Precision
Significant Figures
Significant figures are the number of digits required to express the accuracy of a particular
measurement or the accuracy of a calculation involving measurements. As a general rule one
assumes uncertainty in only the last digit and all other digits are considered to be certain or
definite.
Consider the measurement of the same solid rod using two different rulers. RULER 1 has
graduations (marks) every 1/10 of a cm (i.e. every millimeter). RULER 2 has graduations
every centimeter. Assume the left end of the rod and the metric rulers (not shown in the
figure) are aligned perfectly. The length of the rod in centimeters is obtained by reading the
rulers at the right end of the rod and estimating the last “ digit” . The rulers enlarged for
clarification and are not depicted to scale. That is the centimeter shown is not actually a
centimeter in length.
Let us examine what the reading for length would be for each ruler.
Ruler 1
Ruler 2
1. Length is definitely between 4 and 5 cm.
1. Length is definitely between 4 and 5
cm.
2. Length is definitely between 4.4 and 4.5
cm.
2. Estimate length is around 4.5 cm.
3. Estimate length between 4.47 and 4.49 cm. 3. Report length as 4.5 ± 0.1 cm.
4. Report length as 4.48 ± 0.01 cm
In each measurement there was but one uncertain digit-the last estimated digit. Regardless
RULER 1 produced the more precise measurement. The value of length utilizing RULER 1
has three “ significant figures” while the value of length obtained utilizing RULER 2 has two
“ significant figures” . Generally for a given measurement, the value with the larger number
of “ significant figures” is the more precise value.
Ruler 1 allows measurements to ± 0.01 cm. Consider a different rod again with the left end of
both the rod and ruler perfectly aligned. If the right end of the rod is exactly over the
graduation at 3.8, the length of the rod would be reported as 3.80 ± 0.01 cm. The number of
significant figures is three with uncertainty found only in the last digit which is zero. This
example illustrates the fact that zeros after decimal points are significant. Of course, non-zero
digits are always significant in this type of measurement.
Generally the numerical value of a measured quantity is reported to only one uncertain figure.
The number of digits in the number is called the number of significant figures. One can
assume that the non-zero digits of an experimental measurement are significant. With this
assumption, rules have been devised to understand whether a zero (0) in a number is
significant or not.
Significant Figures in a Measurement
1. The significance of zero and non-zero digits in a number.
a. Assume non-zero digits are significant.
b. Zeros between the decimal and a non-zero digit are not significant.
Example: 0.0448 m - 3 (underlined) significant figures
c. Zeros between non-zero digits are significant
Example: 24,100,386 m - 8 (underlined) significant figures
d. Zeros following non-zero digits are significant if after a decimal.
Example: 23.40 m - 4 (underlined) significant figures
e. Zeros following non-zero digits in a large number are uncertain and not significant
Example: 44,800,000 nm - 3 (underlined) significant figures
2. The last digit is interpreted as ± l unless a more precise statement (as in 4.78 ± 0.014) can
be given.
Our example considered a simple measurement in which all non-zero digits were significant.
Frequently calculators or computer based instruments give values that exceed the know
precision of an experiment. Hence the values need to be rounded to the correct number of
digits. A rule has been established to maintain consistency in the process of rounding. The
rule focuses on the figures that are not significant and are to be dropped from the number.
These digits are called “ superfluous” digits or figures.
Rounding Rule
In dropping superfluous figures, increase the last retained digit by one if the first dropped
figure is greater than 5. If the first dropped figure is exactly five, round off so that the last
digit is an even number.
The rule is relatively simple but applying it practice can be more complex. Let us examine
some examples illustrating the application of the Rounding Rule. All values are rounded to
three significant figures in these examples.
a. 3.468 rounds to 3.47. Comment: the leftmost number to be dropped is 8 (a number greater
than 5) and the 6 must be rounded up to 7.
b. 8.0329 rounds to 8.03. Comment: the leftmost number to be dropped is 2 (a number less
than 5) and the 3 must be left as it is.
c. 5.6350 rounds to 5.64. Comment: the leftmost number to be dropped is exactly 5 and so
the last significant digit (the 3), is rounded up to an even number.
d. 1.245 rounds to 1.24. Comment: the leftmost number to be dropped is exactly 5 but the
last significant digit is an even number (the 4), so it is not changed. The 5 simply is
dropped.
