2.4 Significant figures and rounding

Significant Figures in Values

Significant figures and rounding keep your calculations honest. Every measuring device has limits, and significant figures communicate exactly how precise a measurement actually is. In chemical engineering, where results pass through many calculation steps, sloppy sig fig handling can turn a small measurement uncertainty into a large error in your final answer.

Identifying Significant Figures

Significant figures are the digits in a value that are known with certainty, plus one estimated digit. They tell you (and anyone reading your work) how precise a measurement is. A value with more significant figures is more precise.

For example, a balance reading 4.032 g has four significant figures, while one reading 4.0 g has only two. That difference matters when you carry those numbers into a calculation.

Rules for Determining Significant Figures

These rules can feel arbitrary at first, but they all come back to one question: does this digit carry real information, or is it just holding a place?

• All non-zero digits are always significant. 1.23 has 3 sig figs; 789 has 3 sig figs.
• Zeros between non-zero digits are always significant. 101 has 3 sig figs; 3.0204 has 5 sig figs. These "trapped" zeros clearly represent measured values.
• Leading zeros are never significant. They're just placeholders that show where the decimal point is. 0.0123 has 3 sig figs (the 1, 2, and 3). Writing it as $1.23 \times 10^{-2}$ makes this obvious.
• Trailing zeros are significant only if a decimal point is present. 1.00 has 3 sig figs (the zeros tell you the measurement is precise to the hundredths place). But 1500 is ambiguous and could have 2, 3, or 4 sig figs. Using scientific notation removes the ambiguity: $1.50 \times 10^{3}$ clearly has 3 sig figs.
• Exact numbers have unlimited significant figures. Counted quantities (12 eggs) and defined conversions (1 ft = 12 in) never limit your sig figs because there's no measurement uncertainty.

Rounding to Significant Figures

Rounding Rules

Rounding reduces the number of digits while keeping the value close to the original. Follow these steps:

1. Identify which digit is the last one you want to keep (the last significant figure).
2. Look at the digit immediately to its right.
3. If that digit is less than 5, leave the last significant figure unchanged.
4. If that digit is 5 or greater, increase the last significant figure by 1.
5. Drop all digits to the right of the last significant figure. (If needed, replace them with zeros to maintain the number's magnitude.)

Rounding Examples

• 1.2345 → 4 sig figs → 1.235 (the 5 rounds the 4 up)
• 0.00987 → 2 sig figs → 0.0099 (leading zeros don't count; the 7 rounds the 8 up)
• 45.999 → 3 sig figs → 46.0 (the 9 rounds up, cascading through; the decimal point keeps the trailing zero significant)
• 8,765,000 → 3 sig figs → 8,770,000 (the 5 rounds the 6 up to 7; the trailing zeros are placeholders)

Significant Figures in Calculations

The rules differ for multiplication/division versus addition/subtraction. This trips people up on exams, so pay attention to which operation you're doing.

Multiplication and Division

The result gets the same number of significant figures as the input with the fewest significant figures.

• $3.14 \times 2.7 = 8.478$
  • 3.14 has 3 sig figs; 2.7 has 2 sig figs → round to 2 sig figs → 8.5
• $\frac{12.5}{1.4} = 8.928...$
  • 12.5 has 3 sig figs; 1.4 has 2 sig figs → round to 2 sig figs → 8.9

Addition and Subtraction

The result gets the same number of decimal places as the input with the fewest decimal places. Notice this rule is about decimal places, not sig figs.

• $12.1 + 3.45 + 0.678 = 16.228$
  • 12.1 has 1 decimal place (fewest) → round to 1 decimal place → 16.2
• $1000.0 - 23.47 = 976.53$
  • 1000.0 has 1 decimal place (fewest) → round to 1 decimal place → 976.5

Exact Numbers and Intermediate Results

Exact numbers don't limit sig figs. In the formula $2\pi r$, the 2 is exact (it's a defined multiplier, not a measurement), so it doesn't constrain your answer's precision.

Intermediate results should carry extra digits through multi-step calculations. Only round at the very end. Rounding too early introduces "rounding error" that compounds with each step.

• $\frac{2.3 \times (4.1 + 6.279)}{1.2}$
  • First, add: $4.1 + 6.279 = 10.379$ (keep extra digits for now)
  • Multiply: $2.3 \times 10.379 = 23.8717$
  • Divide: $\frac{23.8717}{1.2} = 19.893...$
  • Now round. The limiting values are 2.3 and 1.2, each with 2 sig figs → 20.

The final answer is 20. (written with a decimal point to show 2 significant figures). If you had rounded the intermediate sum to 1 decimal place before multiplying, you'd risk a slightly different final answer.