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```Scientific Notation
Chemistry deals with very large and very small numbers. Consider this calculation:
(0.000000000000000000000000000000663 x 30,000,000,000) ÷ 0.00000009116
Hopefully you can see, how awkward it is. Try keeping track of all those zeros! In
scientific notation, this problem is:
(6.63 x 10¯31 x 3.0 x 1010) ÷ 9.116 x 10¯8
It is now much more compact, it better represents significant figures, and it is easier to
manipulate mathematically. The trade-off, of course, is that you have to be able to read
scientific notation.
This lesson shows you (1) how to write numbers in scientific notation and (2) how to
convert to and from scientific notation. As you work, keep in mind that a number like
9.116 x 10¯8 is ONE number (0.00000009116) represented as a number 9.116 and an
exponent (10¯8).
Format for Scientific Notation
1. Used to represent positive numbers only.
2. Every positive number X can be written as:
(1 < N < 10) x 10some positive or negative integer
Where N represents the numerals of X with the decimal point after the first nonzero digit.
3. A decimal point is in standard position if it is behind the first non-zero digit. Let X be
any number and let N be that number with the decimal point moved to standard position.
Then:



If 0 < X < 1 then X = N x 10negative number
If 1 < X < 10 then X = N x 100
If X > 10 then X = N x 10positive number
4. Some examples of number three:



0.00087 becomes 8.7 x 10¯4
9.8 becomes 9.8 x 100 (the 100 is seldom written)
23,000,000 becomes 2.3 x 107
5. Some more examples of number three:



0.000000809 becomes 8.09 x 10¯7
4.56 becomes 4.56 x 100
250,000,000,000 becomes 2.50 x 1011
Note that standard position for the decimal place is always just to the right of the first
non-zero digit in the number. Also, it is the first non-zero digit counting from the left of
the number. Another way to remember standard position is that it will always produce a
number between 1 and 10. For example 45.91 x 10¯7 is not in correct scientific notation.
However, the Chemists wishes to stress that it is a correct number, it is just not in
scientific notation.
As a rule of thumb, you generally need not convert numbers where the absolute value
exponent will be 3 or less. However, exceptions do exist and this is only practice in the
Chemists's classroom. Your instructor may have a different standard for you to obey.
Example #1 - Convert 29,190,000,000 to scientific notation.
The answer will be written assuming four significant figures. However, if you are not
sure of significant figures, don't worry - you'll get to it.
The solving process is simply one of factoring this number, but in a particular way. For
example, 5,838,000,000 times 5 is not the correct way, even though this is a correct
factoring of the original number.
First Explanation
Step 1 - start at the decimal point of the original number and count the number of decimal
places you move, stopping to the right of the first non-zero digit. Remember that's the
first non-zero digit counting from the left.
Step 2 - The number of places you moved (10 in this example) will be the exponent. If
you moved to the left, it's a positive value. If you moved to the right, it's negative.
The answer is 2.919 x 1010.
Second Explanation
Step 1 - Write all the significant digits down with the decimal point just to the right of the
first significant digit. Like this: 2.919. Reminder: be aware that this process should
ALWAYS result in a value between 1 and 10.
Step 2 - Now count how many decimal places you would move from 2.919 to recover the
original number of 29,190,000,000. The answer in this case would be 10 places to the
RIGHT. That is the number 10,000,000,000. Written in exponential notation, it would be
1010.
To emphasize the factoring idea, we would have this:
2.919 x 10,000,000,000 = 29,190,000,000
Step 3 - Write 2.919 times the other number, BUT, write the other number as a power of
10. The number of decimal places you counted gives the power of ten. In this example,
that power would be 10 also. The correct answer to this step is:
2.919 x 1010
Please note the the value of the exponent is positive, because you counted to the RIGHT
in step 2.
It may help to think of scientific notation as simply factoring a number, only you are
following rules which dictate how to write the two factors. The first factor is always
between one and ten, while the second factor is always some power of 10.
Example 2 - Write 0.00000000459 in scientific notation.
Step 1 - Write all the significant digits down with the decimal point just to the right of the
first significant digit. Like this: 4.59. Please be aware that this process should ALWAYS
result in a value between 1 and 10.
Step 2 - Now count how many decimal places you would move from 4.59 to recover the
original number of 0.00000000459. The answer in this case would be 9 places to the
LEFT. That is the number 0.000000001. Be aware that this number in exponential
notation is 10¯9.
To emphasize the factoring idea, we would have this:
4.59 x 0.000000001 = 0.00000000459
Step 3 - Write 4.59 times the other number, BUT, write the other number as a power of
10. The number of decimal places you counted gives the power of ten. In this example,
that power would be 9. The correct answer to this step is:
4.59 x 10¯9
Please note the the value of the exponent is negative, because you counted to the LEFT in
step 2.
