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Stoichiometry - Relative atomic mass and relative molecular mass

Grade 11IGCSEChemistry

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

🔑Concepts

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

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The standard for relative mass is the Carbon-12 isotope (12C^{12}C), chosen because it is stable and its mass can be measured accurately.

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Relative Molecular Mass (MrM_r) is the sum of the relative atomic masses of the atoms in a molecule (used for covalent substances).

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Relative Formula Mass (also denoted as MrM_r) is the sum of the relative atomic masses of the atoms in the formula unit of an ionic compound.

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Both ArA_r and MrM_r are dimensionless quantities (they have no units) because they are ratios relative to the mass of 12C^{12}C.

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To calculate the ArA_r of an element with multiple isotopes, you must account for the relative abundance of each isotope.

📐Formulae

Ar=∑(isotopic mass×percentage abundance)100A_r = \frac{\sum (\text{isotopic mass} \times \text{percentage abundance})}{100}

Mr=∑(number of atoms×Ar of that element)M_r = \sum (\text{number of atoms} \times A_r \text{ of that element})

Ar=Average mass of one atom of the element112×mass of one atom of 12CA_r = \frac{\text{Average mass of one atom of the element}}{\frac{1}{12} \times \text{mass of one atom of }^{12}C}

💡Examples

Problem 1:

Calculate the Relative Atomic Mass (ArA_r) of Chlorine, given that it consists of 75%75\% 35Cl^{35}Cl and 25%25\% 37Cl^{37}Cl.

Solution:

Ar=(35×75)+(37×25)100=2625+925100=35.5A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = \frac{2625 + 925}{100} = 35.5

Explanation:

The average mass is weighted by the natural abundance of each isotope. Chlorine-35 is more abundant, so the final ArA_r value is closer to 35 than to 37.

Problem 2:

Calculate the Relative Molecular Mass (MrM_r) of sulfuric acid, H2SO4H_2SO_4. (Given: ArA_r of H=1,S=32,O=16H=1, S=32, O=16)

Solution:

Mr=(2×1)+(1×32)+(4×16)M_r = (2 \times 1) + (1 \times 32) + (4 \times 16) Mr=2+32+64=98M_r = 2 + 32 + 64 = 98

Explanation:

To find MrM_r, multiply the number of atoms of each element by their respective ArA_r and sum the results.

Problem 3:

Calculate the Relative Formula Mass (MrM_r) of Calcium Hydroxide, Ca(OH)2Ca(OH)_2. (Given: ArA_r of Ca=40,O=16,H=1Ca=40, O=16, H=1)

Solution:

Mr=40+[2×(16+1)]M_r = 40 + [2 \times (16 + 1)] Mr=40+(2×17)=40+34=74M_r = 40 + (2 \times 17) = 40 + 34 = 74

Explanation:

The subscript 22 outside the brackets applies to everything inside the brackets (one oxygen and one hydrogen atom).

Relative atomic mass and relative molecular mass Revision - Grade 11 Chemistry IGCSE