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We already began stoichiometry by learning about balanced chemical equations. Stoichiometry is intimately connected to balancing equations. It is central to all of chemistry and is essential for many calculations in chemistry. It is where we finally are able to determine the relationship between the amount of each reactant used and the amount of each product formed. In order to succeed at stoichiometry you must understand chemical formulas and molecular weight (molar mass), balanced chemical equations, converting grams to moles and moles to grams, mole ratios in balanced equations, and dimensional analysis (unit conversions).
Stoichiometry uses the mole ratios of balanced chemical reactions to determine the amount of product produced from a known amount of reactant.
The grams of reactants can be converted to moles of reactants using the conversion factors found from the periodic table. The moles of reactants can be converted to the moles of product using the mole ratios in the balanced chemical equation. The moles of product can then be converted to grams of product using the conversion factors found from the periodic table for the products.
In this section you will learn to determine the moles of products formed when moles of reactants are known, determine the moles of products formed when grams of reactants are known, and determine the grams of products formed when grams of reactants are known.
Consider the following representation of the chemical reaction between hydrogen and oxygen:2H2(g) + O2(g) → 2H2O(g)
The balanced chemical equation can be interpreted as two moles of hydrogen react with one mole of oxygen to form two moles of water.
Example #1 If it is known that two moles of hydrogen and one mole of oxygen are initially present, the balanced chemical equation requires that two moles of water will be formed. This is the theoretical yield, or the maximum amount of product that can be formed from these reactants.
What if there were initially four moles of hydrogen and two moles of oxygen? The first reaction is to notice that the reactant coefficients have been doubled and to then assume that the products would also be doubled and that four moles of water would be produced. That turns out to be correct, but how would you do it if the numbers weren't so nice? How would you do it mathematically?
To mathematically calculate the number of moles of water formed, you could use the ratio between one of the reactants and the product in a way consistent with dimensional analysis.
Or:
(4 mole H2)[
(2 mole H2O)
(2 mole H2)] = 4 mole H2O
(2 mole O2)[
(2 mole H2O)
(1 mole O2)] = 4 mole H2O Example #2 What if you started with 4 g of H2 and 32 g of O2, how many moles of water would be produced in that case?
Since the balanced chemical equation gives mole ratios, the grams of reactants must be converted to moles of reactants. The conversion factor is found from the periodic table.
For H2:Since there are 2 moles of H2 and one mole of O2, the balanced chemical equation can be used to calculate that there are 2 moles of water produced (as shown above).
(4 g H2)[
(1 mole H2)
(2 g H2)] = 2 mole H2 For O2:
(32 g O2)[
(1 mole O2)
(32 g O2)] = 1 mole O2 Example #3
How many grams of water would be produced when 8 g H2 is combined with 64 g O2?
In this case convert the grams of reactants to moles of reactants using the periodic table, determine the moles of water produced using the mole ratios from the balanced chemical equation, and then convert the moles of water produced to grams of water using the periodic table. The result is 72 g of water. Did you get it?
In this section, three examples of stoichiometric calculations have been illustrated. In the first case the moles of reactants were known and the mole ratio from the balanced chemical equation was used to calculate the moles of product (water) produced. In the second case the grams of reactant were given and it required a two step process where the grams of reactant were converted to moles of reactant and then the mole ratio from the balanced chemical equation was used to calculate the moles of product produced. In the last case the grams of reactant was converted to moles of reactant, the mole ratio from the balanced chemical equation was used to determine the moles of product produced, and then the moles of product was converted to grams of product.
Complete Example
Question: How many grams of NaCl will be produced when 34.5 g Na reacts with 53.2 g Cl2?
Write the balanced equation: 2Na + Cl2 → 2NaCl
Convert grams of reactants to moles of reactants:
(34.5 g Na)[
(1 mole Na)
(23 g Na)] = 1.5 mole Na Determine moles and then grams of product (NaCl):
(53.2 g Cl2)[
(1 mole Cl2)
(71 g Cl2)] = 0.75 mole Cl2
(1.5 mole Na)(
(2 mole NaCl)
(2 mole Na))(
(58.5 g NaCl)
(1 mole NaCl)) = 87.8 g NaCl or
(0.75 mole Cl2)(
(2 mole NaCl)
(1 mole Cl2))(
(58.5 g NaCl)
(1 mole NaCl)) = 87.8 g NaCl