Reacting masses and percentage compositionIB MYP Chemistry: Revision notes
Section 1
Using a balanced equation to find masses
A balanced equation gives the mole ratio of the substances. To find an unknown mass:
- Write the balanced equation.
- Convert the known mass to moles (mass ÷ Mr).
- Use the mole ratio from the equation to find the moles of the substance you want.
- Convert to mass (moles × Mr).
Worked example. What mass of water is made from 4.0 g of hydrogen? 2H₂ + O₂ → 2H₂O
Moles of H₂ = 4.0 ÷ 2 = 2.0 mol. The ratio H₂ : H₂O is 2 : 2 = 1 : 1, so 2.0 mol of H₂O forms. Mass = 2.0 × 18 = 36 g.
Comparing masses directly with the equation numbers. The numbers in front of formulae give a ratio of moles, not of grams.
Use the numbers in front of the formulae only for the ratio step, after you have converted to moles.
Section 2
Percentage by mass of an element
The percentage by mass of an element in a compound shows how much of the mass of the compound is that element.
percentage = (number of atoms × Ar ÷ Mr) × 100
Worked example. Iron(III) oxide, Fe₂O₃ (Fe = 56, O = 16). Mr = 2 × 56 + 3 × 16 = 160.
Percentage of iron = (112 ÷ 160) × 100 = 70%. Percentage of oxygen = (48 ÷ 160) × 100 = 30%.
The percentages of all the elements in a compound add up to 100%.
Using the Ar of one atom when the formula has several, such as using 56 instead of 112 for the iron in Fe₂O₃.
Section 3
Limiting reactants
In many reactions one reactant is used up before the other. The reactant that runs out first is the limiting reactant: it decides how much product can be made. The other reactant is in excess, so some is left over.
Analogy: to make sandwiches you need 2 slices of bread and 1 slice of cheese. With 10 slices of bread and 8 slices of cheese you can make only 5 sandwiches, so bread is limiting.
To find the limiting reactant:
- Convert the mass of each reactant to moles.
- Compare with the ratio in the equation.
- The reactant that gives the smaller amount of product is limiting.
- Use the limiting reactant to calculate the mass of product.
Worked example. 2Mg + O₂ → 2MgO. 12 g Mg (0.50 mol) reacts with 4.0 g O₂ (0.125 mol). 0.50 mol Mg needs 0.25 mol O₂, but only 0.125 mol is present, so oxygen is limiting. Moles of MgO = 2 × 0.125 = 0.25 mol, mass = 0.25 × 40 = 10 g.
Always divide by the equation number as well when comparing, for example 2Mg needs only half as many moles of O₂.
Must Know
- Convert to moles, use the mole ratio, convert back to mass
- Percentage by mass = (atoms × Ar ÷ Mr) × 100
- The limiting reactant runs out first and decides the amount of product
- The reactant in excess is left over
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Reacting masses and percentage composition
- A flare manufacturer in South Korea burns magnesium ribbon in oxygen to make a bright white light. The equation for the reaction is 2Mg + O₂ → 2MgO. Relative atomic masses: Mg = 24, O = 16.Calculate the percentage by mass of oxygen in magnesium oxide.2 marks
- A lime works in Morocco heats limestone, calcium carbonate, in a kiln: CaCO₃ → CaO + CO₂. A technician tests a 200 g sample of pure calcium carbonate. Relative atomic masses: Ca = 40, C = 12, O = 16.Calculate the mass of carbon dioxide made from the 200 g sample.2 marks
- A student heats different masses of copper powder strongly in an open crucible so that the copper reacts with oxygen from the air to form black copper(II) oxide: 2Cu + O₂ → 2CuO. She weighs the copper before heating and the copper oxide after cooling. The air supply is always more than enough. Mass of copper and mass of copper oxide formed: 0.64 g and 0.80 g; 1.28 g and 1.60 g; 1.92 g and 2.40 g; 2.56 g and 3.00 g. Relative atomic masses: Cu = 64, O = 16.State a testable hypothesis for this investigation, give a scientific reason for it, and state one control variable.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).