Balancing Chemical Equations


In the notes on Chemical Nomenclature, you learned how to write chemical formulas for numerous compounds.  In the notes on Counting Atoms in Chemical Formulas, you learned how to determine the number of atoms of each element in these formulas.  It is now time to put our knowledge of chemical formulas to use in chemical equations.

Chemistry is primarily concerned with the changes matter undergoes.  Some of these changes are merely physical, such as when ice melts to form water.  But the most interesting changes in chemistry are those in which new substances are formed from "original" substances.  These are called chemical reactions or chemical changes.  In many chemical reactions, the new substances that are formed have properties that are quite different from those of the substances we started with.  But regardless of how dramatic the reaction, or how different the properties of the new substances are from those of the original substances, we know that at the atomic level, a chemical reaction is just a rearrangement of atoms.  Just as the same letters of the alphabet can be be rearranged to form two very different words, so too can the same atoms often by put together in different ways to form substances with very different properties.

We write a chemical equation for a reaction by writing the substances we start with, separating them by plus signs (+) if there is more than one, using an arrow to represent the process of change, and then writing the substances we end up with, again separating them with plus signs.  In order to write a chemical equation, we must know three things:
 

  1. what substances we start with

  2.  
  3. what substances we end up with

  4.  
  5. the chemical formulas of the substances in items 1 and 2 above
In practice, this information must be determined experimentally, but for solving problems in a general chemistry lecture course, this information will usually just be given to you.

Let's now look at some examples of writing and balancing chemical equations.

Problem:  Hydrogen (H2) burns in oxygen (O2) to form water (H2O).  Write a balanced chemical equation for this reaction, showing all substances as gases.

Solution:

Lets begin by writing a word description of the problem.  From the information given above, we have

hydrogen   +   oxygen   ---------->   water

I guess the above is not absolutely necessary, but if you are a beginner, it may help you to see your way.  The next step is to replace each name with its corresponding chemical formula.  The formulas were all given in the problem, so we have

H2(g)   +   O2(g)   ---------->   H2O(g)

The letter g in parentheses after each formula in the equation above indicates that the substance is a gas.  Other phase labels we will encounter are l for liquid, s for solid and aq for aqueous.  The term aqueous means dissolved in water.

Looking at the equation we have so far, we can see it is not balanced.  Keep in mind that a chemical reaction is just a rearrangement of atoms, so every atom you start out with has to be somewhere in the end.  And you can't end up with any atoms you didn't have in the beginning.  The equation above does not meet these requirements.  Using the information in the notes on Counting Atoms in Chemical Formulas, we can see that the left side of this equation has 2 atoms of hydrogen and 2 atoms of oxygen, while the right hand side has 2 atoms of hydrogen and only 1 atom of oxygen.  At this point, hydrogen is ok, but oxygen is not.  In its present form, the equation suggests that an atom of oxygen has magically disappearred.  A pictorial representation of the present form of this equation is shown in line A of Figure 1, where H atoms are red dots and O atoms are blue dots.  Notice that we start with 2 blue dots and end up with only 1.  This of course, is impossible.

Beginners are tempted to solve problems such as the one we now face by adjusting the subscripts, but that is NEVER the thing to do!  For example, one incorrect approach someone might try is to drop the subscript of 2 in O2 and write

H2(g)   +   O(g)   ---------->   H2O(g)

The pictorial representation of this is shown in line B of Figure 1.  This DOES balance the equation, but that's not our only objective.  We also want our equation to be a correct description of what really happens in nature.  Oxygen does not exist as a monatomic gas.  It is known that the oxygen atoms are paried up in groups of 2 in naturally occurring oxygen.  Therefore, to have a correct description of the chemical system, we MUST represent oxygen as O2 and not as O.

Another approach -- also incorrect -- that someone might try -- is to balance oxygen by adding a subscript 2 to the oxygen in water and write the equation as

H2(g)   +   O2(g)   ---------->   H2O2(g)

The pictorial representation of this is shown in line C of Figure 1.  Again, we have a balanced equation, but things are still not right.  Our original objective was to describe the formation of WATER, and H2O2 is not water.  Often, when you start changing subscripts, you get some non-sense formula that does not exist.  In this case we do get a real substance.  H2O2 is known as hydrogen peroxide.  You might have a 3% solution of it in your medicine cabinet.  It's that aniseptic in the dark brown bottle.  Still, we have not achieved the original objective with this equation, so it's not the one we want, even though it shows the formation of a real substance.

