OpenStax Chemistry Chapter 3 Study Guide: Composition, Moles, Solutions, and Concentration Units

Fundamental Chemical Quantities and Molar Mass

  • Formula Mass:

    • Defined as the sum of all atomic masses of the atoms present in a chemical formula unit or molecule.

  • Avogadro's Number:

    • 1 mol=6.022×1023 particles1\,\text{mol} = 6.022 \times 10^{23}\,\text{particles}.
    • The conversion constant for particles per mole is 6.022×1023 particles/mol6.022 \times 10^{23}\,\text{particles/mol}.

  • Molar Mass Calculation:

    • Determined by summing the standard atomic weights of all constituent atoms in a chemical compound as given on the periodic table.

Mass, Mole, and Particle Conversions

  • Conversion Pathway:

    • Interconversions follow the sequence: grams→moles→particles\text{grams} \rightarrow \text{moles} \rightarrow \text{particles}.

  • Grams to Moles:

    • mol=gmolar mass\text{mol} = \frac{g}{\text{molar mass}}

  • Moles to Grams:

    • g=mol×molar massg = \text{mol} \times \text{molar mass}

  • Moles to Particles:

    • N=mol×(6.022×1023)N = \text{mol} \times (6.022 \times 10^{23})
    • Where NN represents the total number of particles (atoms, molecules, or formula units).

Quantitative Chemical Percent Composition and Formula Determination

  • Percent Composition Formula:

    • Percent Composition=element masscompound mass×100\text{Percent Composition} = \frac{\text{element mass}}{\text{compound mass}} \times 100

  • Empirical Formula Determination Procedure:

    • Step 1: Convert elemental mass percentages directly to grams (%→g\% \rightarrow g) by assuming a 100 g100\,\text{g} sample.
    • Step 2: Convert mass in grams to moles (g→molg \rightarrow \text{mol}) using elemental molar masses.
    • Step 3: Divide each molar quantity by the smallest calculated mole value (divide\text{divide}).\n * Step 4: Multiply obtained ratios by integers if necessary to achieve small whole numbers (\text{whole numbers}).).

  • Distinction Between Empirical and Molecular Formulas:

    • Empirical Formula: Represents the simplest, lowest whole-number molar ratio of elements in a compound.
    • Molecular Formula: Represents the actual, exact number of atoms of each element present in a single molecule of a compound.

  • Molecular Formula Multiplier (nn):

    • n=molecular massempirical massn = \frac{\text{molecular mass}}{\text{empirical mass}}
    • The empirical formula subscripts are multiplied by nn to obtain the molecular formula.

Solution Molarity and Dilution Calculations

  • Molarity (MM):

    • Defined as the concentration unit measuring moles of solute per liter of solution:
    • M=molLM = \frac{\text{mol}}{L}

  • Volume Conversion Rule:

    • Always convert volume from milliliters (mL\text{mL}) to liters (LL) before applying the molarity formula (mL→L\text{mL} \rightarrow L).

  • Dilution Equation:

    • M1V1=M2V2M_1 V_1 = M_2 V_2
    • Where M1M_1 and V1V_1 represent initial molarity and volume, and M2M_2 and V2V_2 represent final molarity and volume.

Solution Concentration Expressions: Percent and Parts per Notation

  • Mass Percent (Mass %):

    • Mass %=msolutemsolution×100\text{Mass}\,\% = \frac{m_{\text{solute}}}{m_{\text{solution}}} \times 100

  • Volume Percent (Volume %):

    • Volume %=VsoluteVsolution×100\text{Volume}\,\% = \frac{V_{\text{solute}}}{V_{\text{solution}}} \times 100

  • Parts per Million (ppm):

    • ppm=fraction×106\text{ppm} = \text{fraction} \times 10^6
    • Alternatively expressed as msolutemsolution×106\frac{m_{\text{solute}}}{m_{\text{solution}}} \times 10^6

  • Parts per Billion (ppb):

    • ppb=fraction×109\text{ppb} = \text{fraction} \times 10^9
    • Alternatively expressed as msolutemsolution×109\frac{m_{\text{solute}}}{m_{\text{solution}}} \times 10^9