Lesson Explainer: Units and Equations Chemistry

In this explainer, we will learn how to define SI and derived units and write and manipulate algebraic equations.

Units of measurement were not standardized before the French Revolution. The length of the foot unit used to be different in England and France. The French foot, or pied unit, could even change as you moved from one French town to another. The lack of consistency was problematic and it prompted scientists to devise a set of new units that would be the same throughout the whole world. They proposed the metre (m) for measuring length and used the metre to define the gram (g) for measuring mass and the litre (L) for measuring volume. This set the basis of the metric system.

The metric system would then be revised about onehundred years later to create the SystΓ¨me Internationale d’UnitΓ©s or International System of Units, commonly abbreviated as SI. The SI is a set of seven standardized base units that are determined from precisely measured fundamental constants, such as the speed of light in a vacuum.

Definition: International System of Units (SI)

It is a set of seven standardized base units derived from the seven fundamental constants. The SI units are the metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), and candela (cd).

The table below lists the seven SI base quantities and base units.

Base QuantityBase Unit
Time (𝑑)Second (s)
Length (𝑙)Metre (m)
Mass (π‘š)Kilogram (kg)
Electric current (𝐼)Ampere (A)
Thermodynamic temperature (𝑇)Kelvin (K)
Amount of substance (𝑛)Mole (mol)
Luminous intensity (𝐼)Candela (cd)

Notice that volume is not included as an SI base quantity. This is because volume can be derived from the length.

Quantities, like volume, that can be derived by multiplying and dividing the base quantities are called derived quantities and the unit of the derived quantity is the SI derived unit. The following table shows some common derived quantities and derived units that are used by chemists to quantify different physical properties.

Derived QuantityDerived Unit
Volume (𝑉)m3
Density (𝜌)kgβ‹…mβˆ’3 or kgm
Concentration (𝑐)molβ‹…mβˆ’3 or molm
Energykgβ‹…m2β‹…sβˆ’2 or kgmsβ‹…οŠ¨οŠ¨ or joule (J)
Pressure (𝑃)kgβ‹…mβˆ’1β‹…sβˆ’2 or kgmsβ‹…οŠ¨ or pascal (Pa)
Molar heat capacitykgβ‹…m2β‹…sβˆ’2β‹…molβˆ’1β‹…Kβˆ’1 or kgmsmolKβ‹…β‹…β‹…οŠ¨οŠ¨ or JmolKβ‹…

There are many instances where the magnitude of the base unit is not the most appropriate for the dimension being measured. For example, while we may measure the height of the Eiffel Tower in metres (324 m), it is impractical to measure the radius of a hydrogen atom (0.00000000012 m) or the distance between Earth and Pluto (3β€Žβ€‰β€Ž207β€Žβ€‰β€Ž400β€Žβ€‰β€Ž000β€Žβ€‰β€Ž000 m) with the same unit.

Prefixes can be added to the SI base units to create new units that are larger or smaller than the original base unit. For example, the prefix kilo- means one thousand. By placing the prefix kilo- in front of the base unit metre, we form the unit kilometre (km), which is 1β€Žβ€‰β€Ž000 times larger than the metre. The prefix centi- means one hundredth. By placing the prefix centi- in front of the base unit metre, we form the unit centimetre (cm), which is 100 times smaller than the metre.

There is a sequential series of prefixes that can be used to increase or decrease the size of base units by factors of ten. Common SI prefixes and their relationship to the base units are shown in the table below.

PrefixSymbolMultiplication Factor
Tera-T1β€Žβ€‰β€Ž000β€Žβ€‰β€Ž000β€Žβ€‰β€Ž000β€Žβ€‰β€Ž00010
Giga-G1β€Žβ€‰β€Ž000β€Žβ€‰β€Ž000β€Žβ€‰β€Ž00010
Mega-M1β€Žβ€‰β€Ž000β€Žβ€‰β€Ž00010
Kilo-k1β€Žβ€‰β€Ž00010
Hecto-h10010
Deca-da1010
Base Unit110
Deci-d0.110
Centi-c0.0110
Milli-m0.00110
Micro-πœ‡0.000 00110
Nano-n0.000 000 00110
Pico-p0.000 000 000 00110

We can use the mnemonic The great mighty King Henry died by drinking chocolate milk under-neath pier to remember the SI prefixes in order.

