Mole Calculations (College Board AP® Chemistry): Revision Note

Calculate the number of moles in 19.0 g of O₂

Answer:

• Step 1: Calculate the molar mass of O₂

M = molar mass of O₂

M = 2 × mass of O

M = 2 × 16.00 g/mol

M = 32.00 g/mol

• Step 2: Calculate the number of moles of O₂

n = m/M

n = 19.0 g ÷ 32.00 g/mol

n =  0.594 mol of O₂

• Convert every mass to moles before you use the coefficients

  • The coefficients give a mole ratio, not a mass ratio

• Write the substance with every unit ("mol of O₂", "mol of H₂O") so the units cancel and the ratio cannot end up upside down

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

$\frac{2\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}}{1\mathrm{mol}\mathrm{of}{\mathrm{O}}_2}$

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

 $6.7\mathrm{mol}\mathrm{of}{\mathrm{O}}_2\times \frac{2\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}}{1\mathrm{mol}\mathrm{of}{\mathrm{O}}_2}=13\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}$

6.7 mol of O₂ reacting with excess H₂ produces 13 mol of H₂O (13.4 rounded to 2 significant figures, because 6.7 has 2)

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

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

M = molar mass of H₂O

M = 1 × mass of O + 2 × mass of H

M = 16.00 + 2 × 1.008

M = 18.016 g/mol

n = m/M

n = 350.0 g ÷ 18.016 g/mol

n =  19.4272 mol of H₂O

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

$19.4272\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}\times \frac{1\mathrm{mol}\mathrm{of}{\mathrm{H}}_2}{1\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}}=19.4272\mathrm{mol}\mathrm{of}{\mathrm{H}}_2$

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 = molar mass of H₂

M = 2 × mass of H

M =  2 × 1.008

M = 2.016 g/mol

n = m/M

m = n × M

m = 19.4272 mol × 2.016 g/mol

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

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 limiting reactant calculation (one route for each reactant)

• The initial quantities are always given by the statement

1. H₂:

mass of H₂ → moles of H₂ (n = m/M) → moles of H₂O (multiply by 2 mol H₂O / 2 mol H₂)

2. O₂:

mass of O₂ → moles of O₂ (n = m/M) → moles of H₂O (multiply by 2 mol H₂O / 1 mol O₂)

3. The smaller amount of H₂O identifies the limiting reactant

4. Moles of H₂O (from the limiting reactant) → mass of H₂O (m = n × M)

5. The mass of H₂O is also known as the theoretical yield

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

M = molar mass of H₂

M = 2 × mass of H

M = 2 × 1.008

M = 2.016 g/mol

n = m/M

n = 200.0 g ÷ 2.016 g/mol

n =  99.206 mol of H₂

$99.206\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\times \frac{2\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}}{2\mathrm{mol}\mathrm{of}{\mathrm{H}}_2}=99.206\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}$

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

M = molar mass of O₂

M = 2 × mass of O

M = 2 × 16.00

M = 32.00 g/mol

n = m/M

n = 200.0 g ÷ 32.00 g/mol

n =  6.25 mol of O₂

$6.25\mathrm{mol}\mathrm{of}{\mathrm{O}}_2\times \frac{2\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}}{1\mathrm{mol}\mathrm{of}{\mathrm{O}}_2}=12.5\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}$

• 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 = molar mass of H₂O

M = (1 × mass of O) + (2 × mass of H)

M = 16.00 + (2 × 1.008)

M = 18.016 g/mol

n = m/M

m = n × M

m = 12.5 mol × 18.016 g/mol

m =  225.2 g of H₂O

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

  • Both masses are given to 4 significant figures, so the answer must be written down with 4 significant figures

  • The theoretical yield of water is 225.2 g of H₂O

• Find the limiting reactant from moles: compare the moles of product each reactant can make, as in Step 3.

  • Never choose the reactant with the smaller mass, the smaller molar mass or the smaller coefficient

• Here the two masses are equal (200.0 g), yet O₂ is limiting: equal masses are not equal moles

• To compare the masses of two substances, use both the mole ratio and the molar masses

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

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

 $\mathrm{percent}\mathrm{yield}=\frac{\mathrm{actual}\mathrm{yield}}{\mathrm{theor}e\mathrm{tical}\mathrm{yield}}\times 100\%$

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

$\mathrm{percent}\mathrm{yield}=88.06\%$