Mole Calculations | College Board AP Chemistry Revision Notes 2022

Mole Calculations Using Balanced Equations

Mass to moles calculations

Chemical amounts are measured in moles
The mass contained in one mole of a substance is called the molar mass (M)
The molar mass is is the same as the relative atomic mass in elements, or the relative molecular mass in compounds
- The molar mass can be calculated using the periodic table of the elements
The unit for the molar mass is g mol^-1
The amount of moles (n) in a mass of sample (m) can be calculated with this formula

$n = \frac{m}{M}$

Worked example

Calculate the number of moles in 19.0 g of O₂

Answer:

Step 1: Calculate the molar mass of O₂

M = relative molecular mass of O₂

M = 2 × mass of O

M = 2 × 16.00 g mol^-1

M = 32.00 g mol^-1

Step 2: Calculate the number of moles of O₂

n = m/M

n = 19.0 g/ 32.00 g mol^-1

n = 0.594 mol of O₂

Exam Tip

It is a good practice for your exam to work using the units during the whole calculation. This can help you to keep track if your workings are giving sensible answers. E.g. The number of moles calculated is not usually greater than 20 mol

Reading balanced chemical equations

A balanced chemical equation works like a recipe to that show the moles of reactants and products involved in the reaction
The coefficients (numbers before each chemical formula) are equivalent to the number of moles
E.g Burning hydrogen to produce water is read like this: 2 mol of hydrogen react with 1 mol of oxygen to produce 2 mol of water

2H₂ + O₂ → 2H₂O

Moles to moles calculations

Since the chemical equation is a recipe, the ratio between the moles of products and reactants is the same no matter the amount of moles
Multiple ratios can be written down for the reaction between hydrogen and oxygen to produce water

$\frac{2 mol of H_{2}}{1 mol of O_{2}}$ $\frac{2 mol of H_{2}}{2 mol of H_{2} O}$ $\frac{1 mol of O_{2}}{2 mol of H_{2} O}$

The ratios work as conversion factors and they can also be inverted depending on the practice problem
The ratios are used to calculate unknown moles of reactants and products

Worked example

How many moles of water (H₂O) are produced when 6.7 moles of O₂ react with enough moles of H₂?

2H₂ + O₂ → 2H₂O

Answer:

Step 1: Use the quantity given by the statement to start a new mathematical multiplication

6.7 mol of O₂ ……………..

Step 2: Choose the best ratio from the chemical equation. It must have the following units: the quantity that you already have at the bottom, and the quantity that you need on top

Quantity that you have = mol of O₂

Quantity that you need = mol of H₂O

$\frac{2 mol of H_{2}}{1 mol of O_{2}}$

Step 3: Use the conversion factor to obtain what you need.

$6.7 mol of O_{2} \times \frac{2 mol of H_{2}}{1 mol of O_{2}} = 13.4 mol of H_{2} O$

The reaction of 6.7 moles of O₂ with enough moles of H₂ produced 13.4 moles of H₂O

Mass to mass calculations

One of the most assessed calculations in exams are mass to mass calculations
The general process to perform this calculation is summarized in the next figure:

Mass to mass calculations summary

mass-to-mass-calculations-summary

General process for a mass to mass calculations showing the formula or the ratio that must be used for each step

Worked example

How many g of hydrogen gas (H₂) react with enough O₂ to produce 350.0 g water H₂O?

2H₂ + O₂ → 2H₂O

Answer:

Step 1. Set up the steps using the general process for a mass to mass calculation. The initial quantity is always given by the statement

mass-to-mass-calculation-worked-examplemass-to-mass-calculation-worked-example

Step 2: Calculate the moles of water using its mass and molar mass

M = relative molecular mass of H₂O

M = 1 × mass of O + 2 × mass of H

M = 16.00 + 2 × 1.008

M = 18.016 g mol^-1

n = m/M

n = 350.0 g/ 18.016 g mol^-1

n = 19.4272 mol of H₂O

Step 3: Calculate the moles of hydrogen gas (H₂) needed using the ratio from the chemical equation

$19.4272 mol of H_{2} O \times \frac{1 mol of H_{2}}{1 mol of H_{2} O} = 19.4272 mol of H_{2}$

Since the ratio is 1:1 the moles of hydrogen needed must be the same

Step 4: Calculate the mass of hydrogen gas (H₂)

M = relative molecular mass of H₂

M = 2 × mass of H

M = 2 × 1.008

M = 2.016 g mol^-1

n = m/M

m = n × M

m = 19.4272 mol × 2.016 g mol^-1

m = 39.165 g

Step 5: Write the answer with the correct number of significant figures

Since the only quantity given by the statement has 4 significant figures, the answer must be written down with 4 significant figures