Keep in mind two important ideas when converting to scientific notation: how many
decimal places did you move and in what direction. Both of these affect the power of ten.
Also keep in mind that your answer in scientific notation will always equal the original
value. Suprising as it may seem, students in the Chemists classroom make this
elementary mistake.
Now, convert both of these to scientific notation, and then click the value to see the
35,800,000,000,000
The answer is 3.58 x 1013
1. Write all significant figures as a number between 1 and 10. Do this by putting a
decimal point just to the right of the first non-zero digit.
2. Count the number of decimal places needed to get back to the original starting number,
while noting the direction you must move to do that.
3. Write step one times step two, but write step two in exponential notation. The number
of decimal places moved gives the exponent value and the direction moved gives the sign
(left = positive; right = negative).
Note that the exponent indicates HOW MANY places the decimal point was moved and
the sign of the exponent indicates the DIRECTION of the movement.
There were then two graphics showing with arrows the movement of the decimal point.
In this problem, the decimal point was moved 13 places to the LEFT (positive exponent).
0.00000000821
The answer is 8.21 x 10¯9
1. Write all significant figures as a number between 1 and 10. Do this by putting a
decimal point just to the right of the first non-zero digit.
2. Count the number of decimal places needed to get back to the original starting number,
while noting the direction you must move to do that.
3. Write step one times step two, but write step two in exponential notation. The number
of decimal places moved gives the exponent value and the direction moved gives the sign
(left = positive; right = negative).
Note that the exponent indicates HOW MANY places the decimal point was moved and
the sign of the exponent indicates the DIRECTION of the movement.
In this problem, the decimal point was moved 9 places to the RIGHT (negative
exponent).
Suppose the number to be converted looks something like scientific notation, but it really
isn't. For example, look carefully at the example below. Notice that the number 428.5 is
not a number between 1 and 10. Although writing a number in this fashion is perfectly
OK, it is not in standard scientific notation. What would it look like when converted to
standard scientific notation?
Example #3 - Convert 428.5 x 109 to scientific notation.
Step 1 - convert the 428.5 to scientific notation. (The lesson up to this point has been
covering how to do just this step). Answer = 4.285 x 102.
Step 2 - write out the new number. Answer = 4.285 x 102 x 109.
Step 3 - combine the exponents according to the usual rules for exponents. Answer =
4.285 x 1011.
You don't know the rules for exponents.
Exponent Rules
1) When exponents are multiplied, you add them.
103 x 102 = 105
2) When exponents are divided, you subtract them.
103 ÷ 102 = 101
3) When parenthesis are involved, as in the example, you multiply.
(103)2 = 106
Example #4 - convert 208.8 x 10¯11 to scientific notation.
Step 1 - convert the 208.8 to scientific notation. Answer = 2.088 x 102.
Step 2 - write out the new number. Answer = 2.088 x 102 x 10¯11.
Step 3 - combine the exponents according to the usual rules for exponents. Answer =
2.088 x 10¯9.
Now, convert both of these to scientific notation, then click the value to see the answer
and an explanation.
0.000531 x 1014
Convert 0.000531 x 1014 to scientific notation.
Step 1 - convert the 0.000531 to scientific notation. Answer = 5.31 x 10¯4.
Step 2 - write out the new number. Answer = 5.31 x 10¯4 x 1014.
Step 3 - combine the exponents according to the usual rules for exponents. Answer = 5.31
x 1010.
0.00000306 x 10¯17
Convert 0.00000306 x 10¯17 to scientific notation.
Step 1 - convert the 0.00000306 to scientific notation. Answer = 3.06 x 10¯6.
Step 2 - write out the new number. Answer = 3.06 x 10¯6 x 10¯17.
Step 3 - combine the exponents according to the usual rules for exponents. Answer = 3.06
x 10¯23.
1. When converting a number greater than one (the 428.5 and the 208.8 in the previous
examples), the resulting exponent will become more positive (11 is more positive than 9
while -9 is more positive than -11).
2. When converting a number less than one (the 0.000531 and the 0.00000306 in the
previous examples), the resulting exponent will always be more negative (10 is more
negative than 14 and -23 is more negative than -17).
Another way to put it:
If the decimal point is moved to the left, the exponent goes up in value (becomes more
positive).
If the decimal point is moved to the right, the exponent goes down in value (becomes
more negative).
