Learning Objectives To describe a chemical reaction. To calculate the quantities of compounds produced or consumed in a chemical reaction What happens to matter when it undergoes chemical changes? The Law of conservation of mass says that " Atoms are neither created, nor destroyed, during any chemical reaction ." Thus, the same collection of atoms is present after a reaction as before the reaction. The changes that occur during a reaction just involve the rearrangement of atoms. In this section we will discuss stoichiometry (the "measurement of elements"). Chemical Equations As shown in Figure \(\PageIndex{1}\), 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. This reaction is described with a chemical equation, an expression that gives the identities and quantities of the substances in a chemical reaction. Chemical reactions are represented on paper by chemical equations . For example, hydrogen gas (H 2 ) can react (burn) with oxygen gas (O 2 ) to form water (H 2 O). The chemical equation for this reaction is written as: \[\ce{2H_2 + O_2 \rightarrow 2H_2O} \nonumber \] Chemical formulas and other symbols are used to indicate the starting materials, or reactants, which by convention are written on the left side of the equation, and the final compounds, or products, which are written on the right. An arrow points from the reactant to the products. The chemical reaction for the ammonium dichromate volcano in Figure \(\PageIndex{1}\) is \[ \underbrace{\ce{(NH_4)_2Cr_2O_7}}_{ reactant } \rightarrow \underbrace{\ce{Cr_2O_3 + N_2 + 4H_2O}}_{products }\label{3.1.1} \] The arrow is read as “yields” or “reacts to form.” Equation \(\ref{3.1.1}\) indicates 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 follows: \[\ce{ (NH4)2Cr2O7(s) \rightarrow Cr2O3(s) + N2(g) + 4H2O(g)} \label{3.1.2} \] Equation \(\ref{3.1.2}\) is identical to Equation \(\ref{3.1.1}\) 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 Equations \(\ref{3.1.1}\) and \(\ref{3.1.2}\) . Each side of the reaction 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. Equations \(\ref{3.1.1}\) and \(\ref{3.1.2}\) 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. A chemical reaction changes only the distribution of atoms, not the number of atoms. Introduction to Chemical Reaction Equations: Introduction to Chemical Reaction Equations, YouTube(opens in new window) [youtu.be] 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 (\(C_7H_{16}\)), an important component of gasoline: \[ \ce{C_7H_{16} (l) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g) } \label{3.1.3} \] The complete combustion of any hydrocarbon with sufficient oxygen always yields carbon dioxide and water. Equation \(\ref{3.1.3}\) 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, the coefficients of the reactants and products must be adjusted 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, 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 \(\PageIndex{3}\) . Balancing Combustion Reactions: Balancing Combustions Reactions, YouTube(opens in new window) [youtu.be] 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 Identify the most complex substance. 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. Balance polyatomic ions (if present) as a unit. 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. Check your work by counting the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced. Example \(\PageIndex{1A}\): Combustion of Heptane To demonstrate this approach, let’s use the combustion of n-heptane ( Equation \(\ref{3.1.3}\) ) as an example. Identify the most complex substance . The most complex substance is the one with the largest number of different atoms, which is \(\ce{C_7H_{16}}\). We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance. 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 CO 2 molecules, each of which contains 1 carbon atom, on the right side: \[ \ce{C_7H_{16} + O_2 \rightarrow 7CO_2 + H_2O } \label{3.1.4} \] Balance polyatomic ions as a unit . There are no polyatomic ions to be considered in this reaction. 