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Math Review
1
Scientific Notation
Any quantity can be expressed using a power of ten. As you move the decimal point, you
multiply by 10 as many times as necessary to make the numbers equal. Consider the
following examples:
325
325
325
=
=
=
= 3.25 x 101
= 3.25 x 102
= 3.25 x 103
32.5 x 10
3.25 x 10 x 10
0.325 x 10 x 10 x 10
Because 100 = 1 we can also express 325 as 325 x 100.
A number in scientific notation has two parts. The number in front of the “x 10” is called
the coefficient. The power to which 10 is raised is called the exponent.
3.25 x 103
Exponent
Coefficient
The coefficient must have one and only one digit in front of the decimal point. The
exponent will always be a whole number in chemistry, although the exponent can be
fractional number in other disciplines.
There are three rules for using scientific notation:
Rule 1: To express a number in scientific notation, you move the decimal point to
the position such that there is one nonzero digit to the left of the decimal point.
Rule 2: If the decimal point is moved to the left, the exponent is positive.
Rule 3: If the decimal point is moved to the right, the exponent is negative.
A mnemonic that my help you remember how to keep the signs straight is “Registered
Nurses Love Patients”, or RN, LP, which stands for right --- negative, left --- positive. If
you move the decimal point to the right, the exponent is negative. If you move the decimal
point to the left, the exponent is positive.
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Scientific Notation Example 1:
Express 0.0003821 in scientific notation.
Solution:
0.0003821 = 3.821 x 10-4
The decimal point was moved to the right four spaces. If the decimal point moves
to the right, the exponent is negative. Since it moved four spaces, the exponent is a
–4. Notice that the decimal point was moved until one and only one nonzero digit
was in front of the decimal point.
Scientific Notation Example 2:
Express 56,873 in scientific notation.
Solution:
5.6873 x 104
The decimal point was moved to the left four spaces. If the decimal point moves to
the left, the exponent is positive. Since it moved four spaces, the exponent is a +4.
Notice that the decimal point was moved until one and only one nonzero digit was in
front of the decimal point.
Sometimes it is necessary to change a number from one power of ten to another power of
ten. When changing to a new power of ten, we must consider the sizes of both the
coefficient and the power of ten. In the number 3.2 x 105, the coefficient is 3.2 and the
power of ten is 5. If we wish to express this as another power of 10 that is equal in value
to 3.2 x 105, we must increase the power of ten if the coefficient decreases, and decrease
the power of ten if the coefficient increases. This is shown in the illustration below.
Scientific Notation Example 3:
Fill in the blank: 6.79 x 104 = ________ x 106
Solution:
The exponent has gotten larger. This means that we have multiplied the coefficient
by 10 two more times. Because the exponent has gotten larger by 2 places, the
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coefficient must get smaller by two places. We must change the 6.79 to 0.0679,
which is the value that goes in the blank.
6.79 x 104 = 0.0679 x 106
Although you will probably use your calculator to perform arithmetic operations on
numbers written in scientific notation, you should know the rules that govern these
processes. The rules are presented below.
Addition and Subtraction
1. Express the numbers as the same power of 10.
2. Perform the mathematical operation on the coefficients (add or subtract).
3. Bring down the power of 10 unchanged.
4. Express the answer in proper scientific notation.
Scientific Notation Example 4:
Subtract 4.72 x 102 from 5.32 x 103
Solution:
Make the exponents equal before we subtract the numbers. We can make both of
the exponents 102 or we can make them both 103. The result will be the same no
matter which common exponent we choose. We’ll make them both 102. Therefore
5.32 x 103 = 53.2 x 1024. Now we can subtract the two numbers. Since the final
answer has more than one nonzero digit in the coefficient, we need to rewrite the
number in proper scientific notation.
53.20 x 102
- 4.72 x 102
48.48 x 102 or 4.848 x 103
Multiplication and Division
To multiply two numbers expressed in scientific notation, multiply the coefficients and add
the exponents on the powers of 10.
To divide two numbers expressed in scientific notation, divide the coefficients and subtract
the exponents on the powers of 10.
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Scientific Notation Example 5:
What is the product of 5.1 x 10-4 and 3.2 x 10-2?
Solution:
Multiply 5.1 by 3.2 to get 16. Add the exponents to get –6. The answer is 16 x 10-6.
Change this number to proper scientific notation. The answer is 1.6 x 10-5.
Hints for Using Scientific Notation:
Is it necessary to use scientific notation?
Not every calculation requires scientific notation, but it is a very convenient system for
handling the very large and very small numbers commonly encountered in chemistry.
Scientific notation shows up very frequently in chemistry, so mastering it now will pay off in
the future.
Can I use a calculator to solve problems with number in scientific notation?
Yes. You will need to find out how to input scientific notation numbers into your calculator.
Check your calculator manual for directions. The most common mistake in this chapter
is entering a number in scientific notation incorrectly into a calculator!
5
Rounding
It is often necessary to round numbers after performing calculations. This is usually
necessary when the calculated answer has more significant figures than are allowed. The
rules for rounding numbers are presented below.
1. If the last digit to be kept is followed by a number less than 5, round by dropping
all digits to the right of it.
2. If the last digit to be kept is followed by a number equal to or greater than five,
round up by adding 1 to the last digit to be kept and drop the rest.
Rounding Example 1:
Round the following number to the indicated number of significant digits.
54.674873
0.8765
5.875
rounded to four digits
rounded to two digits
rounded to three digits
Solution:
The last digit to be kept in 54.674873 is a seven. The seven is followed by a
number less than five, so drop all of the digits after the seven in blue. The answer
is 54.67.
The last digit to be kept in 0.8765 is a seven. The seven is followed by a number
greater than or equal to 5, so round up. The answer is 0.88.
The last digit to be kept in 5.875 is a seven. The seven is followed by a number
greater than or equal to 5, so round up. The answer is 5.88.
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Conversion Factors
Conversion factors are ratios of one object to another object. A ratio is a way of comparing
two quantities. The quantities can be compared in three different ways: a to b, a:b, or a/b.
At the local grocery store, a case of soda contains 24 cans. We can express the ratio in
three forms:
(a to b) 24 cans of soda to each case of soda
(a:b)
24 cans of soda: 1 case of soda
(a/b)
24 cans of soda/1 case of soda
This chapter uses conversion factors frequently to solve problems. Conversion factors are
ratios written in the fraction form (a/b).
Some of the conversion factors you will see in this chapter include metric-to-metric
conversion factors such as:
1 gram of silver is 1000 mg of silver
1 liter of water is 1 x 106 µL of water
The most common way that chemists represent ratios like these is in a fraction form. The
conversion factors would look like this in the fraction form:
1 gram silver
1000 mg silver
1 liter water
1 x 106 µL water
Each of these conversion factors can be written in the inverse form. The first ratio we
examined involved cans of soda. We expressed that ratio as 24 cans of soda in 1 case of
soda. The inverse form of this ratio is 1 case of soda contains 24 cans of soda. The key
idea here is that either the original ratio or its inverse form is true. It is true that one
case of soda contains 24 cans and it is true that there are 24 cans of soda in one case.
Conversion factors in chemistry can be written as shown above or in the inverse form.
Regardless of which form is used, the comparison the conversion factor makes is true.
For example, one gram of silver weigh 1000 mg and 1000 mg of silver weighs one gram.
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Here are the conversion factors written in the inverse form:
1000 mg silver
1 gram silver
1 x 106 µL water
1 liter water
The obvious question at this point is: Which form of the conversion factor should you use
in a problem? You use the form that allows the units to cancel. Units cancel when the
unit that appears on the top also appears on the bottom of a fraction somewhere in the
solution. This style of problem solving, where you arrange the conversion factors so that
the units cancel, is called unit analysis.
Conversion Factor Example 1
Which is the correct solution for the question: How many cases of soda can be made from
1200 cans of soda?
1200 cans of soda x
24 cans of soda = 28800 cans of soda2/case of soda
1 case of soda
1200 cans of soda x
1 case of soda = 50 cases of soda
24 cans of soda
Solution:
In which solution do the units cancel? In the second solution, the unit “cans of
soda” cancels. The correct solution will use conversion factors in a way that units
will cancel. Blue lines are drawn through units that cancel.
Conversion Factor Example 2
How many milligrams of silver are in 2.76 grams of silver
Solution
This question uses the metric-to-metric conversion factor: 1 gram = 1000 mg. Set
up the solution so that the units cancel. Blue lines are drawn through units that
cancel.
2.76 grams silver x 1000 mg silver = 2760 mg silver
1 gram silver
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Conversion Factor Example 3
If 8.35 x 109 µL of water was collected in a flask, how many liters of water was collected?
Solution
You don’t need to concern yourself with whether the conversion factor is inverted or
not. Just make sure that the units cancel. Blue lines are drawn through units that
cancel.
8.35 x 109 µL water
x
1 liter water = 8.35 x 103 liters water
1 x 106 µL
Hints for using conversion factors:
How do you know what quantity to start with when solving problems?
Start with the quantity given in the problem and not with the conversion factor. In general,
the first quantity in your unit analysis will not have a denominator. In the last example, we
started with the number of microliters of water:
8.35 x 109 µL water
x
1 liter water = 8.35 x 103 liters water
1 x 106 µL
Notice that the first quantity is a whole number and not a fraction. If you start with the
conversion factor, it is much harder to decide whether to use the original or inverse form of
the conversion factor.
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Should I enter all of the numerator numbers into the calculator before I enter the
denominator numbers, or should I enter each conversion factor into the calculator
before I move on to the next conversion factor?
Either style of entering the numerical quantities will work. You should make a decision to
always use one style of entering numbers into a calculator. This helps you avoid skipping
a number when you are calculating an answer.
In the example calculation below, you could enter the numbers in either of these orders
and obtain the same answer.
16 x 2.54 x 50 = 75.2
3
9
Enter the numbers into your calculator as:
16 x 2.54 ÷ 3 x 50 ÷ 9 =
or
16 x 2.54 x 50 ÷ 3 ÷ 9 =
You don’t need to press the ENTER key after each mathematical.
10
Averages and Weighted Averages
The average of a set of numbers is simply the sum of the values divided by the number of
values in the set.
Average value =
sum of all values___
# of values being summed
For example, the average of 5, 10, and 12 is equal to 9.
5 + 10 + 12
3
= 9
Sometimes, however, the values in a set of numbers do not contribute equally to the set.
For example, in determining your grade point average (GPA) in school, if you earned an A
in one three credit course, and a C in another three credit course, your average will be
exactly a B. However, if you earned an A in a two credit course, and a C in a four credit
course, your average will be lower than a B. This is because the A will count half as much
toward your GPA as the C, because the course in which you earned the A is worth half as
many credit hours.
To determine the average in this case, we must use the fact that the two values contribute
differently to the average, and we say that we calculate the weighted average.
Weighted average = (factor for value 1)(value 1) + (factor for value 2)(value 2) + …
Sum of all values
Averages and Weighted Averages Example 1:
Determine the average of two course grades when the two courses are equal in credit
hours. The quality points assigned to each grade are as follows:
A = 4.0 points
B = 3.0 points
C = 2.0 points
D = 1.0 point
The grade in the first class was an A and the grade in the second class was a C.
Solution:
Average value =
sum of all values___
# of values being summed
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Average of one A and one C = 4.0 + 2.0 = 6.0 = 3.0
2
2
This result is exactly what we expect. An A averaged with a C will yield a B
average.
Averages and Weighted Averages Example 2:
Determine the average of two course grades when the two courses are NOT equal in
credit hours. The quality points assigned to each grade are as follows:
A = 4.0 points
B = 3.0 points
C = 2.0 points
D = 1.0 point
The grade in a two credit class was an A and the grade in a three credit class was a C.
Solution:
Weighted average = (factor for value 1)(value 1) + (factor for value 2)(value 2)
Sum of the weighting factors
Weighted average = (2)(4.0) + (3)(2.0)
2+ 4
=
14
5
= 2.8
You will encounter calculating weighed averages when you calculate the average atomic
weights of elements.
Averages and Weighted Averages Example 3:
Chlorine has two naturally occurring isotopes. Isotope A has a mass of 34.97, and
represents 75.8% of naturally occurring chlorine. Isotope B has a mass of 36.97, and
represents 24.2% of naturally occurring chlorine. What is the average atomic mass of
chlorine?
Solution:
The weighting factors are the percentages, and the values are the masses.
Weighted average = (factor for value 1)(value 1) + (factor for value 2)(value 2)
Sum of the weighting factors
Average atomic mass = (75.8)(34.97) + (24.2)(36.97) = 3545 = 35.45
75.8 +
24.2
100
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Hints for Using Averages and Weighted Averages:
How can I tell if I should use a weighted average in a calculation?
If the question has values that contribute equally to the average, you can use a simple
average calculation. However, if the values contribute unevenly to the average, you must
include the weighting factor and do a weighted average calculation. It is fairly easy to tell
which calculation to use since a weighted average question will include more numbers and
information than a simple average question.
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Geometric Figures
Geometric figures and the angles associated with them form an important foundation for
discussing the shapes of molecules and the bond angles in molecules. You should
become comfortable with the geometric names and angles used in this chapter.
There are 360° as you go around a circle.
The angles change as you cut the circle into different numbers of pieces. The angle of a
straight line is 180°.
Molecules whose atoms have a 180° bond angle are said to have linear geometry.
The bond between the small blue atom and the large gray atom is in the same line as the
bond between the two large gray atoms. The angle between these two bonds is 180°. If
you had to describe the orientation of the blue atom to the two gray atoms, you would say
that they are all on the same line or that they have a linear geometry.
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The angles at the intersection of three lines equally spaced around a circle are
120°.
Molecules whose bonds are 120° apart have trigonal planar geometry.
The angle between the single bond and the double bond is 120°. The blue, black, and red
atoms are all in the same plane. If you traced a line from the blue atom to the red atom to
the other blue atom and then back to the first blue atom, you would have a triangular
shape. Thus, we describe the geometry as trigonal planar.
The angles at the intersection of two lines perpendicular and crossing each other are 90°.
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There are no 90° bond angles in common biological molecules. However, there are
many atoms that have four bonds. What angle do the four bonds make? The make 109°
bond angles. Look at the molecules below. Are the bond angles 90° or 109°?
If you look carefully, you will see that the bond angle is more than 90°. You can also see
that the four blue atoms are not in the same plane. Two of the blue atoms are much
smaller than the other two blue atoms because, in space, they are behind the larger blue
atoms. This kind of molecule is NOT planar. If we traced a line around the molecules
going from blue atom to blue atom, we would have a tetrahedral shape. The geometry of
this molecule is tetrahedral.
Geometry Example 1:
The BF3 molecule, shown below, has a shape that is called trigonal planar. What are the
bond angles in the molecule?
Solution:
Since the angles are formed by the intersection of three lines that come together at
the center, the bond angles are 120°.
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Geometry Example 2:
The BeCl2 molecule below has 180° bond angles. What is the geometry of this molecule?
Solution:
180° bond angles give the molecule linear geometry.
Hints for Using Geometries:
What is the most common mistake in molecular geometry?
Students often think of molecules as flat, planar structures. Once an atom makes 4 bonds,
the geometry becomes very three-dimensional. There are NO 90° bond angles in common
biological molecules. Carbon makes four bonds, but the angle between the bonds is 109°
and the geometry is tetrahedral.
The second most common mistake is not memorizing the possible geometries of biological
molecules. You must have the names and angles memorized in order to understand the
chemical and physical properties of organic and biological molecules.
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Fractions, Decimals and Percents
Fractions, decimals, and percentages are all related to each other. Each is used to
indicate some part of a whole quantity. For example, ¾, 0.75, and 75% are all used to
express a quantity representing three parts out of four. The fraction represents the
number of parts of the whole (3 parts of the whole, which is divided into 4 equal sized
parts).
The decimal represents the part of one whole (0.75 of the whole).
The percentage represents the number of parts of a whole that has been divided into 100
parts (75 parts of the whole which has been divided into 100 equal parts).
Divide pie up into
100 segments
The fraction, 3/4, is converted to a decimal by dividing the numerator (the number on top of
the fraction) by the denominator (the number on bottom of the fraction). Three divided by
four is 0.75.
The decimal is converted to the percentage by moving the decimal point two places to the
right. This is equivalent to multiplying the decimal by 100% in order to convert it to the
percentage. In this conversion, 0.75 becomes 75%
To convert a percent to a decimal, we move the decimal point two places to the left. The
decimal point in 75% is understood to be after the five in the percentage. In this
conversion, 75% becomes 0.75.
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In this chapter, we use percentages more often than decimals or fractions. The term
percent means “per cent” or “per 100.” When we say that 10% of people prefer Brand A,
we mean that 10 people out of 100 people prefer Brand A. When someone has 71% of
the votes in an election, it means that 71 people out of 100 people voted for that person. A
chemist might say that a substance is 64% nitrogen. The chemist means that 64 atoms
out of every 100 atoms are nitrogen.
A percentage is another conversion factor. It can be written as a fraction and used in
problem solving (also called unit analysis). Here are the examples of percentages we
used above written as conversion factors.
10 people prefer Brand A
100 people expressed an opinion
71 people voted for candidate
100 people voted
64 atoms of nitrogen
100 atoms present
The secret to using percentages correctly is to write the units correctly. It is very
important that the units for the “per 100” part indicate what substance is being measured or
counted. The numerator or top of the fraction should clearly indicate what small portion of
the whole quantity is being measured or counted. In other words, a percentage should be
written as:
The Part
The Whole
Percentage Example 1:
How would you present each percentage as a fraction?
47% of applicants were women
15% of atoms were sulfur
Solution
47 female applicants (the Part)
100 total applicants (the Whole)
15 atoms sulfur (the Part)
100 atoms total (the Whole)
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Sometimes we are asked to calculate percentages from actual measurements. In this type
of problem, a percent is not given in the problem.
Percentage Example 2:
What percent of people in the class are women if 30 of the 50 students are female?
Solution:
Remember to divide the part by the whole. What is the “whole” class number? 50.
We divide the “part” (which is 30 female students) by the “whole” (which is 50 male
and female students). Then, we multiply by 100 to create the percentage.
30 women x 100 = 60%
50 students
Chemists frequently use percentages in calculations. Remember, it is important that you
include units in your calculations to avoid mistakes.
Percentage Example 3:
A total of 476 people were surveyed. When asked to choose the best tasting pizza, 34.2%
of them chose Brand B. How many people selected Brand B as the tastiest pizza?
Solution:
First, convert the percentage into a conversion factor. Remember that the
denominator is always 100 (from the “per cent” part of the question).
The conversion factor we would write would be:
34.2 of people selected Brand B
100 people surveyed
The total number of people we surveyed was MORE than 100. It was 476. How do
we include this information in our calculations?
476 people surveyed x
34.2 of people selected Brand B
100 people surveyed
= 163 people selected Brand B
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Notice how the units cancel! We can check the answer to see if it is reasonable. If 476
people were surveyed, is it reasonable that less than 476 chose Brand B? Yes. The
percentage of people selecting Brand B was less than 50%. Is 163 less than half of 476?
Yes.
Percentage Example 4:
If 54% of the mass of a mixture of iron and copper is iron, how much iron is there in a
sample that has a mass of 25 grams?
Solution:
What is the conversion factor we would write for the percentage?
54 grams of iron
100 grams mixture
How did we choose the units “grams?” The unit in which the measurement was
done was given in the problem. Now, let’s use the conversion factor that we
obtained from the percentage in the solution.
25 grams mixture x
54 g iron
100 gram mixture
= 14 grams iron
Again, check to see if the answer is reasonable. If about half of the mixture was
iron and the mixture weighed 25 grams, we would expect an answer of about 12.5
grams. Our estimate of the answer agrees with the calculated answer.
Percentage Example 5:
If a mixture of iron and copper was 54% iron, how much of the mixture would you need to
use to have 200. grams of iron (plus the copper impurity in the mixture)?
Solution:
This question uses the same conversion factor as the previous example. But how
should you arrange the conversion factor so that the units cancel?
200. grams iron x 100 grams mixture = 370 g mixture
54 grams iron
Notice that we used the inverse form of the percentage. We know to do this so that
the units cancel. If you set up the problem so that the units cancel, you will be on
the right track to get the correct answer!
21
Hints for using percentages:
What units do you choose when turning the percentage into a conversion factor?
Look at the question to see what unit of measurement is being used or what substance
was being counted. If the question describes centimeters of metal tubing, then both
numerator and the denominator will be in centimeters. If the question involved milliliters of
solution, then both the numerator and the denominator will be in milliliters. Add other
words to the units to help you distinguish “The Part” from “The Whole.”
22
Conversion Factors
Conversion factors are ratios of one object to another object. A ratio is a way of comparing
two quantities. The quantities can be compared in three different ways: a to b, a:b, or a/b.
At the local grocery store, a case of soda contains 24 cans. We can express the ratio in
three forms:
(a to b) 24 cans of soda to each case of soda
24 cans of soda: 1 case of soda
(a:b)
24 cans of soda/1 case of soda
(a/b)
This chapter uses conversion factors frequently to solve problems. Conversion factors are
ratios written in the fraction form (a/b). In this chapter, there are three major conversion
factors we will learn to use:
A universal conversion factor:
1 mole of chemical contains 6.02 x 1023 molecules (or atoms)
A molar mass conversion factor:
1 mole of chemical weighs the molar mass of the chemical
A chemical formula conversion factor:
1 mole of chemical contains some number of moles of atoms
If the chemical in the three conversion factors was CuCl2, then the three conversion factors
would be:
1 mole of CuCl2 contains 6.02 x 1023 molecules of CuCl2
1 mole of CuCl2 weighs 134.6 grams CuCl2 (the molar mass)
1 mole of CuCl2 contains 2 moles of chlorine atoms
23
The most common way that chemists represent ratios is in a fraction form. The three
conversion factors would look like this in the fraction form:
1 mole CuCl2
6.02 x 1023 molecule CuCl2
1 mole CuCl2
134.6 g CuCl2
1 mole CuCl2
2 mole Cl atoms
Each of these conversion factors can be written in the inverse form. The first ratio we
examined involved cans of soda. We expressed that ratio as 24 cans of soda in 1 case of
soda. The inverse form of this ratio is 1 case of soda contains 24 cans of soda. The key
idea here is that either the original ratio or its inverse form is true. It is true that one
case of soda contains 24 cans and it is true that there are 24 cans of soda in one case.
Conversion factors in chemistry can be written as shown above or in the inverse form.
Regardless of which form is used, the comparison the conversion factor makes is true.
For example, one mole of CuCl2 contains 6.02 x 1023 molecules of CuCl2 and there are
6.02 x 1023 molecules of CuCl2 in every one mole of CuCl2.
Here are the three major conversion factors from this chapter written in the inverse form:
6.02 x 1023 molecule CuCl2
1 mole CuCl2
134.6 g CuCl2
1 mole CuCl2
2 mole Cl atoms
1 mole CuCl2
The obvious question at this point is: Which form of the conversion factor should you use
in a problem? You use the form that allows the units to cancel. Units cancel when the
unit that appears on the top also appears on the bottom of a fraction somewhere in the
solution. This style of problem-solving, where you arrange the conversion factors so that
the units cancel, is called unit analysis.
24
Conversion Factor Example 1
Which is the correct solution for the question: How many cases of soda can be made from
1200 cans of soda?
1200 cans of soda x
24 cans of soda = 28800 cans of soda2/case of soda
1 case of soda
1200 cans of soda x
1 case of soda = 50 cases of soda
24 cans of soda
Solution:
In which solution do the units cancel? In the second solution, the unit “cans of
soda” cancels. The correct solution will use conversion factors in a way that units
will cancel. Blue lines are drawn through units that cancel.
Conversion Factor Example 2
2.5 moles of CuCl2 contain how many molecules of CuCl2?
Solution
This question uses the universal conversion factor of moles to molecules. Set up
the solution so that the units cancel. Blue lines are drawn through units that cancel.
2.5 moles of CuCl2 x 6.02 x 1023 molecules CuCl2 = 1.5 x 1024 molecules CuCl2
1 mole CuCl2
Conversion Factor Example 3
The formula for sugar is C6H12O6. How many moles of sugar are present when 20.0 moles
of oxygen are present?
Solution
This question requires the chemical formula conversion factor in the solution. You
don’t need to concern yourself with whether the conversion factor is inverted or not.
Just make sure that the units cancel. Blue lines are drawn through units that cancel.
20.0 moles of oxygen x
1 mole C6H12O6 = 3.33 moles C6H12O6
6 moles oxygen
25
Hints for using conversion factors:
How do you know what quantity to start with when solving problems?
Start with the quantity given in the problem and not with the conversion factor. In general,
the first quantity in your unit analysis will not have a denominator. In the last example, we
started with the number of moles of oxygen:
20.0 moles of oxygen x
1 mole C6H12O6 = 3.33 moles C6H12O6
6 moles oxygen
Notice that the first quantity is a whole number and not a fraction. If you start with the
conversion factor, it is much harder to decide whether to use the original or inverse form of
the conversion factor.
