# Significant Figures

Chapter 1R

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## Counting Sig. Figs.

Note: Given inexact numbers *should* contain all certain digits and the first uncertain digit.

Count

*left-to-right*when counting significant figures.All non-zero digits of a provided number are significant.

\[1\]

\[2.5\]

\[23.9\]

- Zeroes between two other significant digits are significant.

\[203\]

- Zeroes to the right of a non-zero number, and also to the right of a decimal place, are significant

\[123.0\]

- Zeroes that are placeholders are not significant.

\[0.001~3\]

\[13~000\]

Ambiguous (could be 2, 3, 4, or 5). For the purpose of this class, 2 significant figures.

Use of a decimal place at the end would make the trailing zeroes significant.\[13~000.\]

\[1.3\times 10^4\]

\[1.30\times 10^{4}\]

\[1.300\times 10^{4}\]

\[1.300~0\times 10^{4}\]

*Exact*numbers (those obtained by counting) have an infinite number of significant figures. Fractions*can*be exact. Physical constants such as the*molar gas constant*can also be exact if derived from other exact numbers. See NIST to determine if a physical constant is exact or inexact (i.e. has a non-zero uncertainty).

\[30~\mathrm{lemons}\]

*Some*conversion factors are*exact*while others are*inexact*. For example, 1 inch is defined as being exactly 2.54 cm. Therefore, both values in the following conversion factor, 1 in = 2.54 cm, are exact. However, 1 gallon (US) is*approximately*equal to 3.785412 L (inexact). The quantity given for L is inexact and would have 7 significant figures.A mathematical or physical constants has significant figures to its known digits. For example, as of March 2024, π is known to 102 trillion digits, each of which are significant. Constants such as speed of light (

*c*), gas constant (*R*), etc. also fall into this category. Note: Using a rounded off physical constant (e.g. 3.00×10^{8}m s^{–1}instead of 299 792 458 m s^{–1}for speed of light) will limit the number of significant figures for that constant. See NIST to determine if a physical constant is exact or inexact (i.e. has a non-zero uncertainty).

\[c = 299~792~458~\mathrm{m~s^{-1}}\]

\[c = 3.00\times 10^{8}~\mathrm{m~s^{-1}}\]

\[R = 8.314~462~618~153~24~\mathrm{J~mol^{-1}~K^{-1}}\]

\[R = 8.314~\mathrm{J~mol^{-1}~K^{-1}}\]

## Significant Figures in Calculations

- When adding or subtracting numbers, the result contains no significant digits beyond the place of the last significant digit of any datum.

\[\begin{align*} 3.24 + 1.9 + 12.482 &= 17.\bar{6}22 \\[1.5ex] &= 17.6 \\[4ex] 5.421 - 10.138 + 3.41 &= -1.3\bar{0}7 \\[1.5ex] &= -1.31 \\[4ex] 346 - 343.4 &= \bar{2}.6 \\[1.5ex] &= 3 \\[4ex] 25 - 10.1 &= 1\bar{4}.9 \\[1.5ex] &= 15 \\[4ex] 10. + 16.3 &= 2\bar{6}.3 \\[1.5ex] &= 26 \\[4ex] 10.0 + 16.3 &= 26.3 \end{align*}\]

For numbers that clearly have an ambiguous number of significant figures, assume the zeroes to be insignificant (for the purpose of this class).

\[\begin{align*} 210~000 + 61~435 &= 2\bar{7}1~435 \\[1.5ex] &= 270~000 \end{align*}\]

Here, 210 000 has an ambiguous number of significant figures. The last non-zero digit is located in the ten thousands place. The result should be rounded to the ten thousands place.

Here are a few more examples.

\[\begin{align*} 23~100 + 32 &= 23~\bar{13}32 \\[1.5ex] &= 23~100 \\[4ex] 890 + 12 &= 9\bar{0}2 \\[1.5ex] &= 900 \\[4ex] 312 + 300~000 &= \bar{3}00~312 \\[1.5ex] &= 300~000 \\[4ex] 10 + 16.3 &= \bar{2}6.3 \\[1.5ex] &= 30 \end{align*}\]

- In multiplication or division, the number of significant figures in the answer is determined by the value with the
**fewest significant figures.**

\[\begin{align*} 3.24 \times 812.3 &= 2~6\bar{3}1.852 \\[1.5ex] &= 2.63\times 10^{3}\\[4ex] 1.502 \left ( \dfrac{4.90}{2.11} \right ) &= 1.502 \left ( 2.3\bar{2}2\right ) \\[1.5ex] &= 3.4\bar{8}80 \\[1.5ex] &= 3.49 \\[4ex] \left ( 346 - 343.4 \right ) / 8.4 &= \left ( \bar{2}.6 \right ) 8.4 \\[1.5ex] &= 0.3095 \\[1.5ex] &= 0.3 \end{align*}\]

*As per the textbook*, when a number is rounded off, the last digit retained is increased by one (rounded up) only if the following digit is 5 or greater.**NOTE:**The*round-half-to-even*rule is followed by NIST, ANSI, ASTM, etc. where a number only gets rounded up on a 5*if*the resulting number was an*even*number.

*textbook*.

\[\begin{align*} 12.\bar{6}96 &\rightarrow 12.7 \\[1.5ex] 18.\bar{3}49 &\rightarrow 18.3 \\[1.5ex] 14.\bar{9}99 &\rightarrow 15.0 \\[1.5ex] 14.\bar{3}5 &\rightarrow 14.4 \\[1.5ex] 1.1\bar{2}51 &\rightarrow 1.13 \end{align*}\]

**round-half-to-even**rule.

\[\begin{align*} 14.\bar{3}5 &\rightarrow 14.4 \\[1.5ex] 1.1\bar{2}51 &\rightarrow 1.12 \end{align*}\]

- A rounded value should be obtained in one step by direct rounding of the most precise value available and not in two or more successive roundings. For example: 89 490 rounded to the nearest 1 000 is at once 89 000; it would be incorrect to round first to the nearest 100, giving 89 500 and then to the nearest 1 000, giving 90 000.

- In a multi-step calculation, only round the final value. Determine the number of significant figures in the final result by considering each
*step*in the calculation. In intermediate steps, write the number to the proper number of significant figures and keep*at least*one additional digit.

\[\begin{align*} (2.349~4 + 1.345) \times 1.2 &= 3.69\bar{4}~4 \times 1.2 \\[1.5ex] &= 4.\bar{4}3 \\[1.5ex] &= 4.4 \\[4ex] (2.349~4 \times 1.345) + 1.2 &= 3.15\bar{9}~9 + 1.2 \\[1.5ex] &= 4.\bar{3}5 \\[1.5ex] &= 4.4 \end{align*}\]

- Digits in logarithms, ln(
*x*) or log_{10}(*x*), are significant through the*n*-th place after the decimal when*x*has*n*significant figures. A quantity resulting from a logarithmic transformation is dimensionless.

\[\ln(3.46~\mathrm{kPa}) = 1.241\]

\[\log(3.000\times 10^4) = 4.477~1\]

^{4}has 4 significant figures. The result should have 4 places after the decimal.

- Significant digits as a result of exponentials and antilogarithms,
*e*^{x}or 10^{x}, is equal to the place of the last significant digit in*x*after the decimal. A number resulting from an antilog transformation will have dimensions if the original logarithmic value was derived from a quantity with units.

\[e^{1.241} = 3.4\bar{5}9~07 = 3.46\]

\[10^{4.4771} = 29~9\bar{9}8.531~811~9\ldots = 30~000 = 3.000\times 10^{4}\]