Dimensional Analysis is a technique for solving problems. Particularly, this technique is suited to converting one kind of unit into another.
Converting one unit to another changes nothing: it just expresses a distance, a volume, a mass (etc) in different units. In effect, you are multiplying by one. Using it is the easiest way to find out how many miles someone from Canada means when they say that the hockey rink is about 42 km away. Here’s what you do:
1 mi
42 km × ———————— = 26 mi This works because 1 mi = 1.61 km
1.61 km
Multiplying by any conversion factor is just like multiplying by one. You change the number but not the quantity.
The number of miles is directly proportional to the number of kilometers. The algebraic equation that leads to the math above is:
x mi 1 mi ——————— = ——————— which is: 42 km 1.61 km
1 mi
x mi = ——————— × 42 km so x mi = 26 mi
1.61 km
Notice especially the way the units cancel each other out so
that you are left with the units you want and you cancel the units you have.
You can do the same with other units, too.
For example: 365 day 1 yr
365 days = 1 yr can be written as ——————— = 1 or ——————— = 1
1 yr 365 day
These proportions are called unit factors because they are equal to the number one. Remember, multiplying by the number one does not change the numerical value! The unit factor is used to convert one unit to another unit as in the examples on this page. Pick the unit factor you use carefully as the units must cancel out. If the unit to be cancelled is not part of a fraction or is in the numerator then the unit factor must be written with that unit in the denominator. Here are more examples:
How many seconds are in one year?
365 day 24 hr 60 min 60 s
1 yr × ——————— × ——————— × ——————— × ——————— = 31,536,000 s
1 yr 1 day 1 hr 1 min
Notice that the unit you want to cancel goes on the opposite side of the
conversion factor from the unit you want to replace it: we want the unit
‘yr’ to cancel to be replaced by ‘day’.
Express 25 miles per hour as meters per second
mi 1 km 1000 m 1 hr 1 min
25 ——— × ——————— × ——————— × ——————— × ——————— = 11 m/s
hr 0.621 mi 1 km 60 min 60 s
The math for all of this is really quite simple: multiply by the top, divide by the bottom. Do this for each conversion factor, cancelling units as you go. Do not record intermediate values from your calculator: do all calculations as one long series.
When it comes to knowing how to use the Unit Factor Process (also known as dimensional analisys) there are two basic skills. First, building a unit factor. Second, setting up and carrying out the calculation.
One quantity (like how much milk is left in the jug) can be expressed using different units: saying 1 quart is left is the same as saying 2 pints are left or saying 4 cups are left or even 950 mL are left. A unit equality gives the relationship between two units. It can be used to make two unit factors.
For example:
1 lb 0.454 kg
——————— ———————
0.454 kg 1 lb
This works because of some simple algebra:
1 lb 0.454 kg 1 lb
1 lb = 0.454 kg ----------- = ----------- ----------- = 1
0.454 kg 0.454 kg 0.454 kg
divide both sides by 0.454 kg 0.454 kg/0.454 kg = 1!
The math above gives the unit factor on the left, but what about the one on the right? Can you figure out the necessary math to get that one?
Which of the unit factors do you choose to solve a problem? It depends.
If you have a quantity expressed in kg then choose the unit factor on the left.
1 lb
5.0 kg × ——————— = 11 lb
0.454 kg
0.454 kg
11 lb × ——————— = 5.0 kg
1 lb
If you have a quantity expressed in lb then choose the unit factor on the right.
Why do you choose the unit factors as described in the examples above? You choose them that way so that you cancel out the units you are changing from and introduce the units you are changing to.
Here is a bigger example.
Convert 2 years to milliseconds (ms).2 yr 365 day 24 hr 60 min 60 s 1,000 ms ------ x -------- x -------- x -------- x -------- x ---------- = 6.3072 × 1010 ms 1 1 yr 1 day 1 hr 1 min 1s Enter this into your calculator as 2 × 365 × 24 × 60 × 60 × 1,000 = 6.3072 × 1010
If it helps you to keep track of what you are doing then make a small chart like the one at left. The important thing to note in this example is that the unit to be cancelled out is always on the opposite side of the fraction bar. They cancel out like the variable x in the following algebra problem:
Simplify:
3y 3y 9y
3x · ----- → 3x · ----- → -----
2x 2x 2
Here is an example in which you must cancel units both in the
numerator and the denominator:
Convert 2.88 × 104 km/hr to mi/s
2.88×104 km 1 hr 1 min 0.621 mi ------------ x -------- x -------- x -------- = 4.97 mi/s 1 hr 60 min 60 s 1 kmType into your calculator: 2.88e4 ÷ 60 ÷ 60 × 0.621 =
| Linear Measure | ||
| 2.54 cm | = | 1 in |
| 12 in | = | 1 ft |
| 5280 ft | = | 1 mi |
| 1 m | = | 3.28 ft |
| 1 km | = | 0.621 mi |
| 1 furlong | = | 660 ft |
| Weight/Mass | ||
| 1 stone | = | 14 lbs |
| 1 lb | = | 0.454 kg |
| 1 oz | = | 28.3 g |
| 16 oz | = | 1 lb |
| 1 metric ton | = | 1.10 ton |
| 1 metric ton | = | 103 kg |
| Volumetric Measure | ||
| 1 cm3 | = | 1 mL |
| 1 gal | = | 3.78 L |
| 14.79 mL | = | 1 tbsp. |
| 1 m3 | = | 103 L |
| 1 cup | = | 237 mL |
| 1 gal | = | 4 quarts |
| Time | ||
| 60 s | = | 1 min |
| 60 min | = | 1 hr |
| 24 hr | = | 1 day |
| 365 day | = | 1 yr |
| 10 yr | = | 1 decade |
| 100 yr | = | 1 century |
Do the following conversions using the unit factor process. Show your work for each step and cancel out units as you work. See the examples on the first page for how you should show your work! Express your answers in scientific notation when appropriate.
Common units of speed are miles per hour (mi/hr), kilometers per hour (km/hr), meters per second (m/s), centimeters per second (cm/s), kilometers per second (km/s), and miles per second (mi/s). Show work as before and use scientific notation.
Convert each of the following to all of the given units. Show work for each conversion.
For each of the following numbers calculate how long it would take in days and years to count to that number (1, 2, 3, 4, 5, 6, …) if it takes one second to say each number? There will be two answers to each problem! If you read these directions then give yourself a pat on the back and one extra point.
Use dimensional analysis to solve these problems.
The chemical unit of the mole is extremely useful. It allows chemists to count atoms and molecules just by weighing them on a lab balance. This is because there is a special relationship between the mass of the atom, the mole and the mass of a mole of atoms.
The periodic table specifies the average mass of each type of atom. The types are called elements and each element has a number under its symbol in the table which gives the average mass of atoms of that element. The box that contains carbon in the periodic table is shown at right. The number at top is the atomic number and tells how many protons are in that element. The letter(s) show the symbol for the element.
For our purposes the number at the bottom of the box has units of grams per mole (g/mol). That is, one mole of carbon has a mass of 12.01 g. This can be written as a unit equality: 12.01 g C = 1 mol C. A unit equality like this can be written for any element in the periodic table. For example, 16.00 g O = 1 mol O , 22.99 g Na = 1 mol Na, 14.01 g N = 1 mol N and so on. When using the unit g/mol the number at the bottom of the box is called the molar mass. Carbon (C) has a molar mass of 12.01 g/mol, oxygen (O) has a molar mass of 16.00 g/mol and so on.
The mole is a number that you can think of like the chemist’s dozen. A normal dozen has 12 items but a baker’s dozen has 13 items. A chemist’s dozen has a lot more items: 6.02 × 1023. Since the mole is just a number the unit equality for the mole is: 1 mole = 6.02 × 1023. The number does not have a unit but you can think of it this way: 1 mole = 6.02 × 1023 objects of any kind. This is just the same thing as the fact that a dozen is twelve things of any kind. The unit equality for an ordinary dozen is 1 dozen = 12 objects of any kind.
Examples:
Find how many atoms of carbon are in 36.03 g of carbon
The molar mass of carbon is 12.01 g/mol
1 mol 6.02 x 1023
36.03 g C x --------- x ------------- = 18.06 x 1023
12.01 g 1 mol atoms of carbon
Find the mass of oxgyen in grams of 9.04 x 1023 atoms of oxygen
The molar mass of oxygen is 16.00 g/mol
1 mol 16.00 g
9.04 x 1023 atoms of O x ------------- x --------- = 24.0 g
6.02 x 1023 1 mol mass of oxygen
Do these problems as homework.