Value | Relative Uncertainty | Absolute Uncertainty |
125 mL | +/- 1 part in 125 | +/- 1 mL |
9.700 km | +/- 1 part in 9700 | +/- 0.001 km |
6.02 x 1023 atoms | +/- 1 part in 602 | +/- 1 x 1021 atom |
0.0024 g | +/- 1 part in 24 | +/- 0.0001 g |
347.34 min | +/- 1 part in 34734 | +/- 0.01 min |
In multiplication and division, we retain the largest relative uncertainty. But to make things easier, we make an approximation. In the table above, you notice that values with more significant digits have a smaller relative uncertainty. We relate the relative uncertainy to the number of significant digits. For example, we say that all values with 3 significant digits are known with a relative uncertainty of +/- 1 part in 1000. All values with 5 significant digits are known to within +/- 1 part in 100000. So we can rewrite the above table to reflect this:
Value | Approximate Relative Uncertainty | Absolute Uncertainty |
125 mL | +/- 1 part in 1000 | +/- 1 mL |
9.700 km | +/- 1 part in 10000 | +/- 0.001 km |
6.02 x 1023 atoms | +/- 1 part in 1000 | +/- 1 x 1021 atom |
0.0024 g | +/- 1 part in 100 | +/- 0.0001 g |
347.34 min | +/- 1 part in 100000 | +/- 0.01 min |
In multiplication and division, we retain the largest relative uncertainty. The approximation we made makes this easy. The value with the largest relative uncertainty is also the one with the least number of significant digits. Thus, we retain only as many significant digits as the value with the fewest.
Question: How many moles are contained in 0.025 L of a 0.1104 mol/L solution? |
Solution:
0.025 L 0.1104 mol ------- x ---------- = 0.0027600 mol ==> 0.0028 mol L |
Comments: Since the value 0.025 L only contains two significant digits, we round the result to two significant digits. |
Question: How many Joules are needed to raise the temperature of 15 g of H2O 24.2 K? |
Solution:
15 g 24.2 K 4.184 J ---- x ------ x ------- = 1518.7920 J ==> 1500 J g K |
Comments: Since the value 15 g only contains two significant digits, we round the result to two significant digits. |
Question: What is the frequency of the sodium "D" line (used in many street lamps) given it's wavelength of 588.995 nm? |
Solution:
1 109 nm 2.9979256 x 108 m ---------- x ------ x ----------------- = 1.7657632 x 1020 s-1 588.995 nm m s ==> 1.76576 x 1020 s-1 |
Comments: Since the value of the wavelength 588.995 nm is known to six significant digits, the final result should be determined to six significant digits. Thus, the value of the speed of light must have more than six significant digits so it does not limit the number of significant digits in the result. |