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Appendix B – The Factor Label Method

Introduction

Stoichiometry is the science that deals with quantitative relationships between the elements in a compound (substance stoichiometry) or between the compounds and/or elements involved in a chemical reaction (reaction stoichiometry). A typical stoichiometry problem involves the conversion of a given quantity of one substance into an equivalent amount of another substance, and the factor label method is a common way to solve these problems. This section introduces the method.

B.1. The Factor Label Method

Introduction

The factor label method uses factors to convert one quantity into another. The labels (units) on the quantities and factors are used to determine how the factors are used.

Objectives

B.1-1. Factors and Quantities

A quantity has only one unit, while a factor contains two units.

Quantity

Most numerical problems in chemistry involve the conversion of one amount into another equivalent amount. The following are some examples: The factor-label method is a common method to solve these types of problems. In the factor label method, we usually begin with a quantity and use factors to change its units or to change it into an equivalent amount of a different quantity, so it is important to understand the difference between a quantity and a factor. A quantity is simply an amount and is characterized by a single unit. The following are quantities.

Factor

A factor represents the ratio of two equivalent quantities and is characterized by two units. For example, the equality 60 s = 1 min is expressed by the factor 60 s/1 min, which is used to convert seconds ↔ minutes. The following are also factors. Factors can also be given by identifying the two related quantities as separate amounts:

B.1-2. Quantity or Factor Exercises

Exercise B.1:

Indicate whether each statement is giving a quantity or a factor.
    The length of the string is 6.5 cm.
  • quantity
  • factor The number has a single unit, so it is a quantity.
    The density of the solution is 1.2 g/mL.
  • quantity The number has two units (g and mL), so it is a factor. The equality represented by the factor is 1.2 g = 1 mL.
  • factor
    The concentration of salt in the ocean is about 0.5 mol/L.
  • quantity The number has two units (g and mL), so it is a factor. The equality represented by the factor is 0.5 mol salt = 1 L ocean.
  • factor
    A beaker contains 2 g of Pb, and another contains 5 g of Ca.
  • quantity
  • factor The amounts of Pb and Ca are independent of one another and represent two different quantities. For example, 1 g of one metal could be removed without affecting the mass of the other metal.
    A solution contains 2 g of Pb2+ for every 5 g of Ca2+.
  • quantity You are given the ratio of Pb2+ to Ca2+ in a solution, so this is a factor. It does not say that the solution contains 2 g of Pb2+ or 5 g of Ca2+.
  • factor
    Aluminum oxide contains 3 aluminum atoms for every 2 oxygen atoms.
  • quantity The numbers represent the ratio of the atoms not the number of atoms present, so they are part of a factor, not two quantities. The factor would be expressed as (3 Al atoms)/(2 O atoms).
  • factor
    A brass is 67% copper by mass.
  • quantity Percent by mass is a ratio of masses (the mass of one component per 100 g total). The factor in this case would be expressed as (67 g Cu)/(100 g brass).
  • factor
    A solution is made by dissolving 12 g of salt in 145 mL of water.
  • quantity The 12 g and the 145 mL are amounts, but because they are dependent on one another, they can be combined into a factor that converts between mass of salt and volume of water: (12 g salt)/(145 mL water).
  • factor

B.1-3. Single Factor Conversions

Factors are used to convert the units of a given quantity into those of an equivalent quantity with the factor-label method.
In the factor-label method, a quantity is multiplied by a conversion factor to convert it into an equivalent quantity with different units. The way that the factor appears in the multiplication is dictated by the labels (units) of the quantity and the factor. Conversion factors are always written so that the units of their denominators are the same as those of the given quantity, and the units of their numerators are the same as those of the desired quantity.
units cancel
Figure B.1: The Conversion Factor Method
One quantity can be converted to another by canceling the units.
For example, to determine the number of seconds in 3.5 minutes, you would multiply the given quantity (3.5 min) by the factor (60 s/1 min) to cancel the minutes and produce seconds.
3.5 min ×
60 s
1 min
 = 210 s
However, to determine the number of minutes in 231 seconds, you would multiple the given quantity (231 s) by the factor (1 min/60 s) to cancel seconds and produce minutes.
231 s ×
1 min
60 s
 = 3.85 min
When using the factor label method, it is good practice to include the substance in the label.
In order to maximize the utility of the factor label method, it is important to include the substance in the label. For example, 'g' and 'mL' are acceptable labels, but 'g Na' and 'mL solution' are much better. Consider that a solution of sodium chloride that has a concentration of 0.5 g/1 mL has a density of 1 g/1 mL. Written in this way the concentration and density are indistinguishable, which makes them far less useful when doing a problem. However, if we write that the concentration is 0.5 g NaCl/1 mL solution and the density is 1 g solution/1 mL solution, the two factors are readily distinguished and much more useful in doing a problem. Most of the problems we will deal with in this course involve more than one substance, so including the substance in the factors is very important.

