krit.club logo

Stoichiometric Relationships - The mole concept

Grade 11IBChemistry

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

🔑Concepts

•

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

•

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 12C^{12}C. It has no units.

•

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

•

The empirical formula of a compound is the simplest whole-number ratio of the atoms of each element present in a compound.

•

The molecular formula is the actual number of atoms of each element in a molecule. It is a whole-number multiple of the empirical formula: Molecular Formula=n×(Empirical Formula)\text{Molecular Formula} = n \times (\text{Empirical Formula}).

•

Stoichiometry involves using the coefficients in a balanced chemical equation to determine the molar ratios of reactants and products.

📐Formulae

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

N=n×LN = n \times L

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

%by mass of element=Ar×number of atomsMr(compound)×100%\% \text{by mass of element} = \frac{A_r \times \text{number of atoms}}{M_r (\text{compound})} \times 100\%

💡Examples

Problem 1:

Calculate the number of moles of CO2CO_2 present in 22.0 g22.0 \, g of the gas.

Solution:

First, calculate the molar mass (MM) of CO2CO_2 using relative atomic masses (C=12.01,O=16.00C = 12.01, O = 16.00): M(CO2)=12.01+(2×16.00)=44.01 g⋅mol−1M(CO_2) = 12.01 + (2 \times 16.00) = 44.01 \, g \cdot mol^{-1}. Using the formula n=mMn = \frac{m}{M}: n=22.0 g44.01 g⋅mol−1≈0.500 moln = \frac{22.0 \, g}{44.01 \, g \cdot mol^{-1}} \approx 0.500 \, mol.

Explanation:

The mass of the sample is divided by the molar mass to determine the amount in moles. Ensure units are consistent (gg and g⋅mol−1g \cdot mol^{-1}).

Problem 2:

A compound contains 40.00%40.00\% CC, 6.72%6.72\% HH, and 53.28%53.28\% OO by mass. Determine its empirical formula.

Solution:

  1. Assume 100 g100 \, g of substance to convert percentages to masses: m(C)=40.00 gm(C) = 40.00 \, g, m(H)=6.72 gm(H) = 6.72 \, g, m(O)=53.28 gm(O) = 53.28 \, g.
  2. Calculate moles for each: n(C)=40.0012.01=3.33 moln(C) = \frac{40.00}{12.01} = 3.33 \, mol n(H)=6.721.01=6.65 moln(H) = \frac{6.72}{1.01} = 6.65 \, mol n(O)=53.2816.00=3.33 moln(O) = \frac{53.28}{16.00} = 3.33 \, 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: CH2OCH_2O

Explanation:

To find the empirical formula, determine the molar ratio of the elements. If the results are not whole numbers, multiply all ratios by a common factor.

The mole concept - Revision Notes & Key Formulas | IB Grade 11 Chemistry