Practice Problems
Convert to scientific notation:
1) 28,000,000
2) 305,000
3) 0.000000463
4) 0.000201
5) 3,010,000
6) 0.000000000000057
7) 20,100
8) 0.00025
9) 65,000,000,000,000,000
10) 8.54 x 1012
11) 2101 x 10¯16
12) 305.1 x 107
13) 0.0000594 x 10¯16
14) 0.00000827 x 1019
15) 386 x 10¯22
16) 2511 x 1012
17) 0.000482 x 10¯12
18) 0.0000321 x 1012
19) 288 x 105
20) 4.05 x 1011
1) 2.8 x 107
2) 3.05 x 105
3) 4.63 x 10¯7
4) 2.01 x 10¯4
5) 3.01 x 106
6) 5.7 x 10¯14
7) 2.01 x 104
8) 2.5 x 10¯4
9) 6.5 x 1016
11) 2.101 x 10¯13
12) 3.051 x 109
13) 5.94 x 10¯21
14) 8.27 x 1013
15) 3.86 x 10¯20
16) 2.511 x 1015
17) 4.82 x 10¯16
18) 3.21 x 107
19) 2.88 x 107
Math with Scientific Notation
Speaking realistically, the problems discussed below can all be done on a calculator.
However, you need to know how to enter values into the calculator, read your calculator
screen, and round off to the proper number of significant figures. Your calculator will not
do these things for you.
All exponents MUST BE THE SAME before you can add and subtract numbers in
scientific notation. The actual addition or subtraction will take place with the numerical
portion, NOT the exponent.
The student might wish to re-read the above two sentences with emphasis on the
emphasized portions.
It might be advisable to point out again - DO NOT, under any circumstances, add the
exponents.
Example #1: 1.00 x 103 + 1.00 x 102
A good rule to follow is to express all numbers in the problem in the highest power of
ten.
Convert 1.00 x 102 to 0.10 x 103, then add:
1.00 x 103
+ 0.10 x 103
= 1.10 x 103
Example #2: The significant figure issue is sometimes obscured when numbers are in
scientific notation. For example, add the following four numbers:
(4.56 x 106) + (2.98 x 105) + (3.65 x 104) + (7.21 x 103)
When the four numbers are written in the highest power, we get:
4.56
0.298
0.0365
+ 0.00721
= 4.90171
x
x
x
x
x
106
106
106
106
106
The answer upon adding must be rounded to 2 significant figures to the right of the
decimal point, thus giving 4.90 x 106 as the correct answer.
Generally speaking, you can simply enter the numbers into the calculator and let the
calculator keep track of where the decimal portion is. However, you must then round off
the answer to the correct number of significant figures.
Lastly, be warned about using the calculator. Students often push buttons without
understanding the math behind what they are doing. Then, when the teacher questions
their work, they say "Well, that's what the calculator said!" As if the calculator is to
blame for the wrong answer. Remember, it is your brain that must be in charge and it is
you that will get the points deducted for poor work, not the calculator.
Practice Problems
1) (4.52 x 10¯5) + (1.24 x 10¯2) + (3.70 x 10¯4) + (1.74 x 10¯3)
2) (2.71 x 106) - (5.00 x 104)
Reminder: you must have the same exponent on each number of the problem.
1) On the calculator, the answer to this problem is 1.4555 x 10¯2. Rounding to the correct
number of significant figures (two past the decimal point) and using the rule for rounding
with five, the correct answer is 1.46 x 10¯2.
2) 2.66 x 106.
Math with Scientific Notation
Multiplication and Division
Speaking realistically, the problems discussed below can all be done on a calculator.
However, you need to know how to enter values into the calculator, read your calculator
screen, and round off to the proper number of significant figures. Your calculator will not
do these things for you.
In addition, it is more than likely your instructor has good estimation skills and can look
at problems set up in scientific notation and come very close mentally to the correct
answer. The Chemists feels this is an excellent skill to have. Sometimes a student gives
an answer in the Chemists's classroom and it does not feel right. This is because the
Chemists does have these estimation skills. You should too.
(3.40 x 1015) x (8.58 x 10¯10)
How do you go about solving the above problem? How many significant figures will be
in the answer? The issue of significant figures is discussed elsewhere, so if you do not
know what they are right now, that's OK. Later on, when you are more well-versed in
significant figures, you might want to review these words. By the way, the answer to the
above problem to the correct number of significant figures is 2.92 x 106. The calculator
answer (2917200) contains too many significant figures.
Lastly, these instructions are aimed at calculators using algebraic logic. Those that use
Reverse Polish Notation, such as most Hewlett-Packard calculators, will not be covered
below.
When multiplying numbers in scientific notation, enter one number into the calculator
and multiply it by the other. The calculator should give you the number in proper
scientific notation. You will need to read your calculator properly as well as round off the
number to the correct number of significant figures. Here is a sample problem:
(3.05 x 106) x (4.55 x 10¯10)
You should consult your calculator manual for information on how to enter numbers in
scientific notation into the calculator. The right way involves the use of a key usually
marked "EXP" or "EE." A usual wrong way involves using the times key, where the
student presses times then 10 then presses the "EXP" key.
