Grade: Grade 10 Subject: Science Unit: Stoichiometry SAT: ProblemSolving+DataAnalysis ACT: Science

Mole Calculations

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Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. It allows us to predict how much product can be made from given amounts of reactants, or how much of a reactant is needed to produce a desired amount of product.

The Foundation: Mole Ratios

The coefficients in a balanced chemical equation represent the mole ratios between substances. These ratios are the key to stoichiometric calculations.

Example: 2H2 + O2 --> 2H2O

This equation tells us:

  • 2 moles of H2 react with 1 mole of O2
  • 2 moles of H2 produce 2 moles of H2O
  • 1 mole of O2 produces 2 moles of H2O

Mole ratios: 2:1:2 (H2:O2:H2O)

Types of Stoichiometric Calculations

1. Mole-to-Mole Conversions

The simplest type: convert moles of one substance to moles of another using the mole ratio from the balanced equation.

Formula

moles of B = moles of A x (moles of B from equation / moles of A from equation)

2. Mass-to-Mass Conversions

Convert mass of one substance to mass of another. This requires three steps:

  1. Convert mass of A to moles of A (using molar mass of A)
  2. Convert moles of A to moles of B (using mole ratio)
  3. Convert moles of B to mass of B (using molar mass of B)
Step Conversion Tool Used
1 grams A --> moles A Divide by molar mass of A
2 moles A --> moles B Multiply by mole ratio
3 moles B --> grams B Multiply by molar mass of B

3. Mass-to-Volume Conversions (Gases at STP)

For gases at STP, use the molar volume (22.4 L/mol) to convert between moles and liters.

Limiting Reactant

Limiting Reactant

The limiting reactant (or limiting reagent) is the reactant that is completely consumed first in a chemical reaction. It limits the amount of product that can be formed. The other reactant(s) are called excess reactants.

Analogy: Making Sandwiches

If you have 10 slices of bread and 8 pieces of cheese, and each sandwich needs 2 slices of bread and 1 piece of cheese:

  • Bread: 10 slices / 2 slices per sandwich = 5 sandwiches possible
  • Cheese: 8 pieces / 1 piece per sandwich = 8 sandwiches possible

Bread is the limiting "reactant" - you can only make 5 sandwiches before running out of bread, even though you'll have leftover cheese.

Finding the Limiting Reactant

Method 1: Compare mole ratios

  1. Convert all reactant amounts to moles
  2. Divide each by its coefficient in the balanced equation
  3. The smallest result indicates the limiting reactant

Method 2: Calculate product from each reactant

  1. Calculate how much product each reactant could produce (assuming the other is in excess)
  2. The reactant that produces the least product is limiting

Percent Yield

Theoretical, Actual, and Percent Yield

  • Theoretical yield: The maximum amount of product that could be formed based on stoichiometry (calculated)
  • Actual yield: The amount of product actually obtained from the experiment (measured)
  • Percent yield: (Actual yield / Theoretical yield) x 100%

Why is Actual Yield Usually Less Than Theoretical?

  • Side reactions may occur
  • Some product may be lost during purification
  • The reaction may not go to completion
  • Some product may remain in solution or on equipment

SAT/ACT Connection

Science sections may present stoichiometry problems involving mole ratios, limiting reactants, or percent yield. The key is to use the balanced equation to set up conversion factors. Always identify what you're given and what you need to find.

💡 Examples

Example 1: Mole-to-Mole Conversion

Problem: In the reaction N2 + 3H2 --> 2NH3, how many moles of NH3 can be produced from 4.5 moles of H2?

