing a correlation between a compound’s structure and its reactivity, analytical chemists engage in “measuring important chemical things” [Murray, R. W. Anal. Chem. 2007, 79, 1765]. In this section we review briefly the basic units of measurement and the proper use of significant figures.

## Units of Measurement

A measurement usually consists of a unit and a number that expresses the quantity of that unit. We can express the same physical measurement with different units, which creates confusion if we are not careful to specify the unit. For example, the mass of a sample that weighs 1.5 g is equivalent to 0.0033 lb or to 0.053 oz. To ensure consistency, and to avoid problems, scientists use the common set of fundamental base units listed in Table 2.1.1 . These units are called SI units after the Système International d’Unités.

It is important for scientists to agree upon a common set of units. In 1999, for example, NASA lost a Mar’s Orbiter spacecraft because one engineering team used English units in their calculations and another engineering team used metric units. As a result, the spacecraft came too close to the planet’s surface, causing its propulsion system to overheat and fail.

Some measurements, such as absorbance, do not have units. Because the meaning of a unitless number often is unclear, some authors include an artificial unit. It is not unusual to see the abbreviation AU—short for absorbance unit—following an absorbance value, which helps clarify that the measurement is an absorbance value.

Table 2.1.1 : Fundamental Base SI Units

Measurement

Unit

Symbol

Definition (1 unit is...)

mass

kilogram

kg

...the mass of the international prototype, a Pt-Ir object housed at the Bureau International de Poids and Measures at Sèvres, France. (Note: The mass of the international prototype changes at a rate of approximately 1 μg per year due to reversible surface contamination. The reference mass, therefore, is determined immediately after its cleaning using a specified procedure. Current plans call for retiring the international prototype and defining the kilogram in terms of Planck’s constant; see this link for more details.)

distance

meter

m

...the distance light travels in (299 792 458)–1 seconds.

temperature

Kelvin

K

...equal to (273.16)–1, where 273.16 K is the triple point of water (where its solid, liquid, and gaseous forms are in equilibrium).

time

second

s

...the time it takes for 9 192 631 770 periods of radiation corresponding to a specific transition of the 133Cs atom.

current

ampere

A

...the current producing a force of 2 $$\times$$ 10–7 N/m between two straight parallel conductors of infinite length separated by one meter (in a vacuum).

amount of substance

mole

mol

...the amount of a substance containing as many particles as there are atoms in exactly 0.012 kilogram of 12C.

light

candela

cd

...the luminous intensity of a source with a monochromatic frequency of 540 $$\times$$ 1012 hertz and a radiant power of (683)–1 watts per steradian.

There is some disagreement on the use of “amount of substance” to describe the measurement for which the mole is the base SI unit; see “What’s in a Name? Amount of Substance, Chemical Amount, and Stoichiometric Amount,” the full reference for which is Giunta, C. J. J. Chem. Educ. 2016, 93, 583–586.

We define other measurements using these fundamental SI units. For example, we measure the quantity of heat produced during a chemical reaction in joules, (J), where 1 J is equivalent to 1 m kg/s . Table 2.1.2 provides a list of some important derived SI units, as well as a few common non-SI units.

Table 2.1.2 : Derived SI Units and Non-SI Units of Importance to Analytical Chemistry
Measurement Unit Symbol Equivalent SI Units
length angstrom (non-SI) Å 1 Å = 1 $$\times$$ 10–10 m
volume liter (non-SI) L 1 L = 10–3 m3
force newton (SI) N 1 N = 1 m$$\cdot$$kg/s2

pressure

pascal (SI)

atmosphere (non-SI)

Pa

atm

1 Pa = 1 N/m3 = 1 kg/(m$$\cdot$$s2)

1 atm = 101 325 Pa

energy, work, heat

joule (SI)

calorie (non-SI)

electron volt (non-SI)

J

cal

eV

1 J = 1 N$$\cdot$$m = 1 m2$$\cdot$$kg/s2

1 cal = 4.184 J

1 eV = 1.602 177 33 $$\times$$ 10–19 J

power watt (SI) W 1 W = 1 J/s = 1 m2$$\cdot$$kg/s3
charge coulomb (SI) C 1 C = 1 A$$\cdot$$s
potential volt (SI) V 1 V = 1 W/A = 1 m2$$\cdot$$kg/(s3$$\cdot$$A)
frequency hertz (SI) Hz 1 Hz = s–1
temperature Celcius (non-SI) oC oC = K – 273.15

Chemists frequently work with measurements that are very large or very small. A mole contains 602 213 670 000 000 000 000 000 particles and some analytical techniques can detect as little as 0.000 000 000 000 001 g of a compound. For simplicity, we express these measurements using scientific notation; thus, a mole contains 6.022 136 7 $$\times$$ 1023 particles, and the detected mass is 1 $$\times$$ 10–15 g. Sometimes we wish to express a measurement without the exponential term, replacing it with a prefix (Table 2.1.3 ). A mass of $$1 \times 10^{-15}$$ g, for example, is the same as 1 fg, or femtogram.

Writing a lengthy number with spaces instead of commas may strike you as unusual. For a number with more than four digits on either side of the decimal point, however, the recommendation from the International Union of Pure and Applied Chemistry is to use a thin space instead of a comma.

Table 2.1.3 : Common Prefixes for Exponential Notation
Prefix Symbol Factor Prefix Symbol Factor Prefix Symbol Factor
yotta Y 1024 kilo k 103 micro µ 10–6
zetta Z 1021 hecto h 102 nano n 10–9
eta E 1018 deka da 101 pico p 10–12
peta P 1015 100 femto f 10–15
tera T 1012 deci d 10–1 atto a 10–18
giga G 109 centi c 10–2 zepto z 10–21
mega M 106 milli m 10–3 yocto y 10–24

## Uncertainty in Measurements

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.

Figure 2.1.1 : When weighing an sample on a balance, the measurement fluctuates in the final decimal place. We record this sample’s mass as 0.5730 g ± 0.0001 g.

### Significant Figures

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.

##### Example 2.1.1

How many significant figures are in each of the following measurements? Convert each measurement to its equivalent scientific notation or decimal form.

1. 0.0120 mol HCl
2. 605.3 mg CaCO3
3. $$1.043 \times 10^{-4}$$ mol Ag+
4. $$9.3 \times 10^4$$ 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 in Calculations

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.

1. 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.
2. 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.
3. 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.
##### Exercise 2.1.1

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.

$\frac {0.250 \times (9.93 \times 10^{-3}) - 0.100 \times (1.927 \times 10^{-2})} {9.93 \times 10^{-3} + 1.927 \times 10^{-2}} = \nonumber$

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.

$\frac {0.55{\color{Red} 5} \times 10^{-3}} {2.92{\color{Red} 0} \times 10^{-2}} = 1.9 \times 10^{-2} \nonumber$

This page titled 2.1: Measurements in Analytical Chemistry is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey.