DIMENSIONAL ANALYSIS EXPLAINED

The foundational principle of dimensional analysis is that a word (unit) divided by the same word (unit) is the equivalent numerical result as dividing a number (i.e. 24) by itself.

24
24		 =    1 meter
meter		 =    1

In dimensional analysis, conversion factors are organized so that the similar units (i.e. meters in the example above) cancel out and become 1 as a result of one unit being in the numerator and the same unit being in the denominator. A conversion factor is formed from an equality between two different units . . . .

1000 m = 1 km

dividing both sides by 1 km gives . . . .

1000 m
1 km		 = 1 km
1 km

which reduces to . . . .

1000 m
1 km		 = 1

The "conversion factor" is  
1000 m
1 km
 .  If both sides were divided by 1000 m instead of 1 km, the conversion factor is  
1 km
1000 m
 .

Question: When would you use these conversion factors?

        Answer: When converting from km to m, use 
1000 m
1 km
. . . . when converting m to km, use 
1 km
1000 m

Example: Convert 4.5 m to km

        Solution: 
4.5 m  ×  1 km
1000 m		  =   4.5E-3 km 
The m after the 4.5 is in the numerator and cancels out with the m after the 1000 in the denominator.

Explanation: a conversion factor simply multiplies the original quantity (4.5 m) by "1" since 1000 m = 1 km. This means that the original length is unchanged - it is, however, expressed in a different unit (km).

Extension: multiple conversion factors can be arranged to convert a value in one unit into its corresponding value in a unit for which the direct conversion is not known. The conversion of 1.23 miles into cm is accomplished as shown:

1.23 mi  ×  5280 ft
1 mi		  ×  12 in
1 ft		  ×  2.54 cm
1 in		  =   1.98E5 cm 

1.23 miles is simply multiplied by "1" three different times to display the same length of 1.23 miles as 1.98E5 centimeters.


Of course, if you knew that 6.21E-6 mile = 1 cm, you could have simply used one conversion factor . . . .

1.23 mi  ×  1 cm
6.21E-6 mi		  =   1.98E5 cm 

While there are hundreds of "single conversion factors", you would need to look most of them up each time you needed to "convert" a measurement. Instead, scientists memorize a few "essential" conversion factors and then use a series of conversions to arrive at the desired unit.

Listed below are some Common Conversion factors that you should commit to memory.

ENGLISH
Length Volume Mass
12 inches = 1 foot 8 ounces = 1 cup 16 ounces = 1 pound
3 feet = 1 yard 32 ounces = 1 quart 2000 pounds = 1 ton
1760 yards = 1 mile 2 cups = 1 pint
5280 feet = 1 mile 4 cups = 1 quart
2 pints = 1 quart
4 quarts = 1 gallon



METRIC
Length Volume Mass
1 femtometer = 1E-15 meters 1 femtoliter = 1E-15 liters 1 femtogram = 1E-15 grams
1 picometer = 1E-12 meters 1 picoliter = 1E-12 liters 1 picogram = 1E-12 grams
1 nanometer = 1E-9 meters 1 nanoliter = 1E-9 liters 1 nanogram = 1E-9 grams
1 micrometer = 1E-6 meters 1 microliter = 1E-6 liters 1 microgram = 1E-6 grams
1 millimeter = 1E-3 meters 1 milliliter = 1E-3 liters 1 milligram = 1E-3 grams
1 centimeter = 1E-2 meters 1 centiliter = 1E-2 liters 1 centigram = 1E-2 grams
1 kilometer = 1E3 meters 1 kiloliter = 1E3 liters 1 kilogram = 1E3 grams



METRIC ⇆ ENGLISH
Length Volume Mass
2.54 cm = 1 inch 1.057 quarts = 1 liter 453.6 grams = 1 pound
1 cm3 = 1 mL



CHEMISTRY
Length
1 Angstrom (Å) = 1E-10 meters
Number
since atoms and molecules are small and can't be seen, chemists refer to groups of atoms / molecules. The "grouping" of items is a common occurance in the English language: pair (2), week(7), dozen (12), score (20), gross (144), ream (500), grand (1000), etc. For chemists, the "group's name" is mole which has a numerical value of 6.022E23. We can talk about a mole of anything, but generally it is reserved for groups of atoms or molecules. The number, 6.022E23, is called Avogadro's Number.

Chemists do not have a balance that can weigh a single molecule, or even several hundred molecules. However, we can still talk about the mass of a molecule since we can calculate it from its Molar Mass (see below) and Avogadro's Number. When talking about the mass of a molecule, we use the unit amu (atomic mass unit).

6.022E23 amu = 1 g
Density
a conversion factor between a substance's mass and volume. For liquids and solids, g/mL or g/cm3 are common density units . . . for gases, g/L are typical density units.
You are expected to know that the density of H2O(l) is 1 g/mL - all other densities needed to work a problem are given to you in the text of the problem.
Molar Mass
a conversion factor that relates a substance's mass and moles and has units of g/mole. The molar mass is calculated using mass information located in the Periodic Table - a resource always available for in-class (quizzes / exams) and out-of-class work.
Molar Ratio
a conversion factor between moles. This can be the molar ratio of two reactants, a reactant and a product, or two products. The molar ratio is created from the coefficients in a balanced chemical equation.
Molarity
a conversion factor that relates the moles of dissolved solute and liters of solution (solute + solvent). This conversion factor is given to you in the text of the problem.

A COMPLEX DIMENSIONAL ANALYSIS PROBLEM

How many mL of a 3.20 M HCl solution is needed to completely neutralize 1.20E-1 in3 of solid sodium hydroxide (NaOH(s))?    DensityNaOH(s) = 2.13 g/cm3

First, let's write a balanced equation for the reaction between HCl(aq) and NaOH(s) . . . .

HCl(aq) + NaOH(s) → NaCl(aq) + H2O(l)

1.20E-1 in3 NaOH(s)  ×  (2.54)3 cm3 NaOH(s)
1 in3 NaOH(s)		  ×  2.13 g NaOH(s)
1 cm3 NaOH(s)		  ×  1 mole NaOH(s)
39.997 g NaOH(s)		  × 
1 mole HCl(aq)
1 mole NaOH(s)		  ×  1 L HCl(aq)
3.20 mole HCl(aq)		  ×  1 mL HCl(aq)
1E-3 L HCl(aq)		  =   32.7 mL 


You may be thinking . . . .

     "how would I know to start with 1.20E-1 in3 NaOH(s)"?

                     . . . . and then . . . .

     "how would I know the conversion factor 'order' "?

To answer the first question . . . . since the question asks for volume (mL) of HCl which is "1 unit", you would scan the question to find a piece of data that only contains "1 unit" . . . . this would be 1.20E-1 in3 . . . . note that 3.20 M is actually two units → 3.20 moles / 1 L.

If the question asked for an answer that has two units (molar mass, density, Molarity, etc.), then you would use the dimensional analysis approach displayed above in the numerator and in the denominator of a "large fraction". The starting numerator unit is converted into the ending numerator unit and the starting denominator unit is converted into the ending denominator unit.


To answer the second question . . . . you need a map!! . . . . luckily, we specialize in chemical cartography. Find the map you need here.