The bottom line is NEVER ADJUST SUBSCRIPTS!  Every chemical substance has a definite, unchangable chemical formula.  When you change subscrips, what you are doing is changing the chemical formula, which is never the correct course of action.

Rather than adjusting the subscripts, you can adjust the numbers that go in front of the formulas.  These numbers are called coefficients, and they tell us how many of those formula units we have.  The correctly balanced equation for the formation of water from hydrogen and oxygen is

2H2(g)   +   O2(g)   ---------->   2H2O(g)

The pictorial representation of this is seen in line D of Figure 1.  It is the only version of the equation discussed so far that is correct.  Notice that it is balanced.  It has 4 atoms of hydrogen on both sides, and it has 2 atoms of oxygen on both sides.  You should strive to learn how to obtain these atom counts by using the information you learned in the notes on Counting Atoms in Chemical Formulas.  Since you have had limited exposure to these concepts at this point, however, I have included Figure 1 to help you.  You can count the red (H) and blue (O) dots in line D of Figure 1 to confirm the atom counts I have given you.  In addition to being balanced, it also achieves our objective of describing the reaction of hydrogen and oxygen -- as they naturally exist  -- to produce water.

The balanced equation shows that the H2 and O2 molecules can not react in a one-to-one ratio to produce water.  Rather, it takes twice as many molecules of H2 as you have molecules of O2 to end up with only H2O molecules and have no H2 or O2 molecules left over.  We normally balance equations using the smallest integers possible.  This is called balancing in lowest terms.  Once you have a balanced equation, if you multiply it throughout by any integer, you will still have a balanced equation.  For example, if you multiply our balanced equation

2H2(g)   +   O2(g)   ---------->   2H2O(g)

by 2, you get

4H2(g)   +   2O2(g)   ---------->   4H2O(g)

This is still balanced.  There are now 8 H atoms on each side and 4 O atoms on each side.  Practice your skills in counting atoms in chemical formulas with this equation.  If you can't reproduce the 8 and 4 I get in my atom count, then prove it to yourself by counting the red (H) and blue (O) dots in Figure 2.

Although I did present the balanced equation

2H2(g)   +   O2(g)   ---------->   2H2O(g)

some may still be wondering where I got it.  Admittedly, it may seem that I sort of "pulled it out of a hat".  The previous discussion was more focused on what is and is not a correct way to balance an equation, rather than presenting the thought process that one goes through step by step in actually balancing the equation.  So I am going to balance the equation we have been working with again -- this time, not leaving out any steps.

Our starting equation is

H2(g)   +   O2(g)   ---------->   H2O(g)

Let us begin with the hydrogen.  We see that there are 2 hydrogens on the left and two on the right.  So far, so good.  H atoms are balanced.  Then we look at oxygen.  We have 2 O atoms on the left, but only one on the right.  Problem.  We know that if we are going to have 2 O atoms on the left, we must have 2 O atoms on the right also.  We are not allowed to add a subscript to H2O, because that would be changing the formula, so the only solution is to form 2 molecules of H2O instead of only 1.  Our equation now becomes

H2(g)   +   O2(g)   ---------->   2H2O(g)

O atoms are now balanced -- there are 2 O atoms on each side.  But now H atom are messed up.  There are 2 H atoms on the left, but 4 H atoms on the right.  If we are going to end up with 4 H atoms, then we must start out with 4 H atoms.  The way to acheive this is to use 2 molecules of H2 instead of only 1.  Now we have

2H2(g)   +   O2(g)   ---------->   2H2O(g)

At last, the equation is balanced.  It has 4 atoms of H on both sides and 2 atoms of O on both sides.  Your work is not done until ALL atoms are balanced, so be sure to check the status of each element after each balancing operation you do.  Note that when an element does not balance, you try to increase the number of atoms of that element on the side with fewer atoms to make it equal the number of atoms of that element on the side with more atoms.

