Significant Figures and Rules for Rounding in Calculations

**Rules for Counting Significant Figures**
Nonzero integers always count as significant figures.	511	3 s.f.
Leading zeros are zeros that precede all the nonzero digits. These do not count as significant figures.	0.0025	2 s.f.
Captive zeros are zeros between nonzero digits. These always count as significant figures.	1008 1.008	4 s.f 4 s.f.
Trailing zeros are zeros at the right end of the number. In general assume they are significant only if the number contains a decimal point or they are specified as significant in the question.	100 1.00 100. 102.	1 s.f. 3 s.f. 3 s.f. 3 s.f.
Exact numbers are not obtained using measuring devices but are determined by counting or definition (like conversion factors) and do not impact the number of significant figures in the result of a calculation.	10 trials 8 molecules 1 g = 100 cg 1 in = 2.54 cm	infinite s.f.

**Rules for Operations Using Significant Figures**
Multiplication or division. The number of significant figures in the result is the same as the number in the least precise measurement used in the calculation.	$\frac{99.1}{1.11} = 89.198919892. . . = 89.2$	3 s.f.
Addition or subtraction. The result has the same number of decimal places as the least precise measurement used in the calculation.	$99.1 + 1.1543 = 100.2543 = 100.3$	1 decimal

**Rules for Significant Figures in Logarithms and pH**
Logarithm. When you take the logarithm of a number, keep as many significant figures to the right of the decimal point as there are significant figures in the original number. For example, log 4.000 (4 s.f.) = 0.6021(4 s.f. to right of the decimal point).	The pH of a solution with $[H +] = 3.44 M :$ $pH = - \log [H +] = - \log (3.44) = 3.463$	3.44 has three s.f., so 3.463 is reported to three decimal places.
Antilogarithms; pH. Conversely, when you take the antilogarithm of a number, the result should have the same number of significant figures as the number of significant decimal places in the principal value. For example, antilog 0.0334 (3 s.f. to the right of the decimal) = 1.08 (3 s.f.).	The $[H +]$ in a solution with a pH of 2.55: $[H +] = 10^{- 2.55}$ $= 2.8 \times 10^{- 3}$	2.55 has two decimal places, so 2.8 is reported to two significant figures.

**Rules for Rounding Significant Figures in an Answer**
Rounding answers. Use only the first number to the right of the last significant figure. If this digit is 5 or greater, round the last significant figure up (away from zero).	Rounding to two significant figures: $6.597 = 6.6$ $6.547 = 6.5$ $6.019 = 6.0$

**Rules for Carrying Significant Figures through Extended Calculations**
It is important that you postpone rounding until all the calculations are completed. At least one extra digit beyond the significant digits should be carried through all of the computations in order to avoid a rounding error. Round only the final answer to the correct number of significant digits.
How to calculate a question requiring that all digits (or at a minimum one additional digit) be carried through the calculation:
A sample of benzene, $C_{6} H_{6}$ , has a mass of 4.25 grams. The molar mass $= 78.11$ g/mol. How many molecules of $C_{6} H_{6}$ are in the sample? $N = 6.022 \times 10^{23}$ $4.25 g \times \frac{1 mol}{78.11 g} \times \frac{{6.022 \times 10}^{23} molecules}{mol} = 5.441045 \times 10^{- 2} mol \times \frac{{6.022 \times 10}^{23} molecules}{mol}$ $= 3.276597 \times 10^{22} = 3.28 \times 10^{22} molecules$
This example shows how to carry significant figures through an extended calculation where you may need to report rounded intermediate values. (You should carry the full values between steps, but if your calculator does not allow carrying, carry at least one extra digit.)
A sample of benzene, $C_{6} H_{6}$ , has a mass of 4.25 grams. The molar mass $= 78.11$ g/mol. (a) How many moles of $C_{6} H_{6}$ are in the sample? $4.25 g \times \frac{1 mol}{78.11 g} = 5.441045 \times 10^{- 2} mol = 5.44 \times 10^{- 2} mol$ (b) How many molecules of $C_{6} H_{6}$ are in the sample? $N = 6.022 \times 10^{23}$ $5.441045 \times 10^{- 2} mol \times \frac{{6.022 \times 10}^{23} molecules}{mol} = 3.276597 \times 10^{22} = 3.28 \times 10^{22} molecules$ (c) What is the total number of atoms in the sample? Each molecule of $C_{6} H_{6}$ contains 6 C atoms + 6 H atoms = 12 atoms total. $3.276597 \times 10^{22} molecules \times \frac{12 atoms}{1 molecule} = 3.931917 \times 10^{23} = 3.93 \times 10^{23} atoms (3 s . f .)$