Skills Beta

Significant Figures

A measurement records every certain digit plus one estimated digit; these are its significant figures.

Practice 5: Mathematical Routines

Question set for this topic

Part 1 · Hook

Why this matters

Ask a bathroom scale and a lab balance to weigh the same apple and you get 150 g and 151.37 g. Both are honest. The number of digits in a measurement tells you how carefully it was made, and on the exam, reporting the wrong number of digits is the most common way to lose a calculation point.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Which is 0.00450 in scientific notation?

  1. 4.50 × 10⁻³
  2. 4.50 × 10³
  3. 45.0 × 10⁻⁴
  4. 4.5 × 10⁻²
Show the answer

Move the point 3 places right to get 4.50; a small number has a negative exponent.

  • Correct: 4.50 × 10⁻³:
  • 4.50 × 10³:
  • 45.0 × 10⁻⁴:
  • 4.5 × 10⁻²:

2. What is the density of an object with a mass of 20 g and a volume of 5 mL?

  1. 4 g/mL
  2. 0.25 g/mL
  3. 100 g/mL
  4. 25 g/mL
Show the answer

Density = mass ÷ volume = 20 g ÷ 5 mL = 4 g/mL.

  • Correct: 4 g/mL:
  • 0.25 g/mL:
  • 100 g/mL:
  • 25 g/mL:

Part 4 · See it

See it first

A ruler marked every 0.1 cm; an object's edge falls about seven tenths of the way from 3.4 to 3.5 cm, so the reading is 3.47 cm, with 7 the estimated digit.
Reading a scale: every digit you are sure of, plus one estimated digit between the marks. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Every instrument can only be read so finelythe last recorded digit of a measurement is an estimate
  2. The significant figures are the certain digits plus that one estimated digitthey show how precise the measurement is
  3. A calculated answer cannot be more precise than the data it came fromit is rounded to match the least precise measurement
  4. Multiplying and adding spread uncertainty differentlyproducts keep the fewest significant figures, and sums keep the fewest decimal places

Part 6 · Key ideas

Key ideas

  • Count significant figures: all nonzero digits; zeros between them; trailing zeros after a decimal point. Leading zeros never count. 0.004050 has four.
  • Multiply or divide: keep the fewest significant figures. Add or subtract: keep the fewest decimal places.
  • Exact numbers (counts, defined equalities like 1 kg = 1000 g) never limit an answer.
  • Precision is how closely repeated readings agree; accuracy is how close they are to the true value. Round once, at the end.

Part 7 · Misconception

A common mistake

The wrong idea: Writing more digits makes an answer more accurate, so copy everything the calculator shows.

What actually happens: Extra digits claim a precision the measurements do not have. 23.45 g ÷ 12.0 mL is 1.95 g/mL, not 1.954166667 g/mL: the volume was only known to three significant figures.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

1. How many significant figures does 0.004050 g have?

  1. 7
  2. 3
  3. 4
  4. 2
Show the answer

Write it as 4.050 × 10⁻³ g: four significant figures. Leading zeros are placeholders; the captive zero and the trailing zero after the decimal point are measured.

  • 7: This counts every digit, but leading zeros only place the decimal point.
  • 3: The final zero comes after a decimal point, so it was measured and counts.
  • Correct: 4: Right: leading zeros do not count; 4, 0, 5 and the trailing 0 after the decimal point do.
  • 2: Zeros between nonzero digits (the 0 in 405) always count, and so does the trailing zero here.

2. Calculate 23.45 g ÷ 12.0 mL and report the density with the correct significant figures.

Type a number in g/mL.

Show the answer

23.45 ÷ 12.0 = 1.954… The least precise value, 12.0 mL, has three significant figures, so the answer is 1.95 g/mL.

  • Answer: 1.95 g/mL

3. Add 12.52 g + 3.1 g + 0.247 g and report the total with the correct precision.

Type a number in g.

Show the answer

12.52 + 3.1 + 0.247 = 15.867. In addition the answer keeps the fewest decimal places: 3.1 has one, so the sum is 15.9 g.

  • Answer: 15.9 g

4. A student measures 3 identical bolts as 2.157 g each and reports the total of the 3 bolts as 6.5 g "because 3 has one significant figure." What is wrong?