Balancing equations is somewhat of a trial and error method.  It is not easy to give a "recipe" that will work in all situations.  However, there are some general hints that can make the work easier.  First balance those elements that appear in only one formula on each side of the equation.  If a free element appears, try to balance those atoms last.

The equation forming water that we just balanced does not fit those rules too well, because both elements appear in just one formula on each side of the equation, and because both elements appear as free elements on the left hand side.  I will now work several examples -- step by step -- that hopefully, well make the process more clear.

Example 1

Methane (CH4) burns in oxygen (O2) to form carbon dioxide (CO2) and water (H2O).  Write a balanced equation for this reaction, showing all substances as gases.

Solution:

The word equation for this reaction is

methane   +   oxygen   ---------->   carbon dioxide   +   water

Replacing each name in the above equation with its formula gives

CH4(g)   +   O2(g)   ---------->   CO2(g)   +   H2O(g)

Following the hints I gave above, we would not want to start our attempts at balancing this by considering oxygen.  It appears in more than one formula on the right hand side, so it would not be clear which coefficient to addjust.  This leaves carbon and hydrogen as possibilities.  I generally start with carbon, where C, H and O are involved.

Notice that there is 1 C atom on the left and 1 C atom on the right.  For now, at least, C is balanced.  Now let's look at H atoms.  We have 4 H atoms in CH4 but only 2 H atoms in H2O.  We can get 4 H atoms on the right by forming two H2O molecules instead of only one.  Our equation now becomes

CH4(g)   +   O2(g)   ---------->   CO2(g)   +   2H2O(g)

H atoms are now balanced.  There are 4 of them on each side.  C atoms are still balanced (one on each side).  We must now look at oxygen.  We have 2 O atoms on the left in O2 and 4 O atoms on the right -- two of them in CO2 and two of them in H2O (since we are producing 2 of these molecules).  We can get 4 O atoms on the left side of the equation by taking two O2 molecules instead of only one.  We now have

CH4(g)   +   2O2(g)   ---------->   CO2(g)   +   2H2O(g)

This equation is balanced.  The number of atoms of each element is as follows:  1 C atom on each side, 4 H atoms on each side, and 4 O atoms on each side.  Always remember to do a final check before considering your work complete.  ALL atoms must balance.

Example 2

Propane (C3H8) burns in oyxgen (O2) to form carbon dioxide (CO2) and water (H2O).  Write a balanced equation for this reaction, showing all substances as gases.

Solution:

The word equation for the reaction is

propane   +   oxygen   ---------->   carbon dioxide   +   water

Replacing each chemical name with its formula (and showing it as a gas) gives

C3H8(g)   +   O2(g)   ---------->   CO2(g)   +   H2O(g)

Starting with carbon, we see that there are 3 C atoms on the left but only 1 C atom on the right.  We put a coefficient of 3 in front of CO2 to balance carbon.  Our equation now becomes

C3H8(g)   +   O2(g)   ---------->   3CO2(g)   +   H2O(g)

Carbon is now balanced, with 3 C atoms on each side.  We now look at hydrogen.  There are 8 Hatoms on the left (in C3H8) but only 2 H atoms on the right (in H2O).  We can get 8 H atoms on the right by putting a coefficient of 4 in front of H2O.  Our equation now becomes

C3H8(g)   +   O2(g)   ---------->   3CO2(g)   +   4H2O(g)

The final step is to balance oxygen.  We have 2 O atoms on the left (in O2) and 10 O atoms on the right (6 of them are contained in CO2 molecules and 4 of them are contained in H2O molecules).  We can get 10 O atoms on the left (to balance the 10 on the right) by putting a coefficient of 5 in front of O2.  Our equation now becomes

C3H8(g)   +   5O2(g)   ---------->   3CO2(g)   +   4H2O(g)

This equation is balanced.  The number of atoms of each element on both sides of the equation is as follows:  3 atoms of C, 8 atoms of H and 10 atoms of O.  You should always check your final equation to be sure it really is balanced.

Example 3

Butane (C4H10) burns in oxygen (O2) to form carbon dioxide (CO2) and water (H2O).  Write a balanced equation for this reaction showing all substances as gases.