Typically, this is announced by the student confidently saying, "Uh, <substitute name of
teacher here>, your value is off by a factor of 10, the correct exponent should be 7, not
6." Depending on your teacher's mood and personality, the response will range from a
nice explanation of why you are wrong to an insulting one. Double- and triple- check
your work before announcing confidently to the teacher that they are wrong.
The calculator gives 1.3878 x 10¯3 for the above problem. Rounded to three significant
figures, the correct answer is 1.39 x 10¯3. If you were to write all the displayed digits
When you are dividing using the calculator, there is an additional factor to be aware of.
Division is not commutative like multiplication, so the order in which numbers are
entered into the calculator for division is important. Also, it is important to realize that
while there are two phrases which can be used for problems, "divided by" and "divided
into," the calculator uses only the first.
If the phrase is "divided by" as in this problem: 2.4 x 10¯4 divided by 3.4 x 105, then you
enter the numbers into the calculator in the order written. However, if this is the problem:
2.4 x 10¯4 divided into 3.4 x 105, then you reverse the written order when entering the
numbers into the calculator.
Sometimes the calculator will give the answer in the usual way (e.g., 0.00084). Often, a
calculator will have a way to convert between the usual way (technically called floating
point notation) and scientific notation. Consult the manual that came with your calculator
for more information. If it does not have this capacity, you will have to do it manually by
counting decimal places.
The biggest error Chemists students make is in reading the calculator screen and then
writing the answer down on paper. No matter how hard he tries, it seems the Chemists
coach cannot eradicate this error. For example, let us suppose the answer to some
problem is 2.35 x 10¯5. Here is how many calculators represent this number on their
screen:
2.35¯5
What happens is that the student writes what he or she sees on the calculator screen and
never thinks about what is being written. The problem is that 2.35¯5 is NOT the same
numerical value as 2.35 x 10¯5. What has happened is that the student is ignoring the
context the numbers are being shown in.
2.35¯5 is simply the calculator's shorthand way to show 2.35 x 10¯5. In the context of the
calculator screen, 2.35¯5 is understood as shorthand for scientific notation. On a piece of
paper, 2.35¯5 is something completely different.
The calculator screen is not sacred.
This is how the mental estimation referred to above is done.
Multiply the decimal portions and add the exponential portions. (Remember that adding
exponents is how to multiply them.) Here is a sample problem:
(3.05 x 106) x (4.55 x 10¯10)
Here is the rearranged problem:
(3.05 x 4.55) x (106 x 10¯10)
This can be done since multiplication is commutative.
The exponent is easy, since 6 plus negative 10 is negative 4.
3.05 x 4.55 is easy too. You know it will be a bit bigger than 12, so you estimate 13.
You now have 13 x 10¯4.
So the teacher says out loud, "Well it's about 1.3 or 1.4 times ten to the negative three."
And you, the Chemists kid, are astonished. It's like magic! Nah, just a good sense of
numbers and the rules which govern them.
Suppose it was division, rather than multiplication. Using the previous sample problem
numbers set up as division:
(3.05 x 106) ÷ (4.55 x 10¯10)
Here is the rearranged problem:
(3.05 ÷ 4.55) x (106 ÷ 10¯10)
The exponent is easy, since 6 minus negative 10 is positive 16. Remember that to divide
exponents, you subtract them.
3.05 ÷ 4.55 is easy too. You know it will be a bit less than 0.75, so you estimate 0.7.
You now have 0.7 x 1016.
So the teacher says out loud, "Well it's about 7 times ten to the fifteen."
And once again you, the Chemists kid, are astonished. It's like magic! Nah, just a good
sense of numbers and the rules which govern them.
Practice Problems
1) (2.68 x 10¯5) x (4.40 x 10¯8)
2) (2.95 x 107) ÷ (6.28 x 1015)
3) (8.41 x 106) x (5.02 x 1012)
4) (9.21 x 10¯4) ÷ (7.60 x 105)
Reminders
1) Double check each calculator entry for correctness. Once you begin entering the new
number, the old one is removed from the display. Often students seem to automatically
assume the answer on the display is correct.
2) Try for reasonable approximations mentally. For example, in (3) above 8.41 x 5.02 is
around 40 and 106 x 1012 is 1018. So the approximate answer is around 40 x 1018 or 4 x
1019. Try and predict (3) as a division, not multiplication.
3) Remember, if you struggle with entering numbers in scientific notation into your
calculator, check your calculator manual or get some live help.
1) (2.68 x 10¯5) x (4.40 x 10¯8)
The calculator display gives 1.1792 x 10¯12. Rounded off to three significant figures
gives 1.18 x 10¯12 as the answer.
2) (2.95 x 107) ÷ (6.28 x 1015)
The calculator display gives 4.6975 x 10¯9. Rounded off to three significant figures gives
4.70 x 10¯9 as the answer.
