Understandings
Applications and Skills
Guidance
Figure 1. To measure the volume of liquid in this graduated cylinder, you must mentally subdivide the distance between the 21 and 22 mL marks into tenths of a milliliter, and then make a reading (estimate) at the bottom of the meniscus.
The following examples illustrate the application of this rule in rounding a few different numbers to three significant figures:
Rounding Numbers
Round the following to the indicated number of significant figures:
(a) 31.57 (to two significant figures)
(b) 8.1649 (to three significant figures)
(c) 0.051065 (to four significant figures)
(d) 0.90275 (to four significant figures)
(a) 31.57 rounds “up” to 32 (the dropped digit is 5, and the retained digit is even)
(b) 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5)
(c) 0.051065 rounds “down” to 0.05106 (the dropped digit is 5, and the retained digit is even)
(d) 0.90275 rounds “up” to 0.9028 (the dropped digit is 5, and the retained digit is even)
Addition and Subtraction with Significant Figures
Rule: When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (i.e., the least precise value in terms of addition and subtraction).
(a) Add 1.0023 g and 4.383 g.
(b) Subtract 421.23 g from 486 g.
(a) 1.0023g+4.383g = 5.3853g
Answer is 5.385 g (round to the thousandths place; three decimal places)
(b) 486g−421.23g = 64.77g
Answer is 65 g (round to the ones place; no decimal places)
Multiplication and Division with Significant Figures
Rule: When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).
(a) Multiply 0.6238 cm by 6.6 cm.
(b) Divide 421.23 g by 486 mL.
(a) 0.6238 cm × 6.6 cm= 4.11708 cm2→ result is 4.1cm2 (round to two significant figures)
four significant figures×two significant figures→two significant figures answer
(b) 0.867 g/ml
Calculation with Significant Figures
One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters.
Vl×w×d
13.44 dm × 5.920 dm × 2.54 dm = 202.09459… dm3(value from calculator)
202 dm3,or202 L (answer rounded to three significant figures)
Figure 2. (a) These arrows are close to both the bull’s eye and one another, so they are both accurate and precise. (b) These arrows are close to one another but not on target, so they are precise but not accurate. (c) These arrows are neither on target nor close to one another, so they are neither accurate nor precise.
Typical systematic errors include
For example: A grade B 250 ml volumetric flask may have the following legend.
250 ml ± 0.30ml @ 20ºC.
In a titration the following measurements were taken.
This expresses the data point as having a maximum possible value of 17.35 ml and a minimum value of 17.15 ml.
The pipette may have the tolerance written on the side as: ± 0.26 ml @ 20ºC
This means that when the pipette is used exactly as per the manufacturer's instructions at 20ºC there is an in-built inaccuracy of between + 0.26 ml and - 0.26 ml.
The 50 ml burette has no tolerance value written on the side.
The smallest mark on the graduated scale is 0.1 ml. It is reasonable to assume that you can assess the measurement of the liquid level to within half of the smallest graduation, i.e. 0.05 ml. This means that each reading should be given as [your value] ± 0.05 ml.
Don't forget that when a burette is used you must take TWO readings. Each reading has an uncertainty of ± 0.05 ml, therefore the total uncertainty is 0.05 + 0.05 = ± 0.1 ml
It is possible that the actual reading (as measured by the balance) is as low as 27.525 g, in which case it has rounded up the reading so that the last decimal is 3.
It is equally possible that the actual reading as measured by the balance is as high as 27.534 g. In this case the second decimal would appear as 3.
Hence the range of values that the balance could be reading is from 27.525 g to 27.534 g.
This uncertainty is recorded as 27.53 ± 0.005 g
The burette requires two readings, the initial reading and the final reading. The volume delivered is obtained by subtraction of the initial reading from the final reading.
There is an uncertainty of ± 0.05 in each reading, ∴ total absolute uncertainty of ± 0.1 ml
The percentage uncertainty = (0.1/24.2) x 100 = 0.41% uncertainty
Step 1 The mass of solid needed for preparation of the solution is weighed out and then recorded.
Step 2 The solid is dissolved in approximately 100 ml water and transferred to a volumetric flask, which is then made up to the 250 ml mark with distilled water. (the volumetric flask has a tolerance of ± 0.46 ml written on the side)
Calculation: The solution concentration = mass/(relative mass x volume in litres)
= 1.66/(204.23 x 0.25) = 0.052 mol dm-3
Absolute error = final concentration x percentage error/100 = 0.052 x 0.372/100 = 0.0002 mol dm-3
In reality this value is too small to be included, as it is smaller than the number of decimal places in the final concentration. This is to be expected, as preparation of a standard solution should be accurate.