Due to historical reasons, the SI unit kilogram already contains a prefix. When constructing units that are multiples or submultiples of the kilogram, a prefix is attached to the word gram. Thus, a unit that is 10 of a kilogram is a milligram (mg), not a microkilogram (μkg).

There are also a few non-SI units that are widely used in conjunction with the SI. Examples of a few of these units are shown in the table below.

QuantityUnitValue in SI Units
Time (𝑑)Minute (min)1=60mins
Hour (h)1=60=3600hmins
Day (d)1=24=86400dhs
Volume (𝑉)Litre (L)1=1Ldm
EnergyElectron volt (eV)1=1.6022Γ—10β‹…eVkgms

Except for the time quantities, these approved units can also be given SI prefixes. Thus, a unit one-thousandth the size of a litre is a millilitre (mL).

In chemistry, volume will most commonly be reported in millilitres or litres rather than the SI unit cubic metre (m3). It is useful, therefore, to recognize the following relationships: 1=1000=1000=1.dmcmmLL

We will often find it necessary to convert between different units. The relationship between two units is called a conversion factor.

Definition: Conversion factor

It is a relationship between two equal quantities that have different units. Conversion factors are commonly expressed as equations or ratios.

For example, 1 decimetre is 0.01 metres. This relationship can be expressed as an equation or fraction: 1=0.110.10.11.dmmordmmormdm

The process of using one or many conversion factors to convert between two units is called dimensional analysis or the factor-label method.

Definition: Dimensional analysis

It is a problem-solving method used to convert between two units whereby the original value is multiplied by one or more conversion factors.

In this process, the original value and unit are multiplied by one or more conversion factors written as fractions. Any units that appear in both the numerator and denominator will cancel out.

Let us convert 53.5 decimetres into metres. We can set up the following equation in which we multiply the original value and unit by the conversion factor that relates decimetres to metres: 53.5Γ—0.11.dmmdm

We chose to write the conversion factor with decimetres in the denominator so that the units cancel out: 53.5Γ—0.11.dmmdm

This leaves us with the unit metres: 53.5Γ—0.11=5.35.dmmdmm

Example 1: Converting a Mass in Grams into Kilograms

A small sample of osmium was found to have a mass of 22.6 g. What is that value in kilograms?

Answer

The prefix kilo- means one thousand. This indicates that a kilogram is one thousand times larger than a gram. Another way to say this is that 1 kilogram is equal to 1β€Žβ€‰β€Ž000 grams. This relationship is called a conversion factor and can be used to convert between the two units.

Dimensional analysis is the process of converting units using a conversion factor. We can begin this process by writing the original value: 22.6.g

We then multiply the original value by the conversion factor written as a fraction. We need to decide which of the following representations of the conversion factor should be used: 1100010001.kggorgkg

When performing a calculation, like units will cancel out when they appear in both the numerator and denominator. The original value has the unit grams. To cancel this unit out, grams must appear in the denominator of the conversion factor. Thus, we will multiply the original value by the first fraction: 22.6Γ—11000.gkgg

The gram unit will cancel out: 22.6Γ—11000.gkgg

This leaves us with an answer in kilograms: 22.6Γ—11000=0.0226.gkggkg

A sample of osmium with a mass of 22.6 g will have a mass of 0.0226 kg.

Dimensional analysis can make converting complex units easy. The density of copper is 8β€Žβ€‰β€Ž940 kg/m3. We want to know the density of copper in grams per millilitre. To perform this conversion, we must first recognize the appropriate conversion factors: 1=1000,0.01=1,1=1.kggmcmcmmL

We can write the original density as the fraction 89401.kgm

We can then separate out the unit cubic metres into three separate metre terms: 89401β‹…1β‹…1.kgmmm

Then, we multiply by the appropriate number and type of conversion factors so that all of the undesired units are canceled out: 89401β‹…1β‹…1Γ—10001Γ—0.011Γ—0.011Γ—0.011Γ—11=8.94.kgmmmgkgmcmmcmmcmcmmLgmL

This leaves us with the numerical value of 8.94 and units of grams per millilitre. Notice that it is necessary to use the conversion factor 0.011mcm three times in order to cancel the cubic metres in the original value.

