1B.3: Mathematics in Chemistry
- Page ID
- 50462
Learning Objectives
- Express numbers both in scientific notation and standard notation
- Carry out calculations with the correct number of significant Figures
- Convert between various units
Math Review: 4 Ways of Expressing Numbers
There are four common ways to express numbers, and some of these are preferable when the numbers are very large or very small.
- Floating Number
- Scientific Notation
- SI Prefixes
- Logarithms
Floating Numbers
Floating numbers are common numbers where the "magnitude" is determined by the position of the decimal point. Digits to the left of the decimal are larger than 1 and digits to the right are less than one, or a fraction. So 12.121200 and 121212.00 are two numbers of identical digits but the second is 104 times larger than the first . A fraction typically has a zero in front of the decimal, so 0.25 is \(\frac{25}{100}\) or one quarter \(\frac{1}{4}\).
Scientific Notation
Unlike a floating numbers scientific notation has two parts, first a number with only one digit to the left of the decimal and a variable number of digits to the right of the exponent, which is multiplied by 10 to a power, such that it equals the value of the desired number. We call the power of 10 its exponent.
\[\text{_ . _ _ _ } \times 10^{n} \]
Where n is the power and the underscore represents the number of digits in the number. Note, due to this format, scientific notation always has an unambiguous number of significant digits.
Q&A: What is Exponentiation?
Exponentiation is the replication of multiplication the way multiplication is the replication of addition
\[ \begin{align*} 10^0 &= 1 \\[4pt] 10^1 &= 10 \quad\quad &&10^{-1} =\frac{1}{10} = 0.1 \\[4pt] 10^2 &= 10(10) = 100 \quad\quad &&10^{-2} =\frac{1}{10^{2}} =\frac{1}{100}=0.001 \\[4pt] 10^3 &=10(10)(10) = 1000 \quad\quad &&10^{-3} =\frac{1}{10^{3}}=\frac{1}{1000}=0.0001 \end{align*}\]
Q&A: Can the exponent have units?
No, the exponent can not have units, it is the number of times you multiplied or divided the number by 10, but you write the units behind the number expressed in scientific notation.
\[1.333 \times 10^{4} \; L = 13,333 \; L\]
Advantages of Scientific Notation
- Allows Awkwardly Large and Small Numbers to Be Expressed in Term of Compact and Easily Written Numbers
- Allows Accurate Representation of the Number of Significant Figures in a Number, That Is a Measurement’s Precision, the “Certainty” of Our Measurements
SI Prefixes
SI prefixes are similar to scientific notation, where you express a number with between 1 and 3 digits to the left of the decimal point, times 10 to the power of 3, but use the appropriate SI prefix to represent 10 to the power of 3. Note, it is common to express floating numbers with a comma every three decimal places (power of 103). So the number
13,000,000,000g = 13 x109g or 13 gigagrams.
You need to know the SI prefixes from section 1B.1.3.3.
Logarithms
Logarithms are very similar to scientific notation, but the pre-exponential is expressed as 10 to the power of a fraction, and the log of a number, is the power of 10 that equals a number. Logarithms are often not covered in general chemistry 1, but they are extensively used in general chemistry 2.
Deeper Look: Logarithms
This is optional material for general chemistry 1, but you will need to understand this for general chemistry 2.
Let’s consider the number 26794. It can be expressed in scientific notation or as a log
So logarithms are very similar to scientific notation, where the number being multiplied by 10 to a power is expressed as 10 to the power of a fraction. Please see the math review in Chapter 14 prelude for more information on logarithms, where there is a video relating logarithms to scientific notation
Significant Digits, Calculations and Rounding off Answers
It is important to be able to preserve the number of significant Figures in calculations based on the precision of the data used in those calculations, and there are two sets of rules, one for multiplication and division, and the other for addition and subtraction.
Multiplication and Division
The number of significant Figures in the product of multiplication or dividend of division is based on the valued used with the least number of significant digits.
Exercise \(\PageIndex{1}\)
How many significant Figures are in the answer to the following problem?
\[\frac{(3.42)(127)}{1.6}=? \nonumber\]
- Answer
-
Two, you do not even need to do the math to answer this question.
Addition and Subtraction
Here, the number of significant Figures is based on the least precise measured value. In this case you actually have to line the numbers up by the decimal point and do the math, and then limit the answer to the precision of the least precise value.
Consider:
Exercise \(\PageIndex{2}\)
How many significant Figures are in the answer for 99 + 1?.
