It’s nearly the start of the school year, and you’ve gathered 10 friends for an end-of-semester bonfire. What would a bonfire be without s’mores? You pack supplies, making sure you have enough to make one s’more for everyone.
You can think of making a s’more as a chemical equation (see our Chemical Equations module for more on those):
2 Graham crackers + 1 piece of chocolate + 4 mini-marshmallows → 1 s’more
Just as with a chemical equation, the coefficients in front of the “reactants” and “products” show the proportions in which they react to produce the desired product—one s’more.
So, to make 10 s’mores, you would need:
20 Graham crackers + 10 pieces of chocolate + 40 mini-marshmallow → 10 s’mores
Congratulations, you’ve just made it through your first exercise in what chemists call stoichiometry. This mouthful of a term was coined in the 1790s by chemist Jeremias Benjamin Richter, who became fascinated by the proportional mathematics of combining chemicals, convinced that it held clues to the nature of matter (which it does indeed; Dalton drew on this math to devise his early atomic theory, described in depth in our module Early Ideas about Matter: From Democritus to Dalton. Richter combined the Greek words stoicheion, which means “element,” and metron, which means “measure.” In other words, stoichiometry is a way of measuring the amount of each reactant combining in a chemical reaction, in our case, the amounts of reactant (20 graham crackers, 10 pieces of chocolate and 30 mini-marshmallows) that in turn predict the amount of product (10 s’mores), or vice versa.
Stoichiometry may seem like a complicated word, but it’s a fairly straightforward concept when you apply it to chemical equations: the proportions expressed in a chemical equation (the coefficients) can be used to predict how much product will be produced from a given measure of reactants.
Stoichiometry predicts the amount of product produced
For example, we used stoichiometry to determine how many s’more “reactants” we would need to make 10 s’mores. We can also use stoichiometry to predict how much product we’ll get with the amount of each reactant we have. If we have lots and lots of chocolate and marshmallows but only 12 graham crackers, how many s’mores can we make?
Again, our equation is: 2 Graham crackers + 1 piece of chocolate + 4 mini-marshmallows → 1 s’more
If we have 12 graham crackers, that’s enough to make 6 s’mores. It doesn’t matter how much extra chocolate we have, because without the graham crackers, it isn’t a s’more.
So the mole ratio of graham crackers to s’mores produced is:
From s’mores to manufacturing: Real-world stoichiometry
Using the same concept of mole ratios as explained above, stoichiometry is used to figure out how much reactant is needed to make a desired quantity of product in a laboratory or manufacturing facility. An important industrial example is the production of nitrogen-based fertilizer, which provides important nutrients to the soil and allows modern farmers to grow more food per acre.
For centuries, farmers have understood the importance of adding nutrients to the soil in which they grow crops, but prior to the 1900s they were limited to using animal manure or expensive, naturally occurring mineral deposits as fertilizer. In the 1840s, the German chemist Justus von Liebig identified nitrogen as fertilizer’s key ingredient. However, despite the abundance of nitrogen in the atmosphere, there was no easy way to convert nitrogen to a form that could be taken up by plants.
This all changed in the early 1900s when the German chemist Fritz Haber invented a chemical process for converting nitrogen to ammonia (NH3), the compound that often gives household cleaners their characteristic smell, and which plants can use as a source of nitrogen. His initial method was only economical on a small scale, so Haber worked with a German colleague, Carl Bosch, to adapt this process to work at an industrial scale. The Haber-Bosch process is sometimes referred to as one of the most significant inventions of the 20th century, and it led to Haber winning the Nobel Prize in Chemistry in 1918. In its equation form, the Haber-Bosch process is relatively simple:
N2 + 3H2 → 2NH3
The ability to perform this simple reaction on a large scale had important historical consequences. Cheap ammonia provided an avenue to widely available inexpensive fertilizers, which created a boom in agriculture (and an associated increase in population) in the 20th century. And it indirectly prolonged World War I by providing Germany with an inexpensive source of the nitrogen necessary to make gunpowder. Some scientists have more recently questioned whether the Haber-Bosch process is a sustainable practice, given the environmental impact of agriculture and a growing population, as well as the fact that considerable energy is required to generate the hydrogen gas.
Mole Ratios: Applying stoichiometry to the production of fertilizer
Let’s apply our stoichiometry discussion here and imagine that an agricultural company needs to manufacture 1,500 kilograms of NH3 to meet the demand for fertilizer. How much N2 and H2 would they need to start with?
