Chemistry · General chemistry I · Worked example
Count significant figures
How many significant figures are in 0.004050, in 1.050, in 2300 and in 2.30 × 10³?
0.004050: skip the leading zeros
The zeros before the 4 only locate the decimal point, so they do not count. The zero between 4 and 5 counts, and so does the final zero, which trails a number with a decimal point: 4, 0, 5, 0. Four significant figures.
1.050: every digit counts
The zero between 1 and 5 counts, and so does the trailing zero after the decimal point. Four significant figures.
2300: ambiguous
Trailing zeros without a decimal point may be placeholders or measured digits. 2300 could have two, three or four significant figures; the number alone cannot say.
2.30 × 10³: the ambiguity removed
Scientific notation shows exactly which digits were measured. 2.30 × 10³ has three significant figures; 2.3 × 10³ would have two.
See it in a calculation
Leading zeros add no precision: in 0.004050 m × 2.0 m, the measurement 2.0 m has only two significant figures, so the product is 0.0081 m².
Result
0.004050 and 1.050 have four significant figures each, 2300 is ambiguous (two to four), and 2.30 × 10³ has three.
Your turn
How many significant figures are in 0.0300 g and in 6.022 × 10²³?
Show the answer and explanation
Three and four.
In 0.0300, the leading zeros do not count but the two trailing zeros after the decimal point do: 3, 0, 0. In 6.022 × 10²³, every digit of the coefficient counts: 6, 0, 2, 2.
Keep exploring
In a Chemistry box, write 0.004050 m × 2.0 m = 0.008100 m²: the arithmetic reads ✓, with a note that the data support only two significant figures.
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