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Stoichiometric Relationships - Introduction to the particulate nature of matter and chemical change

Grade 12IBChemistry

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

🔑Concepts

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Matter is classified into pure substances (elements and compounds) and mixtures. A pure substance has a constant composition, while a mixture contains two or more substances physically combined in varying proportions.

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The three states of matter are solid (ss), liquid (ll), and gas (gg). Phase changes occur at specific temperatures and pressures: melting (s→ls \to l), boiling (l→gl \to g), sublimation (s→gs \to g), freezing (l→sl \to s), condensation (g→lg \to l), and deposition (g→sg \to s).

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A homogeneous mixture (solution) has a uniform composition throughout, whereas a heterogeneous mixture has distinct phases or boundaries.

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Chemical equations must be balanced to satisfy the Law of Conservation of Mass, meaning the number of atoms of each element must be the same on both the reactant and product sides.

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Stoichiometric coefficients in a balanced equation represent the molar ratios of the reactants and products involved in the chemical change.

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Atom economy is a measure of the efficiency of a chemical reaction, calculated by the ratio of the mass of the desired product to the total mass of all reactants used.

📐Formulae

Percentage atom economy=molar mass of desired product∑molar masses of all reactants×100%\text{Percentage atom economy} = \frac{\text{molar mass of desired product}}{\sum \text{molar masses of all reactants}} \times 100\%

n=NLn = \frac{N}{L} where nn is the amount in moles, NN is the number of particles, and LL is Avogadro's constant (6.02×1023 mol−16.02 \times 10^{23} \text{ mol}^{-1})

Ar=weighted average mass of one atom of an element112 mass of one atom of 12CA_r = \frac{\text{weighted average mass of one atom of an element}}{\frac{1}{12} \text{ mass of one atom of }^{12}C}

💡Examples

Problem 1:

Balance the following chemical equation for the complete combustion of propane: C3H8(g)+O2(g)→CO2(g)+H2O(l)C_3H_8(g) + O_2(g) \rightarrow CO_2(g) + H_2O(l)

Solution:

C3H8(g)+5O2(g)→3CO2(g)+4H2O(l)C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)

Explanation:

To balance the carbon atoms, we place a coefficient of 33 before CO2CO_2. To balance the hydrogen atoms (88 on the left), we place a 44 before H2OH_2O. Finally, there are 3×2+4×1=103 \times 2 + 4 \times 1 = 10 oxygen atoms on the right, so we place a 55 before O2O_2 on the left.

Problem 2:

Calculate the percentage atom economy for the production of FeFe in the following reaction: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g). (Relative atomic masses: Fe=55.85,O=16.00,C=12.01Fe = 55.85, O = 16.00, C = 12.01)

Solution:

Atom Economy=2×55.85(2×55.85+3×16.00)+3×(12.01+16.00)×100%≈45.8%\text{Atom Economy} = \frac{2 \times 55.85}{(2 \times 55.85 + 3 \times 16.00) + 3 \times (12.01 + 16.00)} \times 100\% \approx 45.8\%

Explanation:

Atom economy is the molar mass of the desired product (2×Fe2 \times Fe) divided by the sum of the molar masses of all reactants (Fe2O3Fe_2O_3 and 3CO3CO). The total mass of reactants is 159.70+84.03=243.73 g/mol159.70 + 84.03 = 243.73 \text{ g/mol}. The mass of the desired product is 111.70 g/mol111.70 \text{ g/mol}. (111.70/243.73)×100=45.8%(111.70 / 243.73) \times 100 = 45.8\%.

Introduction to the particulate nature of matter and chemical change Revision - Grade 12 Chemistry…