# High School Chemistry/Using Mathematics in Chemistry

Unit terms are the words following a measurement that tell you on which standard the measurement is based. Every measurement must have a unit term. The unit terms also follow the algebraic rules of exponents and cancellation. Carrying the unit terms through mathematical operations provide an indication as to whether the mathematical operation was carried out correctly. If the unit term of the answer is not correct, it is an indication that the mathematical operation was not done correctly.

## Lesson Objectives

• Use units in problem solving.
• Do problem solving using dimensional analysis.
• Use significant figures in calculations.

## Using Units in Problem Solving

Anytime we have to do a calculation, it is important to include the units along with the actual numbers. One reason is that you can often decide how to solve the problem simply by looking at the units. For example, let's say you are trying to calculate solubility. One of the units used for solubility is grams/liter (g/L). Imagine that you have forgotten how to do the mathematical calculation for this problem, but you have measured how many grams of a solid dissolved into a certain number of liters. Looking at the units of the values that you have (g and L) and at the units of the answer you want to get (g/L), you can figure out the mathematical set-up. The g/L unit allows you to know it needs to be "grams divided by liters".

You will also note that as you do a calculation, you will be working with units in a similar manner as you would a number. Just as with numbers, units can be divided out when that specific unit appears in the numerator as well as the denominator.

As a final note on units, think of them in an "apples and oranges" context. You can't subtract meters from kilometers without first converting the measurements into common units. Always check a measurement's units to make sure that they are appropriate for a given calculation.

## Using Conversion Factors

Conversion factors are used to convert one unit of measurement into another. A simple conversion factor can be used to convert meters into centimeters, or a more complex one can be used to convert miles per hour into meters per second. Since most calculations require measurements to be in certain units, you will find many uses for conversion factors. What always must be remembered is that a conversion factor has to represent a fact; this fact can either be simple or much more complex. For instance, you already know that 12 eggs equal 1 dozen. A more complex fact is that the speed of light is 1.86×105 miles/sec. Either one of these can be used as a conversion factor depending on what type of calculation you might be working with. The following section provides you with more examples of this.

## Dimensional Analysis

When using conversion factors (and for that matter, a lot of other calculations), a process called dimensional analysis is extremely useful. Dimensional analysis allows you to make a number of unit conversions in a single calculation. It will also help you keep the units straight.

 Example A car travels 58.5 miles, using 1.5 gallons of gasoline. How do you express this in kilometers/liter? You know that there are 3.78 liters in a gallon, and a kilometer is 0.62 miles. How would you make this conversion? Solution: You first need to write out a mathematical expression showing all your conversion factors and units: ${\frac {58.5\,{\text{miles}}}{1.5\,{\text{gallons}}}}\times {\frac {1\,{\text{gallon}}}{3.78\,{\text{liters}}}}\times {\frac {1\,{\text{kilometer}}}{0.62\,{\text{miles}}}}$ Next, you need to check for units to divide out: ${\frac {58.5\,{\overline {\text{miles}}}}{1.5\,{\underline {\text{gallons}}}}}\times {\frac {1\,{\overline {\text{gallon}}}}{3.78\,{\text{liters}}}}\times {\frac {1\,{\text{kilometer}}}{0.62\,{\underline {\text{miles}}}}}$ Notice that at this point you are left with kilometers in the numerator and liters in the denominator. Your last step is to multiply your numbers, and your answer will be in kilometers/liter: ${\frac {58.5}{1.5}}\times {\frac {1}{3.78\,{\text{liters}}}}\times {\frac {1\,{\text{kilometer}}}{0.62}}={\frac {16.64106503\,{\text{kilometers}}}{\text{liter}}}$ This is the answer that your calculator will give to you. However it is not the correct answer. For that we have to proceed to the next section.

## Using Significant Figures in Multiplication and Division

Whenever we do a calculation, we need to pay attention to the significant figures. The rule is that your final answer can only be as precise as your least precise measurement. This means that the least precise tool used for any measurement in the calculation will determine how precise the answer will be.

For multiplication and division, first determine the number of significant figures in each of the measurements; the number of significant figures in your answer will be the same as the least number in the calculation. For example, if you multiplied the number 1.02584 by 2.1, your answer can only have two significant figures. The same rule applies for division.

 Example Divide the number 125.688 by 14.01. Express your answer using correct significant figures. Solution: 125.688 has 6 significant figures, and 14.01 has 4. Therefore, your answer can only have 4 significant figures. ${\frac {125.688}{14.01}}=8.971$ An important point to remember is that if you are multiplying or dividing by an exact number, then you treat that number as having an infinite number of significant figures. An exact number is a number that is written without all of its known significant figures. For instance, one meter or one dozen have many significant figures but we just don't write all of them, so these kind of measurements never determine the number of significant figures in a calculation. There is another type of exact number and that is using a measurement such as five people, ten dogs, or one cat. Again, these do not determine the number of significant figures in an answer.

 Example You have one dozen identical objects, with a total mass of 46.011 grams. What is their average mass? Solution: ${\tfrac {46.011}{12}}=3.83425$ This rounds off to 3.8342 (5 significant figures, the same as 46.011)

## Using Significant Figures in Addition and Subtraction

There is a different rule for determining significant figures when adding or subtracting measurements. Now, you will need to look for the measurement with the least number of significant figures to the right of the decimal place; this number of decimal places will determine the number of significant figures to be used in the answer.

 Example What is the sum of 14.3 and 12.887? Solution: 14.3 + 12.887 = 27.187 The number 14.3 only has 1 digit to the right of the decimal point, so our answer is rounded off to 27.2.

## Lesson Summary

• Dimensional analysis aids in problem solving.
• Conversion factors are created by unit analysis.
• Significant figures must be carried through mathematical operations.
• The answer for an addition or subtraction problem must have digits no further to the right than the shortest addend.
• The answer for a multiplication or division problem must have the same number of significant figures as the factor with the fewest significant figures.

## Review Questions

1. Perform the following calculations and give your answer with the correct number of significant figures:
(a) 0.1886 × 12
(b) ${\tfrac {2.995}{0.16685}}$ (c) ${\tfrac {1210}{0.1223}}$ (d) 910 × 0.18945
2. Perform the following calculations and give your answer with the correct number of significant figures:
(a) 10.5 + 11.62
(b) 0.01223 + 1.01 =1.02
(c) 19.85 − 0.0113
3. Do the following calculations without a calculator:
(a) (2.0×103)(3.0×104)
(b) (5.0×10−5)(5.0×108)
(c) (6.0×10−1)(7.0×10−4)
(d) ${\tfrac {(3.0\,\times \,10^{-4})(2.0\,\times \,10^{-4})}{2.0\,\times \,10^{-6}}}$ ## Vocabulary

dimensional analysis
A technique that involves the study of the dimensions (units) of physical quantities. It affords a convenient means of checking mathematical equations.