Mole Calculations Using Balanced Equations
Mass to moles calculations
- Chemical amounts are measured in moles
- The mass contained in one mole of a substance is called the molar mass (M)
- The molar mass is is the same as the relative atomic mass in elements, or the relative molecular mass in compounds
- The molar mass can be calculated using the periodic table of the elements
- The unit for the molar mass is g mol-1
- The amount of moles (n) in a mass of sample (m) can be calculated with this formula
Worked example
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
Exam Tip
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
Reading balanced chemical equations
- A balanced chemical equation works like a recipe to that show the moles of reactants and products involved in the reaction
- The coefficients (numbers before each chemical formula) are equivalent to the number of moles
- E.g Burning hydrogen to produce water is read like this: 2 mol of hydrogen react with 1 mol of oxygen to produce 2 mol of water
2H2 + O2 → 2H2O
Moles to moles calculations
- Since the chemical equation is a recipe, the ratio between the moles of products and reactants is the same no matter the amount of moles
- Multiple ratios can be written down for the reaction between hydrogen and oxygen to produce water
- The ratios work as conversion factors and they can also be inverted depending on the practice problem
- The ratios are used to calculate unknown moles of reactants and products
Worked example
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
- Step 3: Use the conversion factor to obtain what you need.
The reaction of 6.7 moles of O2 with enough moles of H2 produced 13.4 moles of H2O
Mass to mass calculations
- One of the most assessed calculations in exams are mass to mass calculations
- The general process to perform this calculation is summarized in the next figure:
Mass to mass calculations summary
General process for a mass to mass calculations showing the formula or the ratio that must be used for each step
Worked example
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
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
Limiting reactant Calculations
- In chemical reactions with two or more reactants, the most common scenario is the one of the reactants determines when the reaction ends
- The limiting reactant is completely consumed during the chemical reaction and it limits the amount of product formed
- The reactants in excess occur in greater amount than the amount needed to react completely with the limiting reactant
- The amount of product calculated using the chemical reactant is also known as theoretical yield
- The general process to perform this calculation is summarized in the next figure:
Limiting reactant calculations summary for a generic equation
General process for a limiting reactant calculation showing the formula or the ratio that must be used for each step. The generic equation is shown above the diagram
Worked example
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
- 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
- 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
Percent Yield Calculations
- In all the scenarios, external conditions to the experiment will not allow to reach the 100% of yield during a reaction
- The amount of product that is actually produced in laboratory conditions is called actual yield
- The actual yield is always less than the theoretical yield
- A comparison between the theoretical yield and the actual yield can be represented using the percent yield, whose formula is shown below
- Percent yield calculations are usually mixed with limiting reactant calculations
Worked example
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