Should I enter all of the numerator numbers into the calculator before I enter the
denominator numbers, or should I enter each conversion factor into the calculator
before I move on to the next conversion factor?
Either style of entering the numerical quantities will work. You should make a decision to
always use one style of entering numbers into a calculator. This helps you avoid skipping
a number when you are calculating an answer.
In the example calculation below, you could enter the numbers in either of these orders
and obtain the same answer.
16 x 2.54 x 50 = 75.2
3
9
Enter the numbers into your calculator as:
17 x 2.54 ÷ 3 x 50 ÷ 9 =
or
17 x 2.54 x 50 ÷ 3 ÷ 9 =
You don’t need to press the ENTER key after each mathematical.
26
Fractions, Decimals and Percents
Fractions, decimals, and percentages are all related to each other. Each is used to
indicate some part of a whole quantity. For example, ¾, 0.75, and 75% are all used to
express a quantity representing three parts out of four. The fraction represents the
number of parts of the whole (3 parts of the whole, which is divided into 4 equal sized
parts).
The decimal represents the part of one whole (0.75 of the whole).
The percentage represents the number of parts of a whole that has been divided into 100
parts (75 parts of the whole which has been divided into 100 equal parts).
Divide pie up into
100 segments
The fraction, 3/4, is converted to a decimal by dividing the numerator (the number on top of
the fraction) by the denominator (the number on bottom of the fraction). Three divided by
four is 0.75.
The decimal is converted to the percentage by moving the decimal point two places to the
right. This is equivalent to multiplying the decimal by 100% in order to convert it to the
percentage. In this conversion, 0.75 becomes 75%
To convert a percent to a decimal, we move the decimal point two places to the left. The
decimal point in 75% is understood to be after the five in the percentage. In this
conversion, 75% becomes 0.75.
27
In this chapter, we use percentages more often than decimals or fractions. The term
percent means “per cent” or “per 100.” When we say that 10% of people prefer Brand A,
we mean that 10 people out of 100 people prefer Brand A. When someone has 71% of
the votes in an election, it means that 71 people out of 100 people voted for that person. A
chemist might say that a substance is 64% nitrogen. The chemist means that 64 atoms
out of every 100 atoms are nitrogen.
A percentage is another conversion factor. It can be written as a fraction and used in
problem-solving (also called unit analysis). Here are the examples of percentages we
used above written as conversion factors.
10 people prefer Brand A
100 people expressed an opinion
71 people voted for candidate
100 people voted
64 atoms of nitrogen
100 atoms present
The secret to using percentages correctly is to write the units correctly. It is very
important that the units for the “per 100” part indicate what substance is being measured or
counted. The numerator or top of the fraction should clearly indicate what small portion of
the whole quantity is being measured or counted. In other words, a percentage should be
written as:
The Part
The Whole
Percentage Example 1:
How would you present each percentage as a fraction?
47% of applicants were women
15% of atoms were sulfur
Solution
47 female applicants (the Part)
100 total applicants (the Whole)
15 atoms sulfur (the Part)
100 atoms total (the Whole)
28
Sometimes we are asked to calculate percentages from actual measurements. In this type
of problem, a percent is not given in the problem.
Percentage Example 2:
What percent of people in the class are women if 30 of the 50 students are female?
Solution:
Remember to divide the part by the whole. What is the “whole” class number? 50.
We divide the “part” (which is 30 female students) by the “whole” (which is 50 male
and female students). Then, we multiply by 100 to create the percentage.
30 women x 100 = 60%
50 students
Chemists frequently use percentages in calculations. Remember, it is important that you
include units in your calculations to avoid mistakes.
Percentage Example 3:
A total of 476 people were surveyed. When asked to choose the best tasting pizza, 34.2%
of them chose Brand B. How many people selected Brand B as the tastiest pizza?
Solution:
First, convert the percentage into a conversion factor. Remember that the
denominator is always 100 (from the “per cent” part of the question).
The conversion factor we would write would be:
34.2 of people selected Brand B
100 people surveyed
The total number of people we surveyed was MORE than 100. It was 476. How do
we include this information in our calculations?
476 people surveyed x
34.2 of people selected Brand B
100 people surveyed
= 163 people selected Brand B
29
Notice how the units cancel! We can check the answer to see if it is reasonable. If 476
people were surveyed, is it reasonable that less than 476 chose Brand B? Yes. The
percentage of people selecting Brand B was less than 50%. Is 163 less than half of 476?
Yes.
Percentage Example 4:
If 54% of the mass of a mixture of iron and copper is iron, how much iron is there in a
sample that has a mass of 25 grams?
Solution:
What is the conversion factor we would write for the percentage?
54 grams of iron
100 grams mixture
How did we choose the units “grams?” The unit in which the measurement was
done was given in the problem. Now, let’s use the conversion factor that we
obtained from the percentage in the solution.
25 grams mixture x
54 g iron
100 gram mixture
= 14 grams iron
Again, check to see if the answer is reasonable. If about half of the mixture was
iron and the mixture weighed 25 grams, we would expect an answer of about 12.5
grams. Our estimate of the answer agrees with the calculated answer.
Percentage Example 5:
If a mixture of iron and copper was 54% iron, how much of the mixture would you need to
use to have 200. grams of iron (plus the copper impurity in the mixture)?
Solution:
This question uses the same conversion factor as the previous example. But how
should you arrange the conversion factor so that the units cancel?
200. grams iron x 100 grams mixture = 370 g mixture
54 grams iron
Notice that we used the inverse form of the percentage. We know to do this so that
the units cancel. If you set up the problem so that the units cancel, you will be on
the right track to get the correct answer!
30
Hints for using percentages:
What units do you choose when turning the percentage into a conversion factor?
Look at the question to see what unit of measurement is being used or what substance
was being counted. If the question describes centimeters of metal tubing, then both
numerator and the denominator will be in centimeters. If the question involved milliliters of
solution, then both the numerator and the denominator will be in milliliters. Add other
words to the units to help you distinguish “The Part” from “The Whole.”
31
Balancing Chemical Equations
A mathematical equation is simply a sentence that states that two expressions are equal.
One or both of the expressions will contain a variable whose value must be determined by
solving the equation. Linear equations are equations in which the variable being solved for
is raised to a power no higher than 1. Equations in which the variable being solved for is
raised to a power of 2 are called quadratic equations. The problems in this text do not
require you to solve quadratic equations, so this topic will not be covered here. Consider
the following linear equation.
5x - 3 = 2x + 9
To solve the equation for x, the first step is to get all of the terms involving x on one side of
the equation, and all other terms on the other side. We are allowed to perform the
following operations on any equation without changing the equality that the equation
represents.
1. Adding the same quantity to both sides of the equation.
2. Subtracting the same quantity from both sides of the equation.
3. Multiplying both sides of the equation by the same quantity.
4. Dividing both sides of the equation by the same quantity.
The basic rule is: Whatever you do to one side, do to the other side also.
In the equation 5x - 3 = 2x + 9, we will collect all of the x’s on the left side of the
equation and all of the numbers on the right side. We must add 3 to each side to get the
numbers on the right side, and subtract 2x from each side to get the x’ on the left side.
Original equation:
(1) add three to both side:
5x - 3 = 2x + 9
+3
+ 3
5x
= 2x + 12
-2x
-2x
3x
=
12
(2) subtract 2x from both sides:
Now to solve for x, we must divide both sides of the equation by 3.
32
(3) divide both sides by 3:
3x = 12
3
3
Final Answer:
x
= 4
Balancing Equations Example 1
Solve for the value of x in the equation below.
6x + 7 = 4x + 23
Solution:
First, subtract 7 from both sides. Then, subtract 4x from both sides. Finally, divide
both sides by 2 to get the final answer: x = 8. You will be on track to get the right
answer if you remember one simple rule: Whatever you do to one side, do to the
other side also.
Now what does this have to do with balancing equations? You can use basic algebra
rearrangements like we did in the example above to help you balance equations. Look at
this equation.
Ni(s) + HCl(aq)
NiCl2(aq) + H2(g)
A balanced equation must have the same number of each kind of atom on both sides of
the equation. The same number of nickel atoms must appear on both sides of the
equation. The same number of hydrogen atoms must appear on both sides of the
equation. The same number of chlorine atoms must appear on both sides of the equation.
As it is currently written, the equation is not balanced. The nickel atoms are balanced, but
the hydrogen and chlorine atoms are not balanced. Let’s balance the chlorine atoms using
some algebra.
