Dr. Ray's Learning Center

Units, Significant Figures, and Dimensional Analysis

General Chemistry · Measurement, Matter, and Atoms

Dr. Ray’s Learning Center · covers: Units, Significant Figures, and Dimensional Analysis

  • 25–30 min
  • 7 figures
  • 4 worked examples
  • 5 practice sets

Before this lesson:

No earlier lesson is needed: only arithmetic with decimals and powers of ten.

In 1999 NASA lost the Mars Climate Orbiter. One team’s software reported thruster data in pound-force seconds, while the navigation software expected newton-seconds, the SI unit. Nobody converted the numbers, the spacecraft flew far too close to Mars, and it was lost.

Every number in chemistry is a measurement with a unit and a limited precision. This lesson shows how to change units safely with unit factors, and how many digits a measured or calculated value deserves (Figure 1).

  1. Given quantitynumber and unit
  2. × factorsunit factors: old unit cancels
  3. Calculator resultnew unit, extra digits
  4. roundsignificant-figure rules
  5. Answerright unit, right digits

Unit factors are exact: they never limit significant figures.Multiply or divide: fewest significant figures. Add or subtract: fewest decimal places.

Figure 1. The method of this lesson. Unit factors change the unit without changing the amount; the significant-figure rules decide how many digits the answer keeps.

Learning Objectives

After this lesson, you should be able to:

  1. Convert between SI and common units with unit factors.
  2. Apply significant-figure rules to measured and calculated values.

Key Ideas

1. SI Units and Metric Prefixes

Scientists use the SI (International System of Units). The base units you will meet first are the kilogram (kg) for mass, the meter (m) for length, the second (s) for time, the mole (mol) for amount of substance and the kelvin (K) for temperature. The liter (L) is not an SI base unit but is used with SI: 1 L is 1 dm³.

1L=1000mL=1000cm31\,\text{L} = 1000\,\text{mL} = 1000\,\text{cm}^{3}

A prefix multiplies a unit by a power of ten, so the same prefix works for every unit: 1 kg is 1000 g and 1 km is 1000 m (Figure 2).

Metric prefixes and their powers of tenA table of six rows. kilo, symbol k, factor 10 to the 3, examples km, kg, kJ. No prefix, factor 1, examples m, g, L, s. centi, c, 10 to the minus 2, cm. milli, m, 10 to the minus 3, mm, mg, mL. micro, symbol mu, 10 to the minus 6, micrometer and microgram. nano, n, 10 to the minus 9, nm.Metric prefixesPrefixSymbolFactorExampleskilok103​km, kg, kJ(none)–1m, g, L, scentic10−2​cmmillim10−3​mm, mg, mLmicroµ10−6​µm, µgnanon10−9​nmLarger unit at the top; each row is a power of ten.
Figure 2. Each prefix multiplies the unit by a power of ten. The same prefix means the same factor for every unit: 1 kg is 103 g and 1 km is 103 m.

Key idea: the prefix tells you the power of ten. Write the equality with a 1 in front of the larger unit, for example 1m=100cm1\,\text{m} = 100\,\text{cm} and 1m=1000mm1\,\text{m} = 1000\,\text{mm}.

2. Unit Factors Cancel Units

Any equality between two units gives two unit factors. Each factor equals 1, because its top and bottom are the same quantity, so multiplying by it changes the unit but not the amount.

For example, the inch is defined as exactly 2.54 cm:

1in=2.54cm1\,\text{in} = 2.54\,\text{cm}
2.54cm1in and 1in2.54cm\dfrac{2.54\,\text{cm}}{1\,\text{in}} \text{ and } \dfrac{1\,\text{in}}{2.54\,\text{cm}}

Choose the factor that has the unit you want to remove in its denominator; that unit then cancels (Figure 3). This method is called dimensional analysis: if the units of your answer come out right, the setup is right. Several factors can be chained, one for each unit you change.

