Lesson Explainer: Balancing Chemical Equations Chemistry

In this explainer, we will learn how to determine the stoichiometry of a reaction equation by balancing the numbers of atoms in the reactants and products.

Let us consider the reaction of hydrogen gas (H2) with oxygen gas (O2) to produce water vapor: H()+O()HO()222ggg

The chemical equation as written tells us useful information about the reactants, products, and their states of matter. However, this chemical equation is not complete.

Notice that there are two atoms of hydrogen on both sides of the equation, but there are two atoms of oxygen on the reactant side and only one atom of oxygen on the product side:

This reaction equation violates the law of conservation of mass that states that matter can neither be created nor destroyed. We must, when writing the equation, find a way to ensure that the number of hydrogen and oxygen atoms is the same on both sides of the reaction arrow. A reaction equation that does not violate the law of conservation of mass is called a balanced chemical equation.

Definition: Balanced Chemical Equations

A balanced chemical equation is a reaction equation where the number of atoms of each element is the same on the reactant and product sides of the equation.

To balance the equation, we might consider adding a subscript two after the oxygen atom in the water molecule: H()+O()HO()2222ggl

This new equation does adhere to the law of conservation of mass. However, the product of this reaction is hydrogen peroxide, not water. We cannot change the subscript values for any substance on the right- or left-hand sides of the chemical equation as this will change the identity of the compounds involved in the reaction.

Instead, we should adjust the number of molecules or formula units of each substance involved in the reaction until the equation is balanced. We could place the number two in front of the hydrogen molecule and water molecule in the chemical equation, thus doubling the number of hydrogen and water molecules: 2H()+O()2HO()222ggg

There are now four hydrogen atoms and two oxygen atoms on both sides of the chemical equation and the reaction is balanced. The numerals that we placed in front of the hydrogen and water molecules in the equation are known as coefficients or stoichiometric coefficients.

Definition: Stoichiometric Coefficients

Stoichiometric coefficients are numerical values written before a species in a chemical equation in order to balance the overall reaction.

Stoichiometric coefficients indicate the ratio of each species involved in the reaction. Thus, for the formation of water from hydrogen and oxygen, the ratio of hydrogen molecules to oxygen molecules is 2∢1.

Example 1: A Balanced Equation of Hydrogen Iodide Formation

Hydrogen iodide is produced by the reaction of hydrogen and iodine according to the equation π‘₯𝑦𝑧H+IHI22

The coefficients π‘₯, 𝑦, and 𝑧 are whole numbers. What are the smallest possible values of π‘₯, 𝑦, and 𝑧?

Answer

In a balanced chemical equation, the number of atoms of each element is the same on both sides of the reaction. We can see from either the chemical equation or the model below that there are two hydrogen atoms and two iodine atoms on the reactant side and one atom of each hydrogen and iodine on the product side.

For there to be two hydrogen and iodine atoms on the product side, there must be a second hydroiodic acid molecule.

With two molecules of HI, the reaction is balanced.

π‘₯, 𝑦, and 𝑧 in the chemical equation are the stoichiometric coefficients. Stoichiometric coefficients are numerals that indicate the number of molecules or formula units of each species that are necessary to balance a chemical reaction. In this question, π‘₯, 𝑦, and 𝑧 represent the number of molecules of H2, I2, and HI respectively. The reaction is balanced with one molecule of H2, one molecule of I2, and two molecules of HI. Thus, π‘₯=1, 𝑦=1, and 𝑧=2.

Example 2: Identifying Which Element Is Out of Balance in a Chemical Equation

Consider the chemical equation CHCHOH+O2CO+3HO32222

Which element is out of balance in the chemical equation?

Answer

An element is in balance when the total number of atoms of that element is the same on both sides of the reaction arrow. Subscripts indicate the number of atoms in one molecule or formula unit. Numbers in front of a species are called coefficients and indicate the number of molecules or formula units.