  1. The total should be rounded further, to 7 g, to match the one digit in 3
  2. Adding the three masses keeps two decimal places, so 6.47 g
  3. Multiplying a mass by a count of objects gives the wrong unit
  4. The 3 is an exact count, so it does not limit the answer: 6.471 g
Show the answer

Exact numbers, from counting or definitions such as 1 kg = 1000 g, never limit precision. 3 × 2.157 g = 6.471 g, four significant figures like the measurement.

  • The total should be rounded further, to 7 g, to match the one digit in 3: This makes the same mistake more strongly: the count of 3 is exact.
  • Adding the three masses keeps two decimal places, so 6.47 g: Adding 2.157 three times gives 6.471, which keeps three decimal places, not two.
  • Multiplying a mass by a count of objects gives the wrong unit: The unit stays grams, and multiplying the mass of one by the number of identical bolts is exactly how to get the total.
  • Correct: The 3 is an exact count, so it does not limit the answer: 6.471 g: Right: counted numbers are exact and have unlimited significant figures.

Data table

Three students weigh a 50.00 g standard mass

Three students each weigh the same certified 50.00 g mass four times on different balances.

Four readings by each student, in grams
StudentReading 1 (g)Reading 2 (g)Reading 3 (g)Reading 4 (g)
Ana49.1249.1349.1149.12
Ben50.3149.6250.0849.95
Cai50.0149.9950.0050.00

5. Which student's readings are precise but not accurate?

  1. Ben
  2. Cai
  3. Ana
  4. Cai, because readings that agree to 0.01 g are precise
Show the answer

Precision is how closely repeated readings agree; accuracy is how close they are to the true value. Ana is precise but off by about 0.88 g.

  • Ben: Ben's readings spread over 0.69 g, so they are not precise; their average happens to land near 50.00 g.
  • Cai: Cai's readings are both close together and close to 50.00 g: precise and accurate.
  • Correct: Ana: Right: Ana's readings agree within 0.02 g, but all are about 0.88 g below the true 50.00 g.
  • Cai, because readings that agree to 0.01 g are precise: Cai is precise, but also accurate: the readings sit right at 50.00 g.

6. Ana's results suggest which kind of error?

  1. A random error from the student reading the display inconsistently
  2. A significant-figures error from recording too many digits
  3. A systematic error, such as a balance that reads about 0.9 g low
  4. No error, since the readings are so close together
Show the answer

A consistent shift in one direction points to a systematic error, such as a balance that was not zeroed. Averaging more readings will not fix it.

  • A random error from the student reading the display inconsistently: Random errors scatter readings in both directions; Ana's are tightly grouped.
  • A significant-figures error from recording too many digits: Four-digit readings on a 0.01 g balance are recorded correctly.
  • Correct: A systematic error, such as a balance that reads about 0.9 g low: Right: every reading is shifted the same way by about the same amount.
  • No error, since the readings are so close together: Close agreement shows precision, not accuracy; the true mass is 50.00 g.

7. What is the average of Ben's four readings, reported to the correct precision?

Type a number in g.

Show the answer

(50.31 + 49.62 + 50.08 + 49.95) ÷ 4 = 199.96 ÷ 4 = 49.99 g. The sum keeps two decimal places; dividing by the exact count 4 keeps four significant figures.

  • Answer: 49.99 g

8. Why does Cai's balance report readings to 0.01 g but not to 0.001 g?

  1. The standard mass has four significant figures, so the balance shows four
  2. A third decimal place would be zero for this mass, so it is left off
  3. Balances round to two decimal places to make averaging easier
  4. Its last digit is already estimated; a further digit would not be measured
Show the answer

Every measurement has an uncertainty in its last digit. Reporting more digits than the instrument resolves would claim a precision that was not measured.

  • The standard mass has four significant figures, so the balance shows four: The display depends on the balance, not on the object being weighed.
  • A third decimal place would be zero for this mass, so it is left off: The third decimal place is unknown, not zero; writing a zero would claim it was measured.
  • Balances round to two decimal places to make averaging easier: The number of places reflects the balance's uncertainty, not convenience.
  • Correct: Its last digit is already estimated; a further digit would not be measured: Right: a measurement keeps its certain digits plus one uncertain digit.

Part 9 · Summary

Summary

A measurement records every certain digit plus one estimated digit; these are its significant figures. Leading zeros never count. Products and quotients keep the fewest significant figures, sums and differences the fewest decimal places, and exact numbers never limit. Precision is agreement between readings; accuracy is closeness to the true value.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections