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Quantitative Chemistry - The Mole Concept and Avogadro Constant

Grade 9IB

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

🔑Concepts

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The mole (symbol nn) is the SI unit for the amount of substance. One mole of any substance contains exactly 6.02times10236.02 \\times 10^{23} elementary entities (atoms, molecules, or ions).

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Avogadro's Constant (LL or NAN_A) is the proportionality factor that relates the number of particles in a sample to the amount of substance in moles. L=6.02times1023textmol−1L = 6.02 \\times 10^{23} \\text{ mol}^{-1}.

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Relative Atomic Mass (ArA_r) is the weighted average mass of an atom of an element compared to frac112\\frac{1}{12} of the mass of a carbon-12 atom.

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Molar Mass (MM) is the mass of one mole of a substance, expressed in textgmol−1\\text{g mol}^{-1}. It is numerically equal to the relative molecular mass (MrM_r) or relative formula mass of the substance.

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The relationship between mass, molar mass, and moles allows for stoichiometric calculations in chemical reactions using the formula n=fracmMn = \\frac{m}{M}.

📐Formulae

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

N=ntimesLN = n \\times L

m=ntimesMm = n \\times M

Mr=sumArM_r = \\sum A_r

💡Examples

Problem 1:

Calculate the number of moles in 250textg250 \\text{ g} of Calcium Carbonate (CaCO3CaCO_3). (Given ArA_r values: Ca=40,C=12,O=16Ca = 40, C = 12, O = 16)

Solution:

  1. Calculate the Molar Mass (MM) of CaCO3CaCO_3 by summing the relative atomic masses: beginarrayr4012+48hline100endarray\\begin{array}{r} 40 \\\\ 12 \\\\ + 48 \\\\ \\hline 100 \\end{array} M=100textgmol−1M = 100 \\text{ g mol}^{-1}

  2. Use the mole calculation formula: n=fracmMn = \\frac{m}{M} n=frac250100n = \\frac{250}{100} n=2.5textmoln = 2.5 \\text{ mol}

Explanation:

First, the molar mass is calculated by adding the mass of one Calcium atom (4040), one Carbon atom (1212), and three Oxygen atoms (3times16=483 \\times 16 = 48). Then, the total mass (250textg250\\text{g}) is divided by this molar mass to find the number of moles.

Problem 2:

How many molecules are present in 0.5textmoles0.5 \\text{ moles} of Carbon Dioxide (CO2CO_2)?

Solution:

  1. Identify the given values: n=0.5textmoln = 0.5 \\text{ mol} L=6.02times1023textmol−1L = 6.02 \\times 10^{23} \\text{ mol}^{-1}

  2. Use the particle number formula: N=ntimesLN = n \\times L N=0.5times6.02times1023N = 0.5 \\times 6.02 \\times 10^{23} N=3.01times1023textmoleculesN = 3.01 \\times 10^{23} \\text{ molecules}

Explanation:

To find the total number of molecules (NN), multiply the amount of substance in moles (nn) by Avogadro's constant (LL).

Problem 3:

Find the mass in grams of 1.204times10241.204 \\times 10^{24} atoms of pure Carbon (CC). (Given ArA_r of C=12C = 12)

Solution:

  1. First, find the number of moles (nn) using Avogadro's constant: n=fracNLn = \\frac{N}{L} n=frac1.204times10246.02times1023=2textmoln = \\frac{1.204 \\times 10^{24}}{6.02 \\times 10^{23}} = 2 \\text{ mol}

  2. Next, calculate the mass (mm) using the molar mass (M=12textgmol−1M = 12 \\text{ g mol}^{-1}): m=ntimesMm = n \\times M m=2times12m = 2 \\times 12 m=24textgm = 24 \\text{ g}

Explanation:

The number of particles is converted to moles by dividing by Avogadro's constant. This mole value is then multiplied by the molar mass of Carbon to find the mass in grams.