Module 1 Chemistry: Empirical Formulas, Molecular Formulas, and Chemical Equations

Chemical Formulas and Reactions: Study Notes

Empirical Formulas

• An empirical formula represents the simplest whole-number ratio of atoms within a compound.

• The actual chemical formula (molecular formula) is either identical to the empirical formula or a whole-number multiple of it.

Converting Decimal Subscripts to Whole Numbers (for Empirical Formula Determination)

• Detailed steps for determining an empirical formula, typically involving converting masses/percentages to moles, dividing by the smallest mole value, and then converting to whole numbers as outlined above, are fundamental to this concept.

Molecular Formulas

• The molecular formula provides the actual number of atoms of each element found in one molecule of a covalent compound.

Acetylene and Benzene: An Illustrative Example

• Different molecular formulas: Acetylene (C2H2) and Benzene (C6H6).

• Same empirical formula: For both compounds, the simplest whole-number ratio is CH.

• Same percent composition: Due to the identical empirical formula, both compounds share the same elemental percentage makeup by mass.

Relationship Between Empirical and Molecular Formulas

• The molecular formula is always a whole-number multiple of the empirical formula.

• To find the molecular formula from the empirical formula and molar mass, you first calculate the empirical formula mass ($EFM$).

• Then, determine the multiple ($n$) by dividing the molar mass ($MM$) by the empirical formula mass: $\text{n} = \frac{\text{MM}}{\text{EFM}}$.

• Finally, multiply each subscript in the empirical formula by $n$ to get the molecular formula.

*Example Problem:

• What is the molecular formula of fructose if it has an empirical formula of $CH_2O$ and a molar mass of $180.2 \text{ g/mol}$?

  • 1. Calculate Empirical Formula Mass (EFM):

    • C: $1 \times 12.01 \text{ g/mol} = 12.01 \text{ g/mol}$

    • H: $2 \times 1.008 \text{ g/mol} = 2.016 \text{ g/mol}$

    • O: $1 \times 16.00 \text{ g/mol} = 16.00 \text{ g/mol}$

    • $EFM = 12.01 + 2.016 + 16.00 = 30.026 \text{ g/mol}$

  • 2. Determine the Multiple (n):

    • $n = \frac{180.2 \text{ g/mol}}{30.026 \text{ g/mol}} \approx 6$

  • 3. Find the Molecular Formula:

    • Multiply the subscripts of $CH_2O$ by $6$

    • Molecular Formula = $C<em>{(1 \times 6)}H</em>{(2 \times 6)}O<em>{(1 \times 6)} = C</em>6H<em>{12}O</em>6$

Combustion Analysis

• Combustion analysis is a laboratory technique specifically used to determine the empirical formula of compounds that contain Carbon (C), Hydrogen (H), and sometimes Oxygen (O).

Combustion Analysis Apparatus (Figure 3.19)

• A stream of oxygen ($O_2$) is passed into a combustion chamber where the organic sample is burned.

• The products of combustion ($H<em>2O$ and $CO</em>2$) along with any unreacted oxygen ($O_2$) are then passed through two absorbent traps.

• The first trap contains an absorbent material (e.g., $Mg(ClO4)<em>2$) that specifically absorbs water ($H</em>2O$).

• The second trap contains an absorbent material (e.g., concentrated $NaOH$ or $Ba(OH)<em>2$) that specifically absorbs carbon dioxide ($CO</em>2$).

• By measuring the increase in mass of each absorbent trap, the mass of $H<em>2O$ and $CO</em>2$ produced from the combustion can be determined.

Determining Mass of C, H, and O in a Sample

• From the mass of $CO<em>2$ absorbed, the mass of carbon in the original sample can be calculated using the molar mass ratio ($1 \text{ mol C / 1 mol } CO</em>2$).

• From the mass of $H<em>2O$ absorbed, the mass of hydrogen in the original sample can be calculated using the molar mass ratio ($2 \text{ mol H / 1 mol } H</em>2O$).

• If the original sample also contains oxygen, its mass cannot be directly measured. Instead, the mass of oxygen is determined by subtracting the masses of C and H (and any other known elements) from the total mass of the original sample.

• Once the masses of C, H, and O are known, they are converted to moles, and the empirical formula is determined using the steps outlined previously.

Chemical Reactions and Aqueous Solutions (Chapter 4)

Information in a Chemical Equation

• Chemical equations provide a concise way to represent chemical changes.

• Reactants are the starting substances that undergo a chemical transformation, rearranging their existing bonds, to form new substances called products.

• Example: In the reaction of hydrogen and oxygen to form water: $2H<em>2 + O</em>2 \to 2H_2O$

  • $H<em>2$ and $O</em>2$ are the reactants.