Solution:

The word equation for the reaction is

butane   +   oxygen   ---------->   carbon dioxide   +   water

Replacing each substance with its chemical formula gives

C4H10(g)   +   O2(g)   ---------->   CO2(g)   +   H2O(g)

Starting with carbon, we see there are 4 C atoms on the left (in C4H10) but only 1 C atom on the right (in CO2).  We can get 4 C atoms on the right by putting a coefficient of 4 in front of CO2.  Our equation now becomes

C4H10(g)   +   O2(g)   ---------->   4CO2(g)   +   H2O(g)

Carbon is now balanced, with 4 C atoms on each side.  Now let's look at hydrogen.  Thre are 10 H atoms on the left (in C4H10) and only 2 on the right (in H2O) but we can get 10 H atoms on the right if we put a coefficient of 5 in front of H2O.  This gives

C4H10(g)   +   O2(g)  ---------->   4CO2(g)   +   5H2O(g)

Now lets look at oxygen.  There are only 2 O atoms on the left (in O2) but there are 13 on the right (8 of these are in CO2 and 5 of these are in H2O).  Here, we seem to have a problem.  We have an odd number of oxygen atoms on the right side (13) but on the left side, the oxygen atoms can only come in pairs -- that is, they can only produce an even number .  How are we going to get 13 O atoms when they must come in pairs?

We solve this problem by doing something that might seem rather unusual.  We use a fractional coefficient.  If we put a coefficient of 13 / 2 (that's thirteen halves) in front of O2 and write the equation as

C4H10(g)   +   (13 / 2) O2(g)   ---------->   4CO2(g)   +   5H2O(g)

This balances the equation.  There are now 13 atoms of oxygen atoms on the left side just like the right side.  Recall from the notes on Counting Atoms in Chemical Formulas that you multiply the coefficeint by the subscript to get the number of atoms.  For O2 this multiplication is

13   x   2   =   13
 2

And on the right hand side, we have

4 * 2   +   5 * 1   =   8   +   5   =   13

Although balanced, some may find it troubling that a fraction was used as a coefficient.  Later, we will develope the mole concept, and from then on, there will be nothing wrong with fractional coefficients.  Right now, however, we are interpreting a chemical formula to represent one molecule (or "formula unit" for ionic compounds, which don't exist as molecules).  We can't have a fraction of a molecule -- they come only in whole numbers -- like people.

We can easily get an equation balanced using only integers (no fractions) by multiplying ALL the coefficients by 2.  The number 2 was chosen because the denominator of our fraction is 2, and we want to "clear the fraction".  Multiplying all coefficients by 2 gives

2C4H10(g)   +   13O2(g)   ---------->   8CO2(g)   +   10H2O(g)

The above equation is balanced with the smallest possible integers, that is, in lowest terms.  Each element has the same number of atoms on  both sides of the equation.  The numbers are as follows:  8 C atoms, 20 H atoms, and 26 O atoms.

A mistake sometimes made when trying to clear the fraction is to multiply only the fractional coefficient by 2 and write

C4H10(g)   +   13O2(g)   ---------->   4CO2(g)   +   5H2O(g)

This does not balance oxygen.  There are 26 O atoms on the left (in O2) and 13 O atoms on the right (8 in CO2 and 5 in H2O).  Think of the set of coefficients in a balanced equation as the ingredients in a cooking recepie.  If you want to double the recepie, you must double ALL ingredients.

Example 4

Table sugar (C12H22O11), a solid, can be made to burn in oxygen (O2), forming carbon dioxide (CO2) and water vapor (H2O) as products.  Write a balanced equation for this reaction, indicating the correct physical state for all substances involved.