Again, the equation is: N2 + 3H2 → 2NH3
Looking at the equation, we see that the mole ratio of N2 required to produce NH3 is:
(1 mol N2) / (2 mol NH3)
Now we’ll use this mole ratio to determine how much reactant we need to start with to make 1,500 kilograms of ammonia.
First, a reminder: whenever we are calculating amounts of substance in a reaction, we have to convert the mass of each substance into moles. Why? Because the substances involved don’t have equal weights. Think of it in terms of the s’more: 1 piece of chocolate weighs a lot more than 1 mini-marshmallow. If we used mass in the equation instead of number of pieces, we might say that one s’more requires 1 gram of chocolate and 4 grams of mini-marshmallows. But in reality, that would amount to one piece of chocolate and about 50 mini-marshmallows! (For more on converting from grams to moles, see our module about the mole and atomic mass.)
So, we know that we want to make 1500 kg of NH3, Let’s start by converting kilograms to grams as follows
1,500 kg of NH3 x 1000 g/kg = 1,500,000 g of NH3
Then, we need to calculate how many moles that is. To do that, we multiply the molecular mass of NH3 (17 g per mole) by the number of grams, setting the equation up so that the grams cancel and the answer is in moles. We see that:
1,500,000 g NH3 × (1 mol NH3) / (17 g NH3) = 88,325 mol NH3
Next, we can use the mole ratio to figure out how many moles of N2 will be needed. Since we need 1 mole of N2 to produce 2 moles of NH3, we use that mole ratio to determine how many moles of N2 will be needed to produce 88,235 moles of NH3:
88,235 mol NH3 × (1 mol N2) / (2 mol NH3) = 44,117 mol N2
Now, use the molecular mass of N2 to figure out how many grams of N2 are required, then convert the grams of N2 to kg of N2, since that’s the units we want for our answer:
44,117 mol N2 × (28 g N2) / (1 mol N2) = 882,340 g N2, or 882.340 kg
Now that we know how many kilograms of N2 we would need, we can use the mole ratio of the reactants (N2 and H2) to figure out how many moles of H2 are required.
Remember, the equation states:
N2 + 3H2 → 2NH3
The mole ratio for H2 to N2 is 3 to 1. So, for every mole of nitrogen, we’ll need three times as many moles of hydrogen:
(3 mol H2) / (1 mol N2 )
Remember, we need to calculate how many moles of hydrogen are needed, and then convert those moles of hydrogen into grams of hydrogen. Using the moles of nitrogen we calculated above, we can get kg of H2:
44,117 mol N2 x (3 mol H2) / (1 mol N2 ) × (2 g H2) / (1 mol H2) = 264,702 g H2, or 264.702 kg
This is important information for the fertilizer manufacturer. While nitrogen is readily available from the air, hydrogen gas is not. So the manufacturer would likely have to purchase the hydrogen gas, which is expensive to generate, potentially explosive, and difficult to transport and store. Therefore, the manufacturer needs to know precisely how much hydrogen gas is required.
Limiting Reactant: The first reactant used up limits the amount of product
In the case above, the manufacturer will have an unlimited amount of nitrogen gas, but a precise amount of hydrogen gas. Therefore, the amount of hydrogen gas will limit the amount of ammonia that can be made (just like the number of graham crackers can limit the number of s’mores that can be made).
We would say that hydrogen is the limiting reactant, meaning that this is the reactant that will be used up first. As a result, the amount of it will determine how much product is produced. Determining how much reactant is required to produce a specific amount of product is one of the most important applications of stoichiometry.
We’ll illustrate this first with the s’mores. Let’s say you have the following amounts of s’more “reactants”:
120 graham crackers
70 pieces of chocolate
Here, again, is the s’mores equation: 2 Graham crackers + 1 piece of chocolate + 4 mini-marshmallows → 1 s’more
How many s’mores can you make from your reactants? That will depend on the limiting reactant, the one which will run out first. To determine with reactant is limiting, you will first need to calculate how many s’mores you can make with each of the reactants. You can do this using the mole ratio:
Limiting reactant is an important concept in any manufacturing process. A manufacturer knows they want to make a certain amount of a specific product, and will purchase the reactants accordingly. In many cases, it is more economical to make the most expensive reactant be the limiting one, reducing the cost of excess and waste.
