This is “Quantitative Relationships Based on Chemical Equations”, section 5.3 from the book Introduction to Chemistry: General, Organic, and Biological (v. 1.0).
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A balanced chemical equation not only describes some of the chemical properties of substances—by showing us what substances react with what other substances to make what products—but also shows numerical relationships between the reactants and the products. The study of these numerical relationships is called stoichiometryThe study of the numerical relationships between the reactants and the products in a balanced chemical equation.. The stoichiometry of chemical equations revolves around the coefficients in the balanced chemical equation because these coefficients determine the molecular ratio in which reactants react and products are made.
The word stoichiometry is pronounced “stow-eh-key-OM-et-tree.” It is of mixed Greek and English origins, meaning roughly “measure of an element.”
Let us consider a stoichiometry analogy from the kitchen. A recipe that makes 1 dozen biscuits needs 2 cups of flour, 1 egg, 4 tablespoons of shortening, 1 teaspoon of salt, 1 teaspoon of baking soda, and 1 cup of milk. If we were to write this as a chemical equation, we would write
2 c flour + 1 egg + 4 tbsp shortening + 1 tsp salt + 1 tsp baking soda + 1 c milk → 12 biscuits(Unlike true chemical reactions, this one has all 1 coefficients written explicitly—partly because of the many different units here.) This equation gives us ratios of how much of what reactants are needed to make how much of what product. Two cups of flour, when combined with the proper amounts of the other ingredients, will yield 12 biscuits. One teaspoon of baking soda (when also combined with the right amounts of the other ingredients) will make 12 biscuits. One egg must be combined with 1 cup of milk to yield the product food. Other relationships can also be expressed.
We can use the ratios we derive from the equation for predictive purposes. For instance, if we have 4 cups of flour, how many biscuits can we make if we have enough of the other ingredients? It should be apparent that we can make a double recipe of 24 biscuits.
But how would we find this answer formally, that is, mathematically? We would set up a conversion factor, much like we did in Chapter 1 "Chemistry, Matter, and Measurement". Because 2 cups of flour make 12 biscuits, we can set up an equivalency ratio:
$$\frac{12\text{biscuits}}{\text{2cflour}}$$We then can use this ratio in a formal conversion of flour to biscuits:
$$4\overline{)\text{cflour}}\times \frac{12\text{biscuits}}{\text{2}\overline{)\text{cflour}}}=24\text{biscuits}$$Similarly, by constructing similar ratios, we can determine how many biscuits we can make from any amount of ingredient.
When you are doubling or halving a recipe, you are doing a type of stoichiometry. Applying these ideas to chemical reactions should not be difficult if you use recipes when you cook.
A recipe shows how much of each ingredient is needed for the proper reaction to take place.
© Thinkstock
Consider the following balanced chemical equation:
2C_{2}H_{2} + 5O_{2} → 4CO_{2} + 2H_{2}OThe coefficients on the chemical formulas give the ratios in which the reactants combine and the products form. Thus, we can make the following statements and construct the following ratios:
Statement from the Balanced Chemical Reaction | Ratio | Inverse Ratio |
---|---|---|
two C_{2}H_{2} molecules react with five O_{2} molecules | $\frac{2{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}}{5{\text{O}}_{\text{2}}}$ | $\frac{5{\text{O}}_{\text{2}}}{2{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}}$ |
two C_{2}H_{2} molecules react to make four CO_{2} molecules | $\frac{2{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}}{4{\text{CO}}_{\text{2}}}$ | $\frac{4{\text{CO}}_{\text{2}}}{2{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}}$ |
five O_{2} molecules react to make two H_{2}O molecules | $\frac{5{\text{O}}_{\text{2}}}{2{\text{H}}_{\text{2}}\text{O}}$ | $\frac{2{\text{H}}_{\text{2}}\text{O}}{5{\text{O}}_{\text{2}}}$ |
four CO_{2} molecules are made at the same time as two H_{2}O molecules | $\frac{2{\text{H}}_{\text{2}}\text{O}}{4{\text{CO}}_{\text{2}}}$ | $\frac{4{\text{CO}}_{\text{2}}}{2{\text{H}}_{\text{2}}\text{O}}$ |
Other relationships are possible; in fact, 12 different conversion factors can be constructed from this balanced chemical equation. In each ratio, the unit is assumed to be molecules because that is how we are interpreting the chemical equation.
Any of these fractions can be used as a conversion factor to relate an amount of one substance to an amount of another substance. For example, suppose we want to know how many CO_{2} molecules are formed when 26 molecules of C_{2}H_{2} are reacted. As usual with a conversion problem, we start with the amount we are given—26C_{2}H_{2}—and multiply it by a conversion factor that cancels out our original unit and introduces the unit we are converting to—in this case, CO_{2}. That conversion factor is $\frac{4{\text{CO}}_{\text{2}}}{2{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}}\text{,}$ which is composed of terms that come directly from the balanced chemical equation. Thus, we have
$${\text{26C}}_{\text{2}}{\text{H}}_{\text{2}}\times \frac{4{\text{CO}}_{\text{2}}}{2{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}}$$The molecules of C_{2}H_{2} cancel, and we are left with molecules of CO_{2}. Multiplying through, we get
$$\text{26}\overline{){\text{C}}_{\text{2}}{\text{H}}_{2}}\times \frac{4{\text{CO}}_{\text{2}}}{2\overline{){\text{C}}_{\text{2}}{\text{H}}_{2}}}={\text{52CO}}_{\text{2}}$$Thus, 52 molecules of CO_{2} are formed.
This application of stoichiometry is extremely powerful in its predictive ability, as long as we begin with a balanced chemical equation. Without a balanced chemical equation, the predictions made by simple stoichiometric calculations will be incorrect.
Start with this balanced chemical equation.
KMnO_{4} + 8HCl + 5FeCl_{2} → 5 FeCl_{3} + MnCl_{2} + 4H_{2}O + KClSolution
Start with this balanced chemical equation.
$${\text{2KMnO}}_{\text{4}}+{\text{3CH}}_{\text{2}}{\text{=CH}}_{\text{2}}+{\text{4H}}_{\text{2}}\text{O}\to {\text{2MnO}}_{\text{2}}+{\text{3HOCH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{OH}+\text{2KOH}$$Verify that the equation is balanced.
Give 2 ratios that give the relationship between KMnO_{4} and CH_{2}=CH_{2}. (A total of 30 relationships can be constructed from this chemical equation. Can you find the other 28?)
Explain how stoichiometric ratios are constructed from a chemical equation.
Why is it necessary for a chemical equation to be balanced before it can be used to construct conversion factors?
Stoichiometric ratios are made using the coefficients of the substances in the balanced chemical equation.
A balanced chemical equation is necessary so one can construct the proper stoichiometric ratios.
Balance this equation and write every stoichiometric ratio you can from it.
NH_{4}NO_{3} → N_{2}O + H_{2}OBalance this equation and write every stoichiometric ratio you can from it.
N_{2} + H_{2} → NH_{3}Balance this equation and write every stoichiometric ratio you can from it.
Fe_{2}O_{3} + C → Fe + CO_{2}Balance this equation and write every stoichiometric ratio you can from it.
Fe_{2}O_{3} + CO → Fe + CO_{2}Balance this equation and determine how many molecules of CO_{2} are formed if 15 molecules of C_{6}H_{6} are reacted.
C_{6}H_{6} + O_{2} → CO_{2} + H_{2}OBalance this equation and determine how many molecules of Ag_{2}CO_{3}(s) are produced if 20 molecules of Na_{2}CO_{3} are reacted.
Na_{2}CO_{3}(aq) + AgNO_{3}(aq) → NaNO_{3}(aq) + Ag_{2}CO_{3}(s)Copper metal reacts with nitric acid according to this equation:
3Cu(s) + 8HNO_{3}(aq) → 3Cu(NO_{3})_{2}(aq) + 2NO(g) + 4H_{2}O(ℓ)Gold metal reacts with a combination of nitric acid and hydrochloric acid according to this equation:
Au(s) + 3HNO_{3}(aq) + 4HCl(aq) → HAuCl_{4}(aq) + 3NO_{2}(g) + 3H_{2}O(ℓ)Sulfur can be formed by reacting sulfur dioxide with hydrogen sulfide at high temperatures according to this equation:
SO_{2}(g) + 2H_{2}S(g) → 3S(g) + 2H_{2}O(g)Nitric acid is made by reacting nitrogen dioxide with water:
3NO_{2}(g) + H_{2}O(ℓ) → 2HNO_{3}(aq) + NO(g)NH_{4}NO_{3} → N_{2}O + 2H_{2}O; the stoichiometric ratios are $\frac{{\text{1NH}}_{\text{4}}{\text{NO}}_{\text{3}}}{{\text{1N}}_{\text{2}}\text{O}}\text{,}$ $\frac{{\text{1NH}}_{\text{4}}{\text{NO}}_{\text{3}}}{{\text{2H}}_{\text{2}}\text{O}}\text{,}$ $\frac{{\text{1N}}_{\text{2}}\text{O}}{{\text{2H}}_{\text{2}}\text{O}}\text{,}$ and their reciprocals.
2Fe_{2}O_{3} + 3C → 4Fe + 3CO_{2}; the stoichiometric ratios are $\frac{{\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{3C}}\text{,}$ $\frac{{\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{4Fe}}\text{,}$ $\frac{{\text{2Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{{\text{3CO}}_{\text{2}}}\text{,}$ $\frac{\text{3C}}{\text{4Fe}}\text{,}$ $\frac{\text{3C}}{{\text{3CO}}_{\text{2}}}\text{,}$ $\frac{\text{4Fe}}{{\text{3CO}}_{\text{2}}}\text{,}$ and their reciprocals.
2C_{6}H_{6} + 15O_{2} → 12CO_{2} + 6H_{2}O; 90 molecules