Stoichiometry for atoms, molecules and ions

Definitions & Key takeaways

Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. In the context of atoms, molecules, and ions, stoichiometry involves determining the amount of each species that is involved in a chemical reaction. It involves using the balanced chemical equation for a reaction to determine the amount of reactants consumed and the products formed.

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Stoichiometry 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 two H two plus O2 reacts to form 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.
OK. So we know that one mole is equal to 6.02 times 10 to the 23rd 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.
And 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 reaction.
Stoichiometry problems give us information about one substance in the reaction and ask us to find out about one or more other substances in that same reaction.
There are four general types of stoichiometry problems. The first type of stoichiometry problem is a mole to mole conversion.
You're given moles of something and ask to find the moles of something else. Suppose we start with four moles of hydrogen.
And we're 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 chemical reaction.
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.
So we get H two over O2 equals 2/1. Once we have this mole ratio, we want to set up a proportion, we know that we start out with four moles of hydrogen.
So we put this number in the numerator and we're asked to find the number of moles of oxygen. Since this is unknown, we put an X in the denominator.
So we have H two over O2 equals 2/1 equals four over X. Our next step is to solve for X and we can do this using cross multiplication.
We multiply the two times the X and then set it to equal four times one. Solving for X tells us that X equals two.
So we need two moles of oxygen to completely react with four moles of hydrogen. Now, we wanna 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 chemical reaction, there is a 2 to 2 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.
So we get H two over H2O equals 2/2. Now we look at the numbers we are given in our problem.
We know that we start with four moles of hydrogen. So we put four moles in the numerator of our ratio, but we don't know how many moles of water are produced.
So we put an X in the denominator. So we have H two over H2O equals 2/2 equals four over 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're given moles of one substance in the reaction and asked to find grams of another substance in the reaction.
Suppose we're asked to figure out how many grams of oxygen are necessary to react with four moles of hydrogen. We once again use the mole ratio of hydrogen to oxygen, which is 2 to 1 and set up a simple proportion.
We write four into the numerator. And once again, we use X in the denominator to denote the number of moles of oxygen.
So H two over O2 equals 2/1 equals four over X solving this equation tells us that X equals two moles of oxygen. We now need to convert this number to grams.
For this, we need the molar mass of oxygen because a diatomic oxygen molecule consists of two covalently bonded oxygen atoms.
We find the molar mass of an oxygen atom on our periodic table and then multiply it by two. So two times 16 g per mole equals 32 g per mole.
Next, we multiply two moles of oxygen by 32 g per mole, the moles cancel and we get 64 g of oxygen. We can use this same approach whenever we're given moles of one substance in the reaction.
And we want to find grams of something else. The third type of stoichiometry problem is a grams to mole conversion.
Suppose that we have a goal of producing 36 g of water. The question is how many moles of oxygen do we need to accomplish this?
Our first step is to convert everything into moles. The molar mass of water can be calculated by looking at the periodic table which tells us that each hydrogen atom has a molar mass of 1 g per mole.
And that oxygen has a molar mass of 16 g per mole. So the molar mass of water can be calculated as so two times 1 g per mole plus one times 16 g per mole equals 18 g per mole.
Since we wanna produce 36 g of water, we divide this number by the molar mass which tells us that we want to produce two moles of water to find the number of moles of oxygen.
We need to use mole ratios again by writing down the mole ratio of oxygen to water. For our chemical reaction.
Looking at the balanced reaction, we find the mole ratio by putting one the coefficient of oxygen in the numerator. And then putting two, the coefficient of water into the denominator.
We then set this ratio equal to X divided by the two moles of water that we just calculated. So we get O2 over H2O equals 1/2 equals X over two.
Solving for X tells us that one mole of oxygen is needed for a reaction to produce two moles of water. Note that our final answer has three significant figures.
The fourth type of stoichiometry problem is a grams to grams, conversion. We are given grams of one substance in the reaction and were asked to find the grams of something else in the reaction.
This time, we look at a new chemical reaction corresponding to the reaction where potassium chlorate turns into potassium chloride and oxygen gas.
Suppose we start with 4.4 g of potassium chlorate. And we're asked to figure out how many grams of oxygen will produce.
The first step is to convert to moles. Potassium chlorate has the chemical formula K cl three.
So we look up the molar mass of each atom on our periodic table, then multiply it by its subscript in the chemical formula, 39.1 g per mole plus 35.5 g per mole plus three times 16 g per mole equals 122.6 g per mole.
So one mole of potassium chlorate has a mass of 122.6 g. We now divide 4.4 g of potassium chlorate by the molar mass, which tells us that we have 0.0359 moles of potassium chlorate in our reaction.
To figure out the number of moles of oxygen gas involved in our reaction, we need to write the balanced chemical equation.
So two K CL 03 reacts to form two K cl plus 3 O2. Looking at the balanced equation, we write down the mole ratio of potassium chlorate to oxygen.
We write the coefficient of potassium chlorate two in the numerator and the coefficient of oxygen three in the denominator.
So K cl +03 over O2 equals 2/3. We know that we start with 0.0359 moles of potassium chlorate.
So we write down a new ratio with this number in the numerator. Since we don't know how many moles of oxygen result from our reaction.
We put an X in the denominator. So we get K CL 03 over O2 equals 2/3 equals 0.0359 over X.
We are now ready to solve for X using cross multiplication. We first multiply two times X and we set this quantity equal to three times 0.0359.
Solving this equation gives us X equals 0.0538 moles of oxygen produced during this reaction. Now, our last step is to convert 0.0538 moles of oxygen into the mass of oxygen.
We recall from before that diatomic oxygen molecules have a molar mass of 32 g per mole. So we multiply 0.0538 moles by 32 g per mole, which gives us 1.72 g of oxygen.
Note that our final answer has three significant figures. All right, it's a quick recap stoichiometry problems involve the relationships between the quantities of reactants and products in a chemical reaction.
The coefficients in the balanced equation tell us the relative number of moles of reactants and products. Since the balanced equation tells us the relative number of moles.
When we are given the mass of a reactant or product in grams, our first step should always be convert the mass into moles.
Once we find moles, we can use mole ratios to determine the amount of the other substance of interest in the reaction.