B.1-4. Single Factor Exercises

Exercise B.2:

Use the factor-label method to solve the following. Enter each quantity and the abbreviation for its units as given in the problem separated by a single space. Do not include the substance.
A bag of marbles contains 3 red marbles for every 2 blue marbles. How many red marbles are in a bag that contains 18 blue marbles? Use RM and BM for red marbles and blue marbles, respectively.
o_18 BM_s Start with the amount you wish to convert.
×
o_3 RM_s This is the numerator of the last factor, so it has the same units as the answer.
=
o_27 RM_s Calculate the amount using the other values.
18 BM×
3 RM
2 BM
 = 27 RM
o_2 BM_s You must cancel the given units.
The density of a liquid is 0.740 g/mL. What is the volume of 86.0 g of the liquid? Use g and mL units.
o_86.0 g_s Start with the amount you wish to convert.
×
o_1 mL_s Must have units of answer.
=
o_116 mL_s
86.0 g×
1 mL
0.740 g
 = 116 mL
o_0.740 g_s Cancel given units.
The density of a liquid is 0.740 g/mL. What is the mass of 127 mL of the liquid?
o_127 mL_s Start with the amount you wish to convert.
×
o_0.740 g_s Must have units of answer.
=
o_94.0 g_s
127 mL×
0.740 g
1 mL
 = 94.0 g
o_1 mL_s Cancel given units.
What volume of solution would be required to obtain 18 g of NaCl if the solution is prepared by dissolving 35 g of NaCl in enough water to make 450 mL of solution?
o_18 g_s Start with the amount you wish to convert.
×
o_450 mL_s Must have units of answer.
=
o_230 mL_s Only two significant figures!
18 g×
450 mL
35 g
 = 230 mL

Note that the result of the calculation is 231 mL, but the answer is good to only two significant figures, so it is reported as 230 mL.
o_35 g_s Cancel units of given.
A solution is prepared by dissolving 12 g of sugar in enough water to make 285 mL of solution. How many g sugar are in 375 mL of solution?
o_375 mL_s Start with the amount you wish to convert.
×
o_12 g_s Must have units of answer.
=
o_16 g_s
375 mL×
12 g
285 mL
 = 16 g
o_285 mL_s Cancel units of given.

B.1-5. Multiple Factor Conversions

Conversion factors can be strung together, but the denominator of each should be the same as the numerator of the preceding factor.
Conversions cannot always be done conveniently with a single factor. In these instances, more than one factor may be required. However, the product of a quantity and a factor is simply another quantity, which can then be multiplied by another factor to produce yet another quantity. Each successive quantity can be multiplied by yet another factor to produce new quantities until the desired quantity is obtained. Once again, we use the labels of the factors to determine the order and manner in which the factors are used. The guiding rule is that the units of each denominator must be the same as the label of the preceding numerator. If this method is followed, the unit of the produced quantity will always be the same as the numerator of the final factor because all other units will cancel. The following example determines the number of years in 5.6 × 106 seconds.
5.6 × 106 s ×
1 min
60 s
 ×
1 h
60 min
 ×
1 d
24 h
 ×
1 yr
365 d
 = 0.18 yr
The first multiplication converts seconds to minutes, the next converts the minutes to hours, the next converts the hours to days, and the final conversion takes days into years, the desired units. Note that all of the units cancel except the desired units.

B.1-6. Multiple Factor Exercises

Exercise B.3:

Enter each quantity and the abbreviation for its units as given in the problem separated by a single space. Do not include the substance.
What is the total cost of the gasoline required for a 675 mile trip, if gasoline is 4.06 USD/gal and the car gets 25.3 mi/gal? Use mi, gal, and USD. USD = US dollar.
o_675 mi_s Start with the amount you wish to convert.
×
o_1 gal_s This is the numerator of the factor whose denominator has the same units as the given.
×
o_4.06 USD_s These units must be those of the answer.
=
o_108 USD_s
675 mi×
1 gal
25.3 mi
×
4.06 USD
1 gal
 = 108 USD
o_25.3 mi_s Cancel the units of the given.
o_1 gal_s These units must be the same as previous numerator.

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