One inch is 2.54 centimetersA bar labeled 1 inch drawn to scale above a bar of the same length divided into centimeters: 1 cm, 1 cm and 0.54 cm, which is 2.54 cm in total.One length, two units1 in.1 cm1 cm0.54the same length: 2.54 cm in 1 in.so either unit factor leavesthe amount unchangedFigure drawn to scale.The two unit factors from one equalityTwo stacked fraction cards. The first has 2.54 cm on top and 1 in. on the bottom: in. on the bottom, so it cancels in. The second has 1 in. on top and 2.54 cm on the bottom: cm on the bottom, so it cancels cm.Two unit factors2.54 cm1 in.in. on the bottom:cancels in.1 in.2.54 cmcm on the bottom:cancels cm
Figure 3. One equality, two unit factors. Pick the one whose bottom unit matches the unit you want to remove; it cancels, and the unit on top is left.

3. Counting Significant Figures

A measured value is written with all the digits you are sure of plus one estimated digit. These are its significant figures. The rules (Figure 4):

  • Nonzero digits are always significant.
  • Zeros between nonzero digits are significant: 2007 has 4.
  • Leading zeros only place the decimal point and are not significant: 0.00340 has 3.
  • Trailing zeros after a decimal point are significant: 38.0 has 3.
  • Trailing zeros in a whole number with no decimal point are ambiguous: write 24300 in scientific notation, such as 2.43×1042.43 \times 10^{4}, to show 3.
Counting significant figuresFour numbers as digit boxes. 0.00340: the three leading zeros are not significant, 3, 4 and the final 0 are, so 3 significant figures. 2007: all four digits, 4. 38.0: all three including the trailing zero after the decimal point, 3. 24300: 2, 4 and 3 are significant, the two trailing zeros are ambiguous, so 3 to 5.Which digits are significant?0.003403leading zeros: no20074zeros between: yes38.03trailing zero after point: yes243003 to 5no decimal point: ambiguoussignificantnotambiguous
Figure 4. Green digits count, gray zeros only place the decimal point, and dashed zeros are ambiguous until the number is written in scientific notation.

Exact numbers have unlimited significant figures and never limit a result: counted objects (24 bolts) and defined equalities (1in=2.54cm1\,\text{in} = 2.54\,\text{cm}, 1km=1000m1\,\text{km} = 1000\,\text{m}). Every unit factor in this lesson is exact.

4. Significant Figures in Calculations

A calculated value cannot be more precise than the measurements behind it. There are two rules, one for each kind of operation (Figure 5):

  • Multiplying or dividing: keep as many significant figures as the measured value with the fewest significant figures. 4.56×1.4=6.3844.56 \times 1.4 = 6.384 on a calculator; 1.4 has 2 significant figures, so the result is 6.4.
  • Adding or subtracting: keep as many decimal places as the measured value with the fewest decimal places. 12.52+3.1=15.6212.52 + 3.1 = 15.62 on a calculator; 3.1 has 1 decimal place, so the result is 15.6.
Rounding a productMultiplication: 4.56, with 3 significant figures, times 1.4, with 2, gives 6.384 on a calculator. The digits 8 and 4 are dropped: the result is 6.4, with 2 significant figures.Multiply or dividecount significant figures4.563 sig. figs×1.42 sig. figs6.384calculator6.4 (2 sig. figs)Rounding a sumAddition: 12.52, with 2 decimal places, plus 3.1, with 1 decimal place, gives 15.62 on a calculator. The last 2 is dropped: the result is 15.6, with 1 decimal place.Add or subtractcount decimal places12.522 decimal places+3.11 decimal place15.62calculator15.6 (1 decimal place)
Figure 5. Two different rules. For a product the least precise factor limits the significant figures; for a sum the least precise term limits the decimal places. Gray digits are dropped.

To round, look at the first digit you drop. If it is less than 5, keep the last digit; if it is greater than 5, or 5 followed by other nonzero digits, round the last digit up. A dropped digit that is exactly 5 with nothing after it is handled differently by some textbooks, so the problems on this page avoid that case. Round only once, at the end of a calculation.

Key idea: multiply or divide: fewest significant figures. Add or subtract: fewest decimal places. Exact numbers never count.