Solution:

Step 1: Identify the mole ratio from the balanced equation:

3 mol H2 : 2 mol NH3

Step 2: Set up the conversion:

4.5 mol H2 x (2 mol NH3 / 3 mol H2)

Step 3: Calculate:

4.5 x (2/3) = 3.0 mol NH3

Answer: 3.0 moles of NH3

Example 2: Mass-to-Mass Conversion

Problem: How many grams of water are produced when 8.0 g of hydrogen gas reacts with excess oxygen? (2H2 + O2 --> 2H2O)

Solution:

Step 1: Convert grams H2 to moles H2:

Molar mass H2 = 2.02 g/mol

8.0 g / 2.02 g/mol = 3.96 mol H2

Step 2: Use mole ratio (2 mol H2 : 2 mol H2O, which is 1:1):

3.96 mol H2 x (2 mol H2O / 2 mol H2) = 3.96 mol H2O

Step 3: Convert moles H2O to grams:

Molar mass H2O = 18.02 g/mol

3.96 mol x 18.02 g/mol = 71.4 g H2O

Answer: 71.4 grams of water

Example 3: Finding Limiting Reactant

Problem: If 10.0 g of zinc reacts with 15.0 g of hydrochloric acid (Zn + 2HCl --> ZnCl2 + H2), which is the limiting reactant?

Solution:

Step 1: Convert both reactants to moles:

  • Zn: 10.0 g / 65.38 g/mol = 0.153 mol Zn
  • HCl: 15.0 g / 36.46 g/mol = 0.411 mol HCl

Step 2: Divide each by its coefficient:

  • Zn: 0.153 mol / 1 = 0.153
  • HCl: 0.411 mol / 2 = 0.206

Step 3: Compare results. The smaller value indicates the limiting reactant.

0.153 < 0.206

Answer: Zinc (Zn) is the limiting reactant.

Example 4: Calculating Theoretical Yield

Problem: Based on Example 3, what is the theoretical yield of hydrogen gas (H2)?

Solution:

Step 1: Use the limiting reactant (Zn) to calculate moles of H2:

Mole ratio: 1 mol Zn : 1 mol H2

0.153 mol Zn x (1 mol H2 / 1 mol Zn) = 0.153 mol H2

Step 2: Convert moles H2 to grams:

0.153 mol x 2.02 g/mol = 0.309 g H2

Answer: 0.309 grams of H2 (theoretical yield)

Example 5: Calculating Percent Yield

Problem: A student performed the reaction from Examples 3-4 and obtained 0.25 g of hydrogen gas. What is the percent yield?

Solution:

Step 1: Identify the values:

  • Theoretical yield = 0.309 g (from Example 4)
  • Actual yield = 0.25 g (given)

Step 2: Apply the percent yield formula:

Percent yield = (Actual / Theoretical) x 100%

Percent yield = (0.25 g / 0.309 g) x 100%

Step 3: Calculate:

Percent yield = 80.9%

Answer: 80.9% yield

✏️ Practice

1. In the reaction 2Al + 3Cl2 --> 2AlCl3, what is the mole ratio of Al to AlCl3?

A) 2:3

B) 1:1

C) 3:2

D) 2:1

2. How many moles of O2 are needed to completely react with 6 moles of CH4? (CH4 + 2O2 --> CO2 + 2H2O)

A) 3 mol

B) 6 mol

C) 12 mol

D) 2 mol

3. What is the first step in a mass-to-mass stoichiometry problem?

A) Use the mole ratio

B) Convert grams to moles

C) Balance the equation

D) Calculate percent yield

4. The reactant that is completely consumed in a reaction is called the:

A) Excess reactant

B) Product

C) Limiting reactant

D) Catalyst

5. If the theoretical yield is 50 g and the actual yield is 40 g, what is the percent yield?

A) 125%

B) 80%

C) 90%

D) 10%

6. In the reaction 2H2 + O2 --> 2H2O, if you have 4 mol H2 and 3 mol O2, which is limiting?

A) H2

B) O2

C) Neither, they are equal

D) Cannot be determined

7. What mass of CO2 is produced when 12 g of carbon burns completely? (C + O2 --> CO2, molar mass of CO2 = 44 g/mol)

A) 12 g

B) 22 g

C) 44 g

D) 56 g

8. Why is actual yield usually less than theoretical yield?

A) The calculations are always wrong

B) Side reactions, product loss, or incomplete reactions

C) The molar masses are estimates

D) The balanced equation is incorrect

9. In 2NaOH + H2SO4 --> Na2SO4 + 2H2O, how many moles of water form from 0.5 mol H2SO4?

A) 0.25 mol

B) 0.5 mol

C) 1.0 mol

D) 2.0 mol

10. A percent yield of 110% would indicate:

A) An excellent experiment

B) Experimental error (impure product or measurement mistake)

