5.1.2Additional lessons
Reacting Mass Calculations — IB Chemistry SL Revision Notes & Topical Questions
5 min read 6 parts Lesson 60 of 98
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Part 1 of 6The Mole Link
The mole links the mass of a substance to the number of particles in the equation.
Amount
- Amount The amount in moles equals mass in grams divided by molar mass in g mol⁻¹, written n = m/M.
Molar mass
- Molar mass Add the relative atomic masses of all atoms in the formula, using the values in the data booklet.
Mole ratio
- Mole ratio The coefficients of a balanced equation give the ratio in which amounts of substances react and form.
Mass ratio
- Mass ratio Reacting masses are not in the ratio of the coefficients, since different substances have different molar masses.
Part 2 of 6Method
1
Write the balanced equation for the reaction and identify the substance whose mass is given and the one whose mass is required.
2
Convert the given mass to moles using n = m/M.
3
Use the coefficients to find the moles of the required substance by ratio.
4
Convert these moles into a mass using m = n × M.
5
Round the answer to the same number of significant figures as the least precise data and include units.
Part 3 of 6Worked Pattern
Worked Pattern
Given
- Given Always the starting point, whether it is a reactant or a product.
Ratio
- Ratio The relation that makes the calculation specific to the equation, such as 1 : 1 or 2 : 3.
Required
- Required The quantity asked for, found by the last step of converting from moles to mass.
Check
- Check A product mass should not exceed the total mass of reactants, which follows from conservation of mass.
| Step | Formula |
|---|---|
| Moles from mass | n = m ÷ M |
| Moles by ratio | n(B) = n(A) × coefficient B ÷ coefficient A |
| Mass from moles | m = n × M |
Part 4 of 6Typical Contexts
Typical Contexts
Thermal decomposition
- Thermal decomposition Heating a carbonate gives an oxide and CO₂, and the loss in mass can be predicted from the equation.
Metals and acids
- Metals and acids The mass of metal that reacts can be used to find the mass of a salt formed.
Industrial scale
- Industrial scale Chemical manufacturers use these calculations to decide how much raw material to buy.
Experimental analysis
- Experimental analysis Measured masses can be compared with theoretical values to find yield and purity.
Part 5 of 6Worked Example
Question
- Calculate the mass of calcium oxide formed when 25.0 g of calcium carbonate is heated: CaCO₃(s) → CaO(s) + CO₂(g).
- (Mr CaCO₃ = 100.09, Mr CaO = 56.08)
- Amount of CaCO₃ = 25.0 ÷ 100.09 = 0.2498 mol.
- The ratio CaCO₃CaO is 1 : 1, so n(CaO) = 0.2498 mol.
- Mass of CaO = 0.2498 × 56.08 = 14.0 g.
Answer 14.0 g of calcium oxide
Part 6 of 6Exam Checklist & Tips
5.1.2 – Key Points for the Exam
- Molar mass The mass of one mole of a substance, in g mol⁻¹.
- Mole ratio The ratio of coefficients in a balanced equation.
- Amount of substance The quantity measured in moles.
- Theoretical mass The mass predicted from the balanced equation, assuming complete reaction.
Common Mistakes
- Using masses directly in the ratio of the equation rather than converting to moles first.
- Using an unbalanced equation to get the mole ratio.
- Using the wrong molar mass because a subscript or bracket was ignored.
- Giving the answer to too many significant figures.
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