There is one chlorine atom on the left side of the equation and two chlorine atoms on the
right side of the equation. What do you need to multiply the left side of the equation by in
order to make the two sides of the equation have the same number of chlorine atoms? Let
this unknown multiplication factor be “x”. We can set up an algebraic equation for this
question that looks like:
x(1) = 2
33
To solve for x, divide both sides by one. The final answer is x = 2. If we put a coefficient
of two in front of HCl, the equation becomes:
NiCl2(aq) + H2(g)
Ni(s) + 2HCl(aq)
The equation is now balanced.
That was a very simple application of algebra to the problem of balancing an equation.
Let’s look at a more complex example.
Fe(s) + O2(g)
Fe2O3(s)
Let’s balanced the iron atoms first. There is one iron atom on the left side of the equation
and two iron atoms on the right side of the equation. The algebra expression that
represents this situation would be:
x(1) = 2
Solving for x by dividing both sides by one gives us the final answer of: x = 2. We need to
put a coefficient of two in front of Fe(s) on the left to balance the iron atoms.
2 Fe(s) + O2(g)
Fe2O3(s)
There are two oxygen atoms on the left side of the equation and three oxygen atoms on
the right side of the equation. The algebra expression that represents this situation would
need to say that some unknown multiplication factor, y, times the two oxygen atoms on the
left will equal the three oxygen atoms on the right.
y(2) = 3
Solve for y by dividing both sides by two. The final answer is y = 3/2. We need to put a
coefficient of 3/2 in front of the O2 on the left to balance the oxygen atoms. It is easier to
leave this answer in the fractional form rather than convert it to a decimal number. You’ll
see why soon.
2 Fe(s) + 3/2 O2(g)
Fe2O3(s)
The equation is now balanced. However, as often happens, a fraction appears in the
balanced equation. Chemical reactions, except in a few limited circumstances, should not
be balanced with fractions. In order to eliminate the fraction, we need to multiply all of the
coefficients by the number in the denominator. Remember the basic rule: Whatever you
do to one side, do to the other side also.
.
.
2 2Fe(s) + 2 3/2 O2(g)
2 Fe2O3(s)
34
The equation becomes:
4 Fe(s) + 3 O2(g)
2 Fe2O3(s)
Balancing Equations Example 2
Balance this equation:
HF(g) + SiO2(s)
SiF4(g) + H2O(l)
Solution:
If chemical equation is difficult to balance, you can use simple algebra expressions
to make balancing the equation easier. To balance the fluorine atoms, you could
use the algebra expression:
x(1) = 4
Solving for x gives you a final answer of: x = 4. You need a coefficient of four in
front of HF(g). To balance the oxygens, you could use the algebra expression:
2 = y(1)
Solving for y gives you a final answer of: 2 = y. You need a coefficient of two in
front of H2O(l).
The balanced equation is:
4HF(g) + SiO2(s)
SiF4(g) + 2H2O(l)
35
Balancing Equations Example 3
Balance the equation
C2H6(g) + O2(g)
CO2(g) + H2O(l)
Solution
Begin by balancing the carbons. The algebra expression for carbons could be 2 =
x(1). Solving for x gives the final answer: 2 = x. You need a coefficient of two in
front of CO2(g). The algebra expression for hydrogen could be 6 = y(2). Solving for
y gives the final answer: 3 = y. You need a coefficient of 3 in front of H2O(l).
C2H6(g) + O2(g)
2 CO2(g) + 3 H2O(l)
The algebra expression for oxygen is more complex. There are two oxygen atoms
on the left side of the equation and 4 + 3 oxygen atoms on the right side of the
equation. The algebra expression could be:
z(2) = 7
Solving for z gives the final answer: z = 7/2. The balanced equation is:
C2H6(g) + 7/2 O2(g)
2 CO2(g) + 3 H2O(l)
Although the equation is balanced, we need to eliminate the fraction from the
chemical equation. We do this by multiplying coefficients on both sides of the
equation by whatever value is in the denominator. In this case, we would multiply
by two.
.
2 C2H6(g) + 2 7/2 O2(g)
.
The equation becomes
2 C2H6(g) + 7 O2(g)
.
2 2 CO2(g) + 2 3 H2O(l)
4 CO2(g) + 6 H2O(l)
36
Hints for Balancing Equations:
Do I need to use algebra to balance an equation?
No. Most students quickly learn how to balance the simpler chemical equations.
However, if a equation is difficult or tricky to balance, writing out algebra expression for
each different kind of atom is a reliable way to balance an equation.
37
Basic Algebra
Algebra encompasses the mathematical manipulations needed to solve for an unknown.
The unknown is often called the variable. It is most often symbolized with an “x”. In
chemistry, the unknown or variable often is labeled with a different letter. The letter is
chosen to remind the scientist what quantity he is solving for. If the quantity is pressure,
the variable is “P”. If the quantity is temperature, the variable is “T”.
Many different kinds of equations can be solved using algebra. The most common algebra
equation you will have to solve in chemistry has the form:
6x = 24
where a variable appears on one side of the equation with a coefficient. You can solve this
basic algebra equation by dividing BOTH sides of the equation by the coefficient in front of
the variable (in this case, the six).
6x = 24
6
6
On the side with the variable, the six cancels out. That is because 6 divided by 6 is one.
On the other side of the equation (the side without the variable), 24 divided by 6 is four.
6x = 24
6
6
The equation now is: x = 4.
You will see this basic algebra equation frequently in this chapter. Boyle’s Law, where
pressure is compared to volume, is the good example of this algebraic manipulation.
Basic Algebra Example 1:
A gas occupies 3.5 L at 0.80 atm. If the volume is reduced to 1.7 L, what pressure will the
gas experience?
Solution
Boyle’s Law uses the equation P1V1 = P2V2. One of the four variables in the
equation is unknown.
38
Initial Conditions
P1 = 0.80 atm
V1 = 3.5 L
Final Conditions
P2 = ?
V2 = 1.7 L
When we fill in the numbers, the equation becomes:
(0.80 atm)(3.5 L) = (P2)(1.7 L)
(0.80 atm)(3.5 L) = (P2)(1.7 L)
1.7 L
1.7 L
Divide both sides by the coefficient on the side of the equation with the variable P2.
For this question, we divide both sides by 1.7 L. The unit “liter” cancels out on both
sides. The value “1.7” on the right side is canceled out by dividing by “1.7”.
1.6 atm = P2
Hints for solving basic algebra equations:
What does it mean when the question says ”solve for P2”?
This is just a mathematician’s way to tell you to do what we did in example 1.
Can I rearrange the equation so that both pressures are on the same side?
No. The equation P1V1 = P2V2 is NOT the same as P1P2 = V1V2. The units wouldn’t
cancel and you would get a different answer. Although this alternative equation could be
solved in the same manner as the first equation using basic algebra, you cannot change
the equation (or formula) and get the same answer.
I didn’t get the answer in the book, but I know I’m solving the equation using correct
algebra. What’s the problem?
The most common mistake in this chapter is not assigning the initial conditions and final
conditions correctly. If we take the information in the first example, but reverse the initial
and final volumes, the table would look like:
Initial Conditions
Final Conditions
39
P1 = 0.80 atm
V1 = 1.7 L
P2 = ?
V2 = 3.5 L
Instead of the volume getting smaller, the question now has the volume getting larger. The
equation becomes
(0.80 atm)(1.7 L) = (P2)(3.5 L)
0.39 atm = P2
We got a totally different answer because we are asking a different question. Read each
question carefully to determine what quantity is present initially and what quantity is
present at the end.
40
Cross Multiplication and Gas Laws
Cross multiplication is a valuable mathematical solving technique because it allows you to
convert a fraction, which can be tricky to use, into a simpler equation. If, for example,
there were 5 members on a team and you had 7 teams, you could set up an equality using
fractions to find the number of people on 7 teams.
x
5 team members =
1 team
7 teams
To solve this problem with the two fractions more easily, you cross multiply. The term
“cross multiply” was coined because students first drew a big “X” or “cross” to connect the
two numbers that need to be multiplied together to solve the problem.
5 team members
1 team
=
x
7 teams
The arrows tell us to multiply 5 team members by 7 teams and to multiply 1 team by the
unknown. This trick helps us turn a tricky fraction into a common algebra problem.
(5 team members)(7 teams) = (1 team)(x)
In the previous section of this review, we learned how to solve this basic algebra problem:
divide both sides by the term “1 team.” Notice how the unit “team” cancels on both sides
of the equation.
