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Stoichiometric Relationships - The mole concept

Grade 12IBChemistry

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

🔑Concepts

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The mole (nn) is the SI unit for the amount of substance. One mole contains exactly 6.02imes10236.02 imes 10^{23} elementary entities, a value known as the Avogadro constant (NAN_A or LL).

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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 a carbon-12 atom. It is dimensionless.

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Molar mass (MM) is the mass of one mole of a substance, expressed in units of g⋅mol−1g \cdot mol^{-1}.

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The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula is the actual number of atoms of each element in a molecule and is a multiple of the empirical formula.

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Stoichiometry involves using the quantitative relationships (mole ratios) from a balanced chemical equation to calculate the amounts of reactants and products.

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The limiting reactant is the substance that is entirely consumed in a reaction, determining the maximum amount of product that can be formed (theoretical yield).

📐Formulae

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

N=n×NAN = n \times N_A

n=VVmn = \frac{V}{V_m}

PV=nRTPV = nRT

c=nVc = \frac{n}{V}

Percentage Yield=Experimental YieldTheoretical Yield×100%\text{Percentage Yield} = \frac{\text{Experimental Yield}}{\text{Theoretical Yield}} \times 100\%

Percentage Atom Economy=Molar mass of desired productSum of molar masses of all reactants×100%\text{Percentage Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Sum of molar masses of all reactants}} \times 100\%

💡Examples

Problem 1:

Calculate the number of moles in 25.0 g25.0\text{ g} of calcium carbonate (CaCO3CaCO_3).

Solution:

  1. Calculate the molar mass (MM) 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}.
  2. Use the formula n=mMn = \frac{m}{M}: n=25.0 g100.09 g mol−1≈0.250 moln = \frac{25.0\text{ g}}{100.09\text{ g mol}^{-1}} \approx 0.250\text{ mol}.

Explanation:

To find the amount in moles, we divide the given mass by the sum of the relative atomic masses of all atoms in the chemical formula.

Problem 2:

A compound consists of 40.00% C40.00\%\text{ C}, 6.72% H6.72\%\text{ H}, and 53.28% O53.28\%\text{ O} by mass. Determine its empirical formula.

Solution:

  1. Assume 100 g100\text{ g} of the substance: mC=40.00 g,mH=6.72 g,mO=53.28 gm_C = 40.00\text{ g}, m_H = 6.72\text{ g}, m_O = 53.28\text{ g}.
  2. Convert mass to moles: nC=40.0012.01=3.33 moln_C = \frac{40.00}{12.01} = 3.33\text{ mol} nH=6.721.01=6.65 moln_H = \frac{6.72}{1.01} = 6.65\text{ mol} nO=53.2816.00=3.33 moln_O = \frac{53.28}{16.00} = 3.33\text{ mol}
  3. Divide by the smallest value (3.333.33): C:3.333.33=1C: \frac{3.33}{3.33} = 1; H:6.653.33≈2H: \frac{6.65}{3.33} \approx 2; O:3.333.33=1O: \frac{3.33}{3.33} = 1. Empirical formula is CH2OCH_2O.

Explanation:

The empirical formula is found by calculating the mole ratio of the constituent elements and simplifying it to the smallest whole-number ratio.

Problem 3:

Determine the volume of 0.500 mol0.500\text{ mol} of an ideal gas at STP (273 K273\text{ K} and 100 kPa100\text{ kPa}), given the molar volume Vm=22.7 dm3 mol−1V_m = 22.7\text{ dm}^3\text{ mol}^{-1}.

Solution:

Use the formula V=n×VmV = n \times V_m: V=0.500 mol×22.7 dm3 mol−1=11.35 dm3V = 0.500\text{ mol} \times 22.7\text{ dm}^3\text{ mol}^{-1} = 11.35\text{ dm}^3.

Explanation:

According to Avogadro's Law, one mole of any gas at STP occupies the same volume (22.7 dm322.7\text{ dm}^3 under IB standard conditions).

The mole concept - Revision Notes & Key Formulas | IB Grade 12 Chemistry