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Stoichiometry

- Chemical Analysis
- Chemical Reactions
- Acid-Base Reactions
- Acid-Base Titration
- Hydrates
- Percent Composition
- Physical and Chemical Changes
- Redox Reactions
- Redox Titration
- Representing Chemical Reactions
- Single and Double Replacement Reactions
- Skeleton Equation
- Stoichiometry
- Inorganic Chemistry
- Catalysts
- Chlorine Reactions
- Group 2
- Group 2 Compounds
- Halogens
- Ion Colours
- Period 3 Elements
- Period 3 Oxides
- Periodic Table
- Periodic Trends
- Properties of Halogens
- Properties of Transition Metals
- Reactions of Halides
- Reactions of Halogens
- Shapes of Complex Ions
- Test Tube Reactions
- Titrations
- Transition Metal Ions in Aqueous Solution
- Transition Metals
- Variable Oxidation State of Transition Elements
- Ionic and Molecular Compounds
- Bond Hybridization
- Bond Length
- Bonding and Elemental Properties
- Coulomb Force
- Formal Charge
- Interstitial and Substitutional Alloys
- Intramolecular Force and Potential Energy
- Lattice Energy
- Lewis Dot Diagrams
- Limitations of Lewis Dot Structure
- Naming Ionic Compounds
- Polar and Non-Polar Covalent Bonds
- Potential Energy Diagram
- Properties of Covalent Compounds
- Resonance Chemistry
- Saturated Bond
- Sigma and Pi Bonds
- Structure of Ionic Solids
- Structure of Metals and Alloys
- The Octet Rule
- Types of Chemical Bonds
- VSEPR
- Kinetics
- Activation Energy
- Catalysis
- Concentration
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- First Order Reaction
- Multistep Reaction
- Pre-equilibrium Approximation
- Rate Constant
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- Steady State Approximation Example
- The Change of Concentration with Time
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- Nuclear Chemistry
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- Carbon Dating
- Mass Energy Conversion
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- Acylation
- Alcohol Elimination Reaction
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- Amide
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- Avogadro Constant
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- Equilibrium Constant Kp
- Equilibrium Constants
- Examples of Covalent Bonding
- Factors Affecting Reaction Rates
- Finding Ka
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- Gas Constant
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- Ideal Gas Law
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- Intermolecular Forces
- Introduction to Acids and Bases
- Ion and Atom Photoelectron Spectroscopy
- Ionic Bonding
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- Solutes Solvents and Solutions
- Solution Representations
- Solutions and Mixtures
- Specific Heat
- Spectroscopy
- Standard Potential
- States of Matter
- Stoichiometry in Reactions
- Strength of Intermolecular Forces
- The Laws of Thermodynamics
- The Molar Volume of a Gas
- Thermodynamically Favored
- Trends in Ionic Charge
- Trends in Ionisation Energy
- Types of Mixtures
- VSEPR Theory
- Van der Waals Forces
- Vapor Pressure
- Water in Chemical Reactions
- Wave Mechanical Model
- Weak Acid and Base Equilibria
- Weak Acids and Bases
- Writing Chemical Formulae
- pH
- pH Change
- pH Curves and Titrations
- pH Scale
- pH and Solubility
- pH and pKa
- pH and pOH

When baking a cake, you start by gathering all of your ingredients. You check your recipe to see how many eggs, cups of flour, and other ingredients you need. Using the correct number of ingredients is important for a successful cake! Chemistry works the same way! We use a "recipe" (chemical equation) and the quantities are based on stoichiometry. Like a cake, the chemist also uses recipes.

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

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!

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.

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!

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:

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.

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**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

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

More about Stoichiometry

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