# Stoichiometry for atoms, molecules and ions

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## Pre-med

#### Chemistry

#### Content Reviewers:

Rishi Desai, MD, MPH#### Contributors:

Jessica Reynolds, MS, Evan Debevec-McKenney, Sam Gillespie, William GilpinStoichiometry deals with the relationships between the quantities of reactants and products in a chemical reaction.

For example, consider when hydrogen and oxygen react to form water.

In a balanced chemical equation, the coefficients tell us the relative number of molecules involved.

2 H2 + O2 → 2 H2O

The balanced equation tells us that two molecules of hydrogen will react with one molecule of oxygen to produce two molecules of water.

We know that one mole is equal to 6.02 times ten to the twenty-third molecules.

This incredibly large number is called Avogadro’s number.

So if we have one molecule of oxygen and we multiply by Avogadro’s number, we have one mole of oxygen.

If we have two molecules of hydrogen and we multiply that by Avogadro’s number, then we have two moles of hydrogen.

Notice that the one mole of oxygen and two moles of hydrogen correspond to the coefficients in the balanced equation.

So two moles of hydrogen will react with one mole of oxygen to produce two moles of water.

Therefore, the coefficients in a balanced equation tell you the relative number of moles of everything involved in the reaction.

We can use these mole relationships to find out how much product can be made from a given amount of reactant, or to find out how much of a reactant is needed to form a known amount of product.

Stoichiometry problems give us information about one substance in the reaction, and ask us to find out about one or more other substances in the reaction.

There are four general types of stoichiometry problems:

The first type of stoichiometry problem is a mole-to-mole conversion: you are given moles of something and asked to find moles of something else.

Suppose we start with 4 moles of hydrogen, and we are asked to figure out how many moles of oxygen are necessary to completely react with the hydrogen.

To accomplish this, we need to use a mole ratio.

Looking at the coefficients in our balanced equation, we can see that the mole ratio is 2 to 1, hydrogen to oxygen.

We put the coefficient of hydrogen in the numerator, and the coefficient of oxygen in the denominator:

H2 / O2 = 2/1

Once we have this mole ratio, we want to set up a proportion.

We know that we start out with 4 moles of hydrogen.

So we put this number in the numerator.

We are asked to find the number of moles of oxygen.

Since this is an unknown, we put an “x” in the denominator:

H2 / O2 = 2/1 = 4/x

Our next step is to solve for “x”.

We can do this using cross multiplication.

We multiply the 2 times the x, and then set it equal to four times 1

(2)(x) = (4)(1)

Solving for x tells us that x equals 2.

So we need two moles of oxygen to completely react with four moles of hydrogen.

Now we want to know how much water our reaction will produce.

To find this, we once again need to use a mole ratio.

This time, however, we want to find the ratio of our reactant, hydrogen, to our product, water.

Looking at the coefficients of the balanced equation there is a two to two mole ratio of hydrogen to water.

Therefore, we put a two in the numerator for hydrogen, as well as a two in the denominator for water

H2 / H2O = 2/2

Now, we look at the numbers we are given in our problem.

We know that we start with 4 moles of hydrogen, so we put 4 moles in the numerator of our ratio.

We don’t know how many moles of water are produced, so we put an “x” in the denominator.

H2 / H2O = 2/2 = 4/x

We now have an equation, which we can solve for “x” using cross multiplication.

“x” is equal to four.

So four moles of water are produced by our reaction.

The second type of stoichiometry problem is a “mole to gram” conversion.

In this type of problem we are given moles of one substance in the reaction and asked to find grams of another substance in the reaction.

Suppose we are asked to figure out how many grams of oxygen are necessary to react with 4.00 moles of hydrogen.

We once again use the mole ratio of hydrogen to oxygen which is two to one and set up a simple proportion.

We write 4.00 into the numerator and once again use an “x” in the denominator to denote the number of moles of oxygen.

H2 / O2 = 2/1 = 4.00/x

Solving this equation tells us that “x” equals 2.00 moles of oxygen.