2.5: Using Algebra in Chemistry
Lesson Objectives
The student will:
 perform algebraic manipulations to solve equations.
 use the density equation to solve for the density, mass, or volume when two of the quantities in the equation are known.
 construct conversion factors from equivalent measurements.
 apply the techniques of dimensional analysis to solving problems.
 perform metric conversions using dimensional analysis.
Vocabulary
 conversion factor
 dimensional analysis
Introduction
During your studies of chemistry (and physics as well), you will note that mathematical equations are used in a number of different applications. Many of these equations have a number of different variables that you will need to work with. You should also note that these equations will often require you to use measurements with their units. Algebra skills become very important here!
Solving Algebraic Equations
Chemists use algebraic equations to express relationships between quantities. An example of such an equation is the relationship between the density, mass, and volume of a substance: \begin{align*}D~ =~ \frac {m} {V}\end{align*}



\begin{align*}D = \frac {\text{mass}} {\text{volume}} = \frac {67.5 \ \text{g}} {5.00 \ \text{mL}} = 13.5 \ \text{g/mL}\end{align*}
D=massvolume=67.5 g5.00 mL=13.5 g/mL

\begin{align*}D = \frac {\text{mass}} {\text{volume}} = \frac {67.5 \ \text{g}} {5.00 \ \text{mL}} = 13.5 \ \text{g/mL}\end{align*}

You should notice both units and significant figures are carried through the mathematical operations.
Frequently, you may be asked to use the density equation to solve for a variable other than density. For example, you may be given measurements for density and mass and be asked to determine the volume.
Example:
The density of solid lead is \begin{align*}11.34 \ \text{g/mL}\end{align*}



Since \begin{align*}D = \frac {m} {V}\end{align*}
D=mV , then \begin{align*}V = \frac {m} {D}\end{align*}V=mD .

Since \begin{align*}D = \frac {m} {V}\end{align*}




\begin{align*}V = \frac {81.0 \ \text{g}} {11.34 \ \text{g/mL}} = 7.14 \ \text{mL}\end{align*}
V=81.0 g11.34 g/mL=7.14 mL

\begin{align*}V = \frac {81.0 \ \text{g}} {11.34 \ \text{g/mL}} = 7.14 \ \text{mL}\end{align*}

A common equation used in chemistry is \begin{align*}PV = nRT\end{align*}



\begin{align*}P = \frac {nRT} {V} \ \ \ \ \ V = \frac {nRT} {P} \ \ \ \ \ n = \frac {PV} {RT} \ \ \ \ \ R = \frac {PV} {nT} \ \ \ \ \ T = \frac {PV} {nR}\end{align*}
P=nRTV V=nRTP n=PVRT R=PVnT T=PVnR

\begin{align*}P = \frac {nRT} {V} \ \ \ \ \ V = \frac {nRT} {P} \ \ \ \ \ n = \frac {PV} {RT} \ \ \ \ \ R = \frac {PV} {nT} \ \ \ \ \ T = \frac {PV} {nR}\end{align*}

Make sure you recall these skills from algebra. If necessary, you should practice them.
Example:
Use the equation \begin{align*} \frac {A} {B} = \frac {C} {D}\end{align*}



\begin{align*}D = \frac {BC} {A} = \frac {(3.000 \ \text{mL})(326.96 \ \text{g})} {(15.1 \ \text{g})} = 65.0 \ \text{mL}\end{align*}
D=BCA=(3.000 mL)(326.96 g)(15.1 g)=65.0 mL

\begin{align*}D = \frac {BC} {A} = \frac {(3.000 \ \text{mL})(326.96 \ \text{g})} {(15.1 \ \text{g})} = 65.0 \ \text{mL}\end{align*}

