Significant figures are the digits in a measured number that carry real information. Count them with five rules: nonzero digits count, zeros between nonzero digits count, leading zeros never count, trailing zeros count when there is a decimal point, and trailing zeros in a whole number with no decimal point are ambiguous. In calculations, multiplication and division keep the fewest significant figures of the inputs, addition and subtraction keep the fewest decimal places, and you round once, at the end.
The counting rules are quick to learn. Most lost marks come from multi-step problems: a subtraction that quietly cuts the precision, or an answer rounded halfway through that lands one digit off. This guide covers both, with every example worked from the raw numbers.
The Five Rules for Counting Significant Figures
4.37 has 3.2. Zeros between nonzero digits are significant:
1.005 has 4.3. Leading zeros are never significant:
0.0082 has 2.4. Trailing zeros are significant when the number has a decimal point:
6.200 has 4, 0.0530 has 3.5. Trailing zeros in a whole number with no decimal point are ambiguous:
2,500 could have 2, 3 or 4.Why do leading zeros not count? They only place the decimal point. If you write 0.0082 m as 8.2 mm, the zeros vanish and nothing about the measurement changes, so they can't have been telling you anything. A trailing zero after the decimal point is different. Nobody writes 6.200 g by accident: it says the balance read to the thousandth of a gram and showed zeros there, which is a measured result. These conventions match the treatment in OpenStax Chemistry 2e, section 1.5.
How many significant figures are in 0.004050, 30.70, 1,002 and 2.500 × 10³?
0.004050: the three zeros in front are leading zeros (rule 3). Then 4, the captive 0, 5 (rules 1 and 2) and the final 0, which sits after a decimal point (rule 4). That gives 4.
30.70: the 3, the captive 0, the 7 and the trailing 0 after the decimal point. That gives 4.
1,002: both zeros are trapped between the 1 and the 2, so 4.
2.500 × 10³: in scientific notation every digit of the coefficient counts, so 4. The 10³ only sets the size.
4, 4, 4, 4A fast way to count: start at the first nonzero digit and count to the end. If there is no decimal point, stop at the last nonzero digit instead and treat any trailing zeros as unknown.
Why 2,500 Is Ambiguous, and How to Fix It
Suppose a crowd count is reported as 2,500. Was it counted to the nearest hundred, ten or one? The digits cannot say, so rule 5 calls those zeros ambiguous. The cure is scientific notation, where you choose the coefficient to show exactly what you know:
| Written as | Significant figures | Means known to the nearest |
|---|---|---|
2.5 × 10³ | 2 | 100 |
2.50 × 10³ | 3 | 10 |
2.500 × 10³ | 4 | 1 |
The same trick rescues answers. If a calculation should give 2 significant figures and your calculator shows 40, write 4.0 × 10¹. Plain "40" would hide that the zero is real.
Exact Numbers Never Limit Your Answer
The rules above are for measured numbers. Counted and defined numbers are exact: they have unlimited significant figures, so they never decide how far you round. OpenStax gives the standard examples: 1 foot is exactly 12 inches and 1 inch is exactly 2.54 centimeters (OpenStax Chemistry 2e, section 1.5). The same goes for the 2 in C = 2πr, the ½ in ½mv², and "3 trials" when you average three results.
Constants like π are not measured either. Use your calculator's π key, which carries more digits than any classroom measurement, and let the measured data set the rounding.
How to Round to a Set Number of Significant Figures
- Count from the first nonzero digit
Underline the digits you are keeping. To round
0.0456729to 3 significant figures, keep456. - Look at the first digit you are dropping
Less than 5: leave the last kept digit alone. More than 5, or 5 followed by any nonzero digit: add 1 to it. Here the dropped part starts with 7, so 456 becomes 457, giving
0.0457. - Keep the size the same
Dropped digits left of the decimal point turn into zeros or a power of ten, never just disappear.
1,456,000to 2 significant figures is1.5 × 10⁶, not 15.
The one tricky case is a dropped part that is exactly 5 (a 5 with nothing, or only zeros, after it). OpenStax rounds to whichever choice leaves the kept digit even, and rounds up whenever any nonzero digit follows the 5 (OpenStax Chemistry 2e, section 1.5). Under that rule 2.345 to 3 significant figures is 2.34, while 7.2651 is 7.27.
If your course teaches plain round-half-up, 2.345 becomes 2.35 instead. The two conventions only disagree on that exact-5 case, so check which one your teacher uses.
Multiplication and Division: Fewest Significant Figures
Why significant figures here? In a product, a 1% uncertainty in one input becomes about a 1% uncertainty in the result. Significant figures are a rough stand-in for that relative uncertainty, so the least precise input, measured as a fraction of itself, sets the result.
A rectangle measures 12.5 cm by 3.2 cm. Find its area. Then find the average speed of a runner who covers 125.0 m in 9.6 s.
Area: 12.5 cm × 3.2 cm = 40.0 cm² (cm times cm gives cm²). The inputs have 3 and 2 significant figures, so the answer gets 2.
Two significant figures of 40.0 is 40, but "40 cm²" is ambiguous, so write 4.0 × 10¹ cm².
Speed: 125.0 m ÷ 9.6 s = 13.0208… m/s. The inputs have 4 and 2 significant figures, so keep 2: 13 m/s.
4.0 × 10¹ cm² and 13 m/sAddition and Subtraction: Fewest Decimal Places
Adding works on absolute uncertainty, not relative. If one mass is only known to the nearest 0.1 g, the total can't be known to the nearest 0.001 g, however precise the other masses are. The column with the coarsest last digit wins.
Add 12.52 g + 3.1 g + 0.083 g. Then find the volume delivered by a burette that reads 25.37 mL after starting at 25.12 mL.
Sum: 12.52 + 3.1 + 0.083 = 15.703. The inputs have 2, 1 and 3 decimal places. The fewest is 1 (from 3.1 g), so round to the tenths: 15.7 g.
Difference: 25.37 − 25.12 = 0.25. Both have 2 decimal places, so the answer keeps 2: 0.25 mL.
Notice the difference has only 2 significant figures, though each reading had 4. Subtracting two close numbers cancels the leading digits, and the precision goes with them.
15.7 g and 0.25 mLUsing the significant-figure rule for a sum. In Example 3, the inputs have 4, 2 and 2 significant figures, so a student who applies "fewest significant figures" reports 16 g. That throws away a digit the data supports: 3.1 g limits the tenths place, not the whole answer. Ask which operation you are doing before you round, every time.
Multi-Step Calculations: Round Once, at the End
Real problems mix both rules. The method is to track what each step's rule allows without actually throwing digits away:
- Work one operation at a time
Follow the order of operations, and for each step note how many significant figures (or decimal places) the rule allows.
- Carry guard digits
Write the intermediate result with at least one or two extra digits, and mark the last significant one (bold or underline works). Never feed a rounded value into the next step.
- Round once
The final answer gets the significant figures allowed by the last step's rule, applied to the tracked counts.
OpenStax makes the same point: round at the end of a multistep calculation so rounding errors don't pile up (OpenStax Chemistry 2e, section 1.5). The next example shows what goes wrong otherwise.
A flask full of liquid has a mass of 52.38 g. The empty flask has a mass of 41.2 g. The liquid's volume is 12.0 mL. Find the density.
Step 1 (subtract): 52.38 g − 41.2 g = 11.18 g. The rule allows 1 decimal place, so 11.2, which is 3 significant figures. Carry 11.18 g and remember "3 significant figures".
Step 2 (divide): d = 11.18 g ÷ 12.0 mL = 0.93166… g/mL, so grams per milliliter is the unit. The inputs to this step have 3 (tracked) and 3 significant figures, so keep 3.
Round once: 0.932 g/mL.
Compare rounding too early: 11.2 g ÷ 12.0 mL = 0.93333… g/mL, which rounds to 0.933. The last digit is off by one, purely because of where the rounding happened.
d = 0.932 g/mLA pencil is 7.25 in long. Convert it to centimeters. Then find the circumference of a circle of radius 4.25 cm.
Conversion: 7.25 in × 2.54 cm/in = 18.415 cm. The inches cancel. Since 2.54 cm per inch is exact, only 7.25 (3 significant figures) counts: 18.4 cm.
Circumference: C = 2πr = 2 × π × 4.25 cm = 26.703… cm. The 2 is exact and π comes from the calculator key, so the radius sets 3 significant figures: 26.7 cm.