- Answer
-
3, the answer is 100. or 1.00 x 102 because both numbers are known to the "ones" place.
Note: in the above exercise, a number with two sig. figs was added to a number with one sig fig., and the answer had three sig. figs. Unlike multiplication and division, you must do the math to determine the number of sig. figs for addition and subtraction.
Rounding Off Answers
When you have to truncate an answer by removing numbers you may round up or round down. Truncation is removing a digit from the number, and all digits smaller than that digit.
- If truncated digit is less than 5, round down (use value of preceding digit)
- if truncated value is greater than 5, round up (add one to preceding value).
- if truncated value equals 5, look at the next digit.
- if it is even, round down
- if it is odd, round up*
*Note, when the truncated value equals 5 you round in the direction that makes the answer end in an even number
Contradictions of the Rules
Sometimes the rules break down and you need to be aware of it. For example, if you weight two nickels and each nickel weights 5.0 grams, what is the weight of both nickels? Well, multiplication is the repetition of addition and so you could multiply the weight of each nickel by two
\[2\, \text{nickels} \left ( \dfrac{5.0 \,\text{grams}}{\text{nickel}} \right )= 10. \; \text{grams}\]
or you could add the mass of the two nickels
\[5.0\,g +5.0\,g = 10.0\,g\]
So when you multiplied the answer would have two significant Figures because the number with the least was the measured mass (the 2 nickels was an exact counted number), but when you added, they were all known to the tenths position, and so your answer has 3 significant digits.
Converting Floating Decimal Point Numbers to Scientific Notation
A closer look at scientific notation shows that the number is being represented by two parts, the pre-exponential and the exponential (10 to a power). To convert a floating decimal point number to scientific notation you need to multiple or divide the pre-exponential by 10 to a power, until there is just one digit to the left of the decimal point. Because this changes the value of the number, you need to divide or multiple that pre-exponential by 10 to a power (the second term of the notation, noting that division is represented by 10 to a negative exponent), so that they combined expression equals the value of the floating decimal point number.
This is easy to see by looking at the solutions to the next two examples, where you convert a fraction and a number greater than (or equal to) 10 to scientific notation.
Convert the following floating numbers to scientific notation
Exercise \(\PageIndex{3}\)
0.00456 (A number smaller than 1)
- Answer
-
Step 1: Since this number is less than 1 you need identify the power of 10 which you can multiply it by and set one digit to the left of the decimal. Here, we can multiply by 1000 or 103 and force the original number to have one digit to the left of the decimal, but we have changed its value.
0.00456 x 103 = 4.56 (note, which is not equal to the original number)
Step 2: Since we multiplied it by 103 we must also divide by 103 (effectively multiplying the original number by 1 and not changing it's value). Than we express the factor in the denominator as a power of 10.
\[\left ( 0.00456 \right )= \left ( 0.00456 \right )\left ( \frac{10^{3}}{10^{3}} \right )= \left ( \frac{0.00456\times 10^{3}}{10^{3}} \right )= \left ( \frac{4.56}{10^{3}} \right )= 4.56\times 10^{-3}\]
Note, once you get the hang of this you can just count the number of digits you need to move the decimal to bring the original number to one digit to the left of the decimal and multiply that number by 10 to the negative value of those digits. Here we moved the decimal three positions to the right, so we multiply by 10-3 (ie., divide by 103).
Exercise \(\PageIndex{4}\)
456.00, (A number larger than or equal to 10)
- Answer
-
Since this number has more than one digit to the left of the decimal, we must divide it by a factor of 10 to make it smaller so that one digit is to the left of the decimal. Then we must multiply everything by 10 to the same power that we divided by 10 in the first step.
\[\left ( 456.00 \right )= \left ( 456.00 \right )\left ( \frac{10^{2}}{10^{2}} \right )= \left ( \frac{456.00\times 10^{2}}{10^{2}} \right )= \left ( 4.56\times 10^{2} \right )\]
Note, once you get the hang of this you can just count the number of digits you need to move the decimal to bring the original number to one digit to the left of the decimal and multiply that number by 10 to the value of those digits. Here we moved the decimal two positions to the left, making it 100 times smaller, so we multiply by 102 to maintain the same value.
Significant Digits:
How can you express the number 400 to 2 significant digits?
You must use scientific notation.
4.0 x 102
As scientific notation always has a decimal point there is never a problem expressing significant digits.
