Conservation of Matter

The law of conservation of matter summarizes many scientific observations about matter: It states that there is no detectable change in the total quantity of matter present when matter converts from one type to another (a chemical change) or changes among solid, liquid, or gaseous states (a physical change). What does this mean for chemistry? In any chemical change, one or more initial substances change into a different substance or substances. Both the initial and final substances are composed of atoms because all matter is composed of atoms. According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and type of atoms after the chemical change as were present before the chemical change.

Although this conservation law holds true for all conversions of matter, convincing examples are few and far between because, outside of the controlled conditions in a laboratory, we seldom collect all of the material that is produced during a particular conversion. For example, when you eat, digest, and assimilate food, all of the matter in the original food is preserved. But because some of the matter is incorporated into your body, and much is excreted as various types of waste, it is challenging to verify by measurement.

When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a chemical equation. Consider as an example the reaction between one methane molecule (CH₄) and two diatomic oxygen molecules (O₂) to produce one carbon dioxide molecule (CO₂) and two water molecules (H₂O). The chemical equation representing this process is provided in the upper half of Figure 1, with space-filling molecular models shown in the lower half of the figure.

This example illustrates the fundamental aspects of any chemical equation:

1. The substances undergoing reaction are called reactants, and their formulas are placed on the left side of the equation.
2. The substances generated by the reaction are called products, and their formulas are placed on the right side of the equation.
3. Plus signs (+) separate individual reactant and product formulas, and an arrow (⟶) separates the reactant and product (left and right) sides of the equation.
4. The relative numbers of reactant and product species are represented by coefficients (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.

It is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the relative numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios.

Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4, or 3-6-3-6, and so on (Figure 2). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:

• 1 methane molecule and 2 oxygen molecules react to yield 1 carbon dioxide molecule and 2 water molecules.
• 1 dozen methane molecules and 2 dozen oxygen molecules react to yield 1 dozen carbon dioxide molecules and 2 dozen water molecules.
• 1 mole of methane molecules and 2 moles of oxygen molecules react to yield 1 mole of carbon dioxide molecules and 2 moles of water molecules.

Balancing Equations

The chemical equation described above is a balanced equation, meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO₂ and H₂O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is 2 (from CO₂) + 2 (from 2 H₂O)  =  4 product oxygen atoms.  Thus, we’ll need two O₂ molecules as reactants to similarly have 4 reactant oxygen atoms.

The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:

CH₄ + 2 O₂  ⟶  CO₂ + 2 H₂O

A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation:

H₂O  ⟶  O₂ + H₂

Comparing the number of H and O atoms on either side of this equation confirms its imbalance:

The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the coefficients of the equation may be changed as needed. Keep in mind, of course, that the formula subscripts define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H₂O to H₂O₂ would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H₂O to 2.

2 H₂O  ⟶  O₂ + H₂

The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H₂ product to 2.

These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:

2 H₂O  ⟶  O₂ + 2 H₂

Notice that in all these equations, the charge is also balanced on both the reactant and product sides. In the previous example, there is a total of zero charge on both sides of the equation since H₂O, O₂, and H₂ are all neutral molecules. This will become important later when we start writing equations with ions—the total charge on the reactant side must be equal to the total charge on the product side since electrons cannot be lost in a chemical equation.

It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation’s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane (C₂H₆) with oxygen to yield H₂O and CO₂, represented by the unbalanced equation:

C₂H₆ + O₂  ⟶  CO₂ + H₂O

Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:

C₂H₆ + O₂  ⟶  2 CO₂ + 3 H₂O

This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient can be used with the O₂ reactant to yield an odd number, so a fractional coefficient, 7/2, is used instead to yield a provisional balanced equation:

C₂H₆ + O₂  ⟶  2 CO₂ + 3 H₂O

A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:

2 C₂H₆ + 7 O₂  ⟶  4 CO₂ + 6 H₂O

Finally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,

3 N₂ + 9 H₂  ⟶  6 NH₃

the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:

N₂ + 3 H₂  ⟶  2 NH₃

Additional Information in Chemical Equations

The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include s for solids, l for liquids, g for gases, and aq for substances dissolved in water (aqueous solutions are solutions where water is the solvent). These notations are illustrated in the example equation here:

2 Na(s) + 2 H₂O(ℓ)  ⟶  2 NaOH(aq) + H₂(g)

This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).

Stoichiometry

A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction’s stoichiometry, a term derived from the Greek words stoicheion (meaning “element”) and metron (meaning “measure”). Here, the use of balanced chemical equations for various stoichiometric applications is explored.

The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for 1 cup pancake mix, ¾ cup milk, and one egg. The “equation” representing the preparation of pancakes per this recipe is

1 cup mix + ¾ cup milk + 1 egg  →  8 pancakes

If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is

24 pancakes × = 3 eggs

Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive stoichiometric factors that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:

N₂(g) + 3 H₂(g)  →  2 NH₃(g)

This equation shows ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:

or or

These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.

Example 2: Moles of Reactant Required in a Reaction

How many moles of I₂ are required to react with 0.429 mol of Al according to the following equation (see Figure 1)?

2 Al(s) + 3 I₂(s)  ⟶  2 AlI₃(s)

Solution

Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is . The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:

mol I₂ = 0.429 mol Al × = 0.644 mol I₂

Check Your Learning

How many moles of Ca(OH)₂ are required to react with 1.36 mol of H₃PO₄ to produce Ca₃(PO₄)₂ according to the following equation?

3 Ca(OH)₂(aq) + 2 H₃PO₄(aq)  →  Ca₃(PO₄)₂(s) + 6 H₂O(ℓ)

Answer

2.04 mol

These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.