Conversion Factors

Introduction

Now, let's deal with conversion problems. If you are already familiar with doing calculations using conversion factors, feel free to move on to the next topic. Or try the Quick Quiz or Evaluation Quiz for this topic and then decide whether to move on.

A Simple Example

You've done conversion factor calculations before you even came to a chemistry class. For example, how many inches are there in two feet? The answer, or course, is 24 inches. I'm sure you could figure that out quickly, almost without thinking about it, simply because you have dealt with that kind of question before.

But now, I'd like you to think about it, not just do it automatically. How is it you come up with 24 inches from 2 feet? Well, you multiply 2 by 12. That gives you 24. But why multiply by 12? You multiply by 12 because there are 12 inches in one foot. This is set up in a rather formal fashion here (and in example 24 in your workbook). Let's go through the process step by step.

starting information conversion factor answer
2 feet x 12 inches
  1 foot
= 24 inches

This is what we call the conversion factor method. It also goes by other names. We start with "2 feet." We are going to change this to "inches," that is find out what 2 feet is equal to in inches. To do this we can multiply by a conversion factor that will change from feet (that goes on the bottom) to inches (that goes on the top). The equivalence between feet and inches goes into the conversion factor. The top part (the numerator) of the conversion factor has to be equivalent to the bottom part (the denominator). The relationship, of course, is that 12 inches is the same as 1 foot. Because the unit "foot" is in the denominator, it will cancel with the feet that we started with, leaving us with the unit "inches." Then calculating 2 x 12 1 tells us that we have 24 inches.

This has been a somewhat ridiculous example. To work out this problem in this way makes a difficult problem out of a simple one. If you can work out a problem simply, do so. The purpose of the example is to use something very familiar to introduce what may be a new technique.

Another Simple Example

Let's go through another similar example.  How many feet are there in 30 inches?   (This is also shown in example 25 in your workbook.)

starting information conversion factor answer
30 inches x    1 foot   
12 inches
= 2.5 feet

Here we want to change inches into feet. So, let's find out how many feet there are in 30 inches. Write down 30 inches first and then the conversion factor with inches on the bottom and feet on the top. The reason for this is that we are changing from inches to feet. Then put the right numbers with the units. So we have to multiply by one foot over 12 inches or by 1/12 in order to get the correct answer. The inches cancel out, and 30 times 1 divided by 12 gives 2.5 feet as the final answer.

Rounding Off

I should point out that I did not round the answer off to one digit even though I only have one digit written here (1 ft.). This relationship between feet and inches is known to be an exact relationship. There is no uncertainty or limits to its precision. Therefore, the answer is rounded off to the same degree of precision as the starting value.

Sometimes conversion factors are exact relationships and sometimes they are limited in their precision. You have to look at each case as it comes along.

Purpose of These Examples

The purpose of these examples has been to show you a method for doing certain types of calculations, even though you might be able to solve these particular examples without using the method.

In these examples a measurement made in one unit (such as feet) was converted into another unit (such as inches). To work these kinds of problems, you need to have a conversion factor--a relationship between the units. In these examples you needed to know that one foot equals 12 inches.

This approach can be used anytime you are converting from one thing to another if they are proportional to one another.

Metric Conversions

In this course you must be able to do conversions from one unit to another within the metric system. For example you would have to be able to change a measurement made in milliliters into liters, or a measurement made in meters into centimeters, or a measurement made in kilograms into grams or milligrams. To do such things, you need to know conversion factors that relate the different units.

Metric Units

In the metric system, all the units are related to one another by factors of ten as indicated by the prefixes. Factors of ten include 10, 100, 1,000, 1 million and so forth, as well as 1/10, 1/100, or 1/1,000. You should have already learned the most common of these prefixes. (They are also listed in example 26-a and 26-b in your workbook.) The unit can be meters, liters or grams.

General Relationships
1 milliunit = 0.001 unit or 1 unit = 1000 milliunits
1centiunit = 0.01 unit or 1 unit = 100 centiunits
1 kilounit = 1000 units or 1 unit = 0.001 kilounit

The definitions of the prefixes give you the conversion relationships you need. One milliliter is 1/1,000th of a liter, or .001 liter. One centimeter is 1/100th of a meter or .01 meter. One kilogram is 1,000 grams. Since people usually talk about how many of the smaller units there are in the larger one, this example also shows that you can look at one liter as 1,000 milliliters, and one meter as 100 centimeters. You can use those relationships and often they will be easier.

You might prefer to think of these prefixes lined up from smallest to largest. Whatever unit you are working with (meters, liters or grams), the prefixes to the left on this scale represent fractions of that unit. Deci- is one-tenth, centi- is one-hundredth, milli- is one-thousandth, and micro- is one-millionth. The prefixes on the right side of this scale represent multiples of that unit. Deka- is ten times, hecto- is one hundred times, kilo- is one thousand times, and mega- is one million times.1042ex26.bmp (481078 bytes)

To make sure you have the relationships figured out correctly, take the following quick quiz on metric units.

***Insert Quick Quiz***

Conversion Factors

The relationships between the different metric units can be used as conversion factors. For example, if you need to change a measured value from mg to g, you can multiply by the conversion factor of 1 g over 1000 mg.