On the reactant side, left of the reaction arrow, we see the following: CHCHOH+OΒ·322

There are no coefficients, meaning that there is one molecule of each species. We can determine the number of carbon, hydrogen, and oxygen atoms by adding the subscripts together for each element. An unwritten subscript is an implied one:

There are two carbon atoms, six hydrogen atoms, and three oxygen atoms on the reactant side.

On the product side, right of the reaction arrow, we see the following: 2CO+3HOΒ·22

The coefficient two in front of the carbon dioxide molecule indicates that there are two molecules of carbon dioxide. Two molecules will have twice as many atoms as one molecule. The number of atoms from a species will be equal to the number of atoms in one molecule times the coefficient:

There are two carbon atoms, six hydrogen atoms, and seven oxygen atoms on the product side. The number of carbon and hydrogen atoms is the same on both sides of the reaction, but the number of oxygen atoms is not. The element that is out of balance in the chemical equation is oxygen.

Balancing a chemical equation is often a process of guess and check, but there are steps we can take to make this process easier.

How To: Balancing a Chemical Equation

  1. Make a list of the number of atoms of each element on both sides of the reaction.
    Tip: If any polyatomic ions are present on both sides of the reaction, treat them as a unit. For instance, CO32– will be counted as one group rather than one carbon and three oxygen atoms.
  2. Identify atoms/ions that are not equal on both sides of the reaction (unbalanced).
  3. Add coefficients in front of any species in the reaction to balance the atoms/ions. The number of atoms/ions of each type in the species will be multiplied by the coefficient.
    Tip: Begin by balancing the atoms or ions in the most complex species.
    Tip: Balance elements in their pure form last, for instance, Fe, O2, S8, and so on.
  4. Continue adding coefficients until all atoms/ions are balanced.

Let us see how the steps for balancing a chemical equation apply in practice. Below is the unbalanced chemical equation for the reaction of sodium phosphate with calcium chloride: NaPO()+CaCl()NaCl()+Ca(PO)()342342aqaqaqs

We will begin by making a list of the atoms of each element on both sides of the reaction. However, we will notice that phosphate (PO43–) appears on both sides of the equation. This means that we can count the phosphate ion as a unit instead of breaking it into phosphorus and oxygen atoms. A table, like the one below, is useful for keeping track of the atoms/ions:

We need to pay special attention when counting ions. Notice that there is one phosphate ion in each sodium phosphate unit, not four, and that there are two phosphate ions in each calcium phosphate unit, not 12.

From the table, we can see that all of the atoms and ions in the equation are currently unbalanced. We can now begin to balance the equation by adding coefficients.

We start by identifying the most complex species in the equation. Calcium phosphate contains the most atoms and ions. Therefore, we can begin by balancing the calcium and phosphate. There are three calcium atoms on the product side and only one calcium atom on the reactant side. We can place a coefficient of three in front of the calcium chloride:

If one formula unit of calcium chloride contains one calcium atom and two chlorine atoms, then three formula units will contain a total of three calcium atoms and six chlorine atoms:

With the calcium atoms balanced, we can balance the phosphate ions by placing a coefficient of two in front of the sodium phosphate on the reactant side:

If one formula unit of sodium phosphate contains three sodium atoms and one phosphate ion, then two formula units will contain six sodium atoms and two phosphate ions:

We can now balance the sodium and chlorine atoms by placing a coefficient of six in front of the sodium chloride:

The complete balanced chemical equation for the reaction between sodium phosphate and calcium chloride is 2NaPO()+3CaCl()6NaCl()+Ca(PO)()342342aqaqaqs

Example 3: A Balanced Equation in the Marsh Test for Arsenic

The Marsh test is a method used historically for the detection of arsenic, a toxic element, in mixed materials. The sample is burned to convert any arsenic present to arsenic trioxide (AsO23) and then treated with nitric acid (HNO3) and metallic zinc. If arsenic is present, a garlic smell is detected due to the production of arsine gas (AsH3). The equation for this reaction is AsO+Zn+HNOAsH+6Zn(NO)+HO2333322π‘Žπ‘π‘π‘‘

The coefficients π‘Ž, 𝑏, 𝑐, and 𝑑 are all whole numbers.

