This is "Unit 7", section 7.1 from the book General Chemistry (v. 1.0).

Has this book helped you? Consider passing it on:
Help Creative Commons
Creative Commons supports free culture from music to education. Their licenses helped make this book available to you.
Help a Public School
DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators.

7.1 Chemical Equations

Learning Objectives

  1. To describe a chemical reaction.
  2. To be able to write balanced chemical equations.

As shown in Figure 7.1(a), applying a small amount of heat to a pile of orange ammonium dichromate powder results in a vigorous reaction known as the ammonium dichromate volcano. Heat, light, and gas are produced as a large pile of fluffy green chromium(III) oxide forms. We can describe this reaction with a chemical equationAn expression that gives the identities and quantities of the substances in a chemical reaction. Chemical formulas are used to indicate the reactants on the left and the products on the right. An arrow points from reactants to products., an expression that gives the identities and quantities of the substances in a chemical reaction. Chemical formulas and other symbols are used to indicate the starting material(s), or reactant(s)The starting material(s) in a chemical reaction., which by convention are written on the left side of the equation, and the final compound(s), or product(s)The final compound(s) produced in a chemical reaction., which are written on the right. An arrow points from the reactant to the products:

Figure 7.1(a) An Ammonium Dichromate Volcano: Change during a Chemical Reaction

By Mikk Mihkel Vaabel (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)], via Wikimedia Commons. The starting material (reddish granules) is solid ammonium dichromate. A chemical reaction transforms it to solid chromium(III) oxide (greenish curls), depicted showing a portion of its chained structure, nitrogen gas, and water vapor. (In addition, energy in the form of heat and light is released.) During the reaction, the distribution of atoms changes, but the number of atoms of each element does not change. Because the numbers of each type of atom are the same in the reactants and the products, the chemical equation is balanced.

Equation 7.1(eq1)

(NH 4 ) 2 Cr 2 O 7 reactant → Cr 2 O 3 + N 2 + 4H 2 O products

The arrow is read as "yields" or "reacts to form." So Equation 7.1(eq1) tells us that ammonium dichromate (the reactant) yields chromium(III) oxide, nitrogen, and water (the products).

The equation for this reaction is even more informative when written as

Equation 7.1(eq2)

(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g)

Equation 7.1(eq2) is identical to Equation 7.1(eq1) except for the addition of abbreviations in parentheses to indicate the physical state of each species. The abbreviations are (s) for solid, (l) for liquid, (g) for gas, and (aq) for an aqueous solution, a solution of the substance in water.

Consistent with the law of conservation of mass, the numbers of each type of atom are the same on both sides of these equations. Since the number of atoms of each element is the same on both sides of the arrow, the mass is the same on both sides of the arrow. Each side has two chromium atoms, seven oxygen atoms, two nitrogen atoms, and eight hydrogen atoms. In a balanced chemical equation, both the numbers of each type of atom and the total charge are the same on both sides. Equation 7.1(eq1) and Equation 7.1(eq2) are balanced chemical equations. What is different on each side of the equation is how the atoms are arranged to make molecules or ions. A chemical reaction represents a change in the distribution of atoms but not in the number of atoms. In this reaction, and in most chemical reactions, bonds are broken in the reactants (here, Cr-O and N-H bonds), and new bonds are formed to create the products (here, O-H and N≡N bonds). If the numbers of each type of atom are different on the two sides of a chemical equation, then the equation is unbalanced, and it cannot correctly describe what happens during the reaction. To proceed, the equation must first be balanced.

Note the Pattern

A chemical reaction changes only the distribution of atoms, not the number of atoms.

Balancing Simple Chemical Equations

When a chemist encounters a new reaction, it does not usually come with a label that shows the balanced chemical equation. Instead, the chemist must identify the reactants and products and then write them in the form of a chemical equation that may or may not be balanced as first written. Consider, for example, the combustion of n-heptane (C7H16), an important component of gasoline:

Equation 7.1(eq3)

C7H16(l) + O2(g) → CO2(g) + H2O(g)

The complete combustion of any hydrocarbon with sufficient oxygen always yields carbon dioxide and water (Figure 7.1(b) ).

Figure 7.1(b) An Example of a Combustion Reaction

The wax in a candle is a high-molecular-mass hydrocarbon, which produces gaseous carbon dioxide and water vapor in a combustion reaction. When the candle is allowed to burn inside a flask, drops of water, one of the products of combustion, condense at the top of the inner surface of the flask. Image by Nanda93 (Own work) [Public domain], via Wikimedia Commons

Equation 7.1(eq3) is not balanced: the numbers of each type of atom on the reactant side of the equation (7 carbon atoms, 16 hydrogen atoms, and 2 oxygen atoms) is not the same as the numbers of each type of atom on the product side (1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms). Consequently, we must adjust the coefficients of the reactants and products to give the same numbers of atoms of each type on both sides of the equation. Because the identities of the reactants and products are fixed, we cannot balance the equation by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described.

