Moles, Relative Formula Mass & Calculations
Quantitative chemistry uses relative formula masses (Mr) and the mole concept to calculate reacting masses, gas volumes, and solution concentrations. Applying the core relationship moles = mass ÷ Mr allows you to determine limiting reactants, percentage yields, and titration concentrations.
- Relative Atomic Mass (Ar)
- The average mass of an atom of an element compared to 1/12th the mass of an atom of carbon-12.
- Mole
- The amount of substance containing 6.02 x 10^23 particles.
- Limiting Reactant
- The reactant that is completely used up in a reaction, limiting the amount of product formed.
Moles, Mass & Solution Concentration Step-by-Step Calculator
Enter any mass, relative formula mass (Mr), and solution volume in cm³ to see the exact step-by-step calculation.
Step 1: Calculate Moles (n = m ÷ Mᵣ):
n = 11.7 g ÷ 58.5 g/mol = 0.2000 mol
Step 2, Convert Volume from cm³ to dm³ (÷ 1000):
V = 250 cm³ ÷ 1000 = 0.250 dm³
Step 3: Calculate Concentration (c = n ÷ V):
c = 0.2000 mol ÷ 0.250 dm³ = 0.800 mol/dm³ (or 46.80 g/dm³)
Exam Trap: Always divide cm³ by 1000 to convert into dm³ before using c = n / V!
Exam notes: All 6 Quantitative Chemistry Formulas + Titration Steps
- Every quantitative chemistry calculation starts by finding moles (n). Once you have moles of the known substance, use the big stoichiometric numbers in front of formulas in the balanced equation (the mole ratio) to find moles of the unknown substance.
- Never confuse Concentration in g/dm³ (mass ÷ volume in dm³) with Concentration in mol/dm³ (moles ÷ volume in dm³). To convert from mol/dm³ to g/dm³, multiply by the relative formula mass (M_r).
- At room temperature and pressure (20°C and 1 atm / 101 kPa), 1 mole of any gas occupies 24 dm³ (24,000 cm³).
| Quantity | Formula | Key Unit Conversion / Trap |
|---|---|---|
| Moles from Mass | Moles (n) = Mass (g) ÷ M_r | Use atomic masses (top/bigger number on Periodic Table), never atomic number. |
| Solution Concentration (mol/dm³) | Concentration (c) = Moles (n) ÷ Volume (dm³) | Always divide cm³ by 1,000 to get dm³ first. (25 cm³ = 0.025 dm³) |
| Convert mol/dm³ ↔ g/dm³ | Conc (g/dm³) = Conc (mol/dm³) × M_r | Multiply by M_r to go from moles to grams |
| Gas Volume at RTP | Volume (dm³) = Moles (n) × 24 | If volume is in cm³, divide by 24,000 cm³ instead of 24 dm³ |
| Percentage Yield | % Yield = (Actual mass ÷ Theoretical max mass) × 100 | Always less than 100% due to incomplete reaction, side reactions, or loss in transfer |
| Atom Economy | % Atom Economy = (M_r of desired product ÷ Sum of M_r of all reactants) × 100 | A reaction with only one product always has 100% atom economy. |
Key Concepts & Exam Definitions (8)
- Relative Atomic Mass (Ar)
- The average mass of an atom of an element compared to 1/12th the mass of an atom of carbon-12.
- Mole
- The amount of substance containing 6.02 x 10^23 particles.
- Limiting Reactant
- The reactant that is completely used up in a reaction, limiting the amount of product formed.
- Percentage Yield
- A comparison of the actual mass of product obtained to the theoretical maximum possible mass.
- Atom Economy
- A measure of the proportion of reactant atoms that become part of the desired product in a balanced chemical equation.
- Avogadro Constant (6.02 × 10²³)
- One mole of any substance contains 6.02 × 10²³ particles (atoms, ions, or molecules). The mass in grams of 1 mole of a substance is numerically equal to its relative formula mass (M_r).
- Limiting Reactant
- The reactant that is completely used up first in a chemical reaction; it stops the reaction and determines the maximum theoretical amount of product formed. The other reactant is 'in excess'.