(5 team members)(7 teams) = (1 team)(x)
1 team
1 team
We are left with:
35 team members = x
The total number of team members (unknown x) is 35.
Cross multiplication is used frequently in this chapter to solve for one of the four quantities
that define a gas: pressure, temperature, volume, or moles of gas.
41
Cross Multiplication Example 1
If a gas at 250 K has a pressure of 1.30 atm, what pressure will the gas experience if the
gas is heated to 300 K?
Solution:
This is Gay-Lussac’s Law. The equation to compare pressure and temperature is:
P1
T1
=
P2
T2
Initial Conditions
P1 = 1.30 atm
T1 = 250 K
1.30 atm
250 K
=
Final Conditions
P2 = ?
T2 = 300 K
P2
300 K
Insert values into equation
(1.30 atm)(300 K) = (P2)(250 K)
Convert to multiplication form
(1.30 atm)(300 K) = (P2)(250 K)
250 K
250 K
Isolate the unknown P2
1.56 atm = P2
Solve for unknown P2
With a bit of practice, you will be able to solve this problem with fewer intermediate
steps.
Cross Multiplication Example 2
When the volume of a gas at 325 K is decreased from 5.00 L to 3.00 L, what will the new
temperature of the gas be?
Solution:
This is Charles’s Law. It is no harder to solve for an unknown when it is in the
denominator of the fraction than it is to solve for an unknown when it is in the
numerator of the fraction. Why? Because we convert the fraction to multiplication
during the process of cross multiplication. The equation to compare volume and
temperature is:
42
V1
T1
=
V2
T2
Initial Conditions
V1 = 5.00 L
T1 = 325 K
5.00 L
325 K
=
Final Conditions
V2 = 3.00 L
T2 = ?
3.00 L
T2
Insert values into equation
(5.00 L)(T2) = (3.00 L)(325 K)
Convert to multiplication form
T2 = 195 K
Solve for unknown T2
The step in which both sides were divided by 5.00 L is not commonly written out by
students.
Cross Multiplication Example 3
When a 0.75 L sample of a gas at 100 K and 2.0 atm changes to 200 K and 2.5 atm, what
will the new volume of the gas be?
Solution:
Volume, temperature, and pressure are changing in this question. Because more
than two quantities are changing, we should use the combined gas law to solve the
problem.
P1V1 =
T1
P2V2
T2
Initial Conditions
P1 = 2.0 atm
V1 = 0.75 L
T1 = 100 K
Final Conditions
P2 = 2.5 atm
V2 = ?
T2 = 200 K
43
(2.0 atm)(0.75 L)
100K
=
(2.5 atm)(V2)
200K
We still cross multiply even though there is more than one quantity in the
numerator. The fraction is converted to:
(2.0 atm)(0.75 L)(200K) = (2.5 atm)(V2)(200K)
Divide both sides by 2.5 atm and by 200 K. The units “atm” and “K” will cancel.
0.60 L = V2
Hints for using Cross Multiplication:
Does it matter which pressure is designated P1 and which is designate P2?
YES! It matters very much. Your answer will be different if you reverse the pressures.
Read the question very carefully to determine what the initial pressure is and what the final
pressure is. Reversing initial and final conditions is one of the most common mistakes in
this chapter.
Is it harder to solve for a denominator unknown than for a numerator unknown?
No. Cross multiplication is a wonderful mathematical trick to turn a fraction into a simpler
multiplication. Therefore, it doesn’t matter where the unknown appears in the equation. It
will always be simple to solve for because of cross multiplication.
If the unknown in the denominator really bothers you, you can reverse the initial equation.
In the second example, we compared volume and temperature using the equation. The
unknown was T2
V1
T1
=
V2
T2
We could have reversed the equation and then solved for T2 (now in the numerator).
T1
V1
=
T2
V2
44
Manipulating Negative Numbers
Both positive and negative numbers are used to describe quantities or properties in
chemistry. Certain properties can have only positive values, such as the number of atoms
in a molecule. Some variables, however, can have negative values, such as the
temperature in degrees Fahrenheit or Celsius.
Positive numbers are written with no sign in front of them, but negative numbers will
always have a negative sign in front of them. For example, a temperature of 20 degrees
below zero Fahrenheit will be expressed as –20° F, but a temperature of 20 degrees
above zero Fahrenheit will be expressed as 20° F.
In chemistry, we will have to perform arithmetic operations on both positive and negative
numbers. Although you will usually be performing these operations with your calculator, it
is helpful to be able to predict the answer so that you will not be dependent on the
calculator.
In this chapter, we see signed numbers when we determine the temperature of gases.
Temperatures are measured in the laboratory in the Celsius scale. The Celsius scale
includes both positive and negative values. In calculations, however, temperatures must
be in Kelvin. The Kelvin scale does not have any negative values.
Converting a Temperature to Kelvin Example 1:
What is the temperature in Kelvin when the temperature of a gas is 62 oC?
Solution:
When the temperature is a positive number, you add 273 to the number to obtain
the temperature in the Kelvin scale. The difference between any Celsius
temperature and the Kelvin temperature is always 273.
Temperature in Kelvin = Temperature in Celsius + 273
Temperature in Kelvin = 62 oC + 273
Temperature in Kelvin = 335 K
The temperature in Kelvin cannot be negative. Temperatures in Celsius are
identified with the symbol “oC” which is read “degree Celsius.” Temperatures in
Kelvin do not include the degree symbol. Temperatures in Kelvin have the unit “K”
only.
45
Converting a Temperature to Kelvin Example 2:
What is the temperature in Kelvin when the temperature of a gas is - 47 oC?
Solution:
When the temperature is a negative number, you add 273 to the number to obtain
the temperature in the Kelvin scale just like you did when the temperature was
positive.
Temperature in Kelvin = Temperature in Celsius + 273
Temperature in Kelvin = - 47oC + 273
Temperature in Kelvin = 226 K
The temperature in Kelvin must be less than 273 since the Celsius temperature was
negative. Use the “plus/minus” key on your calculator to enter the negative
sign, not the “subtract” key.
Converting a Temperature to Celsius Example 3:
What is the temperature in Celsius when the temperature of a gas is 238 K?
Solution:
Temperature in Kelvin = Temperature in Celsius + 273
238 K = Temperature in Celsius + 273
238 K – 273 = Temperature in Celsius
- 35 oC = Temperature in Celsius
Since the temperature in Kelvin was less than 273, we expect the Celsius
temperature to be negative.
46
Hints for adding or subtracting negative numbers:
How do I enter negative numbers in my calculator?
The most common error when using negative numbers is that a student uses the “subtract”
key on their calculator instead of the “plus/minus” key. In example two, we needed to
calculate the temperature in Kelvin of a gas that is below zero in the Celsius scale.
Temperature in Kelvin = - 47oC + 273
You could enter this into your calculator as:
47 plus/minus key
addition key
273 equal or enter key
How can I predict the new temperature?
If the temperature in Celsius is negative, the temperature in Kelvin will be less than 273. A
temperature of 273K is the same temperature as 0 oC in the laboratory. A temperature in
Kelvin greater than 273 K indicates a gas at a temperature above zero.
47
Percentages in Solutions
Fractions, decimals, and percentages are all related to each other. Each is used to
indicate some part of a whole quantity. For example, ¾, 0.75, and 75% are all used to
express a quantity representing three parts out of four. The fraction represents the
number of parts of the whole (3 parts of the whole, which is divided into 4 equal sized
parts).
The decimal represents the part of one whole (0.75 of the whole).
The percentage represents the number of parts of a whole that has been divided into 100
parts (75 parts of the whole which has been divided into 100 equal parts).
Divide pie up into
100 segments
The fraction, 3/4, is converted to a decimal by dividing the numerator (the number on top of
the fraction) by the denominator (the number on bottom of the fraction). Three divided by
four is 0.75.
The decimal is converted to the percentage by moving the decimal point two places to the
right. This is equivalent to multiplying the decimal by 100% in order to convert it to the
percentage. In this conversion, 0.75 becomes 75%
To convert a percent to a decimal, we move the decimal point two places to the left. The
decimal point in 75% is understood to be after the five in the percentage. In this
conversion, 75% becomes 0.75.
