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Chemistry: Stoichiometry - Relative Atomic Mass and Molecular Mass

Grade 8IB

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

🔑Concepts

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Relative Atomic Mass (ArA_r) is the weighted average mass of an atom of an element relative to 112\frac{1}{12} of the mass of an atom of Carbon-12 (12C^{12}C).

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Carbon-12 is used as the standard reference because its mass is exactly 1212 atomic mass units (uu).

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Relative Atomic Mass is a ratio and therefore has no units.

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Relative Molecular Mass (MrM_r) is the sum of the Relative Atomic Masses of all the atoms present in a molecule.

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For ionic compounds, the term 'Relative Formula Mass' is used instead of 'Molecular Mass', but the calculation method remains the same.

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The Law of Conservation of Mass states that the total mass of reactants must equal the total mass of products in a chemical reaction, which is verified using MrM_r values.

📐Formulae

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

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

💡Examples

Problem 1:

Calculate the Relative Molecular Mass (MrM_r) of water (H2OH_2O). Given: ArA_r of H=1H = 1 and ArA_r of O=16O = 16.

Solution:

Mr(H2O)=(2×Ar(H))+(1×Ar(O))M_r(H_2O) = (2 \times A_r(H)) + (1 \times A_r(O)) Mr(H2O)=(2×1)+(1×16)=2+16=18M_r(H_2O) = (2 \times 1) + (1 \times 16) = 2 + 16 = 18

Explanation:

The molecule of water contains 22 atoms of Hydrogen and 11 atom of Oxygen. We multiply the number of atoms by their respective Relative Atomic Masses and sum them up.

Problem 2:

Find the Relative Formula Mass of Sulfuric Acid (H2SO4H_2SO_4). Given: Ar(H)=1A_r(H) = 1, Ar(S)=32A_r(S) = 32, Ar(O)=16A_r(O) = 16.

Solution:

Mr(H2SO4)=(2×1)+(1×32)+(4×16)M_r(H_2SO_4) = (2 \times 1) + (1 \times 32) + (4 \times 16) Mr(H2SO4)=2+32+64=98M_r(H_2SO_4) = 2 + 32 + 64 = 98

Explanation:

To calculate the formula mass, we identify the quantity of each element: 22 Hydrogens, 11 Sulfur, and 44 Oxygens, then add their total atomic masses.

Problem 3:

Calculate the Relative Formula Mass of Calcium Carbonate (CaCO3CaCO_3). Given: Ar(Ca)=40A_r(Ca) = 40, Ar(C)=12A_r(C) = 12, Ar(O)=16A_r(O) = 16.

Solution:

Mr(CaCO3)=(1×40)+(1×12)+(3×16)M_r(CaCO_3) = (1 \times 40) + (1 \times 12) + (3 \times 16) Mr(CaCO3)=40+12+48=100M_r(CaCO_3) = 40 + 12 + 48 = 100

Explanation:

Summing the atomic masses: 1 atom of Ca1 \text{ atom of } Ca (4040), 1 atom of C1 \text{ atom of } C (1212), and 3 atoms of O3 \text{ atoms of } O (3×16=483 \times 16 = 48) gives a total of 100100.