Worked Example 1: A One-Step Metric Conversion

Problem: A vitamin C tablet contains 845 mg of ascorbic acid. Express this mass in g.

Step 1. Write the equality between the two units.

1g=1000mg1\,\text{g} = 1000\,\text{mg}

Step 2. Choose the unit factor with mg in the denominator, so mg cancels.

1g1000mg\dfrac{1\,\text{g}}{1000\,\text{mg}}

Step 3. Multiply and cancel the units.

845mg×1g1000mg=0.845g845\,\text{mg} \times \dfrac{1\,\text{g}}{1000\,\text{mg}} = 0.845\,\text{g}

Final Answer: 0.845 g

The unit factor is exact, so the answer keeps the 3 significant figures of 845 mg. 1 g is bigger than 1 mg, so the number got smaller: a quick check that the factor is the right way up.

Your turn: a one-step metric conversion

Same three steps as Worked Example 1: write the equality, choose the factor that cancels the given unit, multiply.

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Worked Example 2: A Chain of Unit Factors

Problem: A delivery truck travels at 65.0 km/h. What is this speed in m/s?

Step 1. Write the unit equalities you need. km must become m, and h must become s.

  • 1km=1000m1\,\text{km} = 1000\,\text{m}
  • 1h=3600s1\,\text{h} = 3600\,\text{s}

Step 2. Arrange each unit factor so the unit you want to remove is in the denominator, then multiply. km is on top in km/h, so km goes on the bottom of the first factor; h is on the bottom, so h goes on top of the second factor.

65.0km/h×1000m1km×1h3600s=18.06m/s65.0\,\text{km/h} \times \dfrac{1000\,\text{m}}{1\,\text{km}} \times \dfrac{1\,\text{h}}{3600\,\text{s}} = 18.06\,\text{m/s}
18.06m/s≈18.1m/s18.06\,\text{m/s} \approx 18.1\,\text{m/s}

Final Answer: 18.1 m/s

  1. Speed65.0km/h65.0\,\text{km/h}in km/h
  2. ×1000m1km\times \frac{1000\,\text{m}}{1\,\text{km}}km cancels
  3. Speed6.50×104m/h6.50 \times 10^{4}\,\text{m/h}in m/h
  4. ×1h3600s\times \frac{1\,\text{h}}{3600\,\text{s}}h cancels
  5. Speed18.06m/s≈18.1m/s18.06\,\text{m/s} \approx 18.1\,\text{m/s}Final answer to 3 significant figures
Figure 6. Worked Example 2 one factor at a time. Each unit factor removes one unwanted unit; the answer is rounded only at the end.

Both factors are exact, so the answer has 3 significant figures like 65.0 km/h. The calculator shows 18.0555…; writing one extra digit (18.06) and rounding once at the end keeps the last digit honest.

Your turn: a chain of unit factors

Same method as Worked Example 2: list the equalities, line up the factors so every unwanted unit cancels, multiply, then round to 3 significant figures.

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Worked Example 3: Significant Figures When Multiplying

Problem: A rectangular metal plate measures 15.24 cm by 3.8 cm. What is its area? Give the answer with the correct number of significant figures.

Step 1. Count the significant figures in each value.

  • 15.24 cm: 4 significant figures
  • 3.8 cm: 2 significant figures

Step 2. Multiply. Write the full calculator result.

A=15.24cm×3.8cm=57.912cm2A = 15.24\,\text{cm} \times 3.8\,\text{cm} = 57.912\,\text{cm}^{2}

Step 3. Round to 2 significant figures, the fewest in the measured values. The first dropped digit is 9, so the 7 rounds up (Figure 7).

57.912cm2≈58cm257.912\,\text{cm}^{2} \approx 58\,\text{cm}^{2}

Final Answer: 58 cm²

Rounding the area in Worked Example 3The calculator result 57.912 shown as digit boxes: 5 and 7 are kept, the decimal point and 9, 1, 2 are dropped. The first dropped digit is 9, so the 7 rounds up: the answer is 58 square centimeters.Rounding 57.912 to 2 significant figures57.912keep 2dropfirst dropped digit is 9, so round the 7 up58 cm2​
Figure 7. Step 3 of Worked Example 3. The width has only 2 significant figures, so only the first two digits of 57.912 survive, and the 9 after them rounds the 7 up.

