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Quantitative Chemistry - Relative Formula Mass and Percentage Composition

Grade 9IB

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 an atom of an element compared to 112\frac{1}{12} of the mass of an atom of Carbon-12.

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Relative Formula Mass (MrM_r) is the sum of the relative atomic masses of all the atoms present in the chemical formula of a substance.

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The MrM_r is a dimensionless quantity (it has no units) because it is a ratio relative to a standard mass.

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In a balanced chemical equation, the total MrM_r of the reactants is always equal to the total MrM_r of the products, following the Law of Conservation of Mass.

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Percentage Composition by mass identifies what proportion of the total mass of a compound is made up of a specific element.

📐Formulae

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

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

💡Examples

Problem 1:

Calculate the Relative Formula Mass (MrM_r) of Magnesium Hydroxide, Mg(OH)2Mg(OH)_2. Given Ar:Mg=24,O=16,H=1A_r: Mg = 24, O = 16, H = 1.

Solution:

Mr=24+2×(16+1)M_r = 24 + 2 \times (16 + 1) Mr=24+2×(17)M_r = 24 + 2 \times (17) Mr=24+34=58M_r = 24 + 34 = 58

Explanation:

The formula contains one Magnesium atom and two 'OH' groups. We sum the ArA_r of Magnesium with twice the sum of Oxygen and Hydrogen.

Problem 2:

Calculate the percentage by mass of Nitrogen in Ammonium Nitrate, NH4NO2NH_4NO_2. Given Ar:N=14,H=1,O=16A_r: N = 14, H = 1, O = 16.

Solution:

Mr=(2×14)+(4×1)+(2×16)M_r = (2 \times 14) + (4 \times 1) + (2 \times 16) Mr=28+4+32=64M_r = 28 + 4 + 32 = 64 Total mass of Nitrogen=2×14=28\text{Total mass of Nitrogen} = 2 \times 14 = 28 % Nitrogen=2864×100\% \text{ Nitrogen} = \frac{28}{64} \times 100 % Nitrogen=43.75%\% \text{ Nitrogen} = 43.75\%

Explanation:

First, find the total MrM_r of the compound. Then, identify the total mass contributed by Nitrogen atoms (there are two in the formula). Divide the Nitrogen mass by the total MrM_r and multiply by 100.

Problem 3:

Calculate the percentage of water of crystallization in hydrated Copper(II) Sulfate, CuSO4⋅5H2OCuSO_4 \cdot 5H_2O. Given Ar:Cu=63.5,S=32,O=16,H=1A_r: Cu = 63.5, S = 32, O = 16, H = 1.

Solution:

Mr(CuSO4)=63.5+32+(4×16)=159.5M_r(CuSO_4) = 63.5 + 32 + (4 \times 16) = 159.5 Mr(5H2O)=5×((2×1)+16)=5×18=90M_r(5H_2O) = 5 \times ((2 \times 1) + 16) = 5 \times 18 = 90 Total Mr=159.5+90=249.5\text{Total } M_r = 159.5 + 90 = 249.5 % Water=90249.5×100≈36.07%\% \text{ Water} = \frac{90}{249.5} \times 100 \approx 36.07\%

Explanation:

For hydrated salts, treat the water molecules (H2OH_2O) as a single unit or sum all atoms. The total MrM_r includes the salt and the five water molecules.