MISC 476 Interpreting and Balancing Chemical Equations
Prepared by M. L. Gillette, Indiana University Kokomo and H. A. Neidig, Lebanon
Valley College
Purpose of the Experiment: Interpret and balance chemical equations.
Background Information
Imagine models of an apartment building and a recreation center, built out of toy
blocks. Suppose that we disassembled those structures and used the same blocks to
construct models of three houses and two garages, with no blocks left over. We could
summarize the process using the expression shown in Equation 1.
apartment building + recreation center  3 houses + 2 garages
(Eq. 1)
The process of starting with one arrangement of building blocks, disassembling the
arrangement, and reassembling the blocks into a different arrangement is like the
process that occurs during a chemical reaction. We can represent a chemical reaction by
writing an expression called a chemical equation, as shown in Equation 2.
reactants  products
(Eq. 2)
In chemical equations, we represent the starting arrangements, or structures, called the
reactants, and the final structures, called the products, using chemical formulas. We
indicate the rearrangement process using an arrow (), which we read as either
"yields" or "produces."
Interpreting Chemical Equations
If we heat metallic nickel (Ni), a silvery solid, with antimony (Sb), a silvery-white solid,
we produce nickel antimonide (NiSb), a light copper-red solid. We can describe this
process using the chemical equation shown in Equation 3.
Ni(s, silver) + Sb(s, silver-white) "heat
"# NiSb(s, copper-red)
(Eq. 3)
In Equation 3, the information in parentheses following the elemental symbols and
formulas indicates the physical!states (solid, liquid, or gas) and colors of the reactants
and products. The "s" stands for solid, "g" for gas, and "l" for liquid. If the substance is
dissolved in water, we write "aq," which stands for aqueous. Frequently, we omit the
color descriptions in order to make the notation simpler. The information above the
reaction arrow describes reaction condition's, such as the application of heat.
Write the chemical equations for each of the descriptions in Problem Set 1.
Balancing Equations
The disassembly and reassembly process represented by Equation 1 illustrates a
principle called the conservation of matter: no blocks were added to make the product
structures, and none were left over. When substances are combined in the correct
proportions, based on the appropriate chemical equation, every reactant atom should be
incorporated into a product; no extra atoms should be added, and none should be left
over. We establish the correct proportions for a reaction by balancing the equation for
the reaction. Balancing equations is a bookkeeping procedure that helps us in counting
the number of atoms, molecules, and moles involved in a chemical reaction.
First, we need to recognize the number of atoms of each element that are in a unit of a
compound. Potassium telluride has the formula K2Te. A subscript after a symbol
indicates how many of that atom are in a unit of the compound. The subscript after the
symbol K indicates that there are two K atoms in one unit of K2Te. There is no subscript
for Te. It is understood that if there is no subscript after the symbol for an atom the
subscript is one. Hence, there is only one atom of Te in a unit of K2Te.
In a unit of potassium disulfide (K2S4). there are two atoms of potassium and four atoms
of sulfur. In a molecule or unit of cyclohexane (C6H12), there are six carbon atoms and
twelve hydrogen atoms.
In the formula for some compounds, we find that a group of atoms is enclosed in
parentheses and that there is a subscript following the parenthesis. An example is
radium iodate, Ra(IO3)2. The subscript three after the oxygen indicates that in the iodate
ion (IO3-) there are three oxygen atoms. However, the subscript two appearing after the
parentheses indicates that there are two iodate ions in the compound. Therefore, there
are two iodine atoms in a unit of Ra(IO3)2 while the number of oxygen atoms in a unit of
Ra(IO3)2 is six.
Indicate the number of atoms in a unit of each of the compounds in Problem Set 2.
Sometimes, an equation is already balanced when it is written in its simplest form. For
example, Equation 3 is already balanced because one Ni atom and one Sb atom are on
each side of the equation. In most cases, however, we have to do additional work to
balance an equation. A chemical equation is quantitatively useful only when it is
properly balanced. eonsider the equation for the reaction of sodium (Na) with Sb to
produce sodium antimonide (Na3Sb), shown in Equation 4.
