Chapter 5: Chemical Accounting – Study Notes for CHEM 110

Chemical Accounting

Introduction to Chemical Accounting

Chemical accounting refers to the quantitative methods used to track and calculate the amounts of substances involved in chemical reactions. These calculations are essential in chemistry, biology, and medicine, ensuring that reactions proceed with the correct proportions of reactants and products.

• Key Concept: Just as baking requires precise amounts of ingredients, chemical reactions require exact quantities of reactants for proper outcomes.

• Applications: Used in laboratory preparations, industrial processes, and biological systems.

5.1 Chemical Sentences: Equations

Vocabulary of Chemistry

Chemistry uses a symbolic language to describe substances and reactions:

• Symbols: Represent elements (e.g., H for hydrogen, O for oxygen).

• Formulas: Represent compounds (e.g., H2O for water).

• Equations: Represent chemical reactions, showing how reactants transform into products.

Chemical equations are the sentences in the language of chemistry, communicating chemical changes using symbols and formulas.

Describing Chemical Reactions

• Word Description: Carbon reacts with oxygen to form carbon dioxide.

• Chemical Equation: $\mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2$

• The arrow ($\rightarrow$) means "yields" or "reacts to produce".

Reactants and Products

• Reactants: Species present before the reaction (written to the left of the arrow).

• Products: Species present after the reaction (written to the right of the arrow).

• General format: Reactants → Products

States of Matter in Equations

• (s): solid

• (l): liquid

• (g): gas

• (aq): aqueous solution (dissolved in water)

Law of Conservation of Matter

The law states that matter can neither be created nor destroyed in a chemical reaction. The total mass and number of atoms must remain constant.

• Example: $2\mathrm{H}_2 (g) + \mathrm{O}_2 (g) \rightarrow 2\mathrm{H}_2\mathrm{O} (l)$

• 2 molecules of hydrogen + 1 molecule of oxygen produce 2 molecules of water.

Balancing Chemical Equations

Coefficients and Subscripts

Coefficients are numbers placed before formulas to balance chemical equations. Subscripts indicate the number of atoms in a molecule and must not be changed when balancing equations.

• Each type of atom must have the same number on both sides of the equation.

• The coefficient multiplies everything in the formula.

• Only coefficients can be changed to achieve balance; changing subscripts alters the compound's identity.

Steps for Balancing Equations

1. Write correct formulas for all reactants and products.

2. Balance atoms that appear only once on each side first.

3. Balance free elements last.

4. Count atoms of each kind in reactants and products to ensure balance.

Example of Balancing

• Unbalanced: $\mathrm{CH}_4 (g) + \mathrm{O}_2 (g) \rightarrow \mathrm{CO}_2 (g) + \mathrm{H}_2\mathrm{O} (g)$

• Balanced: $\mathrm{CH}_4 (g) + 2\mathrm{O}_2 (g) \rightarrow \mathrm{CO}_2 (g) + 2\mathrm{H}_2\mathrm{O} (g)$

Volume Relationships in Chemical Equations

Law of Combined Volumes

When measured at the same temperature and pressure, the volumes of gaseous reactants and products are in small whole-number ratios.

• Example: $3\mathrm{H}_2 (g) + \mathrm{N}_2 (g) \rightarrow 2\mathrm{NH}_3 (g)$

• Volume ratio: 3:1:2

Avogadro's Hypothesis

Equal volumes of gases, at the same temperature and pressure, contain the same number of molecules.

• Avogadro's Number: $6.02 \times 10^{23}$ particles per mole

Mole and Molar Mass

Definition of Mole

A mole (mol) is the amount of substance containing $6.02 \times 10^{23}$ particles (atoms, molecules, ions, or formula units).

Formula Mass and Molar Mass

• Formula Mass: The sum of atomic masses in a chemical formula, usually in atomic mass units (u).

• Molar Mass: The mass of one mole of a substance, expressed in grams per mole (g/mol).

• Example: Molar mass of Na = 23.0 g/mol; CO2 = 44.0 g/mol; (NH4)2SO4 = 132.1 g/mol

Calculating Mass and Moles

• To convert moles to grams: $\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}$

• To convert grams to moles: $\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

• Example: $0.400$ mol N2 $\times \frac{28.0\,\text{g}}{1\,\text{mol}} = 11.2$ g N2

Stoichiometry

Stoichiometric Relationships

Stoichiometry involves the quantitative relationships between reactants and products in a balanced chemical equation. The coefficients represent moles.

• Molecular: 2 molecules of NO react with 1 molecule of O2 to form 2 molecules of NO2.

• Molar: 2 mol NO react with 1 mol O2 to form 2 mol NO2.

• Mass: 60.0 g NO react with 32 g O2 to form 92 g NO2.

Stoichiometric Calculations

1. Write and balance the chemical equation.

2. Determine molar masses of substances involved.

3. Convert given mass to moles.

4. Use coefficients to convert moles of given to moles of desired substance.

5. Convert moles of desired substance to grams.

• Example: Calculate mass of O2 needed to react with 10.0 g C:

  1. Equation: $\mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2$

  2. Molar masses: C = 12.0 g/mol, O2 = 32.0 g/mol

  3. Convert mass of C to moles: $10.0\,\text{g} \times \frac{1\,\text{mol}}{12.0\,\text{g}} = 0.833\,\text{mol}$

  4. Convert moles of C to moles of O2: $0.833\,\text{mol C} \times \frac{1\,\text{mol O}_2}{1\,\text{mol C}} = 0.833\,\text{mol O}_2$

  5. Convert moles of O2 to grams: $0.833\,\text{mol O}_2 \times \frac{32.0\,\text{g}}{1\,\text{mol}} = 26.7\,\text{g O}_2$

Solutions and Concentration

Definitions

• Solution: Homogeneous mixture of two or more substances.

• Solute: Substance being dissolved.

• Solvent: Substance doing the dissolving (often water).

• Aqueous Solution: Solution where water is the solvent.

Concentration of Solutions

• Concentration: Amount of solute in a given amount of solvent.

• Dilute Solution: Contains a small amount of solute.

• Concentrated Solution: Contains a large amount of solute.

Molarity (M)

Molarity is defined as the moles of solute per liter of solution.

• Formula: $M = \frac{\text{moles of solute}}{\text{liters of solution}}$

• Example: Calculate the molarity of a solution containing 10.0 mol NaOH in 2.0 L:

  • $M = \frac{10.0\,\text{mol}}{2.0\,\text{L}} = 5.0\,\text{M}$

• Example: 3.50 mol NaCl in 2.00 L:

  • $M = \frac{3.50\,\text{mol}}{2.00\,\text{L}} = 1.75\,\text{M}$

• Example: 18.23 g HCl in 2.0 L:

  • Molar mass HCl = 36.46 g/mol

  • Moles HCl = $\frac{18.23\,\text{g}}{36.46\,\text{g/mol}} = 0.5000\,\text{mol}$

  • $M = \frac{0.5000\,\text{mol}}{2.0\,\text{L}} = 0.25\,\text{M}$

Preparing Solutions

• Dissolve the proper amount of solute and dilute to the desired volume.

• Alternatively, dilute a concentrated solution to the required concentration.

Calculating Mass for Solution Preparation

• Example: Prepare 0.50 L of 0.15 M NaCl solution:

  • Moles NaCl needed: $0.50\,\text{L} \times 0.15\,\text{mol/L} = 0.075\,\text{mol}$

  • Mass NaCl: $0.075\,\text{mol} \times 58.4\,\text{g/mol} = 4.38\,\text{g}$

Percent Concentrations

• Percent by Volume: $\text{Percent by volume} = \frac{\text{volume of solute}}{\text{volume of solution}} \times 100\%$

• Example: 120 mL oil in 4000 mL gasoline: $\frac{120}{4000} \times 100\% = 3.0\%$

• Percent by Mass: $\text{Percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%$

• Example: 25.5 g NaCl in 425 g water: $\frac{25.5}{25.5 + 425} \times 100\% = 5.66\%$

Summary Table: States of Matter Symbols

Symbol	State of Matter
(s)	Solid
(l)	Liquid
(g)	Gas
(aq)	Aqueous Solution

Summary Table: Steps in Stoichiometric Calculations

Step	Description
1	Write and balance the chemical equation
2	Determine molar masses of substances
3	Convert given mass to moles
4	Use coefficients to convert moles of given to moles of desired substance
5	Convert moles of desired substance to grams

Additional info: Some steps and examples have been expanded for clarity and completeness, following standard introductory chemistry textbook conventions.