Mathematical Operations with Scientific Notation
Multiplication and Division
Try plugging the following number into your calculator. Odds are you will get an error message, and you need to know how to solve calculations that are bigger or smaller than your calculator can handle.
\[\frac{\left ( 9.47\times 10^{598} \right )\left ( 7.39\times 10^{-98} \right )}{\left ( 1.432\times 10^{-645} \right )\left ( 2.46\times 10^{4} \right )}\]
The trick is to break the above equation into two parts, one dealing with the pre-exponential, and one dealing with the powers of 10. The following video shows how to do this, and the above problem is solved in the exercise below the video.
Example \(\PageIndex{1}\): multiplication and division with scientific notation
Solve the following problem
\[\frac{\left ( 3.00\times 10^{480} \right )\left ( 4.0\times 10^{-20} \right )}{\left ( 2.00\times 10^{-500} \right )\left ( 1.50\times 10^{70} \right )}\nonumber\]
Solution
(4.0x10890)
Exercise \(\PageIndex{5}\)
Solve the following problem.
\[\frac{\left ( 9.47\times 10^{598} \right )\left ( 7.39\times 10^{-98} \right )}{\left ( 1.432\times 10^{-645} \right )\left ( 2.46\times 10^{4} \right )} \nonumber \]
- Answer
-
\[= \left ( \frac{9.47x7.39}{1.432x2.46} \right )\left ( \frac{10^{598}\times 10^{-98}}{10^{-645}\times 10^{4}} \right ) \nonumber\]
\[= \left ( 19.86\times 10^{\left ( 598-98+645-4 \right )} \right )\nonumber \]
\[= \left ( 19.9\times 10^{1141} \right )= \left ( 1.99\times 10^{1142} \right )\nonumber \]
Addition and Subtraction
This can easily be solved with your calculator, the challenge is how do you determine the number of significant Figures in the answer. To do this you need to line them up by multiplying everything by the same power of 10.
Consider the following problem
4.860 x 1012 + 9.7 x 1010 + 3.68x1011
Would you believe the answer has four significant Figures? (The solution to this problem is below the following video.)
Example \(\PageIndex{2}\): Addition and subtraction with scientific notation
Solve the following problem and report answer to the correct number of significant Figures:
3 x 1023 + 3.00 x 1025
Solution
(3.00 x 1025)
Exercise \(\PageIndex{4}\)
4.860 x 1012 + 9.7 x 1010 + 3.68x1011
- Answer
-
You first need to express all numbers to the same power so you can line up the decimal point. It is suggested that you choose the largest power and make everything else a fraction.
4.860 x 1012
+0.097 x 1012
+0.368 x 1012
5.325 x 1012
Unit Conversions
As chemists have different ways of expressing measurements they need to be able to convert between different units. Central to this is the concept of an equivalence statement which says two ways of representing the same thing are equivalent. For example 12 in = 1 foot is an equivalence statement. An equivalence statement allows you to convert from one unit to the other, and this is done by creating a conversion factor. A conversion factor is simply the ration of one part of the equivalence statement to the other, where the numerator has the unit you want to convert to, and the denominator has the unit you want to transform from.
12 in = 1 foot
has two conversion factors
- to convert from feet to inches multiply by the conversion factor of \[ \left (\frac{12 \; in}{ 1 \; foot} \right)\]
- to convert from inches to feet multiply by the conversion factor of \[\left (\frac{1 \; foot}{12 \;inches} \right)\]
There are three fundamental types of conversions that you will need to be able to perform. The first involve different ways of expressing numbers within the same scale, which include transforming floating numbers to SI prefixes and scientific notation. The second involves converting from one scale to another, like the metric system to the English [read: American] system, and you are expected to know that 2.54 cm=1 inch [read: memorize that]. The third involves physical constants of specific substance, where you convert from one type of physical measurement to another. Density is such a physical constant, and it allows you to convert mass measurements to volume.
What is consistent with each of these three different types is that they all represent an equivalence statement, and from the equivalence statement you can generate a conversion factor.
Three Basic Types Conversions
A. Different prefixes of same base scale (cm -> Km)
B. Conversions between scales (cm -> in)
C. Physical Constants d=M/V
Conversions involving SI Prefixes
2 Basic Techniques:
1. Basic Technique : Use two equivalence statements
2. Advanced Technique: Use one equivalent statement
Basic Technique
How many micrograms are in a kilogram?
Set up conversion factors based on an equivalence statements.
