Grade 10

Advanced

10 min

Lesson 8 of 12

Mass-to-Mass Stoichiometry

Not Started

Solving multi-step problems to determine the actual weight of substances involved in a reaction.

Imagine you are a rocket scientist: if you load even a few grams too little fuel, the mission fails; too much, and the rocket is too heavy to lift off. How do engineers calculate the exact mass of fuel needed for a specific amount of oxygen?

Learning Objectives

1.Execute three-step stoichiometry calculations (Mass A to Mole A to Mole B to Mass B)
2.Apply the law of conservation of mass to verify stoichiometric results
3.Solve real-world laboratory scenarios involving gram-to-gram conversions

The Mole Bridge

In chemistry, we cannot compare grams of one substance directly to grams of another. Why? Because atoms have different weights! A gram of Hydrogen contains many more atoms than a gram of Lead. To solve this, we use the Mole as a universal bridge. The balanced chemical equation tells us the ratio of particles (moles) needed for a reaction, not the ratio of mass. To move from the mass of substance A to the mass of substance B, we must follow a specific three-step path: 1. Convert Mass A to Moles A using Molar Mass, 2. Convert Moles A to Moles B using the Mole Ratio, and 3. Convert Moles B back to Mass B.

Quick Check

Why can we not use the coefficients in a balanced equation to relate the masses of substances directly?

Think about the difference in weight between a dozen feathers and a dozen bricks.

Answer

Coefficients represent the ratio of moles (particles), and different substances have different molar masses.

How many grams of water ( $H_2O$ ) are produced from $4.04\text{ g}$ of Hydrogen gas ( $H_2$ )? Equation: $2H_2 + O_2 \rightarrow 2H_2O$

1. Mass to Moles: Convert

4.04\text{ g } H_2

to moles.

4.04\text{ g } H_2 \times \frac{1\text{ mol } H_2}{2.02\text{ g } H_2} = 2.00\text{ mol } H_2

2. Mole to Mole: Use the ratio from the equation (

2:2

2.00\text{ mol } H_2 \times \frac{2\text{ mol } H_2O}{2\text{ mol } H_2} = 2.00\text{ mol } H_2O

3. Mole to Mass: Convert moles of

H_2O

to grams (

18.02\text{ g/mol}

2.00\text{ mol } H_2O \times \frac{18.02\text{ g } H_2O}{1\text{ mol } H_2O} = 36.04\text{ g } H_2O

The Power of the Mole Ratio

The second step of our journey is the most critical: the Mole Ratio. This ratio comes directly from the coefficients of the balanced chemical equation. It acts as a conversion factor that 'switches' our calculation from substance A to substance B. If your equation is not balanced, your mole ratio will be wrong, and your entire calculation will fail. This is where we apply the Law of Conservation of Mass, which states that atoms are neither created nor destroyed. The total mass of your reactants must equal the total mass of your products in a closed system.

If $44.1\text{ g}$ of Propane ( $C_3H_8$ ) burns, what mass of Oxygen ( $O_2$ ) is consumed? Equation: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

1. Moles of Propane:

44.1\text{ g } C_3H_8 \times \frac{1\text{ mol }}{44.1\text{ g}} = 1.00\text{ mol } C_3H_8

2. Mole Ratio: The ratio is

1:5

1.00\text{ mol } C_3H_8 \times \frac{5\text{ mol } O_2}{1\text{ mol } C_3H_8} = 5.00\text{ mol } O_2

3. Mass of Oxygen:

5.00\text{ mol } O_2 \times \frac{32.00\text{ g }}{1\text{ mol }} = 160.0\text{ g } O_2

Quick Check

If a balanced equation shows $2A + B \rightarrow C$ , what is the mole ratio used to convert from moles of B to moles of A?

The substance you are going TO goes on top of the fraction.

Answer

The ratio is 2 moles of A / 1 mole of B.

Verification and Real-World Lab Scenarios

In a professional laboratory, stoichiometry is used to determine theoretical yield—the maximum amount of product you can make. Advanced problems often require you to verify your results by checking the Law of Conservation of Mass. If you calculate the mass of all reactants used and all products formed, the sums should be identical. Discrepancies usually point to a calculation error in molar mass or a failure to use the correct stoichiometric coefficients.

Calculate the mass of $O_2$ produced from the decomposition of $245.2\text{ g}$ of $KClO_3$ and verify the Law of Conservation of Mass. Equation: $2KClO_3 \rightarrow 2KCl + 3O_2$

1. **Moles of

KClO_3

**:

245.2\text{ g} \div 122.6\text{ g/mol} = 2.00\text{ mol } KClO_3

2. Moles of Products: -

KCl

2.00\text{ mol } KClO_3 \times (2/2) = 2.00\text{ mol } KCl

O_2

2.00\text{ mol } KClO_3 \times (3/2) = 3.00\text{ mol } O_2

3. Mass of Products: -

KCl

2.00\text{ mol} \times 74.6\text{ g/mol} = 149.2\text{ g}

O_2

3.00\text{ mol} \times 32.0\text{ g/mol} = 96.0\text{ g}

4. Verification:

149.2\text{ g} (KCl) + 96.0\text{ g} (O_2) = 245.2\text{ g}

. This matches the starting mass!

Key Takeaways

1The 3-step sequence is: Mass A $\rightarrow$ Moles A $\rightarrow$ Moles B $\rightarrow$ Mass B.
2The Mole Ratio from the balanced equation is the only way to switch between different substances.
3Molar mass (from the Periodic Table) is the conversion factor between grams and moles.
4The Law of Conservation of Mass ensures that the total mass of reactants equals the total mass of products.

Quick Check

Multiple Choice

What is the correct order of operations for a mass-to-mass stoichiometry problem?

Multiple Choice

In the reaction $N_2 + 3H_2 \rightarrow 2NH_3$ , what is the mole ratio of $H_2$ to $NH_3$ ?

True/False

The Law of Conservation of Mass states that the number of moles of reactants must always equal the number of moles of products.

What's Next?

Review Tomorrow

In 24 hours, try to sketch the 'Stoichiometry Map' from memory, showing the arrows between Mass, Moles, and Particles.

Practice Activity

Find a recipe for bread and try to 'balance' it like a chemical equation—if you double the flour, what happens to the yeast? This is the same logic as stoichiometry!

Browse

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Mass-to-Mass Stoichiometry

Learning Objectives

The Mole Bridge

The Power of the Mole Ratio

Verification and Real-World Lab Scenarios

Key Takeaways

Quick Check

What's Next?

Mass-to-Mass Stoichiometry

Learning Objectives

The Mole Bridge

The Power of the Mole Ratio

Verification and Real-World Lab Scenarios

Key Takeaways

Quick Check

What's Next?