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: \[ \ce{C_7H_{16} + O_2 \rightarrow 7CO_2 + 8H_2O} \label{3.1.5} \] 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, O 2 , on the reactant side:\[ \ce{C_7H_{16} (l) + 11O_2 (g) \rightarrow 7CO_2 (g) + 8H_2O (g)} \label{3.1.6} \] 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.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. Example \(\PageIndex{1B}\): Combustion of Isooctane Consider, for example, a similar reaction, the combustion of isooctane (\(\ce{C8H18}\)). Because the combustion of any hydrocarbon with oxygen produces carbon dioxide and water, the unbalanced chemical equation is as follows: \[ \ce{C_8H_{18} (l) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g) } \label{3.1.7} \] Identify the most complex substance . Begin the balancing process by assuming that the final balanced chemical equation contains a single molecule of isooctane. Adjust the coefficients . The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO 2 molecules in the products:\[ \ce{C_8H_{18} + O_2 \rightarrow 8CO_2 + H_2O} \label{3.1.8} \] Balance polyatomic ions as a unit . This step does not apply to this equation. Balance the remaining atoms . Eighteen hydrogen atoms in isooctane means that there must be 9 H 2 O molecules in the products:\[ \ce{C_8H_{18} + O_2 \rightarrow 8CO_2 + 9H_2O } \label{3.1.9} \]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, O 2 , but because there are 2 oxygen atoms per O 2 molecule, we must use a fractional coefficient (25/2) to balance the oxygen atoms: \[ \ce{C_8H_{18} + 25/2 O_2 \rightarrow 8CO_2 + 9H_2O} \label{3.1.10} \] Equation \(\ref{3.1.10}\) 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: \[ \ce{2C_8H_{18} (l) + 25O_2 (g) \rightarrow 16CO_2 (g) + 18H_2O (g) }\label{3.11} \] Check your work . The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side. Balancing Complex Chemical Equations: Balancing Complex Chemical Equations, YouTube(opens in new window) [youtu.be] 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 \(\PageIndex{1C}\): Hydroxyapatite The reaction of the mineral hydroxyapatite (\(\ce{Ca5(PO4)3(OH)}\)) with phosphoric acid and water gives \(\ce{Ca(H2PO4)2•H2O}\) (calcium dihydrogen phosphate monohydrate). Write and balance the equation for this reaction. Given : reactants and product Asked for : balanced chemical equation Strategy : Identify the product and the reactants and then write the unbalanced chemical equation. 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 that phosphoric acid is H 3 PO 4 ). The initial (unbalanced) equation is as follows: \[ \ce{ Ca5(PO4)3(OH)(s) + H_3PO4 (aq) + H_2O_{(l)} \rightarrow Ca(H_2PO_4)_2 \cdot H_2O_{(s)} } \nonumber \] 1. B Identify the most complex substance . We start by assuming that only one molecule or formula unit of the most complex substance, \(\ce{Ca5(PO4)3(OH)}\), 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 \(\ce{Ca5(PO4)3(OH)}\) contains 5 calcium atoms, so we need 5 Ca(H 2 PO 4 ) 2 •H 2 O on the right side: \[ \ce{Ca5(PO4)3(OH) + H3PO4 + H2O \rightarrow 5Ca(H2PO4)2 \cdot H2O} \nonumber \] 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 (PO 4 3 − ), shows up in three places. In H 3 PO 4 , the phosphate ion is combined with three H + ions to make phosphoric acid (H 3 PO 4 ), whereas in Ca(H 2 PO 4 ) 2 • H 2 O it is combined with two H + ions to give the dihydrogen phosphate ion. Thus it is easier to balance PO 4 as a unit rather than counting individual phosphorus and oxygen atoms. There are 10 PO 4 units on the right side but only 4 on the left. The simplest way to balance the PO 4 units is to place a coefficient of 7 in front of H 3 PO 4 : \[ \ce{Ca_5(PO_4)_3(OH) + 7H_3PO_4 + H_2O \rightarrow 5Ca(H_2PO_4)_2 \cdot H_2O } \nonumber \] 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, H 2 O, by placing a coefficient of 4 in front of H 2 O on the left side, giving a total of 4 H 2 O molecules: \[ \ce{Ca_5(PO_4)_3(OH) (s) + 7H_3PO_4 (aq) + 4H_2O (l) \rightarrow 5Ca(H_2PO_4)_2 \cdot H_2O (s) } \nonumber \] The