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Stoichiometry - Relative masses

Grade 12IGCSEChemistry

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

🔑Concepts

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Relative Atomic Mass (ArA_r): The weighted average mass of naturally occurring atoms of an element on a scale where an atom of Carbon-12 (12C^{12}C) 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, this is referred to as Relative Formula Mass.

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The Mole: One mole is the amount of substance that contains the same number of particles as there are atoms in 12.0012.00 g of 12C^{12}C. This number is known as the Avogadro constant (LL or NAN_A), which is 6.02imes1023 mol−16.02 imes 10^{23} \text{ mol}^{-1}.

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Molar Mass (MM): The mass of one mole of a substance, expressed in g⋅mol−1g \cdot mol^{-1}. Numerically, it is equal to the ArA_r or MrM_r of the substance.

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Stoichiometric relationships: In a balanced chemical equation, the coefficients represent the molar ratio in which reactants combine and products form.

📐Formulae

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

N=n×LN = n \times L

Mr=∑(Atomic Mass×Number of Atoms)M_r = \sum (\text{Atomic Mass} \times \text{Number of Atoms})

% by mass of an element=n×ArMr×100%\% \text{ by mass of an element} = \frac{n \times A_r}{M_r} \times 100\%

💡Examples

Problem 1:

Calculate the number of moles in 22.022.0 g of Carbon Dioxide (CO2CO_2). Given ArA_r of C=12C = 12 and O=16O = 16.

Solution:

  1. Calculate Mr(CO2)M_r(CO_2): 12+(2×16)=44.0 g/mol12 + (2 \times 16) = 44.0 \text{ g/mol}.
  2. Use the formula n=mMn = \frac{m}{M}: n=22.044.0=0.50 moln = \frac{22.0}{44.0} = 0.50 \text{ mol}.

Explanation:

First, the relative molecular mass is determined by summing the atomic masses. Then, the mass is divided by this molar mass to find the number of moles.

Problem 2:

Calculate the mass of 3.01×10223.01 \times 10^{22} molecules of water (H2OH_2O). Given ArA_r of H=1H = 1, O=16O = 16, and L=6.02×1023 mol−1L = 6.02 \times 10^{23} \text{ mol}^{-1}.

Solution:

  1. Find the number of moles (nn): n=NL=3.01×10226.02×1023=0.05 moln = \frac{N}{L} = \frac{3.01 \times 10^{22}}{6.02 \times 10^{23}} = 0.05 \text{ mol}.
  2. Find Mr(H2O)M_r(H_2O): (2×1)+16=18 g/mol(2 \times 1) + 16 = 18 \text{ g/mol}.
  3. Find mass (mm): m=n×M=0.05×18=0.90 gm = n \times M = 0.05 \times 18 = 0.90 \text{ g}.

Explanation:

To find the mass from the number of particles, we first convert particles to moles using the Avogadro constant, and then convert moles to mass using the molar mass.

Relative masses - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry