Stoichiometry

Stoichiometry

Student Learning Outcomes

• After studying this chapter, students will be able to:

  • State the formulae of common elements and compounds.

  • Define molecular formula of a compound as the number and type of different atoms in one molecule.

  • Define empirical formula of a compound as the simplest whole number ratio of different atoms in a molecule.

  • Deduce the formula and name of binary ionic compounds from ions with relevant information.

  • Deduce the formula of a molecular substance from the given structure of molecules.

  • Use the relationship: amount of substance = mass / molar mass to calculate:

  • number of moles,

  • mass,

  • molar mass,

  • relative mass (atomic/molecular/formula),

  • number of particles.

  • Define mole as the amount of substance containing Avogadro's number (6.02×10²³) of particles.

  • Explain the relationship between a mole and Avogadro's constant.

  • Construct chemical equations and ionic equations to show reactants forming products, including state symbols.

  • Deduce the symbol equation with state symbols for a chemical reaction given relevant information.

Introduction to Stoichiometry

• Stoichiometry is an important concept in chemistry that helps to calculate the amounts of reactants and products using a balanced chemical equation.

• It is based on the law of conservation of mass, which states that matter can neither be created nor destroyed. Thus, the total mass of all the reactants must equal the total mass of all the products.

• The stoichiometric coefficients used to balance a chemical equation provide the mole ratio between reactants and products.

4.1 Chemical Formula

• Elements exist in different forms:

  • Some elements are represented by their symbols (e.g., Na, Ca, C, Fe).

  • Some elements exist as discrete molecules (e.g., O₂, N₂, H₂).

  • Example: Ozone (O₃) consists of three bonded oxygen atoms.

• Chemical Compounds:

  • Compounds like sodium chloride (NaCl) exist as ions bonded in a crystal lattice. The 1:1 ratio of sodium to chloride ions leads to a formula unit of NaCl.

  • Other examples include CaCl₂, KBr, BaCl₂, etc.

  • Covalent compounds exist in discrete molecules (e.g., water (H₂O)).

  • Example: Ammonia is represented by NH₃ and methane by CH₄.

• A chemical compound is represented by a molecular formula, showing all types of atoms in one molecule of that compound.

• Stoichiometry in Industry: Estimation of raw materials required to produce desired products is a vital application.

Exercise

• Differentiate between the chemical formula of an element and that of a compound. Provide examples. List the names of ionic and covalent compounds from this article.

4.2 Empirical Formula

• The Empirical Formula of a compound indicates the minimum ratio between atoms present.

  • All ionic compounds are represented by empirical formulas, showing the simplest ratio of ions.

  • Example: Empirical formula of calcium fluoride is CaF₂, reflecting the ratio of calcium to fluoride ions in the crystal.

• For covalent compounds that exist as molecules, empirical formulas may differ from molecular formulas.

  • Example: Hydrogen peroxide (H₂O₂) has an empirical formula of HO.

  • Example: Benzene (C₆H₆) has an empirical formula of CH.

  • Water (H₂O) has both molecular and empirical formulas the same (H₂O).

• Some compounds can share the same empirical formula.

  • Example: Benzene (C₆H₆) and acetylene (C₂H₂) both have the empirical formula CH.

Exercise

• Provide two examples of compounds that share the same empirical and molecular formulas.

4.3 Chemical Formula of Binary Ionic Compounds

• To write the formula of an ionic compound, identify the cations and anions along with their charges, then combine them to form an electrically neutral compound.

• If the binary ionic compound's name is known, its formula can be derived:

  • Example: Lithium oxide

  • Lithium cation: Li⁺

  • Oxygen anion: O²⁻

  • Using the crisscross method:

    • Formula = Li₂O

  • Example: Aluminum oxide

  • Aluminum cation: Al³⁺

  • Oxygen anion: O²⁻

  • Formula = Al₂O₃

  • Example: Magnesium nitride

  • Magnesium cation: Mg²⁺

  • Nitride anion: N³⁻

  • Formula = Mg₃N₂

Interesting Information

• Chemical products like shampoos, perfumes, soaps, and fertilizers are made using stoichiometric calculations. Stoichiometry is essential for the chemical industry's operations.

4.4 Chemical Formula of Compounds

• Knowledge of the empirical formula allows the determination of a molecular formula. To calculate the empirical formula:

  • Determine the simplest whole-number ratio of atoms in the compound using experimental mass percent composition.

• Relationship to find molecular formula:

  • $\text{Molecular formula} = n \times \text{Empirical Formula}$

  • Where $n = \frac{\text{Molar Mass}}{\text{Empirical Formula mass}}$.

