The formation of water from hydrogen and oxygen, is shown as a recipe instead of a chemical equation. Study Smarter Original

In this article, we will learn all about **stoichiometry **and how to use it!

- This article is about
**stoichiometry** - First, we'll use stoichiometry to balance chemical equations
- Next, we'll use it to calculate
**yield** - We'll use the same technique to calculate reactant amounts from a given product
- Lastly, we'll look at the
**ideal gas law**and how it is used to calculate yields for gas product

## Stoichiometry Definition

Let's begin by stating the definition of stoichiometry.

**Stoichiometry **is the ratio between products and reactants in a chemical reaction.

Stoichiometry is based on the **law of conservation of mass**. This states that, in a closed system (no outside forces), the mass of the products is the same as the mass of the reactants. **Stoichiometry **is used to balance reactions so that they obey this law. It is also used to calculate the mass of products and/or reactants.

In order for an equation to be balanced, the number of elements needs to be equal on the left (reactants) and right (products) sides of the equation. We balance equations by using **stoichiometric coefficients.**

**Stoichiometric coefficients **are the numbers before an element/compound which indicate the number of moles present. They show the ratio between reactants and products.

Let's start with an example: the reaction of aluminum metal and sulfuric acid.

The unbalanced equation is:

$$Al + H_2SO_4 \rightarrow H_2 + Al_2(SO_4)_3$$

The first step is to count how many of each element we have. On our reactants side, we have 1 mole of aluminum, 2 moles of hydrogen, 1 mole of sulfur, and 4 moles of oxygen. On our product side, we have 2 moles of hydrogen, 2 moles of aluminum, 3 moles of sulfur, and 12 moles of oxygen. For compounds like SO_{4}, it is important to remember to multiply each element by the subscript (the little number outside the parenthesis).

The next step is to pick the first element to balance. For this example, we will use aluminum. Since we only have 1 mole of Al on the reactant side, we multiply it by 2, so we now have:

$$2Al + H_2SO_4 \rightarrow H_2 + Al_2(SO_4)_3$$

Next, let's balance sulfur. Since coefficients are for the entire molecule, we multiply sulfuric acid by 3 to get:

$$2Al + 3H_2SO_4 \rightarrow H_2 + Al_2(SO_4)_3$$

Since the coefficient affects the whole molecule, oxygen is also balanced. However, now we have 6 moles of hydrogen on the left and only 2 on the right. So lastly, we multiply the moles of hydrogen gas by 3 to get the fully balanced equation of:

$$2Al + 3H_2SO_4 \rightarrow 3H_2 + Al_2(SO_4)_3$$

One thing to remember is that stoichiometric coefficients are always whole numbers, no fractions! Also, we always want our coefficients to be as simplified as possible, so if all of them are divisible by the same number, then you should divide them all by that number. Always remember to check if your equation is balanced *first, *or else all of your calculations may be wrong!

## Stoichiometry: Calculating Yield

After we balance a reaction, we can also utilize stoichiometry to determine the reaction's **yield. **

The **yield **of a reaction is the amount of product made given a certain amount of reactant(s). The **expected yield **is calculated using the balanced equation, while the **actual or experimental yield **is calculated by measuring the actual product made.

Determining a reaction's yield can be critical for an experimenter. If the actual yield is much different from the expected yield, there might have been some significant errors in the experiment that need to be fixed. If the product(s) are needed for another reaction, we can make sure that we are making the amount we need.

We can compare the actual to the expected yield by calculating the **percent error. **The formula for percent error is: $$\text{Percent error}=\frac{|\text{actual-expected}|}{\text{expected}}*100\%$$

All experiments will have some errors, but a very high percent error may mean something went really wrong!

Before we learn how to calculate yield, let's first learn how to convert from grams and/or milliliters to moles.

### Converting Mass/volume to Moles

When performing an experiment, we typically measure our reactants using grams (solids) and milliliters (liquids). However, our chemical equations use molar ratios. So to utilize our equation, we need to convert these measurements to moles.