The calculatordetermined value for this arithmetic may yield \begin{align*}64.956954 \ \text{mL}\end{align*}
Conversion Factors
A conversion factor is a factor used to convert one unit of measurement into another unit. A simple conversion factor can be used to convert meters into centimeters, or a more complex one can be used to convert miles per hour into meters per second. Since most calculations require measurements to be in certain units, you will find many uses for conversion factors. What must always be remembered is that a conversion factor has to represent a fact; because the conversion factor is a fact and not a measurement, the numbers in a conversion factor are exact. This fact can either be simple or complex. For instance, you probably already know the fact that 12 eggs equal 1 dozen. A more complex fact is that the speed of light is \begin{align*}1.86 \times 10^5\end{align*}
Dimensional Analysis
Frequently, it is necessary to convert units measuring the same quantity from one form to another. For example, it may be necessary to convert a length measurement in meters to millimeters. This process is quite simple if you follow a standard procedure called dimensional analysis (also known as unit analysis or the factorlabel method). Dimensional analysis is a technique that involves the study of the dimensions (units) of physical quantities. It is a convenient way to check mathematical equations. Dimensional analysis involves considering the units you presently have and the units you wish to end up with, as well as designing conversion factors that will cancel units you don’t want and produce units you do want. The conversion factors are created from the equivalency relationships between the units. For example, one unit of work is a newton meter (abbreviated N \begin{align*} \cdot \end{align*}
Suppose you want to convert 0.0856 meters into millimeters. In this case, you need only one conversion factor that will cancel the meters unit and create the millimeters unit. The conversion factor will be created from the relationship \begin{align*}1000 \ \text{mL} = 1 \ \text{m}\end{align*}



\begin{align*}(0.0856 \ \text{m}) \cdot ( \frac {1000 \ \text{mm}} {1 \ \text{m}}) = (0.0856 \ \cancel{\text{m}}) \cdot ( \frac {1000 \ \text{mm}} {1 \ \cancel{\text{m}}}) = 85.6 \ \text{mm}\end{align*}
(0.0856 m)⋅(1000 mm1 m)=(0.0856 m)⋅(1000 mm1 m)=85.6 mm

\begin{align*}(0.0856 \ \text{m}) \cdot ( \frac {1000 \ \text{mm}} {1 \ \text{m}}) = (0.0856 \ \cancel{\text{m}}) \cdot ( \frac {1000 \ \text{mm}} {1 \ \cancel{\text{m}}}) = 85.6 \ \text{mm}\end{align*}

In the above expression, the meter units will cancel and only the millimeter unit will remain.
Example:
Convert 1.53 g to cg.
The equivalency relationship is \begin{align*}1.00 \ \text{g} = 100 \ \text{cg}\end{align*}



\begin{align*}(1.53 \ \text{g}) \cdot ( \frac {100 \ \text{cg}} {1 \ \text{g}}) = 153 \ \text{cg}\end{align*}
(1.53 g)⋅(100 cg1 g)=153 cg

\begin{align*}(1.53 \ \text{g}) \cdot ( \frac {100 \ \text{cg}} {1 \ \text{g}}) = 153 \ \text{cg}\end{align*}

Example:
Convert 1000. in. to ft.
The equivalency between inches and feet is \begin{align*}12 \ \text{in.} = 1 \ \text{ft}\end{align*}



\begin{align*}(1000. \ \text{in.}) \cdot ( \frac {1 \ \text{ft}} {12 \ \text{in.}}) = 83.33 \ \text{ft}\end{align*}
(1000. in.)⋅(1 ft12 in.)=83.33 ft

\begin{align*}(1000. \ \text{in.}) \cdot ( \frac {1 \ \text{ft}} {12 \ \text{in.}}) = 83.33 \ \text{ft}\end{align*}