3) (8.41 x 106) x (5.02 x 1012)
The calculator display gives 4.2218 x 1019. Rounded off to three significant figures gives
4.22 x 1019 as the answer.
When done as a division, the answer to this problem is 1.68 x 10¯6.
4) (9.21 x 10¯4) ÷ (7.60 x 105)
The calculator display gives 1.2118 x 10¯9. Rounded off to three significant figures gives
1.21 x 10¯9 as the answer.
Significant Figure Rules
There are three rules on determining how many significant figures are in a number:
1. Non-zero digits are always significant.
2. Any zeros between two significant digits are significant.
3. A final zero or trailing zeros in the decimal portion ONLY are significant.
Focus on these rules and learn them well. They will be used extensively throughout the
remainder of this course. You would be well advised to do as many problems as needed
to nail the concept of significant figures down tight and then do some more, just to be
sure.
Please remember that, in science, all numbers are based upon measurements (except for a
very few that are defined). Since all measurements are uncertain, we must only use those
numbers that are meaningful. A common ruler cannot measure something to be
22.4072643 cm long. Not all of the digits have meaning (significance) and, therefore,
should not be written down. In science, only the numbers that have significance (derived
from measurement) are written.
If you're not convinced significant figures are important, you may want to read this
Significant Figure Fable.
Rule 1: Non-zero digits are always significant.
Hopefully, this rule seems rather obvious. If you measure something and the device you
use (ruler, thermometer, triple-beam balance, etc.) returns a number to you, then you have
made a measurement decision and that ACT of measuring gives significance to that
particular numeral (or digit) in the overall value you obtain.
Hence a number like 26.38 would have four significant figures and 7.94 would have
three. The problem comes with numbers like 0.00980 or 28.09.
Rule 2: Any zeros between two significant digits are significant.
Suppose you had a number like 406. By the first rule, the 4 and the 6 are significant.
However, to make a measurement decision on the 4 (in the hundred's place) and the 6 (in
the unit's place), you HAD to have made a decision on the ten's place. The measurement
scale for this number would have hundreds and tens marked with an estimation made in
the unit's place. Like this:
Rule 3: A final zero or trailing zeros in the decimal portion ONLY are significant.
This rule causes the most difficulty with Chemists students. Here are two examples of
this rule with the zeros this rule affects in boldface:
0.00500
0.03040
Here are two more examples where the significant zeros are in boldface:
2.30 x 10¯5
4.500 x 1012
What Zeros are Not Discussed Above
Zero Type #1: Space holding zeros on numbers less than one.
Here are the first two numbers from just above with the digits that are NOT significant in
boldface:
0.00500
0.03040
These zeros serve only as space holders. They are there to put the decimal point in its
correct location. They DO NOT involve measurement decisions. Upon writing the
numbers in scientific notation (5.00 x 10¯3 and 3.040 x 10¯2), the non-significant zeros
disappear.
Zero Type #2: the zero to the left of the decimal point on numbers less than one.
When a number like 0.00500 is written, the very first zero (to the left of the decimal
point) is put there by convention. Its sole function is to communicate unambiguously that
the decimal point is a deciaml point. If the number were written like this, .00500, there is
a possibility that the decimal point might be mistaken for a period. Many students omit
that zero. They should not.
Zero Type #3: trailing zeros in a whole number.
200 is considered to have only ONE significant figure while 25,000 has two.
This is based on the way each number is written. When whole number are written as
above, the zeros, BY DEFINITION, did not require a measurement decision, thus they
are not significant.
However, it is entirely possible that 200 really does have two or three significnt figures.
If it does, it will be written in a different manner than 200.
Typically, scientific notation is used for this purpose. If 200 has two significant figures,
then 2.0 x 102 is used. If it has three, then 2.00 x 102 is used. If it had four, then 200.0 is
sufficient. See rule #2 above.
How will you know how many significant figures are in a number like 200? In a problem
like below, divorced of all scientific context, you will be told. If you were doing an
experiment, the context of the experiment and its measuring devices would tell you how
many significant figures to report to people who read the report of your work.
Zero Type #4: leading zeros in a whole number.
00250 has two significant figures. 005.00 x 10¯4 has three.
Exact Numbers
Exact numbers, such as the number of people in a room, have an infinite number of
significant figures. Exact numbers are counting up how many of something are present,
they are not measurements made with instruments. Another example of this are defined
numbers, such as 1 foot = 12 inches. There are exactly 12 inches in one foot. Therefore, if
a number is exact, it DOES NOT affect the accuracy of a calculation nor the precision of
the expression. Some more examples:
There are 100 years in a century.
2 molecules of hydrogen react with 1 molecule of oxygen to form 2 molecules of water.
There are 500 sheets of paper in one ream.
Interestingly, the speed of light is now a defined quantity. By definition, the value is
299,792,458 meters per second.