The width, 3.8 cm, is known only to 2 significant figures, so the area is too. The extra digits of 57.912 are not real information.

Your turn: significant figures when multiplying

Same method as Worked Example 3. These answers are checked exactly: type the rounded result with the right number of significant figures.

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Worked Example 4: Significant Figures When Subtracting

Problem: A beaker containing some salt reads 60.3 g on a top-loading balance. The empty beaker, weighed earlier on an analytical balance, reads 52.67 g. What is the mass of the salt? Give the answer with the correct number of significant figures.

Step 1. Count the decimal places in each value.

  • 60.3 g: 1 decimal place
  • 52.67 g: 2 decimal places

Step 2. Subtract. Write the full calculator result.

msalt=60.3g−52.67g=7.63gm_{\text{salt}} = 60.3\,\text{g}-52.67\,\text{g} = 7.63\,\text{g}

Step 3. Round to 1 decimal place, the fewest in the measured values.

7.63g≈7.6g7.63\,\text{g} \approx 7.6\,\text{g}

Final Answer: 7.6 g of salt

The top-loading balance reads only to 0.1 g, so the difference is known only to 0.1 g. Notice that the answer has just 2 significant figures although the measurements have 3 and 4: subtracting two similar numbers can lose significant figures.

Lab note: weigh by difference on the same balance whenever you can; mixing balances throws away the precision of the better one.

Your turn: significant figures when adding or subtracting

Same method as Worked Example 4: count decimal places, add or subtract, then round to the fewest decimal places. Answers are checked exactly.

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Common Mistakes

Common mistake: Turning the unit factor upside down. 845mg×1000mg1g845\,\text{mg} \times \frac{1000\,\text{mg}}{1\,\text{g}} gives 845000 with units of mg2g\frac{\text{mg}^{2}}{\text{g}}, which is nonsense. If the units do not cancel, flip the factor.

Common mistake: Using the wrong power of ten. 1m=100cm1\,\text{m} = 100\,\text{cm} but 1m=1000mm1\,\text{m} = 1000\,\text{mm}, and 1g=1000mg1\,\text{g} = 1000\,\text{mg} (not 100). Using 100 for milli in Worked Example 1 gives 8.45 g, ten times too big.

Common mistake: Leaving out a unit factor. Stopping after the first factor in Worked Example 2 gives 6.50×104m/h6.50 \times 10^{4}\,\text{m/h}: the hours were never converted. Check that the units of your answer are the units asked for.

Common mistake: Copying the calculator, or rounding too far. 57.912 cm² claims far more precision than a 2-significant-figure width allows, and rounding to 60 cm² throws away a real digit. Round once, to the number the rule gives.

Common mistake: Using the wrong rule. Rounding the area in Worked Example 3 to the fewest decimal places gives 57.9 cm² (wrong: multiplication counts significant figures). Rounding the salt mass in Worked Example 4 to 3 significant figures gives 7.63 g (wrong: subtraction counts decimal places).

Common mistake: Letting an exact number limit the answer. 24 bolts of 3.47 g each have a total mass of 83.28 g, which rounds to 83.3 g (3 significant figures, from 3.47 g). The count 24 is exact, so 83 g is wrong.

Key Takeaway

Change units by multiplying by unit factors arranged so the units cancel; the factors are exact. Then round the calculator result once: to the fewest significant figures for multiplication and division, or to the fewest decimal places for addition and subtraction.

Practice: randomized problems

Each problem below uses fresh numbers every time: work it out, check your answer, then open the worked solution to compare your steps.

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Your results

This summary collects every practice problem you try on this page and suggests what to review next.

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Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. Unit definitions: BIPM, The International System of Units (SI Brochure, 9th ed., 2019) and NIST Special Publication 811 (2008). The practice problems are generated and checked by an audited engine; see the lesson audit log.