OML Search

Stoichiometry (Calculations from Chemical Equations)

Related Topics:
More Lessons for Chemistry

Math Worksheets

Stoichiometry is the calculation of quantitative relationships of the reactants and products in chemical reactions. Given enough information, we can use stoichiometry to calculate the moles and masses within a chemical equation.

In this lesson, we will look into some examples of stoichiometry problems.

  • What a chemical equation tells you?
  • Does the total mass change during a chemical reaction?
  • How to calculate masses from equations?

What a chemical equation tells you?

The equation C (s) + O2 (g) -> CO2 (g) tells you that:

  • 1 carbon atom reacts with 1 molecule of oxygen to give 1 molecule of carbon dioxide
  • If there was 1 mole of carbon atoms then

1 mole of carbon atom reacts with 1 mole of oxygen to give 1 mole of carbon dioxide

  • The moles can be changed to grams using the relative atomic mass (Ar) and the relative molecular mass (Mr)

The Ar values are: C = 12, O = 16

The Mr values are: O2 = 32, CO2 = (12 + 32) = 44

We can write: 12g of carbon reacts with 32g of oxygen to give 44g of carbon dioxide.

This also means that: 6g of carbon reacts with 16g of oxygen to give 22g of carbon dioxide and so on.

The masses of each substance taking part in the reaction are always in the same ratio.

In general, a chemical equation tells you:

  • how many moles of each substance were involved
  • how many grams of each substance were involved.

The following video shows how to calculate the following stoichiometry problem:

“A solution containing acetic acid is mixed with calcium carbonate. How much acetic acid is consumed in a reaction which produces 0.400 mol CO2?”

Does the total mass change during a chemical equation?

Take note of what happens to the total mass, during the above reaction:

mass of carbon and oxygen at the start: 12 g + 32 g = 44 g

mass of carbon dioxide at the end: 44 g

The total mass has not changed, during the reaction. This is because no atoms have disappeared. They have just been rearranged.

This is one of the basic laws of chemistry:

The total mass does not change during a chemical reaction.

Calculating masses from equations


Hydrogen burns in oxygen to form water. What mass of oxygen is needed to burn 1 gram of hydrogen, and what mass of water is obtained?


Step 1: Write the balanced equation for the reaction.

2H2 (g) + O2 (g) → 2H2O (l)

Step 2: Write down the relative atomic mass (Ar) and the relative molecular mass (Mr), for each substance in the equation.

Ar: H = 1, O = 16

Mr: H2 = 2, O2 = 32, H2O = 18

Step 3: Using Ar or Mr, change the moles in the equation to grams.

Step 4: Find the actual masses.

4 g of hydrogen reacts with 32 g of oxygen to give 36 g of water.

The given question states thatwe need to burn 1 g of hydrogen.

So, 1 g of hydrogen reacts with 32 ÷ 4 = 8 g of oxygen to give 36 ÷ 4 = 9 g of water.

The following video shows an example of calculating mass from equation.


Solving Stoichiometry Problems

In this video, we will look at the steps to solving stoichiometry problems.

Stoichiometry problems

Example 1: How many moles of ammonia are produced by 2.8 mol of hydrogen?

Example 2: Determine the number of moles of oxygen needed to react with 0.56 mol of vanadium to form vanadium (V) oxide?
Example: 3: Carbon dioxide reacts with lithium hydroxide to produce lithium carbonate and water. What mass of lithium htdroxide would be required to react with 1.00 x 103 grams of carbon dioxide?
Example 4: How many moles of silver chloride forms when 2.6 mol of KCl reacts with excess silver nitrate in solution?

Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.

You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

OML Search

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

[?] Subscribe To This Site

follow us in feedly
Add to My Yahoo!
Add to My MSN
Subscribe with Bloglines