Practice Problems
Identify the number of significant figures:
1) 3.0800
2) 0.00418
3) 7.09 x 10¯5
4) 91,600
5) 0.003005
6) 3.200 x 109
7) 250
8) 780,000,000
9) 0.0101
10) 0.00800
1) 3.0800 - five significant figures. All the rules are illustrated by this problem. Rule one
- the 3 and the 8. Rule Two - the zero between the 3 and 8. Rule three - the two trailing
zeros after the 8.
2) 0.00418 - three significant figures: the 4, the 1, and the 8. This is a typical type of
problem where the student errs by giving five significant figures as the answer.
3) 7.09 x 10¯5 - three significant figures. When a number is written in scientific notation,
only significant figures are placed into the numerical portion. If this number were taken
out of scientific notation, it would be 0.0000709.
4) 91,600 - three significant figures. The last two zeros are not considered to be
significant (at least normally). Suppose you had information that showed the zero in the
tens place to be significant. How would you show it to be different from the zero in the
ones place, which is not significant? The answer is scientific notation. Here is how it
would be written: 9.160 x 104. This CLEARLY indicates the presence of four significant
figures.
5) 0.003005- four significant figures. No matter how many zeros there are between two
significant figures, all the zeros are to be considered significant. A number like
70.000001 would have 8 significant figures.
6) 3.200 x 109 - four significant figures. Notice the use of scientific notation to indicate
that there are two zeros which should be significant. If this number were to be written
without scientific notation (3,200,000,000) the significance of those two zeros would be
lost and you would - wrongly - say that there were only two significant figures.
7) 2
8) 2
9) 3
10) 3
Rules for Rounding Off
Now that "everyone" has a calculator that will give a result to six or eight (or more)
figures, it is important that we know how to round the answer off correctly. The typical
rule taught is that you round up with five or more and round down with four or less.
THIS RULE IS WRONG!
act!
The problem lies in rounding "up" (increasing) the number that is followed by a 5. For
example, numbers like 3.65 or 3.75, where you are to round off to the nearest tenth.
OK, let's see if I can explain this. When you round off, you change the value of the
number, except if you round off a zero. Following the old rules, you can round a number
down in value four times (rounding with one, two, three, four) compared to rounding it
upwards five times (five, six, seven, eight, nine). Remember that "rounding off" a zero
does not change the value of the number being rounded off.
Suppose you had a very large sample of numbers to round off. On average you would be
changing values in the sample downwards 4/9ths of the time, compared to changing
values in the sample upward 5/9ths of the time.
This means the average of the values AFTER rounding off would be greater than the
average of the values BEFORE rounding.
This is not acceptable.
We can correct for this problem by rounding "off" (keeping the number the same) in fifty
percent of the roundings-even numbers followed by a 5. Then, on average, the roundings
"off" will cancel out the roundings "up."
The following rules dictate the manner in which numbers are to be rounded to the number
of figures indicated. The first two rules are more-or-less the old ones. Rule three is the
change in the old way.
When rounding, examine the figure following (i.e., to the right of) the figure that is to be
last. This figure you are examining is the first figure to be dropped.
1. If it is less than 5, drop it and all the figures to the right of it.
2. If it is more than 5, increase by 1 the number to be rounded, that is, the preceeding
figure.
3. If it is 5, round the number so that it will be even. Keep in mind that zero is
considered to be even when rounding off.
Example #1 - Suppose you wish to round 62.5347 to four significant figures. Look at the
fifth figure. It is a 4, a number less than 5. Therefore, you will simply drop every figure
after the fourth, and the original number rounds off to 62.53.
Example #2 - Round 3.78721 to three significant figures. Look at the fourth figure. It is 7,
a number greater than 5, so you round the original number up to 3.79.
Example #3 - Round 726.835 to five significant figures. Look at the sixth figure. It is a 5,
so now you must look at the fifth figure also. That is a 3, which is an odd number, so you
round the original number up to 726.84.
Example #4 - Round 24.8514 to three significant figures. Look at the fourth figure. It is a
5, so now you must also look at the third figure. It is 8, an even number, so you simply
drop the 5 and the figures that follow it. The original number becomes 24.8.
When the value you intend to round off is a five, you MUST look at the previous value
ALSO. If it is even, you round down. If it is odd, you round up. A common question is
"Is zero considered odd or even?" The answer is even.
Here are some more examples of the "five rule." Round off at the five.
3.075
3.85
22.73541
0.00565
2.0495
This last one is tricky (at least for high schoolers being exposed to this stuff for the first
time!). The nine rounds off to a ten (not a zero), so the correct answer is 2.050, NOT
2.05.
Would your teacher be so mean as to include problems like this one on a test? In the
Chemists classroom, the sufferers (oops, I mean students) have learned to shout "YES" in
unison to such easy questions.