equation is now balanced. Even though we have not explicitly balanced the oxygen atoms, there are 41 oxygen atoms on each side. 5. Check your work . Both sides of the equation contain 5 calcium atoms, 10 phosphorus atoms, 30 hydrogen atoms, and 41 oxygen atoms. Exercise \(\PageIndex{1}\): Fermentation 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 (\(CH_3CH_2OH\) and carbon dioxide \(CO_2\). Write a balanced chemical reaction for the fermentation of glucose. Commercial use of fermentation . (a) Microbrewery vats are used to prepare beer. (b) The fermentation of glucose by yeast cells is the reaction that makes beer production possible. Answer \[C_6H_{12}O_6(s) \rightarrow 2C_2H_5OH(l) + 2CO_2(g) \nonumber \] Balancing Reactions Which Contain Polyatomics: Balancing Reactions Which Contain Polyatomics, YouTube(opens in new window) [youtu.be] Interpreting Chemical Equations In addition to providing qualitative information about the identities and physical states of the reactants and products, a balanced chemical equation provides quantitative information. Specifically, it gives the relative amounts of reactants and products consumed or produced in a reaction. The number of atoms, molecules, or formula units of a reactant or a product in a balanced chemical equation is the coefficient of that species (e.g., the 4 preceding H 2 O in Equation \(\ref{3.1.1}\) ). When no coefficient is written in front of a species, the coefficient is assumed to be 1. As illustrated in Figure \(\PageIndex{4}\) , the coefficients allow Equation \(\ref{3.1.1}\) to be interpreted in any of the following ways: Two NH 4 + ions and one Cr 2 O 7 2 − ion yield 1 formula unit of Cr 2 O 3 , 1 N 2 molecule, and 4 H 2 O molecules. One mole of (NH 4 ) 2 Cr 2 O 7 yields 1 mol of Cr 2 O 3 , 1 mol of N 2 , and 4 mol of H 2 O. A mass of 252 g of (NH 4 ) 2 Cr 2 O 7 yields 152 g of Cr 2 O 3 , 28 g of N 2 , and 72 g of H 2 O. A total of 6.022 × 10 23 formula units of (NH 4 ) 2 Cr 2 O 7 yields 6.022 × 10 23 formula units of Cr 2 O 3 , 6.022 × 10 23 molecules of N 2 , and 24.09 × 10 23 molecules of H 2 O. These are all chemically equivalent ways of stating the information given in the balanced chemical equation, using the concepts of the mole, molar or formula mass, and Avogadro’s number. The ratio of the number of moles of one substance to the number of moles of another is called the mole ratio. For example, the mole ratio of \(H_2O\) to \(N_2\) in Equation \(\ref{3.1.1}\) is 4:1. The total mass of reactants equals the total mass of products, as predicted by Dalton’s law of conservation of mass: \[252 \;g \;\text{of}\; \ce{(NH_4)_2Cr_2O_7} \nonumber \] yield \[152 + 28 + 72 = 252 \; g \; \text{of products.} \nonumber \] The chemical equation does not, however, show the rate of the reaction (rapidly, slowly, or not at all) or whether energy in the form of heat or light is given off. These issues are considered in more detail in later chapters. An important chemical reaction was analyzed by Antoine Lavoisier, an 18th-century French chemist, who was interested in the chemistry of living organisms as well as simple chemical systems. In a classic series of experiments, he measured the carbon dioxide and heat produced by a guinea pig during respiration, in which organic compounds are used as fuel to produce energy, carbon dioxide, and water. Lavoisier found that the ratio of heat produced to carbon dioxide exhaled was similar to the ratio observed for the reaction of charcoal with oxygen in the air to produce carbon dioxide—a process chemists call combustion. Based on these experiments, he proposed that “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.” Lavoisier was correct, although the organic compounds consumed in respiration are substantially different from those found in charcoal. One of the most important fuels in the human body is glucose (\(C_6H_{12}O_6\)), which is virtually the only fuel used in the brain. Thus combustion and respiration are examples of chemical reactions. Example \(\PageIndex{2}\): Combustion of Glucose The balanced