What is Stoichiometry?

Hi, and welcome to this video on stoichiometry! It’s a super important concept in chemistry that, once you’ve got it mastered, will help you understand many of the calculations carried out for a chemical reaction.

The word stoichiometry, translated from Greek, literally means “the measure of elements.” The dictionary definition is “the quantitative relationship between two or more substances especially in processes involving physical or chemical change.” Basically, we’re looking to equate all the components in our chemical reaction with each other. So, if you have x grams of propane, you can make y grams of carbon dioxide. While it’s a pretty straightforward concept, it can often feel slippery or vague because we either use the word stoichiometry in so many contexts or because we drop it all together and use other equivalent terms. We’ll go through a lot of examples today, pointing out the different phrases and uses of the word itself, but first, we’ll start with a review of the chemical principles that underlie stoichiometry.

First, to fully understand stoichiometry, we need to review the law of conservation of mass.

The law of conservation of mass is straightforward and fairly intuitive. Basically, we can’t destroy or create mass in a chemical reaction. So whatever atoms we begin with, regardless of their molecular arrangement, we must end with the same type and number of atoms. The arrangement of the atoms will change, molecules will split or recombine, but we always have to account for every atom we begin with.

Stoichiometry Examples

For example, when sodium hydroxide reacts with hydrogen chloride, we get sodium chloride and water. Notice that on both sides of the reaction, we have 1 atom each of sodium, chlorine, and oxygen, and we have 2 atoms of hydrogen. Mass has been conserved in this reaction.

\(NaOH+HCl→NaCl+H_{2}O\)

 

Let’s consider a second reaction, the combustion of propane with oxygen to form carbon dioxide and water.

\(C_{3}H_{8}(g)+O_{2}(g)→CO_{2}(g)+H_{2}O(g)\)

 

Imagine that to write this chemical reaction, we had simply carried out a reaction in the lab. We reacted propane with oxygen and discovered that we made carbon dioxide and water. And that’s how we came to write this chemical equation, because we know that those were the reactants and products. But does this reaction obey the law of conservation of mass? Nope! Notice, for example, that on the reactants side of the equation, we have three atoms of carbon and on the products side only one. But we concluded that these were the only products, so how do we fix this?

We can correct this by balancing the reaction with stoichiometric coefficients. While maintaining the chemistry of the reaction that we observed in the lab, we’re simply going to adjust the number of each molecule in the reaction until the law of mass conservation is obeyed. This is called balancing the reaction.

We won’t go into the details here on the best approach to balancing reactions; for now, we’re simply going to add in the stoichiometric coefficients (which are often simply referred to as “the coefficients”). Keep in mind that the coefficients apply to the entire molecule. So, we have 5 molecules of oxygen, 3 of carbon dioxide, and 4 of water. Let’s look at this visually:

\(C_{3}H_{8}(g)+5O_{2}(g)→3CO_{2}(g)+4H_{2}O(g)\)

 

Importantly, notice that to balance the reaction and obey the law of conservation of mass, we did not adjust the composition of the molecules. For example, we didn’t go in and say, “Well, instead of carbon dioxide being \(CO_{2}\), we’ll make it \(C_{3}O_{2}\)… and voilà, carbon is balanced.”

\(C_{3}H_{8}(g)+5O_{2}(g)→C_{3}O_{2}(g)+4H_{2}O(g)\)

 

This would not only be a completely different reaction but it would also be a violation of another chemical principle, the law of definite proportion.

This is the idea that for any given chemical compound, there is a defined and definite composition of atoms.

Basically, applied to carbon dioxide, any molecule of carbon dioxide is one part carbon and two parts oxygen and nothing else. While from today’s detailed molecular view of chemistry, this seems pretty obvious, back in the day, this was a novel idea.

Okay, so back to stoichiometry itself. With our balanced equation, we can now write out stoichiometric equivalencies or stoichiometric factors. That is to say, we can quantitatively equate each component of the reaction. For example, 1 molecule of propane is stoichiometrically equivalent to 3 molecules of carbon dioxide. This is also referred to as the mole ratio—for every 1 mole of propane, we can form 3 moles of carbon dioxide.

You may also hear that a reactant was added stoichiometrically to a reaction. What this implies is that we added the reactant in the exact amount necessary to react fully with the other reactants. So, back to the propane combustion example, if we were planning on reacting 2 moles of propane, we would add 10 moles of oxygen if we added stoichiometrically. This means we have the exact number of oxygen molecules to react with all our propane molecules.

In many problems, you’ll be asked to convert into grams or volumes rather than moles (depending on the chemical) since we need a measurable amount. So, we might ask, how many grams of oxygen are needed to stoichiometrically react with 10 grams of propane? For these problems, we’ll need to use the molar mass of the compounds in addition to the stoichiometric factor to convert between the grams of one compound to the grams of another. Here’s how this problem would be laid out.

\(10.0\text{ grams}\text{ }C_{3}H_{8}\left ( \frac{\text{mol of}\text{ }C_{3}H_{8}}{44.1\text{ g}\text{ of}\text{ }C_{3}H_{8}} \right )\left ( \frac{5\text{ mol}\text{ of}\text{ O}_{2}}{1\text{ mol}\text{ of}\text{ C}_{3}H_{8}} \right )\left ( \frac{32.0\text{ g}\text{ of}\text{ O}_{2}}{mol\text{ of}O_{2}} \right )\)
\(=36.3\text{ grams}\text{ O}_{2}\)

 

We start by converting grams of propane into moles of propane. Then, we convert moles of propane to moles of oxygen using the stoichiometric factor. And finally, we convert moles of oxygen to grams of oxygen. Easy-peasy.

