Significant Figures Calculator

How to count significant figures

Significant figures, or sig figs, are the digits of a number that its measurement vouches for. Every non-zero digit counts, every zero between non-zero digits counts, zeros before the first non-zero digit never count, and zeros at the end count only when the number has a decimal point. So 0.04070 has 4 significant figures: the 4, the zero between the 4 and the 7, the 7 and the final zero. The two zeros in front only place the decimal point.

Type a number into the calculator exactly as it was written and it marks every digit with the rule that decides it, then gives the count, the decimal places and the scientific notation. It also rounds the number to as many significant figures as you ask for: x rounded to n figures is round(x / 10^k) × 10^k, where k = floor(log₁₀|x|) − n + 1 is the place of the last digit kept. The Calculate option takes a whole calculation, such as (25.36 − 18.2) ÷ 2.45, and rounds the answer by the rules for combining measurements, showing each step.

The five rules, with an example of each

• Non-zero digits always count. 285 has 3 significant figures.
• Captive zeros always count. A zero between non-zero digits was measured like any other digit, so 2.005 has 4 significant figures.
• Leading zeros never count. 0.0052 has 2 significant figures. The zeros only say where the decimal point goes, which is why they vanish when the number is written as 5.2 × 10⁻³.
• Trailing zeros count after a decimal point. 0.200 has 3 significant figures, because nobody writes zeros after the point unless those places were read.
• Trailing zeros in a whole number are ambiguous. 1500 has 2 significant figures by convention, though it may be good to the nearest unit, and the calculator says it could have up to 4.

These are the rules as NCERT’s Class XI Physics (section 1.3) and Chemistry (section 1.4.2) texts and OpenStax Chemistry 2e (section 1.5) set them out.

Trailing zeros, and how to write what was measured

A whole number that ends in zeros cannot say how many of them were measured. NIST’s guide to the SI, Special Publication 811, makes the point in section 7.9 with 1200 m, where it is not possible to tell whether the last two zeros are significant, while 1.200 km says they are. Three ways to remove the doubt:

• Scientific notation. 1.5 × 10³ has 2 significant figures, 1.50 × 10³ has 3 significant figures and 1.500 × 10³ has 4 significant figures. The power of ten never counts. The scientific notation converter rewrites a number in that form, or in E notation, keeping the figures it was typed with.
• A decimal point after the number. 1500. has 4 significant figures in the textbooks that use this mark. Not all do, so scientific notation is the safer choice in anything written for someone else.
• A larger unit. 1500 g measured to the gram is 1.500 kg.

When only the written number is available, the calculator follows the convention NCERT states and OpenStax calls prudent: trailing zeros with no decimal point are not counted. It says so beside the count and gives the range, 2 and up to 4 for 1500.

Rounding to significant figures

To round to n significant figures, keep the first n digits, counting from the first non-zero one, and look at the first digit dropped. Below 5, the kept digits stay as they are. Above 5, or a 5 with any non-zero digit after it, the last kept digit goes up by one. As a formula, the last digit kept sits at the place k = floor(log₁₀|x|) − n + 1, and the rounded value is round(x / 10^k) × 10^k.

A 5 with nothing after it, or only zeros, is exactly halfway, a tie, and the rule for a tie is a convention. NCERT’s physics and chemistry texts, OpenStax Chemistry 2e and NIST SP 811 (section B.7.1) all round a tie to the even digit, so 2.745 to 3 significant figures is 2.74, and 2.735 to 3 significant figures is 2.74 as well. Rounding half up gives 2.75 for the first. The calculator rounds half to even unless you choose half up, and says what the other rule gives whenever the two disagree.

When rounding up carries into a new digit, the figures move with it. 9.96 to 2 significant figures is 1.0 × 10¹, and it has to be written that way, because a plain 10 reads as one significant figure. In the same way 1996 to 3 significant figures is 2.00 × 10³, which 2000 cannot show.

Worked example: counting and rounding 0.04070

• The zero before the point and the zero after it are leading zeros, and neither counts.
• The 4 and the 7 are non-zero and count, the zero between them is captive and counts, and the final zero counts because the number has a decimal point.
• That makes 4 significant figures and 5 decimal places: 4.070 × 10⁻² in scientific notation, known to the nearest 0.00001.
• 0.04070 to 2 significant figures is 0.041. The kept digits are 0.040, and the first digit dropped is 7, above 5, so the last kept digit goes up.
• 0.04070 to 3 significant figures is 0.0407, since the only digit dropped is a zero.

Significant figures in addition, subtraction, multiplication and division

The two kinds of operation have different rules, because they carry uncertainty differently.

• Adding and subtracting keep the fewest decimal places. A sum is only as fine as its coarsest term: 12.11 + 18.0 + 1.012 = 31.122, and 18.0 is known only to the tenths, so the sum is 31.1. The rule is about places, not figures: 0.307 − 0.304 = 0.003 keeps 3 decimal places but only 1 significant figure, which is what subtracting two close measurements does.
• Multiplying and dividing keep the fewest significant figures. A product is only as precise, in proportion, as its least precise factor: 2.5 × 1.25 = 3.125, and 2.5 has 2 significant figures, so the product is 3.1.

All three are NCERT’s own examples. For numbers in scientific notation the addition rule reads as the coarsest last significant place, so 1.2 × 10³ + 3.45 × 10² is 1545 known only to the hundreds, which is 1.5 × 10³.

Worked example: (25.36 − 18.2) ÷ 2.45

A sample’s mass is found by difference, 25.36 g in its container less 18.2 g for the container, and divided by its volume, 2.45 mL.

• 25.36 − 18.2 = 7.16. 18.2 has 1 decimal place, so the difference is good to 7.2, which is 2 significant figures, although neither measurement had fewer than 3.
• 7.16 ÷ 2.45 = 2.92244897959…, with every digit of the difference carried. The fewest significant figures is now 2, from the difference, not the 3 of 2.45.
• Rounded once, at the end, the density is 2.9 g/mL.

Applying the multiplication rule to the three numbers as typed would give 2.92, a figure the data do not support: the subtraction, not the division, decides the precision. The calculator’s working shows each step with the rule that set it.

Exact numbers and π

Counted and defined numbers are exact and have unlimited significant figures, so they never limit an answer: 12 eggs, 100 cm in a metre, the 2 in a circumference of 2πr. Tick them in the calculator’s list of numbers. π itself is always exact, carried to 50 decimal places. With the 2 marked exact, 2 × π × 0.52 is 3.3, set by the 2 significant figures of 0.52. Left unmarked, the 2 counts as a measurement with 1 significant figure and the answer becomes 3.

Carry every digit, round once

Rounding at each step throws information away, and the losses build up. NCERT’s example is the reciprocal of 9.58: 1 ÷ 9.58 to 3 significant figures is 0.104, and 1 ÷ 0.104 to 3 significant figures is 9.62, not 9.58. Kept as 0.1044, one digit longer, its reciprocal is 9.579, which rounds back to 9.58. The 1 of a reciprocal is exact, so tick it in the list of numbers to get these figures from the calculator. It carries every digit of every intermediate result, shows how precisely each is known (“good to 7.2”), and rounds only the answer, which is what NCERT’s advice to keep one extra digit approximates.

Common mistakes

• Counting leading zeros. 0.0045 has 2 significant figures. A change of unit makes the point, 0.0045 g being 4.5 mg: a unit cannot change how precisely something was measured.
• Losing a significant trailing zero. 2.50 has 3 significant figures and 2.5 has 2, yet stored as a value they are the same number. That is why this calculator reads the digits you type rather than the value they make.
• Using the multiplication rule on a sum. 436.32 + 227.2 + 0.301 = 663.821 is 663.8, to the tenths of 227.2, not 664, which is NCERT’s warning example.
• Rounding at every step. As the reciprocal of 9.58 shows, the answer can move in its last figure.
• Treating an exact number as a measurement. The 2 in 2πr, or the 100 that turns a fraction into a percentage, would otherwise cut the answer to 1 significant figure.
• Writing a rounded whole number that hides its figures. 2000 cannot say it has 3 significant figures; 2.00 × 10³ can.
• Trusting a floating-point rounding on a tie. 2.675 cannot be stored exactly in binary floating point, which is why Python’s round(2.675, 2) gives 2.67, as its documentation warns. Read as written, 2.675 to 3 significant figures is 2.68 by either tie rule.

What this calculator does not cover

Significant figures are a shorthand for uncertainty, not a measure of it: 1.0 and 9.9 both have 2 significant figures, yet one unit in the last place is 10 percent of the first and about 1 percent of the second. When a measurement comes with a stated uncertainty, propagate it with the error propagation calculator and round the result to match that uncertainty rather than by counting figures. To compare a result with an accepted value, use the percent error calculator, and for the spread of repeated readings the standard deviation calculator.

The calculator does the four arithmetic operations. A power is a repeated product, so it keeps the significant figures of its base: a cube with a side of 7.203 m has a volume of 373.7 m³, to the 4 significant figures of the side, as in NCERT’s example. A logarithm keeps as many decimal places as its argument has significant figures (OpenStax Chemistry 2e, section 14.2), so a hydronium ion concentration of 1.2 × 10⁻³ M gives a pH of 2.92. Report the pH calculator’s answer to the decimal places that rule allows.