Na(s, silver) + Sb(s, silver-white)  Na3Sb(s, black)
(Eq. 4)
Equation 4 is not balanced. One Na atom and one Sb atom are on the reactant side, the
left side of the equation. Three Na atoms and one Sb atom are on the product side, the
right side of the equation. Due to the conservation of matter, it is not possible to have
one atom of Na as the reactant and three atoms of Na in the product. We need to
balance the equation by determining the correct proportions of reactants and products.
If more than one of any species are involved in a chemical reaction, we write the
number involved immediately preceding the appropriate symbol or formula in the
equation for that reaction. The numbers preceding the symbols or formulas in a
chemical equation are the coefficients of the equation. We insert the appropriate
coefficients to balance chemical equations. When no coefficient precedes the symbol or
formula for a species, a coefficient of 1 is understood to be there. Thus, the coefficients
in Equation 1 are 1, 1, 3, and 2, respectively. To balance Equation 4, we write a
coefficient of 3 in front of Na, as shown in Equation 4a.
3 Na(s, silver) + Sb(s, silver-white)  Na3Sb(s, black) (Eq. 4a)
Now there are three Na atoms and one Sb atom on each side of the equation. Therefore,
the equation is balanced.
It is essential that we do not alter the chemical formula of any reactant or product
species in the process of balancing an equation. For example, if we change the
subscripts in the chemical formula for a species, we alter the identity of that reactant or
product species. The resulting equation may be balanced, but it will represent a
different chemical transformation from the one represented by the original
(unbalanced) equation.
We can interpret coefficients in chemical equations in either of two ways: as the number
of individual units involved in a reaction, or as the number of moles of those units
involved in the reaction. Thus, based on Equation 4a, we can say that three Na atoms
react with one Sb atom to produce one unit of Na3Sb, or we can say that 3 mol of Na
atoms react with 1 mol of Sb atoms to produce 1 mol of Na3Sb units.
When we write a chemical equation, we use chemical symbols to represent the elements
and compounds involved in the form in which they participate in the reaction. For
example, when they are in their elemental, or chemically uncombined forms, metal
atoms are not written as if they are bonded to each other, so they can be considered as
essentially reacting individually. Therefore, we can represent them in a chemical
equation by simply using the appropriate chemical symbol, as shown in Equation 4a for
Na and Sb.
In contrast, some nonmetallic elements exist naturally not as individual atoms, but as
molecules composed of two or more atoms of the element. Molecules are groupings of
atoms held together by shared electron pairs. For example, hydrogen, nitrogen, oxygen,
and the halogens (fluorine, chlorine, bromine, and iodine) occur as diatomic (two atom)
molecules: H2, N2, O2, F2, Cl2, Br2, and I2 respectively. We refer to H2 as molecular
hydrogen and H as an atom of hydrogen. Similarly, phosphorus occurs naturally as P4,
and sulfur occurs as S8. These are the formulas that we use in chemical equations
representing reactions involving these molecular elements. For example, carbon (C)
reads with molecular oxygen (O2) to form carbon(IV) dioxide (CO2), as shown in
Equation 5, which is balanced.
C (s) + O2 (g)  CO2 (g)
(Eq. 5)
Another illustration of the correct use of elemental formulas is the reaction of P4 with O2
to form tetraphosphorus decaoxide (P4O10), shown in Equation 6, which is balanced.
P4 (s) + 5 O2 (g)  P4O10 (s)
(Eq.6)
Notice that five O2 molecules react with one P4 molecule to form one P4O10 molecule.
Hydrogen (H2) and nitrogen (N2) gases react to form ammonia (NH3) gas. The balanced
equation for this reaction is shown in Equation 7.
3 H2 (g) + N2 (g)  2 NH3 (g)
(Eq. 7)
As previously discussed, we can interpret Equation 7 in either of two ways. We can say
that three H2 molecules (a total of six H atoms) react with one N2 molecule (two N
atoms). Two NH3 molecules are produced, containing a total of two N atoms and six H
atoms. Or, we can say that 3 mol of H2 molecules (6 mol of hydrogen atoms) react with
1 mol of N2 molecules (2 mol of N atoms). Two moles of NH3 molecules are produced,
containing a total of 2 mol of N atoms and 6 mol of H atoms. There are other sets of
coefficients that we could use to balance Equation 7. For example, coefficients of 6, 2,
and 4, respectively, also work, as shown in Equation 7a.
