Now you're ready to use what you know about conversion factors to solve some stoichiometric problems in chemistry. Almost all stoichiometric problems can be solved in just four simple steps:
Let's begin our tour of stoichiometry by looking at the equation for how iron rusts:
Fe + O_{2}→Fe_{2}O_{3} |
The constituent parts of a chemical equation are never destroyed or lost: the yield of a reaction must exactly correspond to the original reagents. This fact holds not just for the type of elements in the yield, but also the number. Given our unbalanced equation:
Fe + O_{2}→Fe_{2}O_{3} |
Therefore, we must balance the equation by placing coefficients before the various molecules and atoms to ensure that the number of atoms on the left side of the arrow corresponds exactly to the number of elements on the right.
4Fe +3O_{2}→2Fe_{2}O_{3} |
The process of balancing an equation is basically trial and error. It gets easier and easier with practice. You will likely start to balance equations almost automatically in your mind.
The process of converting given units into moles involves conversion factors. Below we will provide the most common and important conversion factors to convert between moles and grams, moles and volumes of gases, moles and molecules, and moles and solutions. These conversion factors function in the same way as those discussed in the previous section Note also that though these conversion factors focus on converting from some other unit to moles, they can also be turned around, allowing you to convert from moles to some other unit.
Converting from Grams to Moles
The gram formula mass of a compound (or element) can be defined as the mass of one mole of the compound. As the definition suggests, it is measured in grams/mole and is found by summing the atomic weights of every atom in the compound. Atomic weights on the periodic table are given in terms of amu (atomic mass units), but, by design, amu correspond to the gram formula mass. In other words, a mole of a 12 amu carbon atom will weigh 12 grams.
The gram formula mass can be used as a conversion factor in stoichiometric calculations through the following equation:
Moles = |
Converting between Volume of a Gas and Moles
The Ideal Gas law, discussed at length in the Sparknote on Gases, provides a handy means of converting between moles and a gas, provided you know certain qualities of that gas. The Ideal Gas Law is PV = nRT , with n representing the number of moles. If we rearrange the equation to solve for n , we get:
n = |
In those instances when a problem specifies that the calculations are to be made at STP (Standard Temperature and Pressure; P = 1 atm, T = 273 K)), the problem becomes even simpler. At STP, a mole of gas will always occupy 22.4 L of volume. If you are given a volume of a gas at STP, you can calculate the moles in that gas by calculating the volume you are given as a fraction of 22.4 L. At STP, 11.2 L of a gas will be .5 moles; 89.6 L of gas will be 4 moles.
Converting between Individual Particles and Moles
Avogadro's Number provides the conversion factor for moving from number of particles to moles. There are 6.02×10^{23} formula units of particles in every mole of substance, with formula unit describing the substance we are looking at, whether it is a compound, molecule, atom, or ion. A formula unit is the smallest unit of a substance that still retains that substance's properties and is the simplest way to write the formula of the substance without coefficients. Some representative formula units are listed below.
Moles = |
Converting between Solutions and Moles
Solutions are discussed in much greater detail in the series of Solutions SparkNotes. But it is possible, and fairly easy to convert between the measures of solution (molarity and molality) and moles.
Molarity is defined as the number of moles of solute divided by the number of liters of solvent. Rearranging the equation to solve for moles yields:
Moles = molarity × liters of solution
MolaLity is defined as the number of moles of solute divided by the number of kilograms of solvent. Rearranging the equation to solve for moles yields:
Moles = molality × kilograms of solution
Before demonstrating how to calculate how much yield a reaction will produce, we must first explain what the mole ratio is.
The Mole Ratio
Let's look once again at our balanced demonstration reaction:
4Fe +3O_{2}→2Fe_{2}O_{3} |
The coefficients in front of iron, oxygen, and iron (III) oxide are ratios that govern the reaction; in other words, these numbers do not demand that the reaction can only take place with the presence of exactly 4 moles of iron and 3 moles of oxygen, producing 2 moles of iron (III) oxide. Instead, these numbers state the ratio of the reaction: the amount of iron and oxygen reaction together will follow a ratio of 4 to 3. The mole ratio describes exactly what its name suggests, the molar ratio at which a reaction will proceed. For example, 2 moles of Fe will react with 1.5 moles of O_{2} to yield 1 mole of Fe_{2}O_{3} . Alternatively, 20 moles of Fe will react with 15 moles of O_{2} to yield 10 moles of Fe_{2}O_{3} . Each of these examples of the reaction follow the 4:3:2 ratio described by the coefficients.
Now, with a balanced equation, the given units converted to moles, and our understanding of the mole ration, which will allow us to see the ratio of reactants to each other and to their product, we can calculate the yield of a reaction in moles. Step 4 demands that we be able to convert from moles to back to the units requested in a specific problem, but that only involves turning backwards the specific converstion factors described above.
Problem: Given the following equation at STP:
N_{2}(g) + H_{2}(g)→NH_{3}(g) |
Solution:
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Step 1: Balance the equation.
N_{2}(g) + 3H_{2}(g)→2NH_{3}(g) |
= 10 moles of NH_{3}(g) |
= 15 moles H_{2}(g) |
= 336 L H_{2}(g) |
Now for a more challenging problem:
Given the following reaction:
2H_{2}S(g) + O_{2}(g)→SO_{2}(g) + 2H_{2}O(s) |
Solution
Step 1. The equation is partially balanced already, but let's finish the job.
2H_{2}S(g) +3O_{2}(g)→2SO_{2}(g) + 2H_{2}O(s) |
Step 2, convert to moles:
1 formula unit of
H_{2}O
has 2 atoms of H and 1 atom of O
The atomic mass of H is 1 gram/mole
Atomic mass of O = 16 grams/mole
GFM of H_{2}O(s) = + = 18 grams / mole |
×1 mole = 1 mole of H_{2}O(s) |
Step 3, mole ratio:
×3 moles O_{2}(g) = 1.5 moles O_{2}(g) |
Step 4, convert to desired units:
= 9.03×10^{23} molecules O_{2}(g) |
Is this the answer? No. The question asks for ATOMS of oxygen. There are two atoms of oxygen in each molecule of O_{2}(g).
×2 atoms O = 1.806×10^{24} atoms O |
Now we're done. Note how important it was to write out not only your units, but what substance you're currently working with throughout the problem. Only a brief check was needed to ascertain if we were really answering the given question. Always check to make sure you have answered the correct question.