Skip to main content

# 5.1: Chemical Sentences- Equations

Learning Objectives

• Identify the reactants and products in any chemical reaction.
• Convert word equations into chemical equations.
• Use the common symbols, $$\left( s \right)$$, $$\left( l \right)$$, $$\left( g \right)$$, $$\left( aq \right)$$, and $$\rightarrow$$ appropriately when writing a chemical reaction.
• Explain the roles of subscripts and coefficients in chemical equations.
• Balance a chemical equation when given the unbalanced equation.
• Explain the role of the Law of Conservation of Mass in a chemical reaction.

In a chemical change, new substances are formed. In order for this to occur, the chemical bonds of the substances break, and the atoms that compose them separate and rearrange themselves into new substances with new chemical bonds. When this process occurs, we call it a chemical reaction. A chemical reaction is the process in which one or more substances are changed into one or more new substances.

## Reactants and Products

To describe a chemical reaction, we need to indicate what substances are present at the beginning and what substances are present at the end. The substances that are present at the beginning are called reactants and the substances present at the end are called products.

Sometimes when reactants are put into a reaction vessel, a reaction will take place to produce products. Reactants are the starting materials, that is, whatever we have as our initial ingredients. The products are just that, wheat is produced or the result of what happens to the reactants when we put them together in the reaction vessel. If we think about baking chocolate chip cookies, our reactants would be flour, butter, sugar, vanilla, some baking soda, salt, egg, and chocolate chips. What would be the products? Cookies! The reaction vessel would be our mixing bowl.

$\underbrace{\text{Flour} + \text{Butter} + \text{Sugar} + \text{Vanilla} + \text{Baking Soda} + \text{Eggs} + \text{Chocolate Chips}}_{\text{Ingredients = Reactants}} \rightarrow \underbrace{\text{Cookies}}_{\text{Product}} \nonumber$

## Writing Chemical Equations

When sulfur dioxide is added to oxygen, sulfur trioxide is produced. Sulfur dioxide and oxygen, $$\ce{SO_2} + \ce{O_2}$$, are reactants and sulfur trioxide, $$\ce{SO_3}$$, is the product.

$\underbrace{\ce{2 SO2(g) + O2(g) }}_{\text{Reactants}} \rightarrow \underbrace{\ce{2SO3(g)}}_{\text{Products}} \nonumber$

In chemical reactions, the reactants are found before the symbol "$$\rightarrow$$" and the products are found after the symbol "$$\rightarrow$$". The general equation for a reaction is:

$\text{Reactants } \rightarrow \text{Products} \nonumber$

There are a few special symbols that we need to know in order to "talk" in chemical shorthand. In the table below is the summary of the major symbols used in chemical equations. You will find there are others but these are the main ones that we need to know. Table $$\PageIndex{1}$$ shows a listing of symbols used in chemical equations.

Table $$\PageIndex{1}$$ Symbols Used in Chemical Equations
Symbol Description Symbol Description
$$+$$ used to separate multiple reactants or products $$\left( s \right)$$ reactant or product in the solid state
$$\rightarrow$$ yield sign; separates reactants from products $$\left( l \right)$$ reactant or product in the liquid state
$$\rightleftharpoons$$ replaces the yield sign for reversible reactions that reach equilibrium $$\left( g \right)$$ reactant or product in the gas state
$$\overset{\ce{Pt}}{\rightarrow}$$ formula written above the arrow is used as a catalyst in the reaction $$\left( aq \right)$$ reactant or product in an aqueous solution (dissolved in water)
$$\overset{\Delta}{\rightarrow}$$ triangle indicates that the reaction is being heated

Chemists have a choice of methods for describing a chemical reaction.

1. They could draw a picture of the chemical reaction.

2. They could write a word equation for the chemical reaction:
"Two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water vapor."

3. They could write the equation in chemical shorthand.

$2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right)$

In the symbolic equation, chemical formulas are used instead of chemical names for reactants and products and symbols are used to indicate the phase of each substance. It should be apparent that the chemical shorthand method is the quickest and clearest method for writing chemical equations.

We could write that an aqueous solution of calcium nitrate is added to an aqueous solution of sodium hydroxide to produce solid calcium hydroxide and an aqueous solution of sodium nitrate. Or in shorthand we could write:

$\ce{Ca(NO_3)_2} \left( aq \right) + 2 \ce{NaOH} \left( aq \right) \rightarrow \ce{Ca(OH)_2} \left( s \right) + 2 \ce{NaNO_3} \left( aq \right)$

How much easier is that to read? Let's try it in reverse? Look at the following reaction in shorthand and write the word equation for the reaction:

$\ce{Cu} \left( s \right) + \ce{AgNO_3} \left( aq \right) \rightarrow \ce{Cu(NO_3)_2} \left( aq \right) + \ce{Ag} \left( s \right)$

The word equation for this reaction might read something like "solid copper reacts with an aqueous solution of silver nitrate to produce a solution of copper (II) nitrate with solid silver."

To turn word equations into symbolic equations, we need to follow the given steps:

1. Identify the reactants and products. This will help you know what symbols go on each side of the arrow and where the $$+$$ signs go.
2. Write the correct formulas for all compounds. You will need to use the rules you learned in Chapter 5 (including making all ionic compounds charge balanced).
3. Write the correct formulas for all elements. Usually this is given straight off of the periodic table. However, there are seven elements that are considered diatomic, meaning they are always found in pairs in nature. They include those elements listed in the table.
 Element Name Formula Hydrogen Nitrogen Oxygen Fluorine Chlorine Bromine Iodine $$H_2$$ $$N_2$$ $$O_2$$ $$F_2$$ $$Cl_2$$ $$Br_2$$ $$I_2$$

Example $$\PageIndex{1}$$

Transfer the following symbolic equations into word equations or word equations into symbolic equations.

1. $$\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right)$$
2. Gaseous propane, $$\ce{C_3H_8}$$, burns in oxygen gas to produce gaseous carbon dioxide and liquid water.
3. Hydrogen fluoride gas reacts with an aqueous solution of potassium carbonate to produce an aqueous solution of potassium fluoride, liquid water, and gaseous carbon dioxide.

Solution

a. An aqueous solution of hydrochloric acid reacts with an aqueous solution of sodium hydroxide to produce an aqueous solution of sodium chloride and liquid water.

b. Reactants: propane ($$\ce{C_3H_8}$$) and oxygen ($$\ce{O_2}$$)

Product: carbon dioxide ($$\ce{CO_2}$$) and water ($$\ce{H_2O}$$)

$\ce{C_3H_8} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \nonumber$

c. Reactants: hydrogen fluoride and potassium carbonate

Products: potassium fluoride, water, and carbon dioxide

$\ce{HF} \left( g \right) + \ce{K_2CO_3} \left( aq \right) \rightarrow \ce{KF} \left( aq \right) + \ce{H_2O} \left( l \right) + \ce{CO_2} \left( g \right) \nonumber$

Exercise $$\PageIndex{1}$$

Transfer the following symbolic equations into word equations or word equations into symbolic equations.

1. hydrogen gas reacts with nitrogen gas to produce gaseous ammonia
2. $$\ce{HCl} \left( aq \right) + \ce{LiOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right)$$
3. copper metal is heated with oxygen gas to produce solid copper(II) oxide.
Answer a
$$H_2 (g) + N_2 (g) \rightarrow NH_3 (g)$$
Answer b
An aqueous solution of hydrochloric acid reacts with an aqueous solution of lithium hydroxide to produce an aqueous solution of lithium chloride and liquid water.
Answer c
$$Cu (s) + O_2 (g) \rightarrow CuO (s)$$

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 reorganized into different arrangements. In a complete chemical equation, the two sides of the equation must be present on the reactant and the product sides of the equation.

## Coefficients and Subscripts

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 the number of each substance involved in the reaction and may be changed in order to balance the equation. The equation above indicates that one mole of solid copper is reacting with two moles of aqueous silver nitrate to produce one mole of aqueous copper (II) nitrate and two atoms of solid silver.

## Balancing a Chemical Equation

Because the identities of the reactants and products are fixed, the equation cannot be balanced by changing the subscripts of the reactants or the products. To do so would change the chemical identity of the species being described, as illustrated in Figure $$\PageIndex{2}$$.

The simplest and most generally useful method for balancing chemical equations is “inspection,” better known as trial and error. The following is 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(s) that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element(s) on both sides.
3. Balance polyatomic ions (if present on both sides of the chemical equation) 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.

Example $$\PageIndex{2}$$: Combustion of Heptane

Balancing the chemical Equation for the combustion of Heptane ($$\ce{C_7H_{16}}$$)

$\ce{C_7H_{16} (l) + O_2 (g) → CO_2 (g) + H_2O (g) } \nonumber$

Solution

1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is $$C_7H_{16}$$. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.

2. Adjust the coefficients.

a. 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:

$\ce{C7H16 (l) + O2 (g) → } \underline{7} \ce{CO2 (g) + H2O (g) } \nonumber$

Reactants Element/Polyatomic Ion Products
7 C 7

b. 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:

$\ce{C7H16 (l) + O2 (g) → 7 CO2 (g) + } \underline{8} \ce{H2O (g) } \nonumber$

Reactants Element/Polyatomic Ion Products
7 C 7
16 H 16

3. Balance polyatomic ions as a unit.

There are no polyatomic ions to be considered in this reaction.

4. Balance the remaining atoms.

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:

$\ce{C7H16 (l) + }\underline{11} \ce{ O2 (g) → 7 CO2 (g) + 8H2O (g) } \nonumber$

Reactants Element/Polyatomic Ion Products
7 C 7
16 H 16
22 O 22

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.

Example $$\PageIndex{3}$$: Combustion of Isooctane

Combustion of Isooctane ($$\ce{C_8H_{18}}$$)

$\ce{C8H18 (l) + O2 (g) -> CO_2 (g) + H_2O(g)} \nonumber$

Solution

The assumption that the final balanced chemical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start. The combustion of any hydrocarbon with oxygen produces carbon dioxide and water.