Lastly, before we get to the problems. Students, when they learn this rule, like to apply it
across the board. For example, in 2.0495, let's say we want to round off to the nearest
0.01. Many times, a student will answer 2.04. When asked to explain, the rule concerning
five will be cited. However, the important number in this problem is the nine, so the rule
is to round up and the correct answer is 2.05.
Practice Problems
Round the following numbers as indicated.
To four figures:
To the nearest 0.1:
the nearest whole number:
1) 2.16347 x 105
13) 3.64
56.912
2) 4.000574 x 106
14) 4.55
3.4125
3) 3.682417
15) 7.250
251.7817
4) 7.2518
16) 0.0865
112.511
5) 375.6523
17) 0.5182
63.541
6) 21.860051
18) 2.473
7.555
To two figures:
To one decimal place:
Round off the farthest right digit
7) 3.512
19) 54.7421
2.473
8) 25.631
20) 100.0925
5.396
9) 40.523
21) 1.3511
8.235
10) 2.751 x 108
22) 79.2588
3.05
11) 3.9814 x 105
23) 0.9114
8.25
12) 22.494
24) 0.2056
8.65
To nearest 0.01:
To
25) 6.675
37)
26) 0.4203
38)
27) 0.03062
39)
28) 4.500
40)
29) 2.473
41)
30) 7.555
42)
To the nearest 0.001:
31) 5.687524
43)
32) 39.861214
44)
33) 104.97055
45)
34) 41.86632
46)
35) 0.03765
47)
36) 0.0045
48)
To nearest 0.01:
To
25) 6.68
37)
26) 0.42
38)
27) 0.03
39)
28) 4.50
40)
29) 2.47
41)
30) 7.56
42)
Round the following numbers as indicated.
To four figures:
To the nearest 0.1:
the nearest whole number:
1) 2.163 x 105
13) 3.6
57
2) 4.000 x 106
14) 4.6
3
3) 3.682
15) 7.2
252
4) 7.252
16) 0.1
112
5) 375.6
17) 0.5
64
6) 21.86
18) 2.5
8
To two figures:
To one decimal place:
Round off the farthest right digit
7) 3.5
19) 54.7
2.47
8) 26
20) 100.1
5.40
To the nearest 0.001:
31) 5.688
43)
32) 39.861
44)
9) 40
8.24
10) 2.8 x 108
3.0
11) 4.0 x 105
8.2
12) 22
8.6
21) 1.4
33) 104.970
45)
22) 79.2
34) 41.866
46)
23) 0.9
35) 0.038
47)
24) 0.2
36) 0.004
48)
Math With Significant Figures
In mathematical operations involving significant figures, the answer is reported in such a
way that it reflects the reliability of the least precise operation. Let's state that another
way: a chain is no stronger than its weakest link. An answer is no more precise that the
least precise number used to get the answer. Let's do it one more time: imagine a team
race where you and your team must finish together. Who dictates the speed of the team?
Of course, the slowest member of the team. Your answer cannot be MORE precise than
the least precise measurement.
For addition and subtraction, look at the decimal portion (i.e., to the right of the decimal
point) of the numbers ONLY. Here is what to do:
1) Count the number of significant figures in the decimal portion of each number in the
problem. (The digits to the left of the decimal place are not used to determine the number
of decimal places in the final answer.)
2) Add or subtract in the normal fashion.
3) Round the answer to the LEAST number of places in the decimal portion of any
number in the problem.
WARNING: the rules for add/subtract are different from multiply/divide. A very
common student error is to swap the two sets of rules. Another common error is to use
just one rule for both types of operations.
Practice Problems
1) 3.461728 + 14.91 + 0.980001 + 5.2631
2) 23.1 + 4.77 + 125.39 + 3.581
3) 22.101 - 0.9307
4) 0.04216 - 0.0004134
5) 564,321 - 264,321
Remember to check the decimal portions of each number in the problem. For example, in
the second problem, the 23.1 will dictate the significant figures in the decimal portion of
the answer. In other words, NO MATTER how many significant figures there are in the
whole number part (to the left of the decimal place), the answer to problem number 2 will
extend only into the tenth place.
1) 3.461728 + 14.91 + 0.980001 + 5.2631
2) 23.1 + 4.77 + 125.39 + 3.581
In each of these two problems, examine the decimal portion only. Find the number with
the LEAST number of digits in the decimal portion. In problem 1 it is the 14.91 and in
problem 2 it is 23.1.
That means problem 1 will have its answer rounded to the 0.01 place and problem 2 will
have its answer rounded to the 0.1 place. The correct answers are 24.61 and 156.8.