45 mg  x      1 g   
1000 mg
= 0.045 g

To change a value from g to mg multiply by the conversion factor of 1000 mg over 1 g.

3.25 g  x 1000 mg
     1 g
= 3250 mg

As you become familiar with metric units, you will be able to move the decimal point back and forth the proper number of places to make metric conversions, rather than using this longer method of multiplying by 100 or 1,000 or dividing by those numbers.

45 mg = 0.045 g
<<<
3.25 g = 3250 mg
  >>>

Now, try the quick quiz on metric conversions. Then continue with metri-American conversions.

***Insert Qick Quiz***

Conversions Between Metric and American Units

Making conversions between the metric and American units is very similar to making metric conversions or any other type of conversion. For example you might be asked how many kilometers are equal to 15 miles or how many feet are there in 100 meters. These problems are solved in the same way as other conversion problems. One difference from metric conversions is that you will not be required you to memorize the conversion factors. We will give you a relationship between metric and current American units when you need to solve a problem, or have you look it up. A number of these relationships are listed here (and in example 28 in your workbook).

Some useful American-metric conversions
1 mile = 1.6 km 1 km = 0.62 mile
1 quart = 946 mL 1 cm = 0.394 inch
1 quart = 0.946 L 1 L = 1.06 quart
1 inch = 2.54 cm 1 g = 0.035 ounce
1 pound = 454 g 1 kg = 2.2 pounds

Multistep Conversions

Sometimes, the relationship available to you may not have the exact units you want. For example, in relating kilometers to miles,  you might be given the relationship or conversion factor between meters and miles. But you can make adjustments within the metric system fairly easily. It just makes two problems instead of one--change miles to meters and then meters to kilometers.

Consider this example.

How many centimeters are in 4.0 yards?
4.0 yards x 36 inches
1 yard
x 2.54 cm
1 inch
=

If you happened to know how many centimeters there are in a yard, this would not be the way to do this problem. The first conversion factor changes yards to inches. The second conversion factor changes inches to centimeters. In the calculation yards cancel and inches cancel, leaving us with centimeters.

Now take some time to work through the problems in the next quick quiz.

***Insert Quick Quiz***

Density Conversions

Now, let's move on to the last objective for the lesson and work with density again. You have experience in making the necessary measurements (or you will when you have completed the lab work) and calculating the density of materials. One use of density is to help identify materials. It can also be used to determine the concentration of certain mixtures such as battery acid, antifreeze, urine and blood.

Density as a Conversion Factor

Another use for density is as a conversion factor between mass and volume. It can be used as a conversion factor because it relates the two proportional properties of mass and volume. I should point out that mass and volume are proportional for a given material under given conditions. If you change materials or conditions, such as temperature or pressure, the density will change.

Suppose you wanted to know how much space a particular mass of a certain substance would occupy. To convert mass to volume (or vice versa), you must know the relationship between the mass and volume and that relationship is density.

Quick review:  This example shows you how to calculate density. You've already done that. You divide the mass by the volume and that gives you the density.
Mass = 5.0 g
Volume = 2.0 mL
What is the density?
Density =   5.0 g 
2.0 mL
= 2.5 g/mL
Now that we know the density for this material, we can calculate how much 6.0 milliliters of it would weigh as asked for here. Take the 6.0 milliliters and then use the density 2.5 grams per milliliter as a conversion factor. The 2.5 goes on top with the grams and the one milliliter goes on the bottom because you want to cancel out milliliters and end up with grams. The result is 15 grams.
How much would 6.0 mL weigh?
6.0 mL x 2.5 g
1 mL
= 15 g
Here we are asked how much volume would four grams of this material occupy? You start with 4.0 grams and multiply by one milliliter over 2.5 grams and you come up with the answer of 1.6 milliliters.
How much volume would 4.0 g take up?
4.0 g x 1 mL
2.5 g
= 1.6 mL

Algebraic Approach

An algebraic approach can be used for density calculations also. If you are familiar with algebra you know that it involves representing a relationship with an equation and using symbols to represent quantities.

You know that density equals mass divided by volume. That relationship can then be expressed in many ways by carefully rearranging the equation. Mass equals volume times density. Volume equals mass divided by density. Which form of the equation is used depends on what you are trying to figure out. D = m
v
D x v = m   or   m = D x v
v = m
D

This example asks the same question as before, but the approach to the answer is different. First write down the appropriate form of the equation--or write down the standard form and then rearrange it. Next substitute the known values into the equation and carry out the calculations.
Given that the density is 2.5 g/mL, how much would 6.0 mL weigh?
m = D x v

m = 2.5 g/mL x 6.0 mL

m = 15 g

In this example we are asked for the volume. The volume equals the mass over the density. So the volume is going to be equal to 4.0 g divided by 2.5 g/mL. Then carry out the calculations to get the volume.
Given that the density is 2.5 g/mL, how much volume would 4.0 g take up?
v = m
D
v = 4.0 g    
2.5 g/mL
v = 1.6 mL

Next, answer the questions in the following quick quiz using whichever method you prefer.