Example 2: Performing Dimensional Analysis on a Cubic Unit

If 1 dm is equal to 10 cm, which of the following is 1 dm3 equal to?

  1. 10 cm3
  2. 1β€Žβ€‰β€Ž000 cm3
  3. 1β€Žβ€‰β€Ž000 cm
  4. 100 cm3
  5. 1 cm3

Answer

Dimensional analysis is the process of converting units using a conversion factor. In this problem, the conversion factor is 1=10.dmcm

We can begin the process of dimensional analysis by writing the original value: 1.dm

We can then multiply the original value by the conversion factor written as a fraction. We need to decide which of the following representations of the conversion factor should be used: 110101.dmcmorcmdm

When performing a calculation, like units will cancel out when they appear in both the numerator and denominator. The original value has the unit cubic decimetres. To cancel this unit out, decimetres must appear in the denominator of the conversion factor. Thus, we will multiply the original value by the second fraction: 1Γ—101.dmcmdm

However, as the original unit is cubic decimetres, only one instance of decimetres will be able to cancel out. This can easily be seen by expanding the unit cubic decimetres: (1Γ—1Γ—1)Γ—101ο€Ή1Γ—1Γ—1×101.dmdmdmcmdmdmdmdmcmdm

In order to cancel out all the decimetre units, we will need to multiply by the conversion factor two additional times: ο€Ή1Γ—1Γ—1×101Γ—101Γ—101.dmdmdmcmdmcmdmcmdm

Performing the calculation leaves us with the unit centimetres times centimetres times centimetres: 1000Γ—Γ—.cmcmcm

A unit that is multiplied by itself can be rewritten as a single unit with a power. The power indicates the number of times the unit is multiplied by itself. The unit centimetres appears three times. Thus, the unit can be rewritten as cm3.

1 dm3 is equal to 1β€Žβ€‰β€Ž000 cm3; hence, the answer is choice B.

Let us consider the equation for density, 𝜌=π‘šπ‘‰, where 𝜌 is the density, π‘š is the mass, and 𝑉 is the volume. Notice that density is isolated on the left-hand side of the equation. Isolated terms are the variables that are being solved for or the subject of the equation. Subjects may be isolated on either side of the equation but must always be in the numerator.

Definition: Subject

It is an isolated term in a mathematical formula. The subject of the formula is the variable being solved for.

The subject of an equation can be changed by rearranging the formula. To rearrange a formula, we must understand two rules of manipulating equations:

  1. Whatever is done to one side of the equation (addition, subtraction, multiplication, division, etc.) must be done to the other side as well. For example, if we divide the left-hand side of an equation by ten, we must also divide the right-hand side of the equation by ten: 10π‘₯=5010π‘₯10=5010π‘₯=5.
  2. To cancel out or move a quantity or variable, we perform the opposite operation on both sides of the equation. For example, variable 𝑏 is being added to variable π‘Ž in the following equation: π‘Ž+𝑏=𝑐. To isolate variable 𝑏, we perform the opposite operation by subtracting variable π‘Ž from both sides of the equation: π‘Ž+𝑏=π‘βˆ’π‘Žβˆ’π‘Žπ‘=π‘βˆ’π‘Ž.

Returning to the density equation, 𝜌=π‘šπ‘‰, let us set the mass, variable π‘š, as the subject. The mass is divided by the volume. To isolate the mass, we need to perform the opposite operation by multiplying both sides of the equation by the volume: π‘‰Γ—πœŒ=π‘šπ‘‰Γ—π‘‰.

The volume will cancel out on the right-hand side: π‘‰Γ—πœŒ=π‘šπ‘‰Γ—π‘‰.

This leaves us with the isolated mass as the subject: π‘‰Γ—πœŒ=π‘š.

If we wish to set the volume as the subject of the equation 𝜌=π‘šπ‘‰, we will first need to bring the volume into the numerator as subjects cannot be in the denominator. We can accomplish this be performing the opposite operation much like we did when setting the mass as the subject: π‘‰Γ—πœŒ=π‘šπ‘‰Γ—π‘‰π‘‰Γ—πœŒ=π‘š.

We still need to isolate the volume. It is currently being multiplied by the density. Performing the opposite operation, π‘‰πœŒπœŒ=π‘šπœŒ, gives us the rearranged formula: 𝑉=π‘šπœŒ.

Example 3: Rearranging an Algebraic Equation to Change the Subject of the Equation

For an aqueous solution at 25∘C, the given equation can be applied: pHpOH+=14.

  1. What is the correct form of the equation when rearranged to have pH as the subject?
  2. The pH of a solution is found to be 11. What is the value of the pOH?

Answer

Part 1

The subject of a formula is the variable being worked out. It must be in the numerator and isolated on either side of the equation. We need to isolate pH as the subject of the formula pHpOH+=14.

This will require us to rearrange the formula. When rearranging formulas, we must remember the following two points:

  1. Whatever operation is done to one side of the equation must be done to the other.
  2. To move or cancel out a variable, the opposite operation is performed on both sides of the equation.

pH is being added to pOH. To isolate the pH, we must perform the opposite operation by subtracting pOH from both sides of the equation: pHpOHpOHpOHpHpOH+=14βˆ’βˆ’=14βˆ’.

The correct form of the equation rearranged to have pH as the subject is pHpOH=14βˆ’.

Part 2

We are given the value of the pH and are asked to solve for the pOH. pOH is the variable being worked out and should be set as the subject of the equation. We can set the pOH as the subject by subtracting pH from both sides of the equation: pHpOHpHpHpOHpH+=14βˆ’βˆ’=14βˆ’.

Now that the equation has been rearranged to solve for the pOH, we can substitute the pH: pOH=14βˆ’11.

Then, we can determine the pOH: pOH=3.

A solution with a pH of 11 will have a pOH of 3.

We can use an equation to determine the unit of the subject. Let us again consider the equation for density, 𝜌=π‘šπ‘‰.

We can substitute units into an equation in place of the corresponding variables. For example, if the mass is given in grams and the volume is given in millilitres, we can write 𝜌=.gmL

We know that like units cancel out when in both the numerator and denominator. Grams and millilitres are not like units. This tells us that the unit of density (𝜌) will be grams per millilitre.

Consider the density equation rearranged so that mass is the subject: π‘‰Γ—πœŒ=π‘š.

If the density was given in kilogram per cubic metre and the volume was given in cubic metres, we could write mkgmmοŠ©οŠ©Γ—=.

We can see that cubic metres is in the numerator and denominator and will cancel out: mkgmmkgmοŠ©οŠ©Γ—==.

This leaves us with the unit kilograms for the mass.

Consider the density equation rearranged so that volume is the subject: 𝑉=π‘šπœŒ.

If the mass was given in grams and the density was given in grams per litre, we could write 𝑉=.ggL

This results in a complex fraction:

We can simplify a complex fraction by multiplying the numerator and denominator by the inverse or reciprocal of the denominator fraction: 𝑉=Γ—Γ—.gLggLLg

The denominator of the complex fraction will cancel out: 𝑣=ο€½1××𝑣=Γ—.gLggLLggLg

Now we can see that the unit grams can be canceled out: 𝑉=Γ—.gLg

Now the volume unit will be litres: 𝑉=.L

Example 4: Determining the Unit of Rate of Reaction from the Units of Mass and Time

The rate of reaction can be calculated using the following equation: rateofreactionmassofproductformedtimetaken=.

If the mass is in grams (g) and the time is in seconds (s), what is the unit of the rate of reaction?

Answer

We can determine the unit of the rate of reaction by substituting the units of mass and time into the equation: rateofreactiongs=.

Like units will cancel out when they appear in both the numerator and denominator. Grams and seconds cannot cancel out one another. Therefore, the unit of the rate of reaction will be grams per second, g/s.

Units of the same quantity must be the same before solving in order for the units to cancel out correctly. Let us consider Boyle’s law, 𝑃𝑉=𝑃𝑉, where 𝑃 represents the pressure, 𝑉 represents the volume, a subscript of one indicates an initial value, and a subscript of two indicates a final value.

If the initial pressure (𝑃) is 101β€Žβ€‰β€Ž325 pascals (Pa), the initial volume (𝑉) is 0.024 L, and the final volume (𝑉) is 35 mL, what is the final pressure (𝑃)? As we are solving for π‘ƒοŠ¨, we can rearrange the equation to set π‘ƒοŠ¨ as the subject: 𝑃𝑉=𝑃𝑉𝑃𝑉𝑉=𝑃𝑉𝑉𝑃𝑉𝑉=𝑃.

Before we substitute the values into the equation, notice that the initial volume is given in litres, while the final volume is given in millilitres. These units must be the same before the problem can be solved. We can convert between litre and millilitre units with the conversion factor of 1=1000litremillilitres. We will convert the 35 millilitres into litres: 35Γ—1100035Γ—11000=0.035.mLLmLmLLmLmL

Once the units of the same quantity are the same, we can substitute the values with units into the rearranged equation: 101325Γ—0.0240.035=𝑃.PaLL

Notice that the unit litre appears in the numerator and denominator and will cancel out: 101325Γ—0.0240.035=𝑃69480=𝑃.PaLLPa

Example 5: Deriving the Equation for Density and Solving given the Mass and Volume

Fill in the blank: A liquid has a volume of 200 mL and a mass of 4β€Žβ€‰β€Ž000 g. The density of this liquid is kg/L.

Answer

We need to determine the density of a liquid, but we have not been given the formula of density. However, we are told that the unit of density is kilograms per litre, kgL. This unit indicates that a value in kilograms was divided by a value in litres.

The problem does not provide a value in kilograms or litres, but we can convert the values given into the appropriate units. Kilograms is a unit of mass. The prefix kilo- means 1β€Žβ€‰β€Ž000 and indicates that the unit kilogram is 1β€Žβ€‰β€Ž000 times larger than a gram. We can therefore write the following relationship, also known as a conversion factor: 1=1000.kgg

We can then perform dimensional analysis to convert the mass in grams into kilograms by multiplying the original value by the conversion factor written as a fraction: 4000Γ—11000.gkgg

The conversion factor is written so that the unit grams is in the denominator. This is because like units that appear in the numerator and denominator will cancel out: 4000Γ—11000.gkgg

This leaves us with the unit kilograms: 4000Γ—11000=4.gkggkg

Litres and millilitres are both units of volume. The prefix milli- means one-thousandth and indicates that the unit millilitre is 1β€Žβ€‰β€Ž000 times smaller than a litre. We can therefore write the following relationship: 1=1000.LmL

We can then perform dimensional analysis to convert the volume in millilitres into litres: 200Γ—11000200Γ—11000=0.2.mLLmLmLLmLL

Now that the mass and volume are in kilograms and litres, respectively, we can divide the mass by the volume to determine the density: densitykgLdensitykgL=40.2=20.

The density of this liquid is 20 kg/L. We should fill in the blank with 20.

Key Points

  • The SI is a series of seven standardized base units: s, m, kg, A, K, mol, cd.
  • Prefixes can be added to the base unit to create new units that are multiples or submultiples of the base unit.
  • The following mnemonic can be used to remember several of the SI prefixes:
    The great mighty King Henry died by drinking chocolate milk under-neath pier.
  • Dimensional analysis can be used to convert between different units.
  • Formulas can be rearranged to change the subject.
  • Units can be substituted into a formula to determine the unit of the subject.
  • Like units that appear in the numerator and denominator will cancel out.

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