In practice, however, we often add one reactant in excess because reactions never go perfectly or maybe we’re taking advantage of Le Châtelier’s principle. In those scenarios, we need to calculate the limiting reagent, which also requires the use of the stoichiometric factors. Let’s run through another example together.

Consider the reaction of hydrazine, \(N_{2}H_{4}\), with dinitrogen tetroxide, \(N_{2}O_{4}\), to form nitrogen gas, \(N_{2}\) , and water, \(H_{2}O\). How many grams of water are formed when 1,000 grams of hydrazine are reacted with 800 grams of dinitrogen tetroxide?

\(2N_{2}H_{4}(l)+N_{2}O_{4}(l)→3N_{2}(g)+4H_{2}O(g)\)

 

First, check that the reaction is balanced. We can see that this one is.

Next, it was not specified in the question that the reactants were added stoichiometrically, so we need to know how much water each reactant can, in theory, produce. To do this, calculate the number of moles of water you can make from 1,000 grams of hydrazine and 800 grams of dinitrogen tetroxide.

Since this is a very similar process to what we just did with the propane combustion, pause the video and try this out on your own.

Here’s what we came up with.

>\(1000\text{ grams}\text{ N}_{2}H_{4}\left ( \frac{\text{mol of}\text{ N}_{2}H_{4}}{32.05\text{ g}\text{ N}_{2}H_{4}} \right )\left ( \frac{4\text{ mol}\text{ of}\text{ H}_{2}O}{2\text{ mol}\text{ of}N_{2}H_{4}}\right )\)
\(=62.40\text{ moles of H}_{2}O\)

 

From 1,000 grams of hydrazine, you can produce 62.40 moles of water.

\(800.0\text{ grams N}_{2}O_{4}\left ( \frac{\text{ mol of N}_{2}O_{4}}{32.0\text{ g N}_{2}H_{4}}\right )\left (\frac{4\text{ mol of H}_{2}O}{1\text{ mol of N}_{2}O_{4}}\right )\)
\(=34.78\text{ moles of H}_{2}O\)

 

And from 800 grams of dinitrogen tetroxide, you can make 34.78 moles of water.

Aha! We have just identified dinitrogen tetroxide as our limiting reagent. Even though we added enough hydrazine to make 62.40 moles of water, we only added enough dinitrogen tetroxide to make 34.78 moles. This also gives us our theoretical yield. In theory, if everything reacted completely, we would produce 34.78 moles of water vapor. This can then be used to calculate the percent yield. Let’s say we ran this reaction and only collected 28 moles of water. Our percent yield would be 80.5%.

\(\frac{28\text{ moles}}{34.78\text{ moles}}\times 100%=80.5%\)

 

Notice that all of these calculations were only done correctly because we used the stoichiometric factors! Any time you’re doing a reaction, regardless of the type of chemistry, you’ll need to keep stoichiometry in mind. It is essential to correctly converting between the different components of the reaction.

For example, this is critical in neutralization reactions or titrations. Think back to the sodium hydroxide–hydrogen chloride reaction. One mole of hydrogen chloride would neutralize one mole of sodium hydroxide. In other words, the stoichiometric equivalency is 1 to 1. However, if we used magnesium hydroxide instead of sodium hydroxide, we would need 2 moles of hydrogen chloride for every 1 mole of magnesium hydroxide to neutralize the solution.

\(Mg\left ( OH_{2} \right ) \left (s \right )+ 2HCl\left (aq\right )→MgCl_{2}\left (aq\right )+2H_{2}O(l)\)

 

Stoichiometry can also be combined with the ideal gas law to answer more complex questions. Let’s consider the combustion of propane again.

Imagine that we did this reaction in a piston with a massless and frictionless top. Ignoring contributions from an increase in temperature, how much would the volume increase when propane reacts with oxygen?

Notice we weren’t given any quantities, just the balanced reaction. That’s because we can solve this just by knowing the stoichiometry. What we need to recognize is that all of the reactants and products are gases, so we start with a total of 6 moles of gas (1 mole of propane and 5 moles of oxygen) and end with 7 moles of gas (3 of carbon dioxide and 4 of water vapor). Of course, we may not actually be using these quantities, but we’re just setting up the stoichiometric equivalencies with a mole unit. So now we can plug these values into Avogadro’s Law, a variant of the ideal gas law, and calculate that the final volume is 7/6 larger than the initial volume. We know this just from the stoichiometric coefficients. Pretty cool!

\(\frac{V_{i}}{n_{i}}+\frac{V_{f}}{n_{f}}\)

\(\frac{V_{i}}{6\text{ moles}}+\frac{V_{f}}{7\text{ moles}}\) \(V_{f}=\frac{7}{6}V_{i}\)

 

At the end of the day, stoichiometry is simply using the coefficients from the balanced equation to convert between the different components in your reaction.

It’s pretty basic, but incredibly important to get right. As we demonstrated today, stoichiometric factors are needed to calculate limiting reagents, theoretical yields, percent yields, and a lot more. And if you understand stoichiometry, you also appreciate the law of conservation of mass, the law of definite proportions, and ultimately the reaction mechanism itself.

Well, that’s it for this review! Thanks for watching, and happy studying!

 

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by Mometrix Test Preparation | This Page Last Updated: July 14, 2022