  1. Find the value of π‘Ž.
  2. Find the value of 𝑏.
  3. Find the value of 𝑐.
  4. Find the value of 𝑑.

Answer

This question is asking us to balance the chemical equation, one coefficient at a time. A chemical equation is balanced when the number of atoms of each element and/or the number of each polyatomic ion is the same on both sides of the reaction.

We can begin by making a list of the number of atoms of each element on both sides of the reaction. We will notice that NO–3 (nitrate) appears on both sides of the equation. We can keep the nitrate ion as a unit in our list. However, we must take care when counting the oxygen atoms to ensure that we do not count the oxygen atoms in the nitrate ion. We can construct a table like the following to organize our list:

Part 1

Coefficient π‘Ž represents the number of zinc atoms on the reactant side. There are six zinc atoms on the product side and only one zinc atom on the reactant side. Setting coefficient a to six will give us six zinc atoms on the reactant side and balance the zinc atoms overall:

The value of coefficient π‘Ž is six.

Part 2

Coefficient 𝑏 represents the number of nitric acid molecules on the reactant side. A nitric acid molecule contains both hydrogen atoms and nitrate ions. Looking at the product side, we can see that all of the nitrate ions are found in the zinc nitrate unit that already has a coefficient of six. As the number of nitrate ions on the product side will not be able to change, we should balance the nitrate ions in order to determine coefficient 𝑏.

As there is one nitrate ion on the reactant side and 12 nitrate ions on the product side, we can set coefficient 𝑏 to 12 to balance the nitrate ions:

We should also recognize that setting coefficient 𝑏 to 12 will increase the number of hydrogen atoms on the reactant side to 12. The value of coefficient 𝑏 is 12.

Part 3

Coefficient 𝑐 represents the number of arsine gas molecules on the product side. Arsine gas contains arsenic and hydrogen atoms. As the hydrogen atoms on the product side are found in two molecules, we cannot balance the hydrogen atoms to determine coefficient 𝑐. Therefore, we should balance the arsenic atoms.

There are two arsenic atoms on the reactant side and one arsenic atom on the product side. We can set coefficient 𝑐 to two in order to balance the arsenic atoms:

We should also recognize that setting coefficient 𝑐 to two will increase the total number of hydrogen atoms on the product side to eight. The value of coefficient 𝑐 is 2.

Part 4

Coefficient 𝑑 represents the number of water molecules on the product side. Water contains oxygen and hydrogen atoms. As the hydrogen atoms on the product side are found in two molecules, it will be easier to determine coefficient 𝑑 by balancing the oxygen atoms.

There are three oxygen atoms on the reactant side and one oxygen atom on the product side. We can set coefficient 𝑑 equal to three to balance the oxygen atoms.

We should also recognize that setting coefficient 𝑑 to three will increase the total number of hydrogen atoms on the product side to 12, balancing the hydrogen atoms as well. The value of coefficient 𝑑 is three.

Chemical equations can be written by analyzing the wording of a chemical statement. Common phrases used in chemical statements and their relationship to reactants and products are shown in the table below.

Reactant AReacts with
Is mixed with
Combines with
Is added to
Burns in
Reactant B
ReactantsYield
Produce
From
Decompose into
Products
ProductsAre produced by
Are generated from
Reactants

Let us consider the following chemical statement. Calcium hydroxide reacts with hydrochloric acid to produce calcium chloride and water.

This chemical statement indicates that calcium hydroxide (Ca(OH)2) and hydrochloric acid (HCl) are the reactants and should be written on the left-hand side of the reaction arrow. Calcium chloride (CaCl2) and water (HO2) are the products and should be written on the right-hand side of the reaction arrow: Ca(OH)+HClCaCl+HO222

Once the unbalanced chemical equation has been written, we can follow our systematic series of guess-and-check steps to determine the appropriate stoichiometric coefficients.

