Stoichiometry

Chemical Formulas

Chemical formulas show the number of each type of atom present in a molecule.

  • H2O: 2 hydrogen atoms are bonded to 1 oxygen atom
  • CO32-: 3 oxygen atoms are bonded to 1 carbon atom producing a compound with a charge of -2
  • Al2O3: an ionic compound with 2 aluminium atoms for every 3 oxygen atoms

Ionic Formulas

The ratio of atoms in an ionic compound depends on the valency of the atoms. Ions with different charges will not have a 1 to 1 ratio.

Al2O3

Aluminium ions have a charge of 3 while oxygen atoms have a charge of -2. The overall charge of an ionic compound must be equal to 0 (neutral).

Aluminium ions have a charge of 3 while oxygen atoms have a charge of -2. The overall charge of an ionic compound must be equal to 0 (neutral).

  1. Find the lowest common multiple for the charges ∣3×−2∣=6|3\times-2|=6
  2. Divide the multiple by the positive charge to get the number of metallic atoms ∣6÷3∣=2|6\div3|=2
  3. Divide the multiple by the negative charge to get the number of nonmetallic atoms ∣6÷−2∣=3|6\div-2|=3
  4. Confirm that the sum of the metal and nonmetal charges are nullified 3×2+−2×3=03\times2+-2\times3=0

The Mole

1 mole of a substance contains 6.02 × 1023 (Avogadro's Constant) molecules. 1 mole of a molecule equals the molecule's relative atomic mass in grams.

  • 1 mole of 12C = 12 grams of 12C
  • 1 mole of 1H = 1 gram of 1H
  • 1 mole of AX = A grams of AX (hypothetical example)

The relative molecular mass is the sum of the relative atomic masses.

Mr=∑ArM_r=\displaystyle\sum_{}^{}{A_r}

The units for relative molecular mass is grams per mole.

m=Mr×nm=M_r\times n

The mass (g) is equal to the relative atomic mass (g mol-1) times the number of moles (mol).

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The relative atomic mass of an atom can be approximated by doubling its atomic number. However, it is better to consult the periodic table before making any assumptions.

Molar Ratios

Balanced chemical equations have molar ratios. These ratios show how many reactant molecules are required per product molecule.

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The law of conservation of mass states that the mass of reactants = mass of products in any chemical reaction.

Finding the number of moles of 1 substance is enough to find the number of moles of other substances. This can be calculated from mass, volumes, and concentrations.

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Please note that X0X_0 and X1X_1 will be used to represent hypothetical molecules.

The function nn represents the total number of moles of an element while the function NN represents the molar ratio found in a chemcial equation. The ratio of moles is equal to the molar ratio in a chemical equation.

n(X0)n(X1)=N(X0)N(X1)\frac{n(X_0)}{n(X_1)}=\frac{N(X_0)}{N(X_1)}

For example, the reaction of magnesium with hydrochloric acid requires 2 hydrogen chloride molecules per magnesium atom.

Mg + 2HCl → MgCl2 + H2

This means multiplying the number of moles of chlorine by a half will return the number of moles of magnesium required to complete the reaction.

n(Mg)n(Cl2)=N(Mg)N(Cl2)n(Mg)n(Cl2)=12n(Mg)=n(Cl2)×12\frac{n(Mg)}{n(Cl_2)}=\frac{N(Mg)}{N(Cl_2)}\\\frac{n(Mg)}{n(Cl_2)}=\frac{1}{2}\\n(Mg)=n(Cl_2)\times\frac{1}{2}

Some questions require calculations involving masses only. The best way to approach these questions is to convert the mass into moles or use the mass ratio formula.

n(X0)=n(X1)×N(X0)N(X1)m(X0)=Mr(X0)×n(X0)n(X_0)=n(X_1)\times\frac{N(X_0)}{N(X_1)}\\m(X_0)=M_r(X_0)\times n(X_0)

The mass ratio is equal to the molecular mass ratio times the molar ratio.

m(X0)m(X1)=Mr(X0)Mr(X1)×N(X0)N(X1)\frac{m(X_0)}{m(X_1)}=\frac{M_r(X_0)}{M_r(X_1)}\times\frac{N(X_0)}{N(X_1)}

5 parameters are required to solve molar ratio and mass questions.

  • 2 Relative Atomic Masses: found in the periodic table
  • 2 Molecular Coefficients: given in the balanced equation beside the molecular formulas
  • 1 Known Mass: given in the question

It takes 2 carbon monoxide molecules and 1 oxygen molecule to produce 2 carbon dioxide molecules.

2CO + O2 → 2CO2

How many grams of oxygen is required to produce 88 grams of carbon dioxide?

  1. Calculate the relative molecular mass of the known chemical in grams per mole as 44 g mol−144 \textrm{ g mol}^{-1}
  2. Divide the known mass by the relative molecular mass to get the number of moles of CO2 as 2 mol2 \textrm{ mol}
  3. Multiply the number of moles by the number of O2 molecules per CO2 molecules to get the number of moles of O2 as 1 mol1 \textrm{ mol}
  4. Multiply by the relative molecular mass of the unknown chemical by the number of moles to get the mass as 32 g32 \textrm{ g}
Mr(CO2)=44 g mol−188 g44 g mol−1=2 mol2 mol×12=1 molMr(O2)=32 g mol−1m(O2)=32 g mol−1×1 molM_r(CO_2)=44\textrm{ g mol}^{-1}\\\frac{88\textrm{ g}}{44\textrm{ g mol}^{-1}}=2\textrm{ mol}\\2\textrm{ mol}\times\frac{1}{2}=1\textrm{ mol}\\M_r(O_2)=32\textrm{ g mol}^{-1}\\m(O_2)=32\textrm{ g mol}^{-1}\times 1\textrm{ mol}

Molar Volume

The volume of 1 mole of any gas is 24 dm3 at room temperature and pressure.

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The molar volume of any gas at (r.t.p.) is usually given in the exam along with a periodic table. The unit for this constant is decimetres cubed per mole.

Concentration

The concentration of a solution is measured in moles per decimetre cubed.

c=nVc=\frac{n}{V}

Concentration is directly proportional to the number of moles but inversely proportional to the volume of the solution.