krit.club logo

Chemistry: Stoichiometry - Chemical Formulae, Word and Symbol Equations, and Equation Balancing

Grade 8IB

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

🔑Concepts

•

Chemical Formulae: A representation of a substance using chemical symbols and numerical subscripts. The subscript indicates the number of atoms of that element in one molecule or formula unit. For example, in H2SO4H_{2}SO_{4}, there are 2 atoms of Hydrogen, 1 atom of Sulfur, and 4 atoms of Oxygen.

•

Word Equations: A way of describing a chemical reaction using the names of the reactants and products. Example: Methane+Oxygen→Carbon dioxide+Water\text{Methane} + \text{Oxygen} \rightarrow \text{Carbon dioxide} + \text{Water}.

•

Symbol Equations: These use chemical formulae to represent reactions. They provide more detail than word equations by showing the ratio of atoms involved.

•

Law of Conservation of Mass: In a chemical reaction, matter is neither created nor destroyed. Therefore, the total mass of the reactants must equal the total mass of the products: ∑Massreactants=∑Massproducts\sum \text{Mass}_{reactants} = \sum \text{Mass}_{products}

•

Balancing Equations: To satisfy the Law of Conservation of Mass, we add coefficients (numbers in front of formulae) so that the number of atoms for each element is the same on both the reactant and product sides.

•

State Symbols: These indicate the physical state of the substances: (s)(s) for solid, (l)(l) for liquid, (g)(g) for gas, and (aq)(aq) for aqueous (dissolved in water).

📐Formulae

Reactants→Products\text{Reactants} \rightarrow \text{Products}

Number of Atoms=Coefficient×Subscript\text{Number of Atoms} = \text{Coefficient} \times \text{Subscript}

2H2(g)+O2(g)→2H2O(l)2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)

Mr=∑Ar (Sum of relative atomic masses)M_{r} = \sum A_{r} \text{ (Sum of relative atomic masses)}

💡Examples

Problem 1:

Balance the following chemical equation: Fe+O2→Fe2O3Fe + O_{2} \rightarrow Fe_{2}O_{3}

Solution:

4Fe+3O2→2Fe2O34Fe + 3O_{2} \rightarrow 2Fe_{2}O_{3}

Explanation:

Start by balancing the Oxygen atoms. Since there are 2 on the left and 3 on the right, the lowest common multiple is 6. Place a 3 in front of O2O_{2} and a 2 in front of Fe2O3Fe_{2}O_{3}. This gives 4 Iron (FeFe) atoms on the right, so place a 4 in front of FeFe on the left.

Problem 2:

Convert the following word equation into a balanced symbol equation: Magnesium+Hydrochloric acid→Magnesium chloride+Hydrogen gas\text{Magnesium} + \text{Hydrochloric acid} \rightarrow \text{Magnesium chloride} + \text{Hydrogen gas}

Solution:

Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)Mg(s) + 2HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)

Explanation:

Magnesium is MgMg, Hydrochloric acid is HClHCl. Magnesium chloride is formed by Mg2+Mg^{2+} and Cl−Cl^{-} ions, giving MgCl2MgCl_{2}. Hydrogen gas is diatomic (H2H_{2}). To balance the 2 Chlorine and 2 Hydrogen atoms on the right, we add a coefficient of 2 to HClHCl.

Problem 3:

Calculate the total number of atoms in one formula unit of Aluminum Sulfate, Al2(SO4)3Al_{2}(SO_{4})_{3}.

Solution:

2+(1×3)+(4×3)=17 atoms2 + (1 \times 3) + (4 \times 3) = 17 \text{ atoms}

Explanation:

There are 2 AlAl atoms. The subscript 3 outside the parentheses multiplies everything inside: 1×3=31 \times 3 = 3 Sulfur atoms and 4×3=124 \times 3 = 12 Oxygen atoms. Total: 2+3+12=172 + 3 + 12 = 17 atoms.