Lesson 1.5 · Unit 1 · HS-PS1-7
Significant figures, and what a measurement is actually claiming
Every measurement carries an implicit claim about how well it is known, and the digits you write down are that claim. Reporting 15.88 g when the balance can only support 15.9 g asserts a precision you do not have. The rules below are not arbitrary bookkeeping; each one exists to stop a calculation from manufacturing certainty that was never measured.
- Precision is reproducibility; accuracy is closeness to the true value. A miscalibrated balance can be highly precise and badly inaccurate, giving the same wrong answer every time.
- Counting significant figures: all non-zero digits count; zeros between non-zero digits count; leading zeros never count; trailing zeros count only if a decimal point is present.
- So 0.00340 has three significant figures (the leading zeros locate the decimal point, the trailing zero is a measured digit) while 3400 has an ambiguous two, which is the reason scientific notation exists.
- For multiplication and division, the answer takes the fewest significant figures of any value used.
- For addition and subtraction, the answer takes the fewest decimal places. This is a different rule with a different criterion, and mixing them up is the commonest error in the topic.
- Exact numbers impose no limit. Counted objects, defined conversions such as 1000 mL per liter, and integers in formulas are exact and never restrict the answer.
- Round once, at the end. Rounding at each intermediate step accumulates error, so carry extra digits through and round the final value.
Where students lose marks: applying the multiplication rule to a subtraction. Subtracting 31.4 from 47.28 does not give four significant figures, because the decimal-place rule applies and 31.4 has only one decimal place.
The problem. A student finds the density of a liquid by mass difference. The empty beaker reads 31.4 g on a balance that displays one decimal place. The beaker with liquid reads 47.28 g on a different balance that displays two. The liquid was delivered by a pipette as 5.00 mL. Report the density correctly.
Step one: identify which rule each operation needs. Finding the mass of liquid is a subtraction, so the decimal-place rule applies. Finding density is a division, so the significant-figure rule applies. The two steps are governed by different rules, and this is the whole difficulty of the problem.
Step two: do the subtraction. \( 47.28 - 31.4 = 15.88 \) as raw arithmetic.
Step three: apply the decimal-place rule. The two values have two and one decimal places respectively, so the result may carry one. The mass of liquid is 15.9 g, not 15.88 g.
Step four: understand why, rather than just obeying. The 31.4 g reading means the true value lies somewhere near 31.35 to 31.45 g. That uncertainty of about a tenth of a gram carries straight into the difference, so a hundredths digit in the answer is not supported by anything. The subtraction did not improve the worse measurement.
Step five: count significant figures for the division. The mass 15.9 g has three. The volume 5.00 mL has three, because the trailing zeros follow a decimal point and are genuine measured digits. The smaller count is three.
Step six: divide. \( 15.9 \text{ g} \div 5.00 \text{ mL} = 3.18 \text{ g/mL} \), which already has three significant figures and needs no rounding. Report 3.18 g/mL.
Step seven: notice what would have gone wrong. Carrying 15.88 g into the division gives 3.176 g/mL, and a student who then applies the significant-figure rule to four and three figures still reports 3.18. Here the error is invisible. It will not always be, and the habit has to be right before it matters.
Step eight: comment on the setup. Using two balances of different precision has thrown away the better one's advantage entirely, because the final uncertainty is set by the worse measurement. Weighing both masses on the two-decimal balance would have supported four significant figures in the mass and a more precise density from the same procedure.
Questions 1 to 5 are recall. Questions 6 to 10 ask for reasoning.
- How many significant figures are in 0.00340?
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Leading zeros do not count; the trailing zero does because a decimal point is present. Three
- How many significant figures are in 2050.0?
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All digits count here: the zero between non-zero digits, and the trailing zeros because the decimal point is shown. Five
- State the rule for multiplication and division.
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The answer carries the fewest significant figures of any value used
- Calculate \( 4.560 + 12.1 \) and report it correctly.
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Raw sum 16.66; the decimal-place rule allows one decimal place. 16.7
- Calculate \( 3.20 \times 4.1 \) and report it correctly.
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Raw product 13.12; the smaller count is two significant figures. 13
- Explain why 3400 is ambiguous and how scientific notation resolves it.
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Trailing zeros in a number with no decimal point may be measured digits or may only be holding the place value, and the notation cannot distinguish the two. A reading of 3400 g could mean the balance resolved the tens and ones digits and found both to be zero, or that it resolved only to the nearest hundred grams. Scientific notation makes the claim explicit because only the digits written in the coefficient are significant: \( 3.4 \times 10^{3} \) states two significant figures, \( 3.40 \times 10^{3} \) states three, and \( 3.400 \times 10^{3} \) states four, while all three represent the same magnitude. Trailing zeros without a decimal point may or may not be measured; scientific notation shows exactly which digits are claimed
- A student measures a rectangle as 2.0 cm by 3.00 cm and reports an area of
6.00 cm2. Find and explain the error.
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The multiplication rule takes the fewest significant figures of the values used, and 2.0 cm has only two, so the area may carry two: \( 6.0 \text{ cm}^2 \). Reporting 6.00 claims the area is known to the nearest hundredth of a square centimeter, which cannot be true when one side is known only to the nearest tenth of a centimeter. The underlying point is that a calculation cannot be more certain than the least certain measurement it used, and writing extra digits does not make the ruler better. 6.0 cm2; the least precise measurement limits the result
- Explain why counting twelve test tubes does not limit the significant figures in a
calculation that uses the number twelve.
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Because it is not a measurement and carries no uncertainty. Counting discrete objects gives an exact value: there are precisely twelve test tubes, not twelve give or take a tenth, so the number could be written as 12.000000 without making any false claim. Significant figure rules exist to propagate measurement uncertainty through a calculation, and a quantity with no uncertainty cannot contribute any. The same applies to defined conversions such as 1000 milliliters per liter and to the integers in a formula, which is why dividing a measured mass by an exact count leaves the significant figures set by the mass alone. It is an exact count with no uncertainty, so it contributes no limit; only measured values do
- Three students weigh the same object on the same balance and get 4.512 g, 4.511 g
and 4.513 g. The object's certified mass is 4.230 g. Describe the balance in terms
of precision and accuracy, and say which problem is easier to fix.
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The readings agree with each other to within two thousandths of a gram, so the balance is highly precise: it returns the same value under the same conditions. It is badly inaccurate, since every reading is about 0.28 g above the certified value, an error more than a hundred times the scatter. The pattern of a constant offset with small scatter is the signature of a calibration fault rather than a random one. That is the easier problem to fix, because a systematic offset can be removed by recalibrating against a standard mass, whereas poor precision points to an instrument or a technique that would have to be replaced or retrained. Precise but inaccurate, with a systematic offset; recalibration fixes it easily
- A student performs a four-step calculation, rounding to two significant figures
after each step, and their answer differs from the correct one in the second digit.
Explain the cause and the remedy.
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Rounding discards information, and rounding at every step discards it four times, so each subsequent step operates on a value that is already slightly wrong and the errors compound rather than cancel. Rounding to two significant figures can shift a value by nearly half a percent, and four such shifts can easily move the second digit of the final answer. The remedy is to carry the full precision of intermediate results through the whole calculation, either by keeping the value in the calculator or by writing down two or three extra guard digits, and to round only once when the final answer is reported. The significant figure rules describe what may be claimed at the end, not what must be written in the middle. Intermediate rounding compounds; carry full precision and round only the final answer