Each conversion factor is designed specifically for the problem. In the case of the conversion above, we need to cancel inches, so we know that the inches component in the conversion factor needs to be in the denominator.
Example:
Convert 425 klums to piks given the equivalency relationship \begin{align*}10 \ klums = 1 \ pik\end{align*}



\begin{align*}(425 \ klums) \cdot ( \frac {1 \ pik} {10 \ klums}) = 42.5 \ piks\end{align*}
(425 klums)⋅(1 pik10 klums)=42.5 piks

\begin{align*}(425 \ klums) \cdot ( \frac {1 \ pik} {10 \ klums}) = 42.5 \ piks\end{align*}

Sometimes, it is necessary to insert a series of conversion factors. Suppose we need to convert miles to kilometers, and the only equivalencies we know are \begin{align*}1 \ \text{mi} = 5,280 \ \text{ft}\end{align*}
Example:
Convert 12 mi to km.


 \begin{align*}(12 \ \text{mi}) \cdot ( \frac {5280 \ \ \text{ft}} {1 \ \text{mi}}) \cdot ( \frac {12 \ \text{in.}} {1 \ \text{ft}}) \cdot ( \frac {2.54 \ \text{cm}} {1 \ \text{in.}}) \cdot ( \frac {1 \ \text{m}} {100 \ \text{cm}}) \cdot ( \frac {1 \ \text{km}} {1000 \ \text{m}}) = 19 \ \text{km}\end{align*}

In each step, the previous unit is canceled and the next unit in the sequence is produced.
Conversion factors for area and volume can also be produced by this method.
Example:
Convert \begin{align*}1500 \ \text{cm}^2 \ \text{to} \ \text{m}^2\end{align*}.


 \begin{align*}(1500 \ \text{cm}^2) \cdot ( \frac {1 \ \text{m}} {100 \ \text{cm}})^2 = (1500 \ \text{cm}^2) \cdot ( \frac {1 \ \text{m}^2} {10,000 \ \text{cm}^2}) = 0.15 \ \text{m}^2\end{align*}

Example:
Convert \begin{align*}12.0 \ \text{in}^3 \ \text{to} \ \text{cm}^3\end{align*}.


 \begin{align*}(12.0 \ \text{in}^3) \cdot ( \frac {2.54 \ \text{cm}} {1 \ \text{in}})^3 = (12.0 \ \text{in}^3) \cdot ( \frac {16.4 \ \text{cm}^3} {1 \ \text{in}^3}) = 197 \ \text{cm}^3\end{align*}

Lesson Summary
 Conversion factors are used to convert one unit of measurement into another unit.
 Dimensional analysis involves considering both the units you presently have and the units you wish to end up with, as well as designing conversion factors that will cancel units you don’t want and produce units you do want.
Further Reading / Supplemental Links
Visit this website for a video series that reviews topics on measurement.
This website provides a math skills review on dimensional analysis.
Review Questions
 For the equation \begin{align*}PV = nRT\end{align*}, rewrite it so that it is in the form of \begin{align*}T =\end{align*}.
 The equation for density is \begin{align*}D = \frac {m} {V} \end{align*}. If \begin{align*}D\end{align*} is \begin{align*}12.8 \ \text{g/cm}^3\end{align*} and \begin{align*}m\end{align*} is \begin{align*}46.1 \ \text{g}\end{align*}, solve for \begin{align*}V\end{align*}, keeping significant figures in mind.
 The equation \begin{align*}P_1 \cdot V_1 = P_2 \cdot V_2\end{align*}, known as Boyle’s law, shows that gas pressure is inversely proportional to its volume. Rewrite Boyle’s law so it is in the form of \begin{align*}V_1 = \end{align*}.
 The density of a certain solid is measured and found to be \begin{align*}12.68 \ \text{g/mL}\end{align*}. Convert this measurement into \begin{align*}\text{kg/L}\end{align*}.
 In a nuclear chemistry experiment, an alpha particle is found to have a velocity of \begin{align*}14,285 \ \text{m/s}\end{align*}. Convert this measurement into \begin{align*}\text{miles/hour}\end{align*} (\begin{align*}\text{mi/h}\end{align*}).
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