The simplest and most generally useful method for balancing chemical equations is "inspection," better known as trial and error. We present an efficient approach to balancing a chemical equation using this method.

Steps in Balancing a Chemical Equation

  1. Identify the most complex substance.
  2. Beginning with that substance, choose an element that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element on both sides.
  3. Balance polyatomic ions (if present) as a unit.
  4. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients.
  5. Count the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced.

To demonstrate this approach, let's use the combustion of n-heptane (Equation 7.1(eq3)) as an example.

  1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is C7H16. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.

  2. Adjust the coefficients. Try to adjust the coefficients of the molecules on the other side of the equation to obtain the same numbers of atoms on both sides. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side:

    Equation 7.1(eq4)

    C7H16 + O2 → 7CO2 + H2O
  3. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction.

  4. Balance the remaining atoms. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side:

    Equation 7.1(eq5)

    C7H16 + O2 → 7CO2 + 8H2O

    The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side:

    Equation 7.1(eq6)

    C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(g)
  5. Check your work. The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a chemical equation is balanced.

Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly.

Example 7.1-1

Write and balance the chemical equation for each given chemical reaction.

  1. Hydrogen and chlorine react to make HCl.
  2. Ethane, C2H6, reacts with oxygen to make carbon dioxide and water.

Solution

  1. Let us start by simply writing a chemical equation in terms of the formulas of the substances, remembering that both elemental hydrogen and chlorine are diatomic:

    H2 + Cl2 → HCl

    The most complex substance appears to be HCl. There are two hydrogen atoms and two chlorine atoms in the reactants and one of each atom in the product. We can fix this by adjusting the coefficient to 2 on the product side:

    H2 + Cl2 → 2HCl

    We check our work and there are two hydrogen atoms and two chlorine atoms on both sides of the chemical equation, so it is balanced.

  2. Start by writing the chemical equation in terms of the substances involved:

    C2H6 + O2 → CO2 + H2O

    Ethane, C2H6 on the reactants side, appears to be the most complex formula. We have two carbon atoms on the left, so we need two carbon dioxide molecules on the product side, so that each side has two carbon atoms; that element is balanced. We have six hydrogen atoms in the reactants, so we need six hydrogen atoms in the products. We can get this by having three water molecules:

    C2H6 + O2 → 2CO2 + 3H2O

    Turning our attention to the remaining atoms. We now have seven oxygen atoms in the products (four from the CO2 and three from the H2O). That means we need seven oxygen atoms in the reactants. However, because oxygen is a diatomic molecule, we can only get an even number of oxygen atoms on the reactants side at a time. We can make the number of oxygen atoms on the products side even by multiplying the coefficients other than oxygen's by 2:

    2C2H6 + O2 → 4CO2 + 6H2O

    By multiplying everything else by 2, we don't unbalance the other elements, and we now get an even number of oxygen atoms in the product, specifically, 14. We can get 14 oxygen atoms on the reactant side by having 7 oxygen molecules:

    2C2H6 + 7O2 → 4CO2 + 6H2O

    As a check, recount everything to determine that each side has the same number of atoms of each element. This chemical equation is now balanced.

Test Yourself

Write and balance the chemical equation that represents nitrogen and hydrogen reacting to produce ammonia, NH3.

Answer

N2 + 3H2 → 2NH3

Example 7.1-2

The reaction of the mineral hydroxyapatite [Ca5(PO4)3(OH)] with phosphoric acid and water gives Ca(H2PO4)2·H2O (calcium dihydrogen phosphate monohydrate). Write and balance the equation for this reaction.

Given: reactants and product

Asked for: balanced chemical equation

Strategy:

A Identify the product and the reactants and then write the unbalanced chemical equation.

B Follow the steps for balancing a chemical equation.

Solution:

A We must first identify the product and reactants and write an equation for the reaction. The formulas for hydroxyapatite and calcium dihydrogen phosphate monohydrate are given in the problem. Recall from Section 5.2 that phosphoric acid is H3PO4. The initial (unbalanced) equation is as follows:

Ca5(PO4)3(OH)(s) + H3PO4(aq) + H2O(l) → Ca(H2PO4)2·H2O(s)
  1. B Identify the most complex substance. We start by assuming that only one molecule or formula unit of the most complex substance, Ca5(PO4)3(OH), appears in the balanced chemical equation.

