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Chemistry: Stoichiometry - The Mole and Avogadro Constant

Grade 8IB

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

🔑Concepts

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The mole (symbol: molmol) is the SI unit used to measure the amount of a substance. One mole contains exactly 6.022×10236.022 \times 10^{23} elementary entities (atoms, molecules, or ions).

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Avogadro's Constant (NAN_A) is defined as 6.02×1023 mol−16.02 \times 10^{23} \text{ mol}^{-1}. It represents the number of particles in one mole of any substance.

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Molar Mass (MM) is the mass of one mole of a substance, expressed in grams per mole (g⋅mol−1g \cdot mol^{-1}). It is numerically equivalent to the relative atomic mass (ArA_r) or relative formula mass (MrM_r) of the substance.

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The relationship between the mass of a substance (mm), its molar mass (MM), and the number of moles (nn) allows for stoichiometric conversions in chemical reactions.

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Stoichiometry uses the quantitative relationships (ratios) from balanced chemical equations to calculate the amounts of reactants and products.

📐Formulae

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

N=n×NAN = n \times N_A

m=n×Mm = n \times M

NA≈6.02×1023 particles/molN_A \approx 6.02 \times 10^{23} \text{ particles/mol}

💡Examples

Problem 1:

Calculate the number of moles in 44 g44 \text{ g} of Carbon Dioxide (CO2CO_2). (Given ArA_r of C=12C = 12, O=16O = 16)

Solution:

M(CO2)=12+(2×16)=44 g⋅mol−1M(CO_2) = 12 + (2 \times 16) = 44 \text{ g} \cdot \text{mol}^{-1} n=mM=44 g44 g⋅mol−1=1 moln = \frac{m}{M} = \frac{44 \text{ g}}{44 \text{ g} \cdot \text{mol}^{-1}} = 1 \text{ mol}

Explanation:

First, find the molar mass of CO2CO_2 by summing the atomic masses. Then, divide the given mass by the molar mass to find the number of moles.

Problem 2:

How many atoms are contained in 0.5 moles0.5 \text{ moles} of pure Iron (FeFe)?

Solution:

N=n×NAN = n \times N_A N=0.5×(6.02×1023)=3.01×1023 atomsN = 0.5 \times (6.02 \times 10^{23}) = 3.01 \times 10^{23} \text{ atoms}

Explanation:

To find the number of particles (NN), multiply the number of moles by Avogadro's constant (6.02×10236.02 \times 10^{23}).

Problem 3:

What is the mass of 2.5 moles2.5 \text{ moles} of Water (H2OH_2O)? (Given ArA_r of H=1H = 1, O=16O = 16)

Solution:

M(H2O)=(2×1)+16=18 g⋅mol−1M(H_2O) = (2 \times 1) + 16 = 18 \text{ g} \cdot \text{mol}^{-1} m=n×M=2.5 mol×18 g⋅mol−1=45 gm = n \times M = 2.5 \text{ mol} \times 18 \text{ g} \cdot \text{mol}^{-1} = 45 \text{ g}

Explanation:

Calculate the molar mass of water (18 g⋅mol−118 \text{ g} \cdot \text{mol}^{-1}) and multiply it by the given number of moles to find the total mass.