  • $H_2O$ is the product.

• Chemical Formulas: The formulas ($H<em>2, O</em>2, H_2O$) indicate the types of substances involved and their elemental composition, revealing what happens during the reaction.

• Coefficients: The numbers preceding the chemical formulas ($2$ in $2H<em>2$ and $2H</em>2O$, $1$ implicit in $O_2$) tell you the proportions (in moles or molecules) in which the reactants combine and products form.

  • For example, two hydrogen molecules react with one oxygen molecule to form two water molecules.

Physical States

• Information in parentheses after each chemical formula specifies the physical state of the reactants and products under the reaction conditions.

• Common states include:

  • $(s)$: solid

  • $(l)$: liquid

  • $(g)$: gas

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

*Example:

• $2H<em>2(g) + O</em>2(g) \to 2H_2O(l)$

Reaction Conditions

• Special conditions required for a reaction to occur (e.g., temperature, pressure, catalyst, energy input) are sometimes written above or below the arrow in the chemical equation.

Summary of Information from Chemical Equations

Balancing Equations

• Chemical reactions must always obey the law of conservation of mass.

  • This fundamental law states that matter cannot be created or destroyed in a chemical reaction.

• In a balanced chemical equation, the number of atoms of each element on the reactant side (left side of the arrow) must be exactly equal to the number of atoms of that same element on the product side (right side of the arrow).

Example: Reaction of Sulfur Dioxide and Oxygen to Form Sulfur Trioxide

• Unbalanced Equation (Initial Observation):
  $SO<em>2 + O</em>2 \to SO_3$

  • Reactants: 1 S atom, 4 O atoms (2 from $SO<em>2$, 2 from $O</em>2$)

  • Products: 1 S atom, 3 O atoms

  • This is unbalanced because the number of oxygen atoms is not equal on both sides.

• Balanced Equation:
  $2SO<em>2 + O</em>2 \to 2SO_3$

  • Reactants:

    • Sulfur: $2 \times 1 = 2 \text{ S atoms}$

    • Oxygen: $(2 \times 2) + (1 \times 2) = 4 + 2 = 6 \text{ O atoms}$

  • Products:

    • Sulfur: $2 \times 1 = 2 \text{ S atoms}$

    • Oxygen: $2 \times 3 = 6 \text{ O atoms}$

  • The equation is now balanced as the number of S and O atoms is equal on both sides.

Balancing Chemical Equations: Overview and Key Principle

• When balancing chemical equations, you must only change the COEFFICIENTS, NEVER the SUBSCRIPTS of the chemical formulas!

• Changing coefficients alters the amounts (number of molecules or moles) of the compounds involved in the reaction. This is permissible because it does not change the identity of the substances.

• Changing subscripts would fundamentally alter the identities of the compounds involved in the reaction. For example, changing $H<em>2O$ to $H</em>2O_2$ changes water (a stable compound) into hydrogen peroxide (a different compound with different chemical properties). This changes the actual chemistry of the reaction and is incorrect for balancing.

Practice Problem:

• Write a balanced equation for the reaction between solid cobalt (III) oxide and solid carbon to produce solid cobalt and carbon dioxide gas.

  • 1. Write the unbalanced equation with correct chemical formulas and states:
    $Co<em>2O</em>3(s) + C(s) \to Co(s) + CO_2(g)$

  • 2. Balance by adjusting coefficients (typically starting with elements other than H and O, then H, then O, and finally checking all):

    • Balance Co: $Co<em>2O</em>3(s) + C(s) \to 2Co(s) + CO_2(g)$

    • Balance O: There are 3 O on the left, 2 O on the right. To get a common multiple of 6:

      • Multiply $Co<em>2O</em>3$ by 2 and $CO<em>2$ by 3: $2Co</em>2O<em>3(s) + C(s) \to 2Co(s) + 3CO</em>2(g)$

    • Rebalance Co (due to change in $Co<em>2O</em>3$ coefficient):

      • There are $2 \times 2 = 4$ Co on the left, so change $2Co$ to $4Co$:
        $2Co<em>2O</em>3(s) + C(s) \to 4Co(s) + 3CO_2(g)$

    • Balance C: There is 1 C on the left, 3 C on the right (from $3CO_2$).

      • Multiply $C(s)$ by 3:
        $2Co<em>2O</em>3(s) + 3C(s) \to 4Co(s) + 3CO_2(g)$

  • 3. Final Check:

    • Reactants: Co: $2 \times 2 = 4$; O: $2 \times 3 = 6$; C: $3 \times 1 = 3$

    • Products: Co: $4 \times 1 = 4$; O: $3 \times 2 = 6$; C: $3 \times 1 = 3$

    • The equation is balanced.