chemical equation for the combustion of glucose in the laboratory (or in the brain) is as follows: \[ \ce{C_6H_{12}O6(s) + 6O2(g) \rightarrow 6CO2(g) + 6H2O(l)} \nonumber \] Construct a table showing how to interpret the information in this equation in terms of a single molecule of glucose. moles of reactants and products. grams of reactants and products represented by 1 mol of glucose. numbers of molecules of reactants and products represented by 1 mol of glucose. Given : balanced chemical equation Asked for : molecule, mole, and mass relationships Strategy : Use the coefficients from the balanced chemical equation to determine both the molecular and mole ratios. Use the molar masses of the reactants and products to convert from moles to grams. Use Avogadro’s number to convert from moles to the number of molecules. Solution : This equation is balanced as written: each side has 6 carbon atoms, 18 oxygen atoms, and 12 hydrogen atoms. We can therefore use the coefficients directly to obtain the desired information. One molecule of glucose reacts with 6 molecules of O 2 to yield 6 molecules of CO 2 and 6 molecules of H 2 O. One mole of glucose reacts with 6 mol of O 2 to yield 6 mol of CO 2 and 6 mol of H 2 O. To interpret the equation in terms of masses of reactants and products, we need their molar masses and the mole ratios from part b. The molar masses in grams per mole are as follows: glucose, 180.16; O 2 , 31.9988; CO 2 , 44.010; and H 2 O, 18.015. \[ \begin{align*} \text{mass of reactants} &= \text{mass of products} \\[4pt] g \, glucose + g \, O_2 &= g \, CO_2 + g \, H_2O \end{align*} \nonumber \] \[ 1\,mol\,glucose \left ( {180.16 \, g \over 1 \, mol \, glucose } \right ) + 6 \, mol \, O_2 \left ( { 31.9988 \, g \over 1 \, mol \, O_2} \right ) \nonumber \] \[= 6 \, mol \, CO_2 \left ( {44.010 \, g \over 1 \, mol \, CO_2} \right ) + 6 \, mol \, H_2O \left ( {18.015 \, g \over 1 \, mol \, H_2O} \right ) \nonumber \] \[ 372.15 \, g = 372.15 \, g \nonumber \] C One mole of glucose contains Avogadro’s number (6.022 × 10 23 ) of glucose molecules. Thus 6.022 × 10 23 glucose molecules react with (6 × 6.022 × 10 23 ) = 3.613 × 10 24 oxygen molecules to yield (6 × 6.022 × 10 23 ) = 3.613 × 10 24 molecules each of CO 2 and H 2 O. In tabular form: Unnamed: 0 \(C_6H_{12}O_{6\;(s)}\) + \(6O_{2\;(g)}\) → \(6CO_{2\;(g)}\) Unnamed: 6 \(6H_2O_{(l)}\) a. 1 molecule NaN 6 molecules NaN 6 molecules NaN 6 molecules b. 1 mol NaN 6 mol NaN 6 mol NaN 6 mol c. 180.16 g NaN 191.9928 g NaN 264.06 g NaN 108.09 g d. 6.022 × 1023 molecules NaN 3.613 × 1024 molecules NaN 3.613 × 1024 molecules NaN 3.613 × 1024 molecule Exercise \(\PageIndex{2}\): Ammonium Nitrate Explosion Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 1947, a ship loaded with ammonium nitrate caught fire during unloading and exploded, destroying the town of Texas City, Texas. The explosion resulted from the following reaction: \[ 2NH_4NO_{3\;(s)} \rightarrow 2N_{2\;(g)} + 4H_2O_{(g)} + O_{2\;(g)} \nonumber \] Construct a table showing how to interpret the information in the equation in terms of individual molecules and ions. moles of reactants and products. grams of reactants and products given 2 mol of ammonium nitrate. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate. Answer : Unnamed: 0 \(2NH_4NO_{3\;(s)}\) → \(2N_{2\;(g)}\) + \(4H_2O_{(g)}\) +.1 \(O_{2\;(g)}\) a. 2NH4+ ions and 2NO3− ions NaN 2 molecules NaN 4 molecules NaN 1 molecule b. 2 mol NaN 2 mol NaN 4 mol NaN 1 mol c. 160.0864 g NaN 56.0268 g NaN 72.0608 g NaN 31.9988 g d. 1.204 × 1024 formula units NaN 1.204 × 1024 molecules NaN 2.409 × 1024 molecules NaN 6.022 × 1023 molecules Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors, YouTube(opens in new window) [youtu.be] Summary A chemical reaction is described by a chemical equation that gives the identities and quantities of the reactants and the products. 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. The mole ratio of two substances in a chemical reaction is the ratio of their coefficients in the balanced chemical equation.