48
In this chapter, we use percentages more often than decimals or fractions. The term
percent means “per cent” or “per 100.” When we say that 10% of people prefer Brand A,
we mean that 10 people out of 100 people prefer Brand A. When someone has 71% of
the votes in an election, it means that 71 people out of 100 people voted for that person. A
chemist might say that a substance is 4.7 % (m/m%). The chemist means that 4.7 grams
of solute are in every 100 grams of solution.
A percentage is another conversion factor. It can be written as a fraction and used in
problem solving (also called unit analysis). Here are the examples of percentages we
used above written as conversion factors.
10 people prefer Brand A
100 people expressed an opinion
71 people voted for candidate
100 people voted
4.7 gram of solute
100 grams solution
The secret to using percentages correctly is to write the units correctly. It is very
important that the units for the “per 100” part indicate what substance is being measured or
counted. The numerator or top of the fraction should clearly indicate what small portion of
the whole quantity is being measured or counted. In other words, a percentage should be
written as:
The Part
The Whole
Percentage Example 1:
How would you present each percentage as a fraction?
47% of applicants were women
15% of atoms were sulfur
Solution
47 female applicants (the Part)
100 total applicants (the Whole)
15 atoms sulfur (the Part)
49
100 atoms total (the Whole)
Sometimes we are asked to calculate percentages from actual measurements. In this type
of problem, a percent is not given in the problem.
Percentage Example 2:
What percent of people in the class are women if 30 of the 50 students are female?
Solution:
Remember to divide the part by the whole. What is the “whole” class number? 50.
We divide the “part” (which is 30 female students) by the “whole” (which is 50 male
and female students). Then, we multiply by 100 to create the percentage.
30 women x 100 = 60%
50 students
Chemists frequently use percentages in calculations. Remember, it is important that you
include units in your calculations to avoid mistakes.
Percentage Example 3:
A total of 476 people were surveyed. When asked to choose the best tasting pizza, 34.2%
of them chose Brand B. How many people selected Brand B as the tastiest pizza?
Solution:
First, convert the percentage into a conversion factor. Remember that the
denominator is always 100 (from the “per cent” part of the question).
The conversion factor we would write would be:
34.2 of people selected Brand B
100 people surveyed
The total number of people we surveyed was MORE than 100. It was 476. How do
we include this information in our calculations?
476 people surveyed x
34.2 of people selected Brand B
100 people surveyed
= 163 people selected Brand B
50
Notice how the units cancel! We can check the answer to see if it is reasonable. If 476
people were surveyed, is it reasonable that less than 476 chose Brand B? Yes. The
percentage of people selecting Brand B was less than 50%. Is 163 less than half of 476?
Yes.
Percentage Example 4:
If a solution is 25.3 % (m/v%), how many grams of solute are present in 250 mL of
solution?
Solution:
What is the conversion factor we would write for the percentage?
25.3 grams of solute
100 grams solution
How did we choose the units “grams?” The unit of the solution was m/m% or
mass/mass percent. We could choose any unit for mass (grams, milligram,
kilogram), but the most commonly chosen unit is gram.
Now, let’s use the conversion factor that we obtained from the percentage in the
solution.
250 grams solution x
25.3 gram solute
100 gram solution
= 63.3 grams solute
Again, check to see if the answer is reasonable. If each 100 grams of solution
contain about 25 grams of solute, then 200 grams of solution should contain 50
grams of solute. We have a little more than 200 grams of solution, so we should
have more than 50 grams of solute.
Percentage Example 5:
If a solution is 16.4 % (m/v) glucose, how many mL of solution are needed to deliver 25.0
grams of glucose to a patient?
Solution:
The conversion factor is 16.4 grams solute/100 mL of solution. Notice that we have
a unit of mass in the numerator and a unit of volume in the denominator. We chose
the unit “milliliter” because we need a volume in the denominator. The “100” part of
the denominator is understood.
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25.0 grams glucose x 100 mL solution = 152 mL solution
16.4 grams glucose
Notice that we used the inverse form of the percentage. We know to do this so that
the units cancel. If you set up the problem so that the units cancel, you will be on
the right track to get the correct answer!
Hints for using percentages:
What units do you choose when turning the percentage into a conversion factor?
Look at the question to see what unit of measurement is being used or what substance
was being counted. If the question describes % (m/m) solution, the units should be
gram/gram. If the question contains a solution with the units % (m/v), the units should be
grams/mL. The units for the conversion factor for this solution would be mL/mL.
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Cross Multiplication and Molarity
Cross multiplication is a valuable mathematical solving technique because it allows you to
convert a fraction, which can be tricky to use, into a simpler equation. If, for example,
there were 5 members on a team and you had 7 teams, you could set up an equality using
fractions to find the number of people on 7 teams.
x
5 team members =
1 team
7 teams
To solve this problem with the two fractions more easily, you cross multiply. The term
“cross multiply” was coined because students first drew a big “X” or “cross” to connect the
two numbers that need to be multiplied together to solve the problem.
5 team members
1 team
=
x
7 teams
The arrows tell us to multiply 5 team members by 7 teams and to multiply 1 team by the
unknown. This trick helps us turn a tricky fraction into a common algebra problem.
(5 team members)(7 teams) = (1 team)(x)
In the previous section of this review, we learned how to solve this basic algebra problem:
divide both sides by the term “1 team.” Notice how the unit “team” cancels on both sides
of the equation.
(5 team members)(7 teams) = (1 team)(x)
1 team
1 team
We are left with:
35 team members = x
The total number of team members (unknown x) is 35.
Cross multiplication is used frequently in this chapter to solve for one of the three
quantities that define a solution: molarity, moles of solute, or liters of solution.
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Cross Multiplication Example 1
How many moles of NaCl are present in 0.285 L of a 2.50 M NaCl solution?
Solution:
The equation for molarity is:
Molarity = moles of solute
liters of solution
At first, this doesn’t look like cross multiplication. What is the denominator for the
term “molarity” in the equation? The denominator is understood to be “one”. This
lets us rewrite the equation as:
Molarity = moles of solute
1
liters of solution
Insert the values from the question.
2.50 M
1
=
moles NaCl
0.285 L
The equation becomes:
(2.50 M)(0.285 L) = (1)(moles NaCl)
0.71 moles = moles NaCl
Cross Multiplication Example 2
How many liters of a 0.81 M HCl solution are required to provide 2.0 moles of HCl?
Solution:
Insert the values from the question and cross multiply.
0.81 M
1
=
2.0 moles HCl
liters of solution
The equation becomes:
(0.81 M)(liters of solution) = (1)(2.0 moles HCl)
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Divide both sides by 0.81 M to get
0.71 moles = moles NaCl
It is no harder to solve for an unknown that is in the denominator than it is to solve
for an unknown that is in the numerator. Cross multiplication makes manipulating
fractions quite easy.
Hints for using Cross Multiplication:
How do the units for molarity cancel?
The units for molarity are moles of solute/liters of solution. If we look at the molarity
formula with units instead if numbers we see.
Moles of solute/liters of solution = moles of solute
1
liters of solution
When we cross multiply, the equation becomes:
(Moles of solute/liters of solution)(liters of solution) = (1)(moles of solute)
The unit “liters of solution” cancels. It is just not very obvious how the units are canceling
unless you change the unit “molarity” to “moles of solute/liters of solution.”
Is it harder to solve for a denominator unknown than for a numerator unknown?
No. Cross multiplication is a wonderful mathematical trick to turn a fraction into a simpler
multiplication. Therefore, it doesn’t matter where the unknown appears in the equation. It
will always be simple to solve for because of cross multiplication.
55
Significant Figures in Logarithms
The logarithm of a number has two parts. The part in front of the decimal point is called
the characteristic. These digits are not significant.
Log (28.29) = 1.4516
characteristic
It represents the power to which ten must be raised in order to get the number 28.29. If
you wrote 28.29 in scientific notation, the number would be 2.829 x 101. Ten is raised to
the power of one so the characteristic is one.
The part of the logarithm after the decimal point is called the mantissa.
Log (28.29) = 1.4516
mantissa
The numbers in the mantissa are all significant. In this example, the number 28.29 has
four significant figures. In order for the logarithm of 28.29 to have four significant figures, it
must have four decimal places.
The number of decimal places in the logarithm of a number is equal to the number
significant figures in the number.
.
Significant Figures in Logarithms Example 1:
How many decimal places would the logarithm of each number have?