  2. Adjust the coefficients. Because calcium is present in only one reactant and one product, we begin with it. One formula unit of Ca5(PO4)3(OH) contains 5 calcium atoms, so we need 5 Ca(H2PO4)2· H2O on the right side:

    Ca5(PO4)3(OH) + H3PO4 + H2O → 5Ca(H2PO4)2· H2O

  3. Balance polyatomic ions as a unit. It is usually easier to balance an equation if we recognize that certain combinations of atoms occur on both sides. In this equation, the polyatomic phosphate ion (PO43-), shows up in three places.In H3PO4, the phosphate ion is combined with three H+ ions to make phosphoric acid (H3PO4), whereas in Ca(H2PO4)2·H2O it is combined with two H+ ions to give the dihydrogen phosphate ion. Thus it is easier to balance PO4 as a unit rather than counting individual phosphorus and oxygen atoms. There are 10 PO4 units on the right side but only 4 on the left. The simplest way to balance the PO4 units is to place a coefficient of 7 in front of H3PO4:

    Ca5(PO4)3(OH) + 7H3PO4 + H2O → 5Ca(H2PO4)2·H2O

    Although OH- is also a polyatomic ion, it does not appear on both sides of the equation. So oxygen and hydrogen must be balanced separately.

  4. Balance the remaining atoms. We now have 30 hydrogen atoms on the right side but only 24 on the left. We can balance the hydrogen atoms using the least complex substance, H2O, by placing a coefficient of 4 in front of H2O on the left side, giving a total of 4 H2O molecules:

    Ca5(PO4)3(OH)(s) + 7H3PO4(aq) + 4H2O(l) → 5Ca(H2PO4)2·H2O(s)

    The equation is now balanced. Even though we have not explicitly balanced the oxygen atoms, there are 45 oxygen atoms on each side.

  5. Check your work. Both sides of the equation contain 5 calcium atoms, 7 phosphorus atoms, 30 hydrogen atoms, and 45 oxygen atoms.

Exercise

Fermentation is a biochemical process that enables yeast cells to live in the absence of oxygen. Humans have exploited it for centuries to produce wine and beer and make bread rise. In fermentation, sugars such as glucose are converted to ethanol and carbon dioxide. Write a balanced chemical reaction for the fermentation of glucose.

Commercial use of fermentation. Microbrewery vats are used to prepare beer. NOLA Brewing Company, Tchoupitoulas Street, Uptown New Orleans. Brewery tour day.By Infrogmation of New Orleans (Photo by Infrogmation of New Orleans) [GFDL (http://www.gnu.org/copyleft/fdl.html), CC BY-SA 2.0 (http://creativecommons.org/licenses/by-sa/2.0), CC BY-SA 2.5 (http://creativecommons.org/licenses/by-sa/2.5) or CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)], via Wikimedia Commons(left-hand inset) The fermentation of glucose by yeast cells is the reaction that makes beer production possible. Saccharomyces cerevisiae cells in DIC microscopy. By Masur (Own work) [Public domain], via Wikimedia Commons

Answer: C6H12O6(s) → 2C2H5OH(l) + 2CO2(g)

Key Takeaways

  • A chemical equation is a concise description of a chemical reaction.
  • Proper chemical equations are balanced.

Summary

In a chemical reaction, one or more substances are transformed to new substances. A chemical reaction is described by a chemical equation, an expression that gives the identities and quantities of the substances involved in a reaction. A chemical equation shows the starting compound(s)-the reactants-on the left and the final compound(s) -the products-on the right, separated by an arrow. In a balanced chemical equation, the numbers of atoms of each element and the total charge are the same on both sides of the equation. The number of atoms, molecules, or formula units of a reactant or product in a balanced chemical equation is the coefficient of that species.

Numerical Problems

  1. From the statement "nitrogen and hydrogen react to produce ammonia," identify the reactants and the products.

  2. From the statement "sodium metal reacts with water to produce sodium hydroxide and hydrogen," identify the reactants and the products.

  3. From the statement "magnesium hydroxide reacts with nitric acid to produce magnesium nitrate and water," identify the reactants and the products.

  4. From the statement "propane reacts with oxygen to produce carbon dioxide and water," identify the reactants and the products.