6 H2 (g) + 2 N2 (g)  4 NH3 (g)
(Eq. 7a)
To simplify equations as much as possible, we choose the smallest whole number
coefficients we can. Because we can divide 6, 2, and 4 by 2 and yield whole numbers, we
prefer to write the equation in the form shown in Equation 7.
Balance the equations in Problem Sets 3 and 4.
Problem Set 1
Write a chemical equation for each of the following descriptions, indicating the physical
states of the reactants and products.
1. If we react metallic germanium (Ge), a gray solid, with bromine, a red-brown liquid,
we produce germanium(lI) bromide (GeBr2). a colorless solid.
2. If we heat metallic iron (Fe), a silvery solid, with metallic selenium (Se), a gray solid,
we produce iron(ll) selenide (FeSe), a black solid.
3. When we heat metallic magnesium (Mg), a silvery solid, with hydrogen (H2), a
colorless gas, we produce magnesium hydride (MgH2), a white solid.
4. When we react metallic copper (Cu), a red-brown solid, with fluorine (F2), a yellowgreen gas, we produce copper(lI) fluoride (CuF2), a white solid.
5. When we heat magnesium carnonate (MgCO3), a white solid, we produce
magnesium oxide (MgO), a white solid, and carbon(lV) dioxide (CO2), a colorless
gas.
Problem Set 2
Indicate the number of atoms of each element that are in one unit of the following
compounds.
1.
CaO
Ca __________
O __________
2.
Na2S
Na __________
S __________
3.
FeBr3
Fe __________
Br __________
4.
Co2O3
Co __________
O __________
5.
Cu3Se2
Cu __________
Se __________
6.
K2SO4
K __________
S __________
O __________
7.
FeC2O4
Fe __________
C __________
O __________
8.
K2Cr2O7
K __________
Cr __________
O __________
9.
Mg(NO3)2
Mg __________
N __________
O __________
Al __________
S __________
O __________
10. Al2(SO3)3
Problem Set 3
1. Rewrite and balance each of the following chemical equations in the space provided.
a. Na (s) + Cl2 (g)  NaCl (s)
b. Mg (s) + Cl2 (g)  MgCl2 (s)
c. H2 (g) + Cl2 (g)  HCl (g)
d. H2 (g) + O2 (g)  H2O2 (l)
e. H2O (l)  H2 (g) + O2 (g)
f. Bi (s) + F2 (g)  BiF3 (s)
g. Al (s) + O2 (s)  Al2O3 (s)
h. Al (s) + Cl2 (g)  AlCl3 (s)
i. CaCO3 (s) + CaO (s)  CO2 (g)
j.
KClO3 (s)  KCl (s) + O2 (g)
k. N2O5 (g)  NO2 (g) + O2 (g)
l. P4O10 (s) + H2O (l)  H3PO4 (s)
m. Fe2O3 (s) + Al (s)  Fe (s) + Al2O3 (s)
n. Sb2S3 (s) + Fe (s)  Sb (s) + FeS (s)
o. NH3 (g) + SO2 (g)  NO (g) + S (s) + H2O (g)
p. NH3 (g) + O2 (g)  NO (g) + H2O (g)
2. What is the sum of the coefficients used in the balanced form of the following
equations?
Example: The sum of the coefficients for 2 H2 + O2  2 H2O is 5
a. In Question 1.a.
b. In Question 1.o.
Problem Set 4
For each of the following incorrectly written equations, write the correct balanced
equation, using the appropriate coefficients. Do not change any of the subscripts.
1. Fe + Br2  2 FeBr2
2. 2 H2 + 2 I2  4 HI
3. 2 Ni + 4 CO  2 Ni(CO)4
4. 3 K + 2 N2  3 KN3
5. 4 KClO3  4 KCl + 3 O2
6. Si + 4 F2  SiF4