Silver nitrate is a good example. This compound, AgNO3, has been used since ancient times as a disinfectant and wound-healing agent. Today it is used in bandages and other medical applications, as well as water purification. It can be easily made by reacting pure silver with nitric acid, according to the equation:
3Ag + 4HNO3 → 3AgNO3 + 2H2O + NO
Silver is a much more expensive reactant than nitric acid, so someone using it to produce silver nitrate will probably want to make silver the limiting reactant.
By starting with a set amount of each reactant, you can determine not only the limiting reactant but also the mass of product that will be produced and the amount of reactant that remains in excess.
Let’s say we start with 150g of silver and 150 g of nitric acid. How much AgNO3 can we make, and which reactant is the limiting one?
To find the answer takes a few steps:
1) convert each reactant to moles 2) use the mole ratio to determine how many moles of one reactant would be required to use up the other 3) calculate the amount of product based on using up all the limiting reactant.
Step 1: Converting to moles
150g Ag × (1 mole) / (108 g Ag) =1.39 mol Ag
150g HNO3 × (1 mole) / (63 g HNO3) = 2.38 mol HNO3
Step 2: Using the mole ratio to equate moles of Ag to moles of HNO3
(4 mol HNO3) / (3 mol Ag)
Since silver is our expensive reactant, we want to use it all up. We can calculate how many moles of HNO3 is required to react with the whole 1.39 mol of Ag, setting up the equation so that moles of HNO3 cancel:
1.39 mol Ag × (4 mol HNO3) / (3 mol Ag) = 1.85 mol HNO3 required
To use up all the, Ag, we need 1.85 moles of HNO3. Look at our calculations above. How much HNO3 do we have? We have 2.38 moles – more than we need. In other words, if we put all of both reactants together, the silver will be used up first, and there will be HNO3 left over. That makes silver the limiting reactant.
This example shows the importance of converting to moles first. We started with the same mass of each reactant, 150 g. But mass doesn’t tell us how many particles there are. That is what the unit of moles tells us.
Knowing that silver is the limiting reactant, we can go further and determine how many moles of AgNO3 is produced from the 1.39 moles of Ag we are starting with. This time we use the mole ratio between Ag and AgNO3.
In the reaction:
3Ag + 4HNO 3 → 3AgNO3 + 2H2O + NO
There are 3 moles of AgNO3 produced for every 3 moles of Ag used. So:
1.39 mol Ag × (3 mol AgNO3) / (3 mol Ag) = 1.379 mol AgNO3 produced
Calculations such as these are vital to our ability to manufacture and use chemicals efficiently, as well as to our ability to understand the impact of the reactions that take place in our everyday world. For example, an engineer for a paint manufacturer must consider the mole ratios of different chemicals in the paint, which will determine the cost of producing that paint. On a grander scale, stoichiometry plays a role in understanding climate change: if we know the quantities of different types of fossil fuels burned in a year, we can determine how much CO2 has been added to the atmosphere. From planning for s’mores to streamlining manufacturing and generating environmental data, we can use stoichiometry to predict and plan the outcome of many chemical processes.
Stoichiometry is the mathematics of chemistry. Starting with a balanced chemical equation, we make use of the proportional nature of chemical reactions to calculate the amount of reactant needed at the start or predict the amount of product that will be produced. While it may not seem all that “chemical,” stoichiometry is a concept that underlies our ability to understand the impact and implications of many chemical processes. A bandage manufacturer may use mole ratios to determine how much silver is required (and therefor the cost) to treat a batch of bandages with silver nitrate. A fertilizer company might apply the concept of limiting reactant to figure out how much product they can produce with a given amount of hydrogen gas. And so on. Stoichiometry, mole ratios, and limiting reactants are indispensable concepts for fully understanding any chemical process.
Stoichiometry uses the proportional nature of chemical equations to determine the amount of reactant needed to produce a given amount of product or predict the amount that will be produced from a given amount of reactant.
The mole ratio shows the proportion of one reactant or product in a reaction to another, and is derived from the balanced chemical equation. While we may need to adjust the amount of reactants to yield more product, the ratio of reactants to products is always the same as the balanced reaction.
The limiting reactant is the chemical used up first in a reaction. it can be determined by comparing the number of moles of each reactant on hand and the mole ratio between reactants and products in the balanced reaction.
Activate glossary term highlighting to easily identify key terms within the module. Once highlighted, you can click on these terms to view their definitions.
Activate NGSS annotations to easily identify NGSS standards within the module. Once highlighted, you can click on them to view these standards.