Balancing Equations
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C ONCEPT
Concept 1. Balancing Equations
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Balancing Equations
Lesson Objectives
• Demonstrate the Law of Conservation of Matter in a chemical reaction.
• Explain the roles of coefficients and subscripts in a chemical reaction.
• Balance equations using the simplest whole number coefficients.
Introduction
Even though chemical compounds are broken up and new compounds are formed during a chemical reaction, atoms
in the reactants do not disappear nor do new atoms appear to form the products. In chemical reactions, atoms are
never created or destroyed. The same atoms that were present in the reactants are present in the products - they are
merely re-organized into different arrangements. In a complete chemical equation, the two sides of the equation
must be balanced. That is, in a complete chemical equation, the same number of each atom must be present on
the reactants and the products sides of the equation.
Subscripts and Coefficients
There are two types of numbers that appear in chemical equations. There are subscripts which are part of the
chemical formulas of the reactants and products and there are coefficients that are placed in front of the formulas to
indicate how many molecules of that substance is used or produced.
The subscripts are part of the formulas and once the formulas for the reactants and products are determined, the
subscripts may not be changed. The coefficients indicate how many molecules of each substance is involved in the
reaction and may be changed in order to balance the equation. The equation above indicates that one atom of solid
copper is reacting with two molecules of aqueous silver nitrate to produce one molecule of aqueous copper (II)
nitrate and two atoms of solid silver. When you learned how to write formulas, it was made clear that when only
one atom of an element is present, the subscript of "1" is not written - so that when no subscript appears for an atom
in a formula, you read that as one atom. The same is true in writing balanced chemical equations. If only one atom
or molecule is present, the coefficient of "1" is omitted.
Coefficients are inserted into the chemical equation in order to balance it; that is, to make equal the total number
of each atom on the two sides of the equation. Consider the equation representing the reaction that occurs when
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gaseous methane, CH4 , is burned in air (reacted with oxygen gas), and produces gaseous carbon dioxide and liquid
water.
The unbalanced symbolic equation for this reaction is given below.
CH4(g) + O2(g) → CO2(g) + H2 O(L)
Equation 1
It is quickly apparent that equation 1 is not balanced. There are 4 hydrogen atoms in the reactants and only 2
hydrogen atoms in the products. The oxygen atoms in the equation are also not balanced. In order to balance this
equation, it is necessary to insert coefficients in front of some of the substances to make the numbers of atoms of
each element the same on the two sides of the equation. We can begin this process by placing a coefficient of 2 in
front of the water molecule.
CH4(g) + O2(g) → CO2(g) + 2 H2 O(L)
Equation 2
The insertion of this coefficient balanced the hydrogen atoms (there are now 4 on each side) but the equation is still
not completely balanced. The oxygen atoms are not balanced. There are 4 oxygen atoms in the products but only
two in the reactants. We can now insert a coefficient of 2 in front of the oxygen molecule in the reactants.
CH4(g) + 2 O2(g) → CO2(g) + 2 H2 O(L)
Equation 3
When we count up the number of each type of atom on the two sides of the equation now, we see that the equation
is properly balanced. There is one carbon atom on each side, four hydrogen atoms on each side, and four oxygen
atoms on each side. Equation balancing is accomplished by changing coefficients, never by changing subscripts.
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Concept 1. Balancing Equations
The Process of Balancing an Equation
The process of writing a balanced chemical equation involves three steps. As a beginning chemistry student, you
will not know whether or not two given reactants will react or not and even if you saw them react, you would not
be sure what the products are without running tests to identify them. Therefore, for the time being, you will be told
both the reactants and products in any equation you are asked to balance.
Step 1: Know what the reactants and products are, and write a word equation for the reaction.
Step 2: Write the formulas for all the reactants and products.
Step 3: Adjust the coefficients to balance the equation.
You must keep in mind that there are a number elements commonly appearing in equations that, under normal
conditions, exist as diatomic molecules (Table 1.1).
TABLE 1.1: Elements that exist as diatomic molecules under normal conditions
Element
Hydrogen
Oxygen
Nitrogen
Chlorine
Fluorine
Bromine
Iodine
Formula for Diatomic Molecule
H2
O2
N2
Cl2
F2
Br2
I2
Phase Under Normal Conditions
Gaseous
Gaseous
Gaseous
Gaseous
Gaseous
Liquid
Solid
When the names of any of these elements appear in word equations, you must remember that the name refers to the
diatomic molecule and insert the diatomic formula into the symbolic equation. If, under unusual circumstances, it
was desired to refer to the individual atoms of these elements, the text would refer specifically to atomic hydrogen
or atomic oxygen and so on.
Sample Problem 1
Write a balanced equation for the reaction that occurs between chlorine gas and aqueous sodium bromide to produce
liquid bromine and aqueous sodium chloride.
Step 1: Write the word equation (keeping in mind that chlorine and bromine refer to the diatomic molecules).
