1. Basic Mole Concepts #
Atoms and molecules are too small to count one by one, so chemists use moles to measure them. Particles are also too small to weigh on a scale.
So instead, we take a very large nummber of these particles, so that we have enough to try and weigh them. This large number is (6.02 × 1023).
This is billions and billions of particles.
When we have this many particles, we say we have 1 mole of particles. We can then weight this 1 mole to see how many grams it is.
Key Definitions #
- The Mole: A unit for measuring particles that represents 6.02 × 1023 particles. (billions and billions of particles)
- Avogadro’s Constant: This very large number of particles that make up 1 mole (6.02 × 1023) of a substance
- Molar Mass: This number is found on the periodic table. It is the mass of 1 mole of a substance. We measure it as grams per mole (g/mol).
Understanding the Mole #
One mole of any substance contains exactly 6.02 × 1023 particles.
For example:
- 1 mole of Carbon has 6.02 × 1023 atoms of carbob
- 1 mole of Oxygen has 6.02 × 1023 atoms of Oxygen
- 1 mole of Water has 6.02 × 1023 molecules of water
2. Finding Molar Mass Using the Periodic Table #
The periodic table shows you the molar mass of each element.
The number below each element symbol is its Molar Mass (Mr). This number tells us that if you take a large bunch of the element (6.02 × 1023 particles), the total mass/weight will be the number shown in the periodic table.
Examples of some elements from the periodic table #
- Hydrogen (H) has a molar mass of 1
- Carbon (C) has a molar mass of 12
- Oxygen (O) has a molar mass of 16
This means that:
- 1 mole of Hydrogen atoms weighs 1 gram
- 1 mole of Carbon atoms weighs 12 grams
- 1 mole of Oxygen atoms weighs 16 grams
This also means that:
- 1g of Hydrogen atoms are 1 mole
- 12g of Carbon atoms are 1 mole
- 16g of Oxygen atoms are 1 mole
Calculating Molar Mass of Compounds #
To find the molar mass of a compound, add up the molar masses of all its atoms:
Example: Finding the Molar Mass of H2O #
- H2O has 2 hydrogen atoms and 1 oxygen atom
- Molar mass of H2O = (2 × mass of H) + (1 × mass of O)
- Molar mass of H2O = (2 × 1 g/mol) + (1 × 16 g/mol) = 18 g/mol
3. Important Mole Formulas #
These formulas will help you solve mole problems.
n = mMr
Mass = Number of moles × Molar mass
m = n × Mr
Number of particles = Number of moles × Avogadro’s constant
N = n × 6.02 × 1023
n = V24
At room temperature and pressure (RTP), 1 mole of any gas takes up 24 dm3
4. Percentage Composition by Mass #
This shows how much of a compound’s mass comes from each element.
Formula: #
Percentage by mass = Mass of element in compoundMolar mass of compound × 100%
Example: Percentage of Oxygen in H2O #
Step 1: Find the molar mass of H2O
- H (1 g/mol) × 2 = 2 g/mol
- O (16 g/mol) × 1 = 16 g/mol
- Total molar mass of H2O = 18 g/mol
Step 2: Find the mass of oxygen in the compound
- Mass of oxygen = 16 g
Step 3: Calculate the percentage
- Percentage of oxygen = 1618 × 100% = 88.9%
5. Calculations with Chemical Equations #
Using moles with chemical equations helps us predict how much product we’ll get from reactions.
Steps for Solving Reaction Problems: #
- Check that the chemical equation is balanced
- Convert the mass of the starting substance to moles
- Use the equation ratio to find moles of product
- Convert moles of product to mass or volume as needed
Example: Producing Hydrogen Gas #
For the reaction: Mg + 2HCl → MgCl2 + H2
If we react 2.4g of magnesium, how much hydrogen gas is produced?
Step 1: The equation is already balanced
The equation shows that 1 mole of Mg produces 1 mole of H2
Step 2: Calculate moles of magnesium
- Molar mass of Mg = 24 g/mol
- Moles of Mg = MassMolar mass = 2.4 g24 g/mol = 0.1 mol
Step 3: Calculate moles of hydrogen gas
- From the equation: 1 mol Mg → 1 mol H2
- So 0.1 mol Mg will produce 0.1 mol H2
Step 4: Calculate volume of hydrogen
- At RTP, 1 mol of gas occupies 24 dm3
- Volume of H2 = 0.1 mol × 24 dm3/mol = 2.4 dm3
6. Limiting Reactants #
In most reactions, one reactant runs out first. This is the limiting reactant because it limits how much product you can make.
Steps to Find the Limiting Reactant: #
- Calculate the moles of each reactant
- Compare the mole ratios in the balanced equation
- The reactant that makes the smallest amount of product is the limiting reactant
Example: Calcium Reaction #
For the reaction: Ca + 2H2O → Ca(OH)2 + H2
If we have 4.0g of Ca and 5.4g of H2O, which is the limiting reactant?