Example 4: A Balanced Equation for Phosphorous Acid Decomposition

When heated, phosphorous acid (HPO33) decomposes into phosphoric acid (HPO34) and the pungent toxic gas phosphine (PH3). Write a balanced chemical equation for this reaction using the smallest possible whole number coefficients for the reactants and products.

Answer

We should begin this problem by deciphering the chemical statement to identify the reactants and products. The key phrase β€œdecomposes into” indicates that species before this phrase in the sentence are the reactants and the species after this phrase are the products. This means that when writing the chemical equation, phosphorous acid should appear to the left of the reaction arrow while phosphoric acid and phosphine gas should appear to the right of the reaction arrow: HPOHPO+PH33343

This is the unbalanced chemical equation. An equation is balanced when the number of atoms of each element and/or the number of each polyatomic ion is the same on both sides of the reaction.

We can begin by making a list of the number of atoms of each element on both sides of the reaction:

Notice that we separated the polyatomic ions, PO33– and PO43–, into phosphorus and oxygen atoms. This is necessary as neither ion appears on both the reactant and product sides of the chemical equation.

We can now begin to balance the equation by adding coefficients. Coefficients are numerals that can be placed in front of any species in the reaction to balance the equation. We need to decide which element to begin balancing. Notice that the hydrogen and phosphorus atoms appear in two species on the product side. This can make balancing the hydrogen and phosphorus atoms difficult. Therefore, we should begin by balancing the oxygen atoms.

There are three oxygen atoms on the reactant side and four oxygen atoms on the product side. The number of oxygen atoms on the reactant side in this equation will always be equal to the number of oxygen atoms in one phosphorous acid unit times the coefficient: 3Γ—=.oxygenatomscoefficienttotaloxygenatomsonreactantside

For there to be four oxygen atoms on the reactant side, we need coefficientoxygenatomscoefficientΓ—3=4=43.

The coefficient would need to be 43. Using this coefficient will make balancing the overall equation very challenging. Instead of using a fraction for the coefficient, we can look for the least common multiple of three, the number of atoms on the reactant side, and four, the number of atoms on the product side. The least common multiple of three and four is 12. We can then set the total number of oxygen atoms needed on both sides of the reaction to 12:

For there to be 12 oxygen atoms on the reactant side, a coefficient of four must be placed in front of the phosphorous acid:

Placing a coefficient of four in front of the phosphorous acid will also change the number of hydrogen and phosphorus atoms on the reactant side. If one unit of phosphorous acid contains three hydrogen atoms and one phosphorus atom, then four phosphorous acid units will contain 12 hydrogen atoms and four phosphorus atoms:

For there to be 12 oxygen atoms on the product side, a coefficient of three must be placed in front of the phosphoric acid:

Placing a coefficient of three in front of the phosphoric acid will also change the number of hydrogen and phosphorus atoms on the product side. If one unit of phosphoric acid contains three hydrogen atoms and one phosphorus atom, then three phosphoric acid units will contain nine hydrogen atoms and three phosphorus atoms:

With a total of 12 hydrogen atoms and four phosphorus atoms on the reactant side, the hydrogen and phosphorus atoms are now balanced. The complete balanced chemical equation for the decomposition of phosphorous acid is 4HPO3HPO+PH33343

Let us consider the following unbalanced combustion reaction: CH+OCO+HO26222

We can begin to balance this chemical equation by making a list of the number of atoms of each element:

Oxygen exists in its pure form on the reactant side. It is useful to balance elements in their pure form last. We can balance the carbon atoms by placing a coefficient of two in front of the carbon dioxide molecule and the hydrogen atoms by placing a coefficient of three in front of the water molecule. The table below shows the updated totals on the product side after these coefficients have been added:

All that remains is to balance the oxygen atoms. There are two oxygen atoms on the reactant side and a total of seven oxygen atoms on the product side. The number of oxygen atoms on the reactant side of this equation will always be equal to the number of atoms in an oxygen molecule times the stoichiometric coefficient: 2Γ—=.oxygenatomscoefficienttotaloxygenatomsonreactantside

For there to be seven oxygen atoms on the reactant side, we need coefficientoxygenatomscoefficientΓ—2=7=72.

