Factor Label method and Dimensional Analysis:
Using factor label implies that labels can be factored by a multiplying and dividing scheme to achieve conversion from one unit to another. Likewise this factoring out of labels that represent units or dimensions is a form of analysis or dimensional analysis.
Conversion factors are based on equivalents. For example, 1000 m = 1 km. This can be converted into two conversion factors, namely.
( A )
1000 m
1 km
or (B)
1km
1000m . If you were to convert 25 m to km select conversion factor A to eliminate meters.
?
km

25 m x
1km
1000m .
Another equivalent is that 100 cm = 1m. Determine how many km are present in 25 cm.
?
km

25 cm x
1m
x
1km
100 m 1000m
When factor label is expressed as dimensional analysis the equal sign shown above is two vertical lines that separate what we want from the quantities utilized in the calculation itself. The multiplication sign is replaced with a single vertical line separating multiplied quantities.
Dimensional analysis is easier to read mainly due to the way fractions are written. In this class we are using dimensional analysis with occasional use of the factor label method.
Sample Problem:
Problems:
1.
How much time in nanocenturies is a 50-minute class period?
Equivalents: century = 100 yr.: 1 yr = 365.25 days: 24 hrs. = 1day;
60 min = 1hr.; 1 x 10 9 nanocenturies = 1 century.
2. How many yards are in 100 meters?
Equivalents: 1 yd = 3 ft, 1 ft = 12 in, 1 in = 2.54 cm, 100 cm = 1 m.
3. What is the speed of light in furlongs per fortnight? The speed of light = 3.0 x
10 8 m/s
Equivalents: 3600 s = 1 hour; 24 hr = 1 da; 14 da = 1fortnight;
1 furlong = 201.16800 meters.
4. There is an old expression about water that states “The world around a pint is a pound.” Is this true? Starting with 1.0 pint of water convert to pounds of water.
Equivalents: 1 quart = 1.06 L; 2 pints = 1 quart; 1000-mL = 1L; 1-mL (of water) =
1 gram; 454 grams = 1.0 pound

5.
Silver can be electroplated onto other metals. Amperage depends on the amount of charge (in coulombs) per second (I=Q/t). One amp = 1C/s. To electroplate silver one electron must be accepted for each atom of Ag that is plated. The charge on an electron is 1.602 x 10 -19 C (coulombs). If a 1.0 amp current is allowed to electroplate silver for 14 min and 32 s, how many atoms of Ag are plated?
6.
One Faraday (F) is equivalent to 96,485 C. Assume that enough time was allowed so that one Faraday of charge acted on plating Ag. Recall that each electron has a charge of 1.602 x 10 -19 C, and that every electron can plate one atom of silver, how many atoms would be plated? Assume this number of atoms is equivalent to 1.00 mole of atoms. How many moles of Ag are plated?
7. How many grams of Ag metal was plated by1.00 mole of atoms in problem number 5?
8. Zinc can be plated onto Fe to make galvanized metal. For every atom of Zn that is plated 2 electrons are required. How many atoms of Zn would be plated by the action of 1 Faraday? What would the mass of Zn plated if the electroplating were
100% efficient and 1 mole of Zn has a mass of 65.64 grams?