Solution:

The word equation is

sugar   +   oxygen   ---------->   carbon dioxide   +   water

Replacing each substance with its formula gives

C12H22O11(s)   +   O2(g)   ---------->   CO2(g)   +   H2O(g)

Starting with carbon, we note that there are 12 C atoms on the left (in C12H22O11) but only 1 C atom on the right (in CO2).  We can get 12 C atoms on the right by putting a coefficient of 12 in front of CO2.  Our equation now becomes

C12H22O11(s)   +   O2(g)   ---------->   12CO2(g)   +   H2O(g)

C atoms are now balanced, with 12 on each side.  Now let's look at hydrogen.  There are 22 H atoms on the left (in C12H22O11) but only 2 H atoms on the right (in H2O).  We can get 22 H atoms on the right by putting a coefficient of 11 in front of H2O.  Our equation now becomes

C12H22O11(s)   +   O2(g)   ---------->   12CO2(g)   +   11H2O(g)

Now look at oxygen.  Let's consider O2 last, because it's the one we're going to adjust to make everything work.  We have to take stock of what we have elsewhere, then decide what must be done with O2 to bring the O atoms into balance.  On the right side, we have a total of 35 O atoms (24 of these are in CO2 and 11 are in H2O).  We must therefore have 35 O atoms on the left side also.  We have 11 of these required 35 O atoms in C12H22O11.  This leaves 35 - 11 or 24 O atoms that must be supplied by O2.  By putting a coefficient of 12 in front of O2 we obtain the needed 24 O atoms.  Therefore, the balanced equation is

C12H22O11(s)   +   12O2(g)   ---------->   12CO2(g)   +   11H2O(g)

This is the balanced equation.  The number of atoms of each element -- which is the same on both sides of the equation -- is as follows:  12 C atoms, 22 H atoms, and 35 O atoms.

Example 5

Potassium metal, a solid, reacts with liquid bromine to form solid potassium bromide.  Potassium is monatomic and bromine is diatomic.  The formula of potassium bromide should be obvious from the known ion charges for these elements.  Write a balanced equation to describe the reaction.

Solution:

The word equation is

potassium   +   bromine   ---------->   potassium bromide

Replacing the chemical names with the corresponding formulas gives

K(s)   +   Br2(l)   ---------->   KBr(s)

Note that the formula of potassium bromide was arrived at by recognizing that potassium forms K+ ions and bromine forms Br- ions.  In lowest terms, it takes one of each ion to give a neutral formula.  If it is not clear to you how I arrive at KBr for potassium bromide, please review the lectures notes titled Chemical Nomenclature.

Looking at the above equation, we see that K atoms are balanced (1 K atom on each side) but Br atoms are not.  There are 2 Br atoms on the left (in Br2) but only 1 Br atom on the right (in KBr).  We can get 2 Br atoms on the right by putting a coefficient of 2 in front of KBr.  We now have

K(s)   +   Br2(l)   ---------->   2KBr(s)

Br atoms are now balanced -- with 2 on each side.  But now K is no longer balanced.  There is only 1 K atom on the left (in K) but 2 K atoms on the right (in KBr).  We can get 2 K atoms on the left by putting a coefficient of 2 in front of K.  We now have

2K(s)   +   Br2(g)   ---------->   2KBr(s)

This is the the balanced equation.  There are 2 K atoms on both sides and 2 Br atoms on both sides.

Example 6

When aqueous solutions of potassium carbonate and calcium chloride are mixed, they produce solid calcium carbonate and aqueous potassium chloride.  The formulas of the substances in the reaction should be obvious from the known ion charges.  Write a balanced equation for this reaction.

Solution:

The word equation is

potassium carbonate   +   calcium chloride   ---------->   calcium carbonate   +   potassium chloride

The formulas can be figured out using the methods in the lecture notes titled Chemical Nomenclature.  Replacing the chemical names with the corresponding chemical formulas gives

K2CO3(aq)   +   CaCl2(aq)   ---------->   CaCO3(s)   +   KCl(aq)

Notice that the carbonate group (CO32-, except that the charge is not written in the formula because it is cancelled by the cation charge) remains intact during the reaction.  Therefore, rather than balancing C atoms and O atoms separately, we can save some time (and steps) by balancing carbonate as a group.  As of right now, carbonate is balanced.  There is 1 carbonate group on each side of the equation.  If carbonate becomes unbalanced during the course of our efforts balancing the other atoms, we will have to deal with it later.  For now, let's turn our attention to another atom.  Let's look at potassium.  We have 2 K atoms on the left side (in K2CO3) but only one K atom on the right side (in KCl).  We can get 2 K atoms on the right side by putting a coefficient of 2 in front of KCl.  We now have

