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Stoichiometry - Relative masses

Grade 12A LevelChemistry

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 6.02×1023 mol−16.02 \times 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 Grade 12 Notes & Examples