**How many moles of sulfur is 6.0 g?**

To convert from g to mol, we must use the **molar mass. **The molar mass of sulfur is** **32.06 g/mol.

$$\frac{6.0\,\cancel{g}}{32.06\frac{\cancel{g}}{mol}}=0.19\,mol$$

Here we see that grams cancel, so only moles are left

Now for liquids, we need to use the **density** as well.

**How many moles of acetone is 62.0 mL? **

The density of iodine is 0.791 g/mL and the molar mass is 58.08 g/mol.

We need to first multiply the milliliter amount by the density so we are in grams. Then we can divide by the molar mass to get the molar amount.

$$62.0\,\cancel{mL}*\frac{0.791\,g}{\cancel{mL}}=49.0\,g$$

$$\frac{49.0\,\cancel{g}}{\frac{58.08\,\cancel{g}}{mol}}=0.844\,mol$$

So now that we know how to calculate moles, we can now learn how to calculate yield

**How many grams of aluminum sulfate are produced from 12.6g of aluminum?**

The equation is:

$$2Al + 3H_2SO_4 \rightarrow 3H_2 + Al_2(SO_4)_3$$

Firstly, we need to convert from grams to moles:

$$\frac{12.6\,g}{\frac{26.98\,g}{mol}}=0.467\,mol$$

From our equation, we see that 2 mols of aluminum produces 1 mol of aluminum sulfate. So to convert:

$$0.467\,mol\,\cancel{Al}*\frac{1\,mol\,Al_2(SO_4)_3}{2\,mol\,\cancel{Al}}=0.234\,mol\,Al_2(SO_4)_3$$

Lastly, we have to convert back from moles to grams. The molar mass of aluminum sulfate is 342.15 g/mol

$$0.234\,\cancel{mol}*\frac{342.15\,g}{\cancel{mol}}=80.1\,g$$

The ratios in our chemical equation only work with mol, so make sure to convert every time!

## Stoichiometry: Calculating Reactants

We can also use our balanced equation to determine the quantity of reactants we need to gain a specific yield. This is especially useful when materials are expensive or scarce, as we want to avoid wasting reactants!

**Given the balanced equation below, how many mLs of HCl are required to produce 23.2 g of CaCl _{2}? The molar mass of CaCl_{2} is 110.8 g/mol, the density of HCl is 1.2 g/mL, the molar mass of HCl is 36.46 g/mol.**

$$CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$$

We just need to follow the same steps as before:

$$\frac{23.2\,g}{\frac{110.98\,g}{mol}}=0.209\,mol$$

$$0.209\,mol\,CaCl_2*\frac{2\,mol\,HCl}{1\,mol\,CaCl_2}=0.418\,mol\,HCl$$$$0.418\,mol*\frac{36.46\,g}{mol}=15.2\,g$$

Since HCl is a liquid, our last step is to divide by the density.

$$\frac{15.2\,\cancel{g}}{\frac{1.2\,\cancel{g}}{mL}}=13\,mL$$

As we saw before, error is common in experiments, so when calculating the required reactant amounts, you may want to use more than needed just in case of error. Here is an illustration that summarizes these steps:

## Stoichiometry and the Limiting Reactant

When calculating yield, it's important to be aware of the **limiting reactant**

The **limiting reactant **is the reactant that is completely consumed in the reaction. Once this reactant is fully consumed, it stops the reaction and therefore limits the product made.

In previous examples, you have only been given the amount for one reactant, so what happens when the amounts of both reactants are given?

Given the balanced equation below, how many moles of NaCl are produced from 33.5 g of Na_{2}S and 45.0 mL of HCl? The molar mass of Na_{2}S is 78.04 g/mol, the molar mass of HCl is 36.46 g/mol, and the density of HCl is 1.2 g/mL.