3) 22.101 - 0.9307
The answer is 21.170. The first value in the problem, with three significant places to the
right of the decimal point, dictates how many significant places to the right of the decimal
4) 0.04216 - 0.0004134
5) 564,321 - 264,321
This problem is somewhat artifical. The correct answer is 300,000, BUT all of the
significant figures are retained. The most correct way to write the answer would be
3.00000 x 105. Sorry 'bout that!
Math With Significant Figures
Multiplication and Division
In mathematical operations involving significant figures, the answer is reported in such a
way that it reflects the reliability of the least precise operation. Let's state that another
way: a chain is no stronger than its weakest link. An answer is no more precise that the
least precise number used to get the answer. Let's do it one more time: imagine a team
race where you and your team must finish together. Who dictates the speed of the team?
Of course, the slowest member of the team. Your answer cannot be MORE precise than
the least precise measurement.
The following rule applies for multiplication and division:
The LEAST number of significant figures in any number of the problem determines the
number of significant figures in the answer.
This means you MUST know how to recognize significant figures in order to use this
rule.
Example #1: 2.5 x 3.42.
The answer to this problem would be 8.6 (which was rounded from the calculator reading
of 8.55). Why?
2.5 has two significant figures while 3.42 has three. Two significant figures is less precise
than three, so the answer has two significant figures.
Example #2: How many significant figures will the answer to 3.10 x 4.520 have?
You may have said two. This is too few. A common error is for the student to look at a
number like 3.10 and think it has two significant figures. The zero in the hundedth's place
is not recognized as significant when, in fact, it is. 3.10 has three significant figures.
Three is the correct answer. 14.0 has three significant figures. Note that the zero in the
tenth's place is considered significant. All trailing zeros in the decimal portion are
considered significant.
Another common error is for the student to think that 14 and 14.0 are the same thing.
THEY ARE NOT. 14.0 is ten times more precise than 14. The two numbers have the
same value, but they convey different meanings about how trustworthy they are.
Four is also an incorrect answer given by some Chemists students. It is too many
significant figures. One possible reason for this answer lies in the number 4.520. This
number has four significant figures while 3.10 has three. Somehow, the student (YOU!)
maybe got the idea that it is the GREATEST number of significant figures in the problem
that dictates the answer. It is the LEAST.
Sometimes student will answer this with five. Most likely you responded with this answer
because it says 14.012 on your calculator. This answer would have been correct in your
math class because mathematics does not have the significant figure concept.
Example #3: 2.33 x 6.085 x 2.1. How many significant figures in the answer?
Which number decides this?
Why?
It has the least number of significant figures in the problem. It is, therefore, the least
precise measurement.
Example #4: (4.52 x 10¯4) ÷ (3.980 x 10¯6).
How many significant figures in the answer?
Which number decides this?
Answer - the 4.52 x 10¯4.
Why?
It has the least number of significant figures in the problem. It is, therefore, the least
precise measurement. Notice it is the 4.52 portion that plays the role of determining
significant figures; the exponential portion plays no role.
Practice Problems
1) (3.4617 x 107) ÷ (5.61 x 10¯4)
2) [(9.714 x 105) (2.1482 x 10¯9)] ÷ [(4.1212) (3.7792 x 10¯5)]. Watch your order of
operations on this problem.
3) (4.7620 x 10¯15) ÷ [(3.8529 x 1012) (2.813 x 10¯7) (9.50)]
4) [(561.0) (34,908) (23.0)] ÷ [(21.888) (75.2) (120.00)]
A Brief Aside
There might come an occasion in chemistry when you are not exactly sure how many
significant figures are called for. Suppose the textbook mentions 100 mL. You look at
this and see only one significant figure.
However, an experienced chemist would know that 100 mL can be easily measured to 3
or 4 significant figures. Why then doesn't the textbook (or the professor) write 100.0 (for
4 sig figs) or 1.00 x 102 (for 3 sig figs)? The textbook writer or the professor might be
assuming that all in his or her audience understands these matters and so it is no big deal
to simply write 100. Or they are lazy.
So, a brief word of advice. If you haven't a clue as to how many significant figures to use,
try using three or four. These are reasonable numbers of significant figures for most
chemical activities.
Also, look out for the instructor who ignores significant figures in lecture, then makes a
big deal of it on a test. Forewarned is forearmed!
Calculators and Significant Figures
It seems that most high school students treat the numbers on their calculator screen as
Holy Writ. NO, IT IS NOT. Most of the string of numbers on the screen does not belong
These two problems are intended to illustrate the difficulty surrounding calculators and
significant figures.
1) Calculate the area in cm2 of these two samples:
length (cm)
width (cm)
area (cm2)
a.
b.
27.81
27.93
20.49
20.36
569.8269
568.6547
2) Calculate the volume in cm3 of the samples from the given thickness:
a.
b.
thickness (cm) volume (cm3)
0.710
404.57709
0.690
392.37181
Solution - ROUND OFF!
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