18.4 cm and 26.7 cmA lab sheet says 250.0 g of water warms from 21.4 °C to 25.36 °C and gives the specific heat c = 4.184 J/(g·°C). Find the heat absorbed, Q = mcΔT.
Step 1 (subtract): ΔT = 25.36 °C − 21.4 °C = 3.96 °C. One decimal place is allowed, so 3.96 rounds to 4.0: only 2 significant figures. Carry 3.96 °C.
Step 2 (multiply): Q = 250.0 g × 4.184 J/(g·°C) × 3.96 °C = 4142.16 J. Grams cancel grams and °C cancels °C, leaving joules.
The inputs to this step have 4, 4 and 2 (tracked) significant figures, so keep 2: 4.1 × 10³ J.
Rounding too early gives 250.0 × 4.184 × 4.0 = 4184 J, reported as 4.2 × 10³ J. That differs in the last kept digit, again only because of where the rounding happened.
Q = 4.1 × 10³ JExample 6 also shows why the whole answer has only 2 significant figures when the mass and specific heat had 4: the thermometer reading 21.4 °C was the weak link, and the subtraction made it weaker. If you want a sharper result, the fix is a better first reading, not more calculator digits.
Rounding each line of working because the worksheet shows rounded intermediate values. Those printed values are for reading, not for re-entering. Keep the full number in your calculator's memory (or write two extra digits) and round only the line you box as the answer.
Significant Figures in Logarithms and pH
Chemistry adds one more rule. In a logarithm, the digits before the decimal point only record the power of ten, so the significant figures live after it: a pH gets as many decimal places as the concentration has significant figures (Chemistry LibreTexts, pH Calculations).
Find the pH of a solution with [H⁺] = 3.2 × 10⁻⁴ M.
pH = −log(3.2 × 10⁻⁴) = 3.49485…
The concentration has 2 significant figures, so the pH keeps 2 decimal places. The leading 3 just mirrors the 10⁻⁴.
pH = 3.49How to Check Your Answer
- Name the limiting input. Before you round, point to the one number that sets the precision. If you can't, you haven't decided which rule applies.
- Re-count the answer. Apply the five counting rules to what you wrote. An answer meant to have 2 significant figures written as "40" or "4,100" fails this test; scientific notation passes it.
- Re-run without early rounding. Put the raw inputs through in one calculator chain. If the last digit changes, a rounded intermediate leaked in.
- Estimate the size. Round every input to one digit and do the arithmetic in your head. For Example 6,
250 × 4 × 4 = 4000, which agrees with 4.1 × 10³ J.
Once you've worked a problem yourself, Solver AI can scan a printed or handwritten chemistry or physics question and show the steps, so you can compare where each rounding decision was made. Treat it as a second check on your own working, not a replacement for it.
Practice Problems
- How many significant figures are in
0.06070,7.0 × 10⁻³and100.0? Answer: 4, 2 and 4. - Compute
4.218 + 0.33 − 1.2to the correct precision. Answer: the raw result is 3.348; the fewest decimal places is 1, so 3.3. - A stone dropped from rest falls for
1.35 s. Usingg = 9.81 m/s²andd = ½gt², find the distance. Answer: 0.5 × 9.81 × 1.35² = 8.939… m, so 8.94 m (the ½ is exact). - Find the pH when
[H⁺] = 2.50 × 10⁻⁵ M. Answer: −log(2.50 × 10⁻⁵) = 4.60206…, so 4.602 (3 decimal places). - Compute
(12.47 − 11.9) ÷ 0.0350. Answer: the subtraction gives 0.57 but allows only 1 decimal place, so it carries 1 significant figure. Then 0.57 ÷ 0.0350 = 16.285…, rounded to 1 significant figure: 2 × 10¹.
If problem 5 felt harsh, that's the point: one coarse reading in a subtraction can leave a single trustworthy digit. The same tracking works in any multi-step physics problem, such as those built on the kinematic equations.
Count from the first nonzero digit. Multiply or divide: fewest significant figures. Add or subtract: fewest decimal places. Exact numbers never limit you. Track the allowed precision at every step, carry guard digits, and round exactly once.