$$Na_2S + 2HCl \rightarrow 2NaCl + H_2S$$

We need to calculate the yield for *both *reactants. Whichever reactant produces the smaller yield is the limiting reactant. Let's start with Na_{2}S

$$\frac{33.5\,g}{\frac{78.04\,g}{mol}}=0.429\,mol$$

$$0.429\,mol\,Na_2S*\frac{2\,mol\,NaCl}{1\,mol\,Na_2S}=0.858\,mol\,NaCl$$

And now for HCl,

$$45.0\,mL*\frac{1.2\,g}{mL}=54.0\,g$$

$$\frac{54.0\,g}{\frac{36.46\,g}{mol}}=1.48\,mol$$

$$1.48\,mol\,HCl*\frac{2\,mol\,NaCl}{2\,mol\,HCl}=1.48\,mol\,NaCl$$

Since Na_{2}S produces less NaCl, Na_{2}S is the limiting reactant and the yield is 0.858 mol NaCl.

## Gas Stoichiometry

So far, we have used stoichiometry in solid and liquid reactions, but what about gas reactions? These calculations are largely the same, except we need to also use the **ideal gas equation**.

The **ideal gas equation** describes how gases hypothetically behave in an ideal situation. The equation is:

$$PV=nRT$$

Where P=pressure, V=volume, n=moles, R=the ideal gas constant, and T=temperature

It's common for "ideal gas" problems to be calculated at STP. STP stands for Standard Temperature and Pressure. This is considered 273 K (0 °C/ freezing point of water) and 1 atm (average pressure at sea level). Now that we've covered the basics, let's start on a problem!

**Given the reaction below, how many liters of H₂ will 17.3 g of potassium produce at STP?**

$$2K + 2H_2O \rightarrow 2KOH + H_2$$

We are still using the same steps as before, except for the last step

$$\frac{17.3\,g}{\frac{39.1\,g}{mol}}=0.442\,mol$$

$$0.442\,mol\,K*\frac{1\,mol\,H_2}{2\,mol\,K}=0.221\,mol\,H_2$$

Now that we have our moles of gas, we need to use the ideal gas equation to convert to liters. Since this reaction is at STP, T=273 K, and P=1 atm. Also, for these units, the ideal gas constant (R) is 0.08205 L*atm/mol*K

$$PV=nRT$$

$$V=\frac{nRT}{P}$$

$$V=\frac{(0.221\,mol)(0.08205\frac{Latm}{molK})(273\,K)}{1\,atm}V=4.95\,L$$

The important thing to remember when calculating the amount of gas is that you must always use the ideal gas equation to convert from moles to liters. Additionally, if your temperature and pressure variables are not given (and you aren't solving for them), then it is likely that the reaction is being done at STP.

## Stoichiometry - Key takeaways

**Stoichiometry**is the mathematical relationship between products and reactants in a chemical reaction.**Stoichiometric coefficients**are the numbers before an element/compound which indicate the number of moles present. They show the ratio between reactants and products. They are used to balance equations.- Stoichiometry can be used to calculate yield by utilizing the ratio between reactants and products. This same concept is used to calculate needed reactant amounts.
- The
**limiting reactant**is the reactant that is completely consumed in the reaction. Once this reactant is fully consumed, it stops the reaction and therefore limits the product made. It can be determined by calculating the yield for all reactants. - For gas reactions, the
**ideal gas law**must be used to calculate yield. - The equation for the ideal gas law is \(PV=nRT\) where P=pressure, V=volume, n=moles, R=ideal gas constant, and T=temperature

###### Learn with 1 Stoichiometry flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Stoichiometry

What is stoichiometry?

**Stoichiometry **is the ratio between products and reactants in a chemical reaction.

What is stoichiometry used for?

Stoichiometry is used to balance chemical reactions, calculate reaction yield, and calculate needed reactant amounts based on a given product.

Why is stoichiometry important?

Stoichiometry ensures that when we calculate reaction yield and/or reactant amounts, we are getting an accurate number.

How to do stoichiometry?

Stoichiometry can be used to balance equations. Count the number of moles of each element on both sides of the equation. Then balance these mol values by multiplying each element and/or compound by the necessary coefficient.

How to calculate yield using stoichiometry?

By using stoichiometric coefficients (the number before an element/compound), we can balance an equation so that the mols of an element on one side are equal to another. We can use the molar ratio between a reactant and a product to calculate the yield of that product.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more