0.05087
-3.87
285.0
2.0
Solution
0.05087
3.87
285.0
2.0
has 4 significant figures, so logarithm would have 4 decimal places
has 3 significant figures, so logarithm would have 3 decimal places
has 4 significant figures, so logarithm would have 4 decimal places
has 2 significant figures, so logarithm would have 2 decimal places
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Significant Figures in Logarithms Example 2:
How many significant figures are in the number if the logarithm of the number is:
-2.80
0.0883
6.3
Solution
-2.80
0.0883
6.3
has 2 decimal places, so number would have 2 significant figures
has 4 decimal places, so number would have 4 significant figures
has 1 decimal place, so number would have 1 significant figure
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Logarithms and Inverse Logarithms
The common logarithm (abbreviated as “log”) of a number is the power to which you must
raise 10 in order to obtain the number. It sounds more complicated than it is.
If the log of 2 is 0.3, we write the expression:
log 2 = 0.3
This means that if we raised 10 to the 0.3 power, the answer would be 2.
100.3 = 2
Here are a two more examples.
Logarithmic Equation
log 4 = 0.6
log 236 = 2.373
Base 10 Equation
100.6 = 4
102.373 = 236
Raising ten to various powers looks odd because so far in our study of chemistry, we only
have seen ten raised to whole number powers. Another reason raising ten to various
powers looks odd is that we usually have a coefficient in front of the ten. We usually see
numbers written like 3.65 x 107, where ten is raised to a whole number power (7) and a
coefficient (3.65) is written in front.
Let's look at the logarithms of some numbers written in scientific notation. You should
notice an important pattern. When working with a number in scientific notation, if the
coefficient is one, then the logarithm is just the power to which ten is raised.
log
log
log
log
log
Logarithm
Expression
1000
100
10
0.01
0.00001
Log Expression in
Scientific Notation
log 1 x 103
log 1 x 102
log 1 x 101
log 1 x 10-2
log 1 x 10-5
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Answer
3
2
1
-2
-5
What happens when the coefficient is not one? We use a calculator!
The common logarithm of a number can be obtained by pressing the LOG button on your
calculator. On some calculators, you enter the number and then press the LOG button.
On other calculators, you press the LOG button and then enter the number.
The logarithms of several numbers are shown below. See if you get the same answers on
your calculator.
Logarithms
log 4.05
= 0.607
log 0.054
= -1.27
log 12.90
= 1.1106
-4
log 2.33 x 10 = -3.633
log 8.7 x 107 = 7.9
Three important things to remember when working with logarithms:
The logarithm of any number less than one is negative number.
The logarithm of any number greater than 1 is a positive number.
The log of one is zero: log 1 = 0 since 100 = 1
Logarithm Example 1:
What are the logarithms of 3.71 and 0.00739?
Solution
log 3.71 = 0.569
log 0.00739 = -2.131
Remember that the logarithm of a number greater than one is positive and the logarithm of
a number less than one is negative. Watch the significant figures. Remember that the
number of decimal places in the logarithm should equal the number of significant figures in
the number.
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When a logarithm of a number is known, you must take the inverse log, or antilog, to
determine the number. The manner in which you do this depends on the calculator you are
using. Try one of these procedures to obtain an inverse logarithm.
Procedure 1: For calculators with an INV button, enter the number for which you
want to obtain the inverse log. Press the INV button (the inverse button) and then
the LOG button.
Procedure 2: For calculators with a 2nd button, enter the number for which you want
to obtain the inverse log. Press the 2nd button. The 2nd button is usually a different
color from the rest of the buttons on the calculator. The 2nd button allows you to use
the mathematical operations on the calculator that are the same color as the 2nd
button. After pressing the 2nd button, press the button under the 10x symbol. The
10x symbol is usually written above the LOG button.
Procedure 3: Use either procedure 1 or 2, but enter the number after you use the
INV or 2nd button.
Find out how to use your calculator to get the inverse logarithms shown below.
Inverse logarithms
invlog 3.0
= 1 x 103
invlog -4.60 = 2.5 x 10-5
invlog 1.61 = 41
Logarithm Example 2:
What is the inverse logarithm of –8.97
Solution
invlog –8.97 = 1.1 x 10-9
For the logarithm of a number to be negative, as it is in this example, the number must be
less than one. The logarithm –8.97 has two decimal places, so the inverse logarithm
should have two significant figures.
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Hints for using logarithms:
Does it matter if I use the LOG button or the LN button on my calculator?
Yes! Common log (the LOG button) is based on a base-10 system. The natural logarithm
(the LN button) is based on a base-e system. The two types of logarithms can be
connected using the equality:
ln x = 2.303 log x
They are related by multiplying the base-10 log value by 2.303.
I’m not getting the right answers but I’m entering the numbers into my calculator
correctly.
You probably are making a small mistake while calculating the logarithm or inverse
logarithm on your calculator. The most common mistake is using the LN button instead of
the LOG button. The next most common mistake is taking the inverse logarithm when you
mean to take the logarithm, or visa versa. Check how you are using your calculator and
consult the manual to your calculator if you continue to have problems.
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Balancing Nuclear Equations
A mathematical equation is simply a sentence that states that two expressions are equal.
One or both of the expressions will contain a variable whose value must be determined by
solving the equation. Linear equations are equations in which the variable being solved for
is raised to a power no higher than 1. Equations in which the variable being solved for is
raised to a power of 2 are called quadratic equations. The problems in this text do not
require you to solve quadratic equations, so this topic will not be covered here. Consider
the following linear equation.
5x - 3 = 2x + 9
To solve the equation for x, the first step is to get all of the terms involving x on one side of
the equation, and all other terms on the other side. We are allowed to perform the
following operations on any equation without changing the equality that the equation
represents.
1. Adding the same quantity to both sides of the equation.
2. Subtracting the same quantity from both sides of the equation.
3. Multiplying both sides of the equation by the same quantity.
4. Dividing both sides of the equation by the same quantity.
The basic rule is: Whatever you do to one side, do to the other side also.
In the equation 5x - 3 = 2x + 9, we will collect all of the x’s on the left side of the
equation and all of the numbers on the right side. We must add 3 to each side to get the
numbers on the right side, and subtract 2x from each side to get the x’ on the left side.
Original equation:
(1) add three to both side:
5x - 3 = 2x + 9
+3
+ 3
5x
= 2x + 12
-2x
-2x
3x
=
12
(2) subtract 2x from both sides:
Now to solve for x, we must divide both sides of the equation by 3.
62
(3) divide both sides by 3:
3x = 12
3
3
Final Answer:
x
= 4
Balancing Nuclear Equations Example 1
Solve for the value of x in the equation below.
6x + 7 = 4x + 23
Solution:
First, subtract 7 from both sides. Then, subtract 4x from both sides. Finally, divide
both sides by 2 to get the final answer: x = 8. You will be on track to get the right
answer if you remember one simple rule: Whatever you do to one side, do to the
other side also.
Now what does this have to do with balancing nuclear equations? You can use basic
algebra rearrangements like we did in the example above to help you balance equations.
Look at this equation.
218
84
Po
4
2
He + ?
A balanced nuclear equation must have the same atomic number (the subscripted
number) on each side of the equation. A balanced nuclear equation must have the same
total mass number (the superscripted number) on each side of the equation.
We can set up two algebraic equations to describe this chemical reaction. For the mass
number, we can say:
Mass number:
218 = 4 + x
For the atomic number, we can say:
Atomic number:
84 = 2 + y
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Isolate the x or the y to solve the equation. In the mass number equation, subtract four
from both sides of the equation. In the atomic number equation, subtract two from both
sides. The equations become:
Mass Number:
Atomic Number:
214 = x
82 = y
Finally, use the periodic table to identify the element that has 82 protons. The element is
Pb. The complete nuclear equation is:
218
84
Po
4
2
He +
214
82
Pb
Balancing Nuclear Equations Example 2
Balance this nuclear equation:
131
53
I
131
54
Xe + ?
Solution:
Set up two algebraic equations, one for the mass number and one for the atomic number.
Mass Number:
Atomic Number:
131 = 131 + x
53 = 54 + y
Isolate the variable, x or y. The equations become:
Mass Number:
Atomic Number:
0 =x
-1 = y
The balanced nuclear equation is:
131
53
I
131
54
Xe +
0
−1
e
A beta particle was emitted.
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Hints for Balancing Nuclear Equations:
Do I need to use algebra to balance an equation?
No, but it is the most reliable problem solving method. You are much less likely to make a
mistake if you write out an algebraic expression than if you attempt to solve the problem in
your head.
I got a negative numbers or a zero for my answer. What did I do wrong?
Nothing at all. Beta and gamma radiation involves particles with a negative number or a
zero.
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