K2CO3(aq)   +   CaCl2(aq)   ---------->   CaCO3(s)   +   2KCl(aq)

All atoms are now balanced.  Sometimes we have those lucky breaks where balancing one atom balances others at the same time.  If you will look back at the original equation, you will see that Cl atoms were not balanced (2 on the left and 1 on the right).  The balancing step we just carried out was performed with K atoms in mind, but it has balanced Cl atoms as well.  Ca atoms and the CO32- were balanced to begin with, and nothing we did while balancing K (and Cl) affected that balance, so Ca and CO32- remain balanced.  On at atom by atom basis, we have the following number of atoms of each element on both sides of the equation:  2 K atoms, 1 C atom, 3 O atoms, and 2 Cl atoms.

Example 7

Ferric oxide, a solid, reacts with carbon monoxide, a gas, to form solid iron metal and carbon dioxide gas.  Iron is a monatomic element.  The formulas of the other substances in this reaction can be determined from their chemical names.  Write a balanced equation for this reaction.

Solution:

The word equation is

ferric oxide   +   carbon monoxide   ---------->   iron   +   carbon dioxide

Since iron forms the Fe2+ and Fe3+ ions, the ferric ion must be Fe3+, because the suffix -ic refers to the ion having the higher charge.  Since the oxide ion is O2-, the formula must be Fe2O3.  If it not clear how I obtained this formula, please study the Chemical Nomenclature notes.  The Greek prefix mono- means 1 and di- means 2.  Therefore carbon monoxide is CO and carbon dioxide is CO2.  (If no prefix is given for the first element, it is mono- by default).  Having figured out all the formulas, we can now write our first draft of the equation.

Fe2O3(s)   +   CO(g)   ---------->   Fe(s)   +   CO2(g)

This equation is rather difficult to balance on an atom by atom basis.  It tends to lead one down a twisted path.  What would seem like a logical start would be to note that C is already balanced and that putting a 2 in front of Fe will balance Fe.

Fe2O3(s)   +   CO(g)   ---------->   2Fe(s)   +   CO2(g)

The problem occurs when you try to balance oxygen.  There are 4 O atoms on the left (3 in Fe2O3 and one in CO).  There are 2  O atoms on the right, in CO2.  If you put a 2 in front of CO2 to get 4 O atoms on the right like this:

Fe2O3(s)   +   CO(g)   ---------->   2Fe(s)   +   2CO2(g)

you will have messed up the balance in C atoms.  If you then put a 2 in front of CO to restore the C atom balance like this:

Fe2O3(s)   +   2CO(g)   ---------->   2Fe(s)   +   2CO2(g)

you've messed up the O atom balance again.  Situations like this can drive you crazy!

The trick to seeing your way clearly in getting this equation balanced is to "get the big picture".  Let's think about what really happens in this reaction.  The CO becomes CO2 which means each CO molecule that participates in this reaction must pick up 1 O atom.  The Fe2O3 is stripped of all its O atoms, to become just Fe.  So each formula unit of Fe2O3 that participates in this reaction must lose 3 O atoms.  Every O atom that is lost by Fe2O3 must be picked up by a CO molecule, because there is no monatomic oxygen (O) being formed in this reaction.

How can each Fe2O3 formula unit lose 3 O atoms, and each CO molecule gain only one O atom, and yet not have any O atoms left over?  There must be 3 molecules of CO for every 1 formula unit of Fe2O3 in the reaction.  That way, all the O atoms lost by the Fe2O3 have somewhere to go, and are not "left out in the cold".  Of course, those 3 CO molecules will become CO2 molecules when they receive the O atoms, so 3 CO2 molecules will be formed.  The balanced equation is therefore

Fe2O3(s)   +   3CO(g)   ---------->   2Fe(s)   +   3CO2(g)

All atoms are now balanced.  On both sides of the equation, there are 2 Fe atoms, 3 C atoms, and 6 O atoms.

I hope you have found these notes helpful.  Your questions, comments and suggestions for improvement are always welcome.

SUGGESTED READING

Section 3.7, "Writing and Balancing Chemical Equations", pages 100 - 104, in your Hill & Petrucci General Chemistry textbook.

This page was last modified Wednesday October 13, 1999