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 = 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

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

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

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

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

 $6.7\mathrm{mol}\mathrm{of}{\mathrm{O}}_2\times \frac{2\mathrm{mol}\mathrm{of}{\mathrm{H}}_2}{1\mathrm{mol}\mathrm{of}{\mathrm{O}}_2}=13.4\mathrm{mol}\mathrm{of}{\mathrm{H}}_2\mathrm{O}$

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

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 = 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

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

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 = 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

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

• 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

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

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

M = 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

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

• 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 = 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

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

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

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

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