Scientific Notation
In exponential notation a number is expressed in the form N × 10a. There are not constraints
on the values of N or a. In contrast scientific notation is a specific type of exponential
notation. It is utilized to express the number of significant figures in a value and to indicate
the magnitude of a particular quantity. Scientific notation has a specifically designed
structure. The general structure is N × 10a where N is a number with the correct number of
significant digits. Beyond being comprised of only significant digits, N has some additional
restrictions; it must be a number that has a value of at least one but less than ten. The
exponent “ a” usually is a positive or negative integer but when necessary, “ a” can have a
value of zero, since 100 = 1.
Decimal notation also has conventions in place for expressing large and small numbers. For
numbers less than one the convention is that they are written with a zero (“ 0” ) followed by a
decimal as shown for 0.348 and 0.00640. The convention for writing numbers greater than
one in the decimal notation (especially large estimated numbers) is not to place a decimal
after the trailing zeros. As such the numbers 80 and 567,630,000 and 3400 do not have a
decimal after the zeros unless the zeros are significant digits. Thus a decimal after the trailing
zeros indicates that all digits in the number are significant. While the decimal notation does
not show the decimal point for a large number, we can assume it is present for the purpose of
counting places to determine the value of the exponent “ a” when converting decimal into
scientific notation.
right in
b. Light is found to have a wavelength of 3.6 × 10-4 m which converts to 0.00036 m.
Comment: the number has two significant figures. As the exponent is negative (-4), the
decimal is moved 4 places to the left requiring three zeros (not significant) between the
decimal and first significant non-zero digit. The leading zero before the decimal point also
is not significant.
Significant Figures
Question: How many significant figures are there in 1.8 × 1016?
Solution: In scientific notation all digits in the decimal portion are considered significant.
The value of the exponent does not affect the number of significant digits. The answer is 2.
The significant digits are underlined: 1.8 × 1016.
Question: How many significant figures are there in 0.001285.
Solution: The leading zeros in a number less than one are not significant. The answer is 4.
The significant digits are underlined: 0.001285
Question: How many significant figures are there in 112300047.
Solution: The zeros between non-zero digits are significant. The answer is 9.
The significant digits are underlined: 112300047
Question: How many significant figures are there in 744000
Solution: The trailing zeros in a large number without a decimal point are uncertain; hence
they are not significant. The answer is 3.
The significant figures are underlined: 744000
Decimal and Scientific Notation
Question: Convert the number 21.7 × 10-1 into a decimal with the correct number of
significant digits.
Solution: To form a decimal from N × 10-a, move the decimal point “ a” places to the left
within N. The answer is 2.17
Question: Convert the decimal number 382.96 into standard scientific notation with the
correct number of significant digits.
Solution: To express the number in the form N × 10a, move the decimal until N is greater or
equal to 1 but less than 10. For each place moved to the left, add +1 to a. The answer is
3.8296 × 102
Question: Convert the decimal number 0.000598 into standard scientific notation with the
correct number of significant figures.
Solution: To express the number in the form N × 10a, move the decimal until N is greater or
equal to 1 but less than 10. For each place moved to the right, add -1 to a. The answer is 5.98
× 10-4.
Question: Convert the number 37.5 × 105 into a decimal number with the correct number of
significant digits.
Solution: To form a decimal number N (which is equivalent to N × 100) from N × 10a move
the decimal point a places to the right within N. The answer is 3750000 without a decimal
point after the trailing zeros.
Rounding
Question: Round the number 0.0370812 to 2 significant figures and express in standard
scientific notation.
Solution: The number rounds to 0.037. To place the answer in standard scientific notation,
the decimal point needs to be moved 2 places to the right which gemerates a factor of 10-2.
The correct response is 3.7 × 10-2.
Question: Round the number 47.98620 to 1 significant figure, and express in decimal
notation.
Solution: As the answer will contain only 1 significant figure, the leftmost number to be
dropped is 7. As 7 is greater than 5, the 4 must be rounded up by 1. The correct answer is 50
without a decimal point after the zero. If a decimal point is placed after the zero it would
imply 2 significant figures.
Question: Round the number 786.37000 to 2 significant figures, and express in decimal
notation.
Solution: The leftmost digit to be dropped is 6 because the answer must be given in 2
significant digits. As 6 is greater than 5, the number 8 must be rounded up to 9. The correct
answer is 790 without a decimal point after the zero. If a decimal point is placed after the zero
it would imply 3 significant figures.
Question: Round the number 185.98690 to 4 significant figures, and express in decimal
notation.
Solution: For the number to contain 4 significant figures, 8 will be the leftmost digit to be
dropped. As 8 is greater than 5, the 9 will be rounded up to 10. This will round the 5 up to six.
The correct answer is 186.0.
Question: Round the number 31.5030 to 2 significant figures and express in decimal
notation.
Solution: The leftmost digit to be dropped in this case is the 5. As the five is followed by at
least one non-zero digit, the rightmost digit to be retained must be rounded up by 1. The
correct answer is 32.
Question: Round the number 32.5000 to 2 significant figures and express in decimal
notation.
Solution: The leftmost digit to be dropped in this case is the 5. As the 5 is followed by all
zeros and the rightmost digit to be retained is an even number, the rightmost digit will be left
as it is. The correct answer is 32.
Question: Round the number 4.3983 to one significant figure and express the answer in
standard scientific notation.
Solution: The leftmost digit to be dropped in this case is the 3. As the 3 is less than 5, the 4
will be left as it is. As the answer must be stated in scientific notation, the factor of 10, must
be included in the answer. The correct answer is 4 × 100.
Question: Round the number 30.2 to two significant digits and express in decimal notation.
Solution: The leftmost digit to be dropped will be the 2. The answer must contain the decimal
point in order to indicate that the answer contains two significant digits. If it were written
“ 30” without the decimal point it would imply that there was only one significant digit in
the number. The correct answer is “ 30.”
Question: Round the number 12.78081 to one significant figure and express in decimal
notation.
Solution: The leftmost digit to be dropped will be 2. As the 2 is less than 5, the 1 will remain
unchanged. The correct answer is 10 without a decimal point. A decimal point would imply
two (2) significant figures.
Question: Round the number 78485.3000 to 4 significant figures and express in scientific
notation.
Solution: As the answer must be written with 4 significant figures, the leftmost digit to be
dropped will be 5. Since 5 is followed by a non-zero digit 1 is added to 8, which is the right
most digit to be retained. To place the answer in scientific notation, the decimal point needs to
be moved 4 places to the left. The answer will therefore contain a factor of 104. The correct
answer is 7.849 × 104.
Applied Precision
Determining the precision in the answer of a calculation in an exact fashion is a complicated
process. In the calculations found in most beginning science courses a simple but approximate
rule is used to determine the number of significant figures in the answer. The number of
significant figures in the answer should equal the number of significant figures in the piece of
data used in the calculation that has the fewest significant figures. This rule is applied to
calculations involving addition and subtraction in a different fashion than to calculations
involving multiplication and division.
Addition and Subtraction
For addition and subtraction, the number with the least number of decimal places determines
the number of decimal places in the answer.
Multiplication and Division
For multiplication and division, the number with the least number of significant figures
determines the number of significant figures in the answer.
Addition and Subtraction
Question: The initial burette reading for a titration using 0.56 M hydrochloric acid was 0.82
mL. The burette reading at the endpoint was 39.4 mL. Calculate the volume of hydrochloric
acid used in the titration to the correct number of significant figures. Calculate the answer
first, then round it.
Solution: The change in volume is ΔV = Vf – Vi. Subtract the initial burette reading from the
final burette reading to obtain ΔV. Note that the concentration of the solution (0.56 M) has
nothing to do with the solving for the volume change and is not considered in the solution.
ΔV = Vf – Vi. = 39.4 mL - 0.82 mL = 38.58 mL (answer not rounded.)
Round the answer to the correct number of significant figures by using the rule that the
number with the fewest places after the decimal determines the number of decimal places in
the answer for addition or subtraction. In this case the number 39.4 determines the number of
decimal places in the answer, one.
The rounded answer is 38.6 mL.
Question: A sample of iron powder weighing 12.60997 grams and a water ethanol mixture
weighing 159.98621 grams were added to a volumetric flask that weighed 28.06 grams.
Calculate the total mass to the correct number of significant figures. Calculate the final
answer, then round it.
Solution: This problem involves adding up the masses of the iron powder, the water ethanol
mixture and the volumetric flask:
The number 28.06 will determine the number of decimal places in the final, rounded answer
because it has the least number of places after the decimal.
The rounded answer is 200.66 g.