Chlorine + sodium bromide → bromine + sodium chloride
Step 2: Substitute the correct formulas into the equation.
Cl2 + NaBr → Br2 + NaCl
Step 3: Insert coefficients where necessary to balance the equation.
By placing a coefficient of 2 in front of the NaBr, we can balance the bromine atoms and by placing a coefficient of
2 in front of the NaCl, we can balance the chloride atoms.
Cl2 + 2 NaBr → Br2 + 2 NaCl
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A final check (always do this) shows that we have the same number of each atom on the two sides of the equation
and we do not have a multiple set of coefficients so this equation is properly balanced.
Sample Problem 2
Write a balanced equation for the reaction between aluminum sulfate and calcium bromide to produce aluminum
bromide and calcium sulfate. It might be worthwhile to note here that the reason you memorized the names and
formulas for the polyatomic ions is that these ions usually remain as a unit throughout many chemical reactions.
Step 1: Write the word equation.
Aluminum sulfate + calcium bromide → aluminum bromide + calcium sulfate
Step 2: Replace the names of the substances in the word equation with formulas.
Al2 (SO4 )3 +CaBr2 → AlBr3 +CaSO4
Equation 1
Step 3: Insert coefficients to balance the equation.
In order to balance the aluminum atoms, we must insert a coefficient of 2 in front of the aluminum compound in the
products.
Al2 (SO4 )3 +CaBr2 → 2 AlBr3 +CaSO4
Equation 2
In order to balance the sulfate ions, we must insert a coefficient of 3 in front of the CaSO4 in the products.
Al2 (SO4 )3 +CaBr2 → 2 AlBr3 + 3 CaSO4
Equation 3
In order to balance the bromine atoms, we must insert a coefficient of 3 in front of the CaBr2 in the reactants.
Al2 (SO4 )3 + 3 CaBr2 → 2 AlBr3 + 3 CaSO4
Equation 4
The insertion of the 3 in front of the CaBr2 in the reactants also balances the calcium atoms in the CaSO4 in the
products. A final check shows 2 aluminum atoms on each side, 3 sulfur atoms on each side, 12 oxygen atoms on
each side, 3 calcium atoms on each side, and 6 bromine atoms on each side. This equation is balanced.
Sample Problem 3
Balance the following skeletal equation. (The term "skeletal equation" refers to an equation that has the correct
formulas but has not yet had the proper coefficients added.)
Fe(NO3 )3 + NaOH → Fe(OH)3 + NaNO3
(skeletal equation)
We can balance the hydroxide ion by inserting a coefficient of 3 in front of the NaOH on the reactant side.
Fe(NO3 )3 + 3 NaOH → Fe(OH)3 + NaNO3
4
(intermediate equation)
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Concept 1. Balancing Equations
Then we can balance the nitrate ions by inserting a coefficient of 3 in front of the sodium nitrate on the product side.
Fe(NO3 )3 + 3 NaOH → Fe(OH)3 + 3 NaNO3
(balanced equation)
Counting the number of each type of atom on the two sides of the equation will now show that this equation is
balanced.
Simplest Whole Number Coefficients
Chemical equations should be balanced with the simplest whole number coefficients that balance the equation. Here
is the properly balanced equation from the previous section.
Al2 (SO4 )3 + 3 CaBr2 → 2 AlBr3 + 3 CaSO4
Equation 4
You should note that the equation in the previous section would have the same number of atoms of each type on each
side of the equation with the following set of coefficients.
2 Al2 (SO4 )3 + 6 CaBr2 → 4 AlBr3 + 6 CaSO4
Equation 5
You should count the number of each type of atom on each side of the equation to confirm that this equation is
"balanced". While this set of coefficients do "balance" the equation, they are not the lowest set of coefficients
possible that balance the equation. We could divide each of the coefficients in this equation by 2 and get a another
set of coefficients that are whole numbers and also balance the equation. Since it is required that an equation be
balanced with the lowest whole number coefficients, equation 5 is NOT properly balanced. The properly balanced
equation for this reaction is equation 4. When you have finished balancing an equation, you should not only check
to make sure it is balanced, you should also check to make sure that it is balanced with the simplest set of whole
number coefficients possible.
Let’s consider another equation for balancing; the combustion of octane, C8 H18 .
C8 H18 + O2 → CO2 + H2 O
Equation 1
To begin balancing this equation, we note that there are 8 carbon atoms on the reactant side and only 1 carbon on
the product side. We can balance the carbon atoms by inserting a coefficient of 8 in front of the carbon dioxide on
the product side.
C8 H18 + O2 → 8 CO2 + H2 O
Equation 2
Next, we note that there are 18 hydrogen atoms in the reactants and only 2 hydrogen atoms in the products. We can
balance the hydrogen atoms by inserting a coefficient of 9 in front of the water molecule on the product side.