- Acid-alkali titration calculation method (4 steps)
- 1) Convert both volumes from cm³ to dm³ (÷ 1,000). 2) Calculate moles of the known solution: n = c × V. 3) Use the balanced equation mole ratio (e.g. 1 H₂SO₄ : 2 NaOH) to find moles of the unknown. 4) Calculate unknown concentration: c = n ÷ V.
Quick-Fire Flashcards (8)
Click any card to check the answer, then mark it as Known to track your progress.
What is the formula linking mass, moles, and Mr?
What is the value of Avogadro’s constant?
How do you convert cm3 to dm3?
Why is it impossible to get 100% yield?
What does a high atom economy mean?
How do you convert concentration in mol/dm³ into g/dm³?
Give three reasons why the percentage yield of a reaction is almost never 100%.
What volume does 0.25 moles of carbon dioxide gas occupy at room temperature and pressure (RTP)?
Exam-Style Questions & Mark Schemes (5)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Calculate the relative formula mass (Mr) of water (H2O). (Ar of H=1, O=16)
Click “Check Mark Scheme” to compare your answer, (2 x 1) + 16 = 18
- Q2
How many moles are there in 44g of carbon dioxide (CO2)? (Mr of CO2 = 44)
Click “Check Mark Scheme” to compare your answer, Moles = mass / Mr. Moles = 44 / 44 = 1 mole.
- Q3
A solution contains 10g of sodium chloride in 250 cm3 of water. Calculate its concentration in g/dm3.
Click “Check Mark Scheme” to compare your answer, Volume in dm3 = 250 / 1000 = 0.25 dm3. Concentration = 10 / 0.25 = 40 g/dm3.
- Q4
In a titration, 25.0 cm³ of 0.200 mol/dm³ sodium hydroxide (NaOH) was exactly neutralised by 20.0 cm³ of sulfuric acid (H₂SO₄). Given 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O, calculate the concentration of the sulfuric acid in mol/dm³. [4 marks]
Click “Check Mark Scheme” to compare your answer, 1) Volume of NaOH = 25.0 ÷ 1,000 = 0.0250 dm³. Moles of NaOH = c × V = 0.200 × 0.0250 = 0.00500 mol [1 mark]. 2) From the equation, 2 mol NaOH react with 1 mol H₂SO₄ (2:1 ratio), so moles of H₂SO₄ = 0.00500 ÷ 2 = 0.00250 mol [1 mark]. 3) Volume of H₂SO₄ = 20.0 ÷ 1,000 = 0.0200 dm³ [1 mark]. 4) Concentration of H₂SO₄ = n ÷ V = 0.00250 ÷ 0.0200 = 0.125 mol/dm³ [1 mark].
- Q5
Calculate the maximum mass of magnesium oxide (MgO, M_r = 40) that can be produced by burning 6.0 g of magnesium (A_r = 24) in excess oxygen (2Mg + O₂ → 2MgO). [3 marks]
Click “Check Mark Scheme” to compare your answer, 1) Moles of Mg = mass ÷ A_r = 6.0 ÷ 24 = 0.25 mol [1 mark]. 2) Mole ratio from 2Mg → 2MgO is 1:1, so moles of MgO formed = 0.25 mol [1 mark]. 3) Mass of MgO = moles × M_r = 0.25 × 40 = 10.0 g [1 mark].
Common Exam Questions (5)
What is the formula for the number of moles?
Use the equation moles = mass (g) ÷ relative formula mass (Mr). For example, 44 g of CO₂ (Mr = 44) contains 44 ÷ 44 = 1 mole.
How do you find mass from moles?
Rearrange the moles equation to mass (g) = moles × Mr. For example, 2 moles of water (Mr = 18) have a mass of 2 × 18 = 36 g.
What is relative formula mass (Mr)?
Relative formula mass (Mr) is the sum of the relative atomic masses (Ar) of all the atoms shown in a substance's chemical formula; for example, Mr of H₂O = (2 × 1) + 16 = 18.
What is a mole in chemistry?
One mole of any substance is the amount that contains Avogadro's constant (6.02 × 10²³) of particles (atoms, molecules, or ions), and has a mass in grams equal to the substance's relative atomic or formula mass.
How do you calculate concentration?
Concentration is calculated either in g/dm³ (mass of solute in g ÷ volume of solution in dm³) or in mol/dm³ (moles of solute ÷ volume of solution in dm³). Convert volumes from cm³ to dm³ by dividing by 1000.
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