1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is $$\ce{C8H18}$$. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance.

2. Adjust the coefficients.

a. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products:

$\ce{C8H18 (l) + O2 (g) -> }\underline{8} \ce{ CO2 (g) + H2O(g)}\nonumber$

Reactants Element/Polyatomic Ion Products
8 C 8

b. Eighteen hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products:

$\ce{C8H18 (l) + O2 (g) -> 8CO2 (g) + }\underline{9} \ce{ H2O(g)} \nonumber$

﻿

Reactants Element/Polyatomic Ion Products
8 C 8
18 H 18

3. Balance polyatomic ions as a unit.

There are no polyatomic ions to be considered in this reaction.

4. Balance the remaining atoms.

The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional coefficient ($$\dfrac{25}{2}$$) to balance the oxygen atoms:

$\ce{C8H18 (l) + } \underline{ \dfrac{25}{2} } \ce{O2 (g)→ 8CO2 (g) + 9H2O(g) }\nonumber$

Reactants Element/Polyatomic Ion Products
8 C 8
18 H 18
25 O 25

The equation is now balanced, but we usually write equations with whole-number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the chemical equation by 2:

﻿

$\underline{2} \ce{C8H18(l) + } \underline{25} \ce{O2(g) ->} \underline{16} \ce{CO2(g) + } \underline{18} \ce{H2O(g)} \nonumber$

5. Check your work.

The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side.

Reactants Element/Polyatomic Ion Products
16 C 16
36 H 36
50 O 50

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 $$\PageIndex{4}$$: Precipitation of Lead (II) Chloride

Aqueous solutions of lead (II) nitrate and sodium chloride are mixed. The products of the reaction are an aqueous solution of sodium nitrate and a solid precipitate of lead (II) chloride. Write the balanced chemical equation for this reaction.

Solution

1. Identify the most complex substance.

The most complex substance is lead (II) chloride.

$\ce{Pb(NO3)2(aq) + NaCl(aq) → NaNO3(aq) + PbCl2(s)} \nonumber$

2. Adjust the coefficients.

There are twice as many chloride ions in the reactants than in the products. Place a 2 in front of the NaCl in order to balance the chloride ions.

$\ce{Pb(NO3)2(aq) + }\underline{ 2} \ce{NaCl(aq) → NaNO3(aq) + PbCl2(s)} \nonumber$

Reactants Element/Polyatomic Ion Products
1 Pb 1
2 Na 1
2 Cl 2

3. Balance polyatomic ions as a unit.

The nitrate ions are still unbalanced. Place a 2 in front of the NaNO3. The result is:

$\ce{Pb(NO3)2(aq) + 2NaCl(aq) → } \underline {2} \ce{NaNO3(aq) + PbCl2(s)} \nonumber$

Reactants Element/Polyatomic Ion Products
1 Pb 1
2 Na 2
2 Cl 2
2 NO3- 2

4. Balance the remaining atoms.

There is no need to balance the remaining atoms because they are already balanced.

5. Check your work.

$\ce{Pb(NO3)2(aq) + 2NaCl(aq) → 2NaNO3(aq) + PbCl2(s)} \nonumber$

Reactants Element/Polyatomic Ion Products
1 Pb 1
2 Na 2
2 Cl 2
2 NO3- 2

Exercise $$\PageIndex{2}$$

Is each chemical equation balanced?

1. $$\ce{2Hg(ℓ)+ O_2(g) \rightarrow Hg_2O_2(s)}$$
2. $$\ce{C_2H_4(g) + 2O_2(g)→ 2CO_2(g) + 2H_2O(g)}$$
3. $$\ce{Mg(NO_3)_2(s) + 2Li (s) \rightarrow Mg(s)+ 2LiNO_3(s)}$$
Answer a
yes
Answer b
no
Answer c
yes

Exercise $$\PageIndex{3}$$

Balance the following chemical equations.

1. $$\ce{N2 (g) + O2 (g) → NO2 (g) }$$
2. $$\ce{Pb(NO3)2(aq) + FeCl3(aq) → Fe(NO3)3(aq) + PbCl2(s)}$$
3. $$\ce{C6H14(l) + O2(g)→ CO2(g) + H2O(g)}$$
Answer a
N2 (g) + 2O2 (g) → 2NO2 (g)
Answer b
3Pb(NO3)2(aq) + 2FeCl3(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)
Answer c
2C6H14(l) + 19O2(g)→ 12CO2(g) + 14H2O(g)

## Summary

• A chemical reaction is the process in which one or more substances are changed into one or more new substances.
• Chemical reactions are represented by chemical equations.
• Chemical equations have reactants on the left, an arrow that is read as "yields", and the products on the right.
• To be useful, chemical equations must always be balanced. Balanced chemical equations 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.

## Contributors and Attributions

• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.

• Henry Agnew (UC Davis)

• Was this article helpful?