The mass of hydrogen gas needed to produce 350.0 g of water is 39.17 g

Limiting reactant Calculations

In chemical reactions with two or more reactants, the most common scenario is the one of the reactants determines when the reaction ends
The limiting reactant is completely consumed during the chemical reaction and it limits the amount of product formed
The reactants in excess occur in greater amount than the amount needed to react completely with the limiting reactant
The amount of product calculated using the chemical reactant is also known as theoretical yield
The general process to perform this calculation is summarized in the next figure:

Limiting reactant calculations summary for a generic equation

limiting-reactant-calculation-summary

General process for a limiting reactant calculation showing the formula or the ratio that must be used for each step. The generic equation is shown above the diagram

Worked example

Calculate the theoretical yield of water (H₂O) collected, when 200.0 g of hydrogen gas (H₂) react with 200.0 g of oxygen gas (O₂)

2H₂ + O₂ → 2H₂O

Answer:

Step 1. Set up the steps using the general process for a mass to mass calculation. The initial quantities are always given by the statement

limiting-reactant-calculation-worked-example

Step 2.1: Calculate the moles of water produced by the mass of hydrogen gas

M = relative molecular mass of H₂

M = 2 × mass of H

M = 2 × 1.008

M = 2.016 g mol^-1

n = m/M

n = 200.0 g/ 2.016 g mol^-1

n = 99.206 mol of H₂

$99.206 mol of H_{2} \times \frac{2 mol of H_{2} O}{2 mol of H_{2}} = 99.206 mol of H_{2} O$

Step 2.2: Calculate the moles of water produced by the mass of oxygen gas

M = relative molecular mass of O₂

M = 2 × mass of O

M = 2 × 16.00

M = 32.00 g mol^-1

n = m/M

n = 200.0 g/ 32.00 g mol^-1

n = 6.25 mol of O₂

$6.25 mol of O_{2} \times \frac{2 mol of H_{2} O}{1 mol of O_{2}} = 12.5 mol of H_{2} O$

Step 3: Analyze the amount of moles and determine the limiting reactant

The least amount of moles of water was produced with the mass of oxygen gas. Therefore, the limiting reactant is oxygen gas (O₂)

Step 4: Calculate the mass of water (H₂O)

M = relative molecular mass of H₂O

M = (1 × mass of O) + (2 × mass of H)

M = 16.00 + (2 × 1.008)

M = 18.016 g mol^-1

n = m/M

m = n × M

m = 12.5 mol × 18.016 g mol^-1

m = 225.2 g of H₂O

Step 5: Write the answer with the correct number of significant figures

Since both quantities given by the statement have with 4 significant figures, the answer must be written down with 4 significant figures

The theoretical yield of water is 225.2 g of H₂O

Percent Yield Calculations

In all the scenarios, external conditions to the experiment will not allow to reach the 100% of yield during a reaction
The amount of product that is actually produced in laboratory conditions is called actual yield
The actual yield is always less than the theoretical yield
A comparison between the theoretical yield and the actual yield can be represented using the percent yield, whose formula is shown below

$percent yield = \frac{actual yield}{theoritical yield} \times 100 %$

Percent yield calculations are usually mixed with limiting reactant calculations

Worked example

The theoretical yield of water in the previous worked example was 225.2 g of water. If the reaction was carried in the laboratory and 198.3 g of water were collected at the end of the reaction. What is the percent yield of the reaction?

Answer:

Step 1: Determine the actual yield and the theoretical yield

Actual yield = 198.3 g

Theoretical yield = 225.2 g

Step 2: Plug the values inside the percent yield formula

$percent yield = \frac{actual yield}{theoritical yield} \times 100 %$

$percent yield = \frac{198.3 g of water}{225.2 g of water} \times 100 %$

$percent yield = 88.06 %$

Mole Calculations (College Board AP Chemistry)

Revision Note

Mole Calculations Using Balanced Equations

Mass to moles calculations

Worked example

Exam Tip

Reading balanced chemical equations

Moles to moles calculations

Worked example

Mass to mass calculations

Mass to mass calculations summary

Worked example

Limiting reactant Calculations

Limiting reactant calculations summary for a generic equation

Worked example

Percent Yield Calculations

Worked example

You've read 0 of your 0 free revision notes

Get unlimited access

Join the 100,000+ Students that ❤️ Save My Exams

Author: Martín