C8 H18 + O2 → 8 CO2 + 9 H2 O
Equation 3
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Now when we count up the number of oxygen atoms in the products, we find there are 25 oxygen atoms. What
coefficient can you insert in front of the O2 molecule in the reactants to have 25 oxygen atoms? Since the oxygen
molecule has a subscript of 2, the only coefficient for this molecule that will produce 25 oxygen atoms is a coefficient
25
of 25
2 . If we insert 2 as the coefficient for the oxygen molecule, we get a balanced equation but we do not get an
equation that is balanced with the simplest whole number coefficients.
C8 H18 +
25
O2 → 8 CO2 + 9 H2 O
2
Equation 4
In order to have whole number coefficients for this equation, we must go through the equation and multiply all the
coefficients by 2. That produces the coefficients shown in equation 5.
2 C8 H18 + 25 O2 → 16 CO2 + 18 H2 O
Equation 5
Now the equation is balanced with the simplest whole number coefficients and is a properly balanced equation.
Sample Problem
Given the following skeletal (un-balanced) equations, balance them.
(a) CaCO3(s) → CaO(s) +CO2(g)
(b) H2 SO4(aq) + Al(OH)3(aq) → Al2 (SO4 )3(aq) + H2 O(L)
(c) Ba(NO3 )2(aq) + Na2CO3(aq) → BaCO3(aq) + NaNO3(aq)
(d) C2 H6(g) + O2(g) → CO2(g) + H2 O(L)
Solutions
(a) CaCO3(s) → CaO(s) +CO2(g) (sometimes the equation balances with all coefficients of 1)
(b) 3 H2 SO4(aq) + 2 Al(OH)3(aq) → Al2 (SO4 )3(aq) + 6 H2 O(L)
(c) Ba(NO3 )2(aq) + Na2CO3(aq) → BaCO3(aq) + 2 NaNO3(aq)
(d) 2 C2 H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2 O(L)
The Conservation of Mass in Chemical Reactions
Both the numbers of each type of atom and the mass are conserved during chemical reactions. An examination of a
properly balanced equation will demonstrate that mass is conserved. Consider the following reaction.
Fe(NO3 )3 + 3 NaOH → Fe(OH)3 + 3 NaNO3
You should demonstrate that this equation is balanced by counting the number of each type of atom on each side of
the equation.
We can also demonstrate that mass is conserved in this reaction by determining the total mass on the two sides of
the equation.
Fe(NO3 )3 + 3 NaOH → Fe(OH)3 + 3 NaNO3
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Concept 1. Balancing Equations
Reactant Side Mass
1 molecule of Fe(NO3 )3 × molecular weight = (1)(241.9 daltons) = 241.9 daltons
3 molecules of NaOH × molecular weight = (3)(40.0 daltons) = 120. daltons
Total mass for reactants = 241.9 daltons + 120. daltons = 361.9 daltons
Product Side Mass
1 molecule of Fe(OH)3 × molecular weight = (1)(106.9 daltons) = 106.9 daltons
3 molecules of NaNO3 × molecular weight = (3)(85.0 daltons) = 255 daltons
Total mass for products = 106.9 daltons + 255 daltons = 361.9 daltons.
As you can see, both number of atom types and mass are conserved during chemical reactions. A group of 20 objects
stacked in different ways will still have the same total mass no matter how you stack them.
Lesson Summary
•
•
•
•
To be useful, chemical equations must always be balanced.
Balanced chemical equations must have the same number and type of each atom on both sides of the equation.
The coefficients in a balanced equation must be the simplest whole number ratio.
Mass is always conserved in chemical reactions.
Review Questions
1. Explain in your own words why it is essential that subscripts remain constant but coefficients can change.
2. Which set of coefficients will properly balance the following equation?
C2 H6 + O2 → CO2 + H2 O
a.
b.
c.
d.
1, 1, 1, 1
1, 3, 2, 2
1, 3.5, 2, 3
2, 7, 4, 6
3. When properly balanced, what is the sum of all the coefficients in the following chemical equation?
SF4 + H2 O → H2 SO3 + HF
a.
b.
c.
d.
4
7
9
None of the above
4. When the following equation is balanced, what is the coefficient found in front of the O2 ?
P4 + O2 + H2 O → H3 PO4
a.
b.
c.
d.
1
3
5
7
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5. Balance the following equations.
a.
b.
c.
d.
XeF6(s) + H2 O(L) → XeO3(s) + HF(g)
Cu(s) + AgNO3(aq) → Ag(s) +Cu(NO3 )2(aq)
Fe(s) + O2(g) → Fe2 O3(s)
Al(OH)3 + Mg3 (PO4 )2 → AlPO4 + Mg(OH)2
Further Reading / Supplemental Links
Website with lessons, worksheets, and quizzes on various high school chemistry topics.
• Lesson 8-1 is on Balancing Equations.
• http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson81.htm
Vocabulary
law of conservation of matter Matter is neither created nor destroyed in chemical reactions.
skeletal equation a chemical equation before it has been balanced
balanced chemical equation a chemical equation in which the number of each type of atom is equal on the two
sides of the equation
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