Question: The volume of sugar solution in a graduated cylinder was determined to be 87.2
mL. After transferring some of the sugar solution, the volume was 6.62 mL. Determine the
volume of sugar solution transferred and express the answer to the correct number of
significant figures. Calculate the answer, then round it.
Solution: To determine the answer, subtract the remaining volume from the original volume
in the volumetric flask. A volumetric flask is numbered differently than a buret. The buret is
numbered so that ΔV = Vf – Vi is a positive number. The volumetric flask is inversely
numbered. Simply subtract the volume that remains from the initial volume to determine how
much solution was transferred.
The figure 87.2 mL will determine the number of significant figures in the final, rounded
answer because it has the fewest number of digits to the right of the decimal point.
The rounded answer is 80.6 mL
Question: The mass of a watch glass containing a sample of potassium acid phthalate was
determined to be 18.357 g. After transferring a sample it weighed 2.3 g. Determine the mass
of potassium acid phthalate transferred to the correct number of significant figures. Calculate
the answer then round.
Solution: To determine the answer, subtract the weight of the sample after the transfer from
its original weight.
The figure 2.3 g will determine the number of significant figures in the final, rounded answer.
Recall the rounding rule: “ If the leftmost digit to be dropped is 5 followed by at least one
non-zero digit add 1 to the rightmost digit to be retained.”
The final rounded answer is 16.1 g.
Question: A sample of calcium carbonate weighing 29.1078 grams and water weighing
150.5422 grams were added to a bottle that weighed 33.1 grams. Calculate the total mass to
the correct number of significant figures.
Solution: To determine the answer, add all the masses together.
The figure 33.1 g will determine the number of significant figures in the final, rounded
answer. Note the rounding rule: “ If the leftmost digit to be dropped is a 5 which is followed
only by zeros (i.e. 5000), make sure the rightmost digit to be retained is even. If this digit is
already even, leave it as it is. If this digit is odd, make it even by adding 1.”
The final rounded answer is 212.8 g.
Multiplication and Division
Question: Calculate (to the correct number of significant figures) the volume (in mL) which
must be taken from an aqueous solution that contains 30.87 g potassium iodide per 100.00 mL
in order to provide 200 g of potassium iodide.
Solution: The problem is asking how many milliliters of the solution contain 200 grams of
potassium iodide. The key to solving it is to recognize that the term “ 30.87 g potassium
iodide per 100.00 mL “ can be used as a conversion ratio. If we want the mass of potassium
iodide we use the first ratio. If we want the volume of solution we use the second ratio. In this
case we want the volume of solution so the second ration is used. Another way to view the
choice of ratios is to use the ratio that cancel the units of the given quantity and leaves the
units of the desired quantity behind.
Remember that the number with the fewest significant figures determines the number of
significant figures in the answer. The result is reported to one significant figure because 200 g
contains one significant figure while 100.00 mL and 30.87 g contain 5 and 4 significant
figures respectively. There is no decimal point after 200 g.
The rounded answer is 600 mL.
The conversion ratio method is preferred but there is another way to do the problem using
proportionality.
Note the equation that results is exactly the same as the equation obtained from the
conversion ratio.
Question: Calculate (to the correct number of significant figures) the volume (in mL) which
must be taken from an aqueous solution that contains 30.87 g potassium iodide per 100.00 mL
in order to provide 200. g of potassium iodide.
Solution: The problem is asking how many milliliters of the solution contain 200. g of
potassium iodide. In this case there is a decimal point in 200. g. Again use the conversion
ratio approach.
Remember that the number with the fewest significant figures determines the number of
significant figures in the answer. The result is reported to three significant figures because
200. g contains three significant figures while 100.00 mL and 30.87 g contain 5 and 4
significant figures respectively.
The rounded answer is 648 mL.
The conversion ratio method is preferred but there is another way to do the problem using
proportionality.
Note the equation that results is exactly the same as the equation obtained from the
conversion ratio.
Question: A pure 19.990 gram sample of a liquid occupies a volume of 3.5844 mL. Calculate
its density in g/mL to the correct number of significant figures.
Solution: To solve this problem it is important to understand the definition of density.
Density is gram per mL. To obtain this unit divide the mass by the volume. Given that 3.5844
milliliters of the liquid has a mass of 19.990 grams, the mass of one milliliter can be
calculated using the given proportion:
Note: The result was reported to 5 significant figures because 19.990 g and 3.5844 g have 5
significant figures each.
Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.