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Structure 1. Models of the particulate nature of matter - Counting particles by mass: The mole

Grade 11IBChemistry

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 contains exactly 6.02214076×10236.02214076 \times 10^{23} elementary entities (atoms, molecules, ions, etc.). This number is known as the Avogadro constant (LL or NAN_A).

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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. It is a dimensionless quantity.

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Relative molecular mass (MrM_r) is the sum of the relative atomic masses of the atoms in a molecule. Relative formula mass is used for ionic compounds.

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Molar mass (MM) is the mass of one mole of a substance and is expressed in grams per mole (g mol−1g \text{ mol}^{-1}). The numerical value of MM is equal to the MrM_r of the substance.

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The relationship between mass (mm), molar mass (MM), and amount in moles (nn) is given by the ratio of mass to molar mass.

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The number of particles (NN) can be determined by multiplying the amount of substance in moles (nn) by Avogadro's constant (LL).

📐Formulae

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

N=n×LN = n \times L

L=6.02×1023 mol−1L = 6.02 \times 10^{23} \text{ mol}^{-1}

Mr=∑ArM_r = \sum A_r

💡Examples

Problem 1:

Calculate the amount in moles of 4.50 g4.50 \text{ g} of water (H2OH_2O).

Solution:

  1. Find the molar mass of H2OH_2O: M(H2O)=(2×1.01)+16.00=18.02 g mol−1M(H_2O) = (2 \times 1.01) + 16.00 = 18.02 \text{ g mol}^{-1}
  2. Use the formula n=mMn = \frac{m}{M}: n=4.50 g18.02 g mol−1≈0.250 moln = \frac{4.50 \text{ g}}{18.02 \text{ g mol}^{-1}} \approx 0.250 \text{ mol}

Explanation:

First, calculate the molar mass by summing the relative atomic masses of Hydrogen and Oxygen from the periodic table. Then, divide the given mass by this molar mass to find the number of moles.

Problem 2:

Determine the number of molecules present in 0.20 mol0.20 \text{ mol} of glucose (C6H12O6C_6H_{12}O_6).

Solution:

Use the formula N=n×LN = n \times L: N=0.20 mol×6.02×1023 mol−1N = 0.20 \text{ mol} \times 6.02 \times 10^{23} \text{ mol}^{-1} N=1.204×1023 moleculesN = 1.204 \times 10^{23} \text{ molecules}

Explanation:

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

Problem 3:

A sample of CalciumCarbonateCalcium Carbonate (CaCO3CaCO_3) contains 3.01×10223.01 \times 10^{22} atoms of oxygen. Calculate the mass of the sample.

Solution:

  1. Determine the moles of Oxygen atoms: n(O)=3.01×10226.02×1023=0.050 moln(O) = \frac{3.01 \times 10^{22}}{6.02 \times 10^{23}} = 0.050 \text{ mol}
  2. Determine the moles of CaCO3CaCO_3: Since there are 3 oxygen atoms per formula unit, n(CaCO3)=0.0503≈0.0167 moln(CaCO_3) = \frac{0.050}{3} \approx 0.0167 \text{ mol}
  3. Calculate molar mass of CaCO3CaCO_3: M=40.08+12.01+(3×16.00)=100.09 g mol−1M = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \text{ g mol}^{-1}
  4. Calculate mass: m=n×M=0.0167 mol×100.09 g mol−1≈1.67 gm = n \times M = 0.0167 \text{ mol} \times 100.09 \text{ g mol}^{-1} \approx 1.67 \text{ g}

Explanation:

First, convert the number of oxygen atoms to moles. Since each molecule/formula unit of CaCO3CaCO_3 contains 3 oxygen atoms, divide the moles of oxygen by 3 to find the moles of the compound. Finally, multiply by the molar mass of the compound to get the mass.