1 kg = 103 g and 1 µg = 10-6 g
so, this gives two conversion factors
\[\left ( \frac{10^{3}g}{1kg} \right )and \left ( \frac{1\mu g}{10^{-6}g} \right )\]
You have two conversion factors. In the first you will convert kilograms to grams, and in the second you will convert the grams to micrograms.
\[\left ( 1kg \right )\left ( \frac{10^{3}g}{1kg} \right )\left ( \frac{1\mu g}{10^{-6}g} \right )= 10^{3-\left ( -6 \right )}\mu g = 10^{9}\mu g \]
In the advanced technique (below) you will solve this problem with only one conversion factor.
Example \(\PageIndex{3}\)
3x1023fsec = ? Gsec
Solution
(3x10-1 Gsec)
Exercise \(\PageIndex{5}\)
Convert 32.4 ng to Mg.
- Answer
-
\[\left ( 32.4ng \right )\left ( \frac{10^{-9}g}{ng} \right )\left ( \frac{Mg}{10^{6}g} \right )= 32.4\times 10^{\left ( -9-6 \right )}Mg = 32.4\times 10^{-15}Mg = 3.24\times 10^{-14}Mg \]
Advanced Technique
How many micrograms are in a kilogram?
Set up a single conversion factor based on an equivalence statement.
The trick here is to multiple both SI prefixes by the number that makes them equal to one. That is, if a micro is one millionth, than there are a million micro in one, and and likewise if a kilo is a thousand, than a thousandth of a kilo is one, and so a million micrograms equals a thousandth of a kilogram.
\[1\mu g= 10^{-6}g\therefore 1g= 10^{6}\mu g\]
\[1kg= 10^{3}g\therefore 1g= 10^{-3}kg\]
so, this gives two conversion factors
\[\frac{10^{6}\mu g}{10^{-3}kg}=\underbrace{ \frac{10^{9}\mu g}{1kg}}_{\text{converts kg to }\mu g}\leftarrow or\rightarrow \frac{10^{-3}kg}{10^{6}\mu g}= \underbrace{\frac{10^{-9}kg}{1\mu g}}_{converts \; \mu g \; to \; kg} \]
(Use the ratio which has the starting unit in the denominator and ending unit in the numerator.)
See Video \(\PageIndex{4}\) for the advanced technique combined with scientific notation
(next video below)
Exercise \(\PageIndex{6}\)
Convert 32.4 ng to Mg..
- Answer
-
1 g = 10-6 Mg = 109 ng
\[\left ( 32.4ng \right )\left ( \frac{10^{-6}Mg}{10^{9}ng} \right )= \left ( 32.4\times 10^{\left ( -6-9 \right )}Mg \right )= \left ( 32.4\times 10^{-15}Mg \right )= \left ( 3.24\times 10^{-14}Mg \right )\]
SI Prefix Conversions & Scientific Notation
The following video shows you how to do SI conversions of numbers expressed in scientific notation. The trick is to write the SI prefix in terms of its power of 10.
Example \(\PageIndex{1}\)
Convert: 3 x 1023 kiloseconds to nanoseconds.
Solution
(3 x 1035 nsec)
Exercise \(\PageIndex{7}\)
Convert 3.57x1014 ml to ml
- Answer
-
\[\left ( 3.57\times 10^{14}\mu l \right )\left ( \frac{10^{3}ml}{10^{6}\mu l} \right )= \left ( 3.57\times 10^{\left ( 14+3-6 \right )}ml \right )= \left ( 3.57\times 10^{11}ml \right )\]
Conversions Between Different Scales
You need equivalence statements between different units. These are the same techniques as the previous, just that the conversions are between different types of units.
Note
UALR students are required to memorize the following conversion factors
- 2.54 cm=1 in (exact) (Metric length to English Length)
- 1 ml = 1 cm3 (Metric volume to Metric length cubed)
Exercise \(\PageIndex{8}\)
How many dm are in an object which is 2.450 feet long?
given 12 in = 1 ft,
- Answer
-
\[\left ( 2.450\,ft \right )\left ( \frac{12\in}{1\,ft} \right )\left ( \frac{2.54\,cm}{1\,in} \right )\left ( \frac{10^{1}dm}{10^{2}cm} \right )= \left ( 7.4676\,dm \right )= \left ( 7.468\,dm \right )\]
Example \(\PageIndex{5}\)
What is the volume in Gallons of a 1.00 cubic foot object?
(Try this first, the answer is at the bottom of the video. )
Solution
Note, you are required to know a 1 ml =1 cm3, 1 in = 2.54 cm, and of course the SI prefix conversions.