C) The reaction was very efficient

D) More limiting reactant was added

Click to reveal answers
  1. B - The coefficients show 2 Al produces 2 AlCl3, which is a 1:1 ratio.
  2. C - 6 mol CH4 x (2 mol O2 / 1 mol CH4) = 12 mol O2
  3. B - First convert the given mass to moles using molar mass.
  4. C - The limiting reactant is used up first and limits the amount of product.
  5. B - Percent yield = (40/50) x 100% = 80%
  6. A - Divide by coefficients: 4/2 = 2 for H2; 3/1 = 3 for O2. H2 gives smaller value, so H2 is limiting.
  7. C - 12 g C / 12 g/mol = 1 mol C; 1 mol C produces 1 mol CO2; 1 mol x 44 g/mol = 44 g CO2
  8. B - Practical losses include side reactions, incomplete reactions, and product loss during handling.
  9. C - 0.5 mol H2SO4 x (2 mol H2O / 1 mol H2SO4) = 1.0 mol H2O
  10. B - Percent yield over 100% is impossible in reality and indicates error (impurities, incomplete drying, or measurement mistakes).

✅ Check Your Understanding

Question 1: Explain why the mole ratio from a balanced equation is essential for stoichiometric calculations. What would happen if we used an unbalanced equation?

Reveal Answer

The mole ratio from a balanced equation represents the actual proportions in which substances react and are produced. These ratios are determined by the law of conservation of mass - atoms must balance on both sides. If we used an unbalanced equation, the mole ratios would be incorrect, leading to wrong predictions. For example, if we incorrectly wrote H2 + O2 --> H2O (unbalanced), we might think 1 mole of H2 produces 1 mole of H2O, but in reality, 2 moles of H2 are needed. Using unbalanced equations would cause us to use wrong amounts of reactants or predict wrong amounts of products.

Question 2: In a reaction, you determine that Reactant A is limiting. What happens to the excess Reactant B after the reaction is complete?

Reveal Answer

When Reactant A (the limiting reactant) is completely consumed, the reaction stops even though some Reactant B remains. The excess Reactant B is left over, unreacted, in the final mixture. The amount of excess B remaining can be calculated by: (1) determining how much B actually reacted (using stoichiometry with the limiting reactant), then (2) subtracting that from the initial amount of B. The excess reactant doesn't disappear - it just doesn't have any more of the limiting reactant to react with.

Question 3: A student obtains a percent yield of 115% in an experiment. What are possible explanations for this result?

Reveal Answer

A percent yield over 100% is theoretically impossible because you cannot create more product than the limiting reactant allows. Possible explanations include: (1) The product is impure - it contains other substances that add to its mass (contamination, water, unreacted starting material). (2) The product wasn't completely dried before weighing. (3) There was a measurement error (wrong mass recorded, calibration issue with the balance). (4) The initial reactant masses were measured incorrectly. (5) A different or additional reaction occurred. The student should re-examine their procedure and purify/dry their product properly before re-weighing.

Question 4: Describe the complete process for solving this problem: "What mass of silver chloride (AgCl) precipitates when 25.0 mL of 0.200 M silver nitrate reacts with excess sodium chloride?" Include all necessary steps.

Reveal Answer

Step 1: Write and balance the equation:
AgNO3 + NaCl --> AgCl + NaNO3 (already balanced, 1:1:1:1)

Step 2: Calculate moles of AgNO3 (given):
moles = Molarity x Volume = 0.200 mol/L x 0.0250 L = 0.00500 mol AgNO3

Step 3: Use mole ratio to find moles of AgCl:
0.00500 mol AgNO3 x (1 mol AgCl / 1 mol AgNO3) = 0.00500 mol AgCl

Step 4: Convert moles of AgCl to grams:
Molar mass AgCl = 107.87 + 35.45 = 143.32 g/mol
0.00500 mol x 143.32 g/mol = 0.717 g AgCl

Answer: 0.717 g of AgCl precipitates

🚀 Next Steps

  • Review any concepts that felt challenging
  • Move on to the next lesson when ready
  • Return to practice problems periodically for review