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Chemistry - Stoichiometry (Formulae and the Mole concept)

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Relative Atomic Mass (ArA_r): The average mass of naturally occurring isotopes of an element on a scale where the 12C^{12}C atom has a mass of exactly 1212 units.

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Relative Molecular Mass (MrM_r): The sum of the relative atomic masses of the elements in a molecule. For ionic compounds, the term Relative Formula Mass is used.

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The Mole: The amount of substance containing the same number of particles as there are atoms in exactly 12g12g of 12C^{12}C. This number is the Avogadro constant, approximately 6.02×10236.02 \times 10^{23} particles per mole.

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Molar Mass (MM): The mass of one mole of a substance, expressed in g/molg/mol. It is numerically equal to the ArA_r or MrM_r.

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Molar Volume: One mole of any gas at Room Temperature and Pressure (RTP) occupies a volume of 24 dm324\ dm^3 (or 24000 cm324000\ cm^3).

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Empirical Formula: The simplest whole-number ratio of atoms of each element present in a compound.

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Molecular Formula: The actual number of atoms of each element in one molecule of a substance. It is a multiple of the empirical formula: Molecular Formula=n×Empirical Formula\text{Molecular Formula} = n \times \text{Empirical Formula}.

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Concentration: The amount of solute (in moles or grams) dissolved in a specific volume of solvent (usually 1 dm31\ dm^3).

📐Formulae

n=mMn = \frac{m}{M}

n=V24 (for gas at RTP in dm3)n = \frac{V}{24} \text{ (for gas at RTP in } dm^3)

c=nV (where V is in dm3)c = \frac{n}{V} \text{ (where } V \text{ is in } dm^3)

% mass of element=number of atoms×ArMr×100%\% \text{ mass of element} = \frac{\text{number of atoms} \times A_r}{M_r} \times 100\%

n=number of particles6.02×1023n = \frac{\text{number of particles}}{6.02 \times 10^{23}}

💡Examples

Problem 1:

Calculate the number of moles in 88g88g of carbon dioxide (CO2CO_2). Given ArA_r of C=12C = 12 and O=16O = 16.

Solution:

  1. Calculate MrM_r of CO2CO_2: 12+(2×16)=4412 + (2 \times 16) = 44.
  2. Use formula n=mMn = \frac{m}{M}: n=88g44g/mol=2.0 moln = \frac{88g}{44g/mol} = 2.0\ mol.

Explanation:

First, find the molar mass by summing the atomic masses, then divide the given mass by this molar mass to find the number of moles.

Problem 2:

A compound contains 40%40\% Calcium, 12%12\% Carbon, and 48%48\% Oxygen by mass. Determine its empirical formula. (ArA_r: Ca=40,C=12,O=16Ca=40, C=12, O=16)

Solution:

  1. Moles of CaCa: 4040=1\frac{40}{40} = 1.
  2. Moles of CC: 1212=1\frac{12}{12} = 1.
  3. Moles of OO: 4816=3\frac{48}{16} = 3. Ratio is 1:1:31:1:3. Empirical formula is CaCO3CaCO_3.

Explanation:

Divide the percentage of each element by its relative atomic mass to find the molar ratio, then simplify to the smallest whole numbers.

Problem 3:

Calculate the volume occupied by 0.5 mol0.5\ mol of O2O_2 gas at RTP.

Solution:

  1. Use the formula V=n×24V = n \times 24.
  2. V=0.5×24=12 dm3V = 0.5 \times 24 = 12\ dm^3.

Explanation:

At RTP, one mole of any gas occupies 24 dm324\ dm^3. Multiplying the number of moles by this constant gives the total volume.