• Example: Empirical formula for hydrogen peroxide (HO) has a molar mass of 34. The molecular formula is:

  • $n = \frac{34}{17} = 2$

  • Result: $\text{Molecular formula} = (HO)_2 = H_2O_2$.

Exercise

• Name three compounds that have different empirical and molecular formulas.

4.5 Deduction of Molecular Formula from Structural Formula

• To deduce the molecular formula from structural formula:

  1. Write down the structural formula of the compound.

  2. Count the number of atoms of each type in the structural formula.

  3. Write symbols of all elements present.

  4. Write total number of each atom as subscript.

  5. Remove the subscript 1.

• Example: Sulfuric acid's structural formula is H-O-S-O-H; containing 2 H, 1 S, and 4 O atoms results in the molecular formula H₂SO₄.

• Example: Acetic acid's structure gives a molecular formula of C₂H₄O₂.

Exercise

• 1. Determine the molecular formula of phosphoric acid (structural formula: HO-P(OH)₂).

• 2. Identify the molecular formula for n-propyl alcohol (structural formula: CH₃-CH₂-CH₂-OH).

• 3. Calculate the formula of calcium carbonate (structural formula: O=C(=O)O).

4.6 Avogadro's Number (Nₐ)

• Avogadro's number (6.022 x 10²³) is fundamental for relating atomic/molecular mass to weight.

• Example: Reaction of carbon and oxygen:

  • Equation: $2C + O_2 \rightarrow 2CO$,

  • Conversions from atoms to grams involve Avogadro's number.

• Weight ratios between reactants and products help establish calculations in grams for laboratory use.

• Example Calculation:

  • 24g of carbon (2 × 6.022 x 10²³ atoms), 32g of oxygen (6.022 x 10²³ molecules), and 56g of CO (2 × 6.022 x 10²³ molecules).

Interesting Information

• The mole concept is essential as it allows counting incredibly small particles by weighing larger, measurable amounts of matter.

4.7 The Mole and Molar Mass

• Mole: Defined as the quantity containing Avogadro's number (Nₐ = 6.022 x 10²³) of particles.

  • A mole of carbon atoms is 12g.

  • A mole of O₂ weighs 32g.

  • A mole of sodium chloride (NaCl) is 58.5g.

• Molar mass: Mass of one mole of any substance (e.g., H: 1.008 g, H₂: 2.016 g).

Example Problem

• Molar masses of:

  • H₃PO₄ (from atomic weights)

  • Calculate: 2H (2×1) + 1P (1×31) + 4O (4×16) = 98 g mol⁻¹.

4.8 Chemical Equations and Chemical Reactions

• Chemical Equations represent chemical changes:

  • Reactants written on the left; products on the right, joined by an arrow (→).

• Requirements for writing a balanced chemical equation:

  1. Obey law of conservation of mass (no atoms lost/gained).

  2. Correctly written formulas & mole ratios.

  3. Indicate states of matter (s, l, g, aq).

• Example of a balanced equation:

  • $Zn(s) + H₂SO₄(aq) \rightarrow ZnSO₄(aq) + H₂(g)$.

• Reversible Reactions indicate both reactants and products can convert back into each other, represented as:

  • $N₂(g) + 3H₂(g) ⇌ 2NH₃(g)$.

4.9 Calculations Based on Chemical Equations

• A balanced equation reveals mole ratios for reactants and products, allowing calculation of reactant/product masses.

• Example: Reaction of calcium carbonate:

  • $CaCO_3 + 2HCl \rightarrow CaCl_2 + H₂O + CO₂$.

• Calculating Product Mass:

  • If 25g of CaCO₃ reacts, find mass of CaCl₂ made:

  • Molar mass: CaCO₃ = 100 g mol⁻¹, CaCl₂ = 111 g mol⁻¹,

  • Using ratio to find mass produced.

Exercises

1. How many moles of oxygen are required for 1.80 moles of ethyl alcohol?

2. How many grams of oxygen required for 0.3 moles of aluminum?

3. Calculate number of molecules produced from 5g of H₂ reacting with oxygen.

Key Points

1. Molecular formula reflects actual atom counts in a molecule.

2. Empirical formula shows minimum ratios of atoms; applicable to ionic and some covalent compounds.

3. Binary ionic compounds have formulas derived from ion names and charges.

4. Molecular formula can be derived from empirical formula by molar mass ratios.

5. Avogadro's number defines the mole and facilitates calculations in chemistry (Nₐ = 6.022 x 10²³).

6. A chemical equation illustrates the reaction, mole ratios, and must be balanced.

7. Mole ratios allow calculation of mass ratios of reactants and products in reactions.

Exercises for Further Study

Multiple Choice & Short Answer Questions