Exercise \(\PageIndex{9}\)
How many gallons are in a cuboid can that is 2.05 ft wide, 12.4 inches long and 1.30 yards deep?
Given: 4 qt = 1 gal, 1.057qt=1L, 12in = 1 ft, 3 ft = 1 yd:
- Answer
-
\[\left ( 2.05ft \right )\left ( 12.4in \right )\left ( 1.3yd \right )\left ( \frac{3ft}{1yd} \right )\left ( \frac{12in}{1ft} \right )^{2}\left ( \frac{2.54cm}{1in} \right )^{3}\left ( \frac{1mL}{1cm^{3}} \right )\left ( \frac{1L}{1000mL} \right )\left ( \frac{1.057qt}{1L} \right )\left ( \frac{1gal}{4qt} \right )= 61.8gal\]
Conversions Involving Physical Constants
Many substances have physical properties showing a linear relationship that can be described by the equation of a straight line Y=mX+b. If these are extensive properties (section 1.6.1), then the Y intercept (b) equals zero and this takes the form of Y=mX, which is an equivalence statement, and allows one to convert measurements in the unit of X to those of the unit of Y. These are often tabulated by the constant m \( \left (m=\frac{Y}{X} \right )\), which allows conversions of measurements between the units of X and Y.
This can be understood by looking at one such constant, density, which is the linear proportionality constant for the extensive properties of mass and volume, where m = dv (the y intercept is zero because if you have nothing, its mass and volume both equal zero - review extensive properties). So if you know the density of something, you can convert its mass to volume, or its volume to mass. Therefor it is common to compile tables of densities and other physical constants.
It should be stated that although these generate conversion factors, they are really describing relationships, that is, mass and volume describe completely different aspects of matter and density allows you to convert from one to another. This is completely different than definind 12 inches as a foot, where inches and feet are different units that describe the same thing.
Conversions Involving Density
Density is defined as
\[d=\frac{m}{v} \]
This gives the equivalence statement
\[m=dv\]
and there are thus two conversion factors
\[\underbrace{ \frac{m}{d}}_{\text{converts mass to volume }}\leftarrow or\rightarrow \underbrace{dv}_{\text{converts volume to mass }} \]
That is, dividing the mass by the density converts it to its volume, and multiply the volume by the density converts its volume to its mass
Example \(\PageIndex{6}\)
How many liters is the volume of 12.140 kg of Iridium? Use 22.65 g/ml as the density of Iridium
(Please do this before looking up the answer at the bottom of the video.)
Solution
0.5360 L, note, the number of sig figs was determined by the density and not the starting mass.
Exercise \(\PageIndex{1}\)
How many pounds would 61.8 gallons of water weight if at a given temperature the water has a density of 0.98g/mL?
- Answer
-
\[\left ( 61.8gal \right )\left ( \frac{4qt}{1gal} \right )\left ( \frac{1L}{1.057qt} \right )\left ( \frac{1000ml}{1L} \right )\left ( \frac{0.98g}{mL} \right )\left ( \frac{1lb}{453.6g} \right )= 505.27lb= 510lb\]
Note the answer has 2 significant digits because of the density.
Fractions and Percentage
A fraction is a \(\frac{Part}{Whole}\), with a value between 0 and 1. Percentage is simply the fraction x 100, where the 100 is a defined number. Fractions and percentages are commonly used to quantify a multicomponent system or mixture. In chemistry there are sort of two ways of quantifying matter, to count the number of particles or measure the mass of each type of particle. It is common (although not required) to use the fraction to represent the fraction of a sample based on the number of particles (mole fraction), and use % to represent the mass distribution (mass %).
Note: \[\sum fractions = 1 \\ \sum percents=100 \]
Exercise \(\PageIndex{10}\)
What is the mass % water if 1 gram of a salt is added to 99.0 g of water?
- Answer
-
99.0%. You may ask, why three significant digits? \[\frac{99.0}{99.0+1}100=\frac{99.0}{100.}100=99.0%\]
Vocabulary
Test Yourself
Query \(\PageIndex{1}\)
Query \(\PageIndex{1}\)
Query \(\PageIndex{1}\)
Query \(\PageIndex{1}\)
Contributors and Attributions
Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, rebelford@ualr.edu. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:
- Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook
- Ronia Kattoum (Learning Objectives)
- Elena Lisitsyna (H5P interactive modules)
- All images from Wikipedia, WikiMedia Commons or other ChemWiki articles