  5. Write and balance the chemical equation described by Numerical Problem 1.

  6. Write and balance the chemical equation described by Numerical Problem 2.

  7. Write and balance the chemical equation described by Numerical Problem 3.

  8. Write and balance the chemical equation described by Numerical Problem 4. The formula for propane is C3H8.

  9. Balance: ___NaClO3 → ___NaCl + ___O2

  10. Balance: ___N2 + ___H2 → ___N2H4

  11. Balance: ___Al + ___O2 → ___Al2O3

  12. Balance: ___C2H4 + ___O2 → ___CO2 + ___H2O

  13. How would you write the balanced chemical equation in Numerical Problem 10 if all substances were gases?

  14. How would you write the balanced chemical equation in Numerical Problem 12 if all the substances except water were gases and water itself were a liquid?

  15. Balance each chemical equation.

    1. KI(aq) + Br2(l) → KBr(aq) + I2(s)
    2. MnO2(s) + HCl(aq) → MnCl2(aq) + Cl2(g) + H2O(l)
    3. Na2O(s) + H2O(l) → NaOH(aq)
    4. Cu(s) + AgNO3(aq) → Cu(NO3)2(aq) + Ag(s)
    5. SO2(g) + H2O(l) → H2SO3(aq)
    6. S2Cl2(l) + NH3(l) → S4N4(s) + S8(s) + NH4Cl(s)

  16. Balance each chemical equation.

    1. Be(s) + O2(g) → BeO(s)
    2. N2O3(g) + H2O(l) → HNO2(aq)
    3. Na(s) + H2O(l) → NaOH(aq) + H2(g)
    4. CaO(s) + HCl(aq) → CaCl2(aq) + H2O(l)
    5. CH3NH2(g) + O2(g) → H2O(g) + CO2(g) + N2(g)
    6. Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)

  17. Balance each chemical equation.

    1. N2O5(g) → NO2(g) + O2(g)
    2. NaNO3(s) → NaNO2(s) + O2(g)
    3. Al(s) + NH4NO3(s) → N2(g) + H2O(l) + Al2O3(s)
    4. C3H5N3O9(l) → CO2(g) + N2(g) + H2O(g) + O2(g)
    5. reaction of butane with excess oxygen
    6. IO2F(s) + BrF3(l) → IF5(l) + Br2(l) + O2(g)

  18. Balance each chemical equation.

    1. H2S(g) + O2(g) → H2O(l) + S8(s)
    2. KCl(aq) + HNO3(aq) + O2(g) → KNO3(aq) + Cl2(g) + H2O(l)
    3. NH3(g) + O2(g) → NO(g) + H2O(g)
    4. CH4(g) + O2(g) → CO(g) + H2(g)
    5. NaF(aq) + Th(NO3)4(aq) → NaNO3(aq) + ThF4(s)
    6. Ca5(PO4)3F(s) + H2SO4(aq) + H2O(l) → H3PO4(aq) + CaSO4·2H2O(s) + HF(aq)

  19. Balance each chemical equation.

    1. NaCl(aq) + H2SO4(aq) → Na2SO4(aq) + HCl(g)
    2. K(s) + H2O(l) → KOH(aq) + H2(g)
    3. reaction of octane with excess oxygen
    4. S8(s) + Cl2(g) → S2Cl2(l)
    5. CH3OH(l) + I2(s) + P4(s) → CH3I(l) + H3PO4(aq) + H2O(l)
    6. (CH3)3Al(s) + H2O(l) → CH4(g) + Al(OH)3(s)

  20. Write a balanced chemical equation for each reaction.

    1. Aluminum reacts with bromine.
    2. Sodium reacts with chlorine.
    3. Aluminum hydroxide and acetic acid react to produce aluminum acetate and water.
    4. Ammonia and oxygen react to produce nitrogen monoxide and water.
    5. Nitrogen and hydrogen react at elevated temperature and pressure to produce ammonia.
    6. An aqueous solution of barium chloride reacts with a solution of sodium sulfate.

  21. Write a balanced chemical equation for each reaction.

    1. Magnesium burns in oxygen.
    2. Carbon dioxide and sodium oxide react to produce sodium carbonate.
    3. Aluminum reacts with hydrochloric acid.
    4. An aqueous solution of silver nitrate reacts with a solution of potassium chloride.
    5. Methane burns in oxygen.
    6. Sodium nitrate and sulfuric acid react to produce sodium sulfate and nitric acid.

Answers

  1. reactants: nitrogen and hydrogen; product: ammonia

  3. reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water

  5. N2 + 3H2 → 2NH3

  7. Mg(OH)2 + 2HNO3 → Mg(NO3)2 + 2H2O

  9. 2NaClO3 → 2NaCl + 3O2

  11. 4Al + 3O2 → 2Al2O3

  13. N2(g) + 3H2(g) → 2NH3(g)