The coefficient would need to be 72 or 3.5, as seen below:

While a coefficient of 3.5 in front of the oxygen molecule does balance the overall equation, it implies that the reaction consumes three and a half oxygen molecules. This cannot be the case. To alleviate the confusion and maintain a balanced chemical equation, we can multiply all of the coefficients, including the unwritten one in front of CH26, by two:

Thus, the complete balanced chemical equation is: 2CH+O4CO+6HO267222

Example 5: A Balanced Equation for Ammonia Oxidation

The reaction of ammonia (NH3) with oxygen produces nitric oxide (NO) and water as the only products. Write a balanced chemical equation for this reaction using the smallest possible whole number coefficients for the reactants and products.

Answer

We should begin this problem by deciphering the chemical statement to identify the reactants and products. The key word β€œproduces” indicates that the species before this word in the sentence are the reactants and the species after this word are the products. This means that when writing the chemical equation, ammonia and oxygen should appear to the left of the reaction arrow and nitric oxide and water should appear to the right of the reaction arrow. Notice that oxygen is written as O2, not O, in the chemical equation below. This is because oxygen exists as a diatomic molecule in its pure form: NH+ONO+HO322

The equation written is the unbalanced chemical equation. An equation is balanced when the number of atoms of each element is the same on both sides of the reaction.

We can begin balancing the equation by making a list of the number of atoms of each element on both sides of the reaction:

The nitrogen atoms are already balanced. Elements that exist in their pure form in the chemical equation, such as the oxygen, are most easily balanced last. This means that we should begin by balancing the hydrogen atoms.

There are three hydrogen atoms on the reactant side and two hydrogen atoms on the product side. As three is not a multiple of two, these atoms will be most easily balanced by finding their least common multiple. The least common multiple of two and three is six. We can then set the total number of hydrogen atoms on both sides of the reaction to six:

For there to be six hydrogen atoms on the reactant side, a coefficient of two must be placed in front of the ammonia molecule. For there to be six hydrogen atoms on the product side, a coefficient of three must be placed in front of the water molecule:

Placing these coefficients has affected the number of nitrogen atoms on the reactant side and that of oxygen atoms on the product side as well. The hydrogen atoms are now balanced, but we must rebalance the nitrogen atoms. We can accomplish this by placing a coefficient of two in front of the nitric oxide molecule:

Placing a coefficient of two in front of the nitric oxide molecule has balanced the nitrogen atoms and has affected the total number of oxygen atoms on the product side. All that remains to do is balance the oxygen atoms.

The number of oxygen atoms on the reactant side of this equation will always be equal to the number of oxygen atoms in an oxygen molecule times the coefficient: 2Γ—=.oxygenatomscoefficienttotaloxygenatomsonreactantside

For there to be five oxygen atoms on the reactant side, we need coefficientoxygenatomscoefficientΓ—2=5=52.

The coefficient would need to be 52 or 2.5, as seen below:

The chemical equation is now balanced. However, the question asked for the equation to be balanced using the smallest possible whole number coefficients. We can maintain a balanced chemical equation and change the decimal coefficient in front of the oxygen molecule to a whole number by multiplying all of the coefficients by two:

The balanced chemical equation for the reaction of ammonia with oxygen using whole number coefficients is 4NH+5O4NO+6HO322

Key Points

  • A balanced chemical equation is a reaction equation where the number of atoms of each element is the same on the reactant and product sides of the equation.
  • Stoichiometric coefficients are placed in front of species in a chemical equation to balance the reaction.
  • Balancing equations is a process of trial and error.
  • Balancing equations can be made easier by balancing polyatomic ions as a unit and waiting to balance pure elements until last.

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