Step 1: Calculate moles of each reactant
- Molar mass of Ca = 40 g/mol
- Moles of Ca = 4.0 g40 g/mol = 0.1 mol
- Molar mass of H2O = 18 g/mol
- Moles of H2O = 5.4 g18 g/mol = 0.3 mol
Step 2: Check the ratio in the equation
- The equation shows: 1 mol Ca reacts with 2 mol H2O
- For 0.1 mol Ca, we need 0.1 × 2 = 0.2 mol H2O
Step 3: Determine the limiting reactant
- We have 0.3 mol of H2O, but only need 0.2 mol
- So H2O is in excess
- Therefore, Ca is the limiting reactant
7. Practice Questions #
-
Define the following
- the mole
- the Avogadro constant
-
Iron(III) oxide, Fe2O3, in iron ore is converted to iron when it reacts with carbon monoxide, CO, in the blast furnace.
a) Calculate the percentage by mass of iron in iron(III) oxide, Fe2O3.
-
Copper(II) oxide is formed when copper(II) nitrate, Cu(NO3)2, is heated.
- Calculate the mass of 0.0200 moles of Cu(NO3)2.
- Calculate the total volume of gas, in dm3 at r.t.p., produced when 0.0200 moles of Cu(NO3)2 is heated.
-
Which two of the following contain the same number of molecules? Show how you arrived at your answer.
- 2.0g of methane, CH4
- 8.0g of oxygen, O2
- 2.0g of ozone, O3
- 8.0g of sulfur dioxide, SO2
-
4.8g of calcium is added to 3.6g of water. The following reaction occurs. Ca + 2H2O → Ca(OH)2 + H2
- The number of moles of Ca = _______
- The number of moles of H2O = _______
- Which reagent is in excess? Explain your choice.
8. Worked Solutions #
Solution 1: #
- The mole is a unit that represents 6.02 × 1023 particles.
- The Avogadro constant is the number of particles in one mole of a substance (6.02 × 1023 particles per mole).
Solution 2: #
a) To calculate the percentage by mass of iron in Fe2O3:
Step 1: Find the molar mass of Fe2O3
- Molar mass of Fe = 56 g/mol
- Molar mass of O = 16 g/mol
- Molar mass of Fe2O3 = (2 × 56) + (3 × 16) = 112 + 48 = 160 g/mol
Step 2: Find the mass of iron in the compound
- Mass of iron in Fe2O3 = 2 × 56 = 112 g
Step 3: Calculate the percentage
- Percentage of iron = 112160 × 100% = 70%
Solution 3: #
a) To calculate the mass of 0.0200 moles of Cu(NO3)2:
Step 1: Find the molar mass of Cu(NO3)2
- Molar mass of Cu = 63.5 g/mol
- Molar mass of N = 14 g/mol
- Molar mass of O = 16 g/mol
- Molar mass of Cu(NO3)2 = 63.5 + (2 × 14) + (6 × 16) = 63.5 + 28 + 96 = 187.5 g/mol
Step 2: Calculate the mass
- Mass = Number of moles × Molar mass = 0.0200 × 187.5 = 3.75 g
b) When Cu(NO3)2 is heated, it decomposes according to the equation:
2Cu(NO3)2 → 2CuO + 4NO2 + O2
Step 1: Find the ratio of gases produced
- From the equation: 2 moles of Cu(NO3)2 produce 4 moles of NO2 and 1 mole of O2
- Total gas moles = 4 + 1 = 5 moles of gas from 2 moles of Cu(NO3)2
Step 2: Calculate moles of gas from 0.0200 moles of Cu(NO3)2
- Moles of gas = 52 × 0.0200 = 0.0500 moles
Step 3: Calculate volume of gas at RTP
- Volume = 0.0500 × 24 = 1.20 dm3
Solution 4: #
To find which two samples contain the same number of molecules, we need to calculate the number of moles in each:
a) 2.0g of methane, CH4:
- Molar mass of CH4 = 12 + (4 × 1) = 16 g/mol
- Moles = 2.0g16 g/mol = 0.125 mol
b) 8.0g of oxygen, O2:
- Molar mass of O2 = 2 × 16 = 32 g/mol
- Moles = 8.0g32 g/mol = 0.25 mol
c) 2.0g of ozone, O3:
- Molar mass of O3 = 3 × 16 = 48 g/mol
- Moles = 2.0g48 g/mol = 0.0417 mol
d) 8.0g of sulfur dioxide, SO2:
- Molar mass of SO2 = 32 + (2 × 16) = 64 g/mol
- Moles = 8.0g64 g/mol = 0.125 mol
Comparing the results, we can see that (a) 2.0g of methane and (d) 8.0g of sulfur dioxide both contain 0.125 moles, which means they have the same number of molecules.
Solution 5: #
For the reaction: Ca + 2H2O → Ca(OH)2 + H2
a) Number of moles of Ca:
- Molar mass of Ca = 40 g/mol
- Moles of Ca = 4.8g40 g/mol = 0.12 mol
b) Number of moles of H2O:
- Molar mass of H2O = 18 g/mol
- Moles of H2O = 3.6g18 g/mol = 0.2 mol
c) Which reagent is in excess?
- From the balanced equation, 1 mol of Ca reacts with 2 mol of H2O
- So 0.12 mol of Ca would need 0.12 × 2 = 0.24 mol of H2O
- We only have 0.2 mol of H2O, which is less than required
- Therefore, H2O is the limiting reactant and Ca is in excess
