Notes on Chemical Formulas and Molar Masses

Chemical Formulas

  • Indicate the elements present in a compound.

  • Show relative number of atoms/ions of each element in a compound.

  • Allow calculation of:

    • Formula mass.

    • Molar mass.

    • Percentage composition.

Formula Masses

  • Formula mass = sum of average atomic masses of all atoms in the formula.

    • Example (Water, H₂O):

    • H: 1.01 amu, O: 16.00 amu

    • Average mass of H₂O molecule = 18.02 amu

  • The mass of molecules (like water) is termed molecular mass.

  • For ionic compounds (like NaCl), use formula mass instead of molecular mass.

Sample Problem: Calculating Formula Mass of Potassium Chlorate (KClO₃)

  1. Calculate:

    • K: 39.10 amu

    • Cl: 35.45 amu

    • O (3 O atoms): 3 × 16.00 amu = 48.00 amu

  2. Total formula mass = 39.10 + 35.45 + 48.00 = 122.55 amu

Molar Masses

  • Molar mass = mass of one mole of a substance in grams; approximately 6.022×10236.022 \times 10^{23} particles.

    • Example (Calcium, Ca): Molar mass = 40.08 g/mol.

  • Calculation of molar mass involves adding the masses of elements in one mole of a compound.

    • Molar mass of water (H₂O): 18.02 g/mol.

  • Note: Molar mass = Formula mass in numeric value.

Sample Problem: Molar Mass of Barium Nitrate (Ba(NO₃)₂)

  1. One mole Ba: 137.33 g/mole

  2. Two moles N: 2 × 14.01 g/mole = 28.02 g/mole

  3. Six moles O: 6 × 16.00 g/mole = 96.00 g/mole

  4. Total molar mass = 137.33 + 28.02 + 96.00 = 261.35 g/mol

Molar Mass as a Conversion Factor

  • Molar mass relates moles to grams:
    Amount in moles×Molar mass (g/mol)=Mass in grams\text{Amount in moles} \times \text{Molar mass (g/mol)} = \text{Mass in grams}

Sample Problem: Mass of 2.50 mol Oxygen Gas (O₂)

  • Given: 2.50 mol O₂

  • Unknown: Mass of O₂ in grams

  • Calculate molar mass (O₂): 2 × 16.00 g/mol = 32.00 g/mol.

  • Convert:
    Mass of O₂=2.50 mol×32.00 g/mol=80.00 g\text{Mass of O₂} = 2.50 \text{ mol} \times 32.00 \text{ g/mol} = 80.00 \text{ g}

Converting Between Amount in Moles and Particles

  1. Determine whether given quantity is moles or particles.

  2. Use conversion factors based on whether going from:

    • Amount to number of particles (left to right).

    • Number of particles to amount (right to left).

    1 mol=6.022×1023 particles1 \text{ mol} = 6.022 \times 10^{23} \text{ particles}

Sample Problem: Ibuprofen (C₁₃H₁₈O₂)

  1. Given 33 g of ibuprofen, molar mass = 206.31 g/mol.

  2. a. Moles of C₁₃H₁₈O₂:

    • moles=gramsmolar mass=33.0 g206.31 g/mol\text{moles} = \frac{\text{grams}}{\text{molar mass}} = \frac{33.0 \text{ g}}{206.31 \text{ g/mol}}

  3. b. Calculate molecules from moles.

  4. c. Total mass of Carbon:

    • mass of Carbon=moles C₁₃H₁₈O₂×12.01 g/mol\text{mass of Carbon} = \text{moles C₁₃H₁₈O₂} \times 12.01 \text{ g/mol}

Percentage Composition

  • Used to find the mass percentage of an element in a compound.

  • Calculated as:
    Percentage composition=(mass of element in 1 molemolar mass of compound)×100\text{Percentage composition} = \left( \frac{\text{mass of element in 1 mole}}{\text{molar mass of compound}} \right) \times 100

  • Independent of sample size; same for any sample.

Sample Problem: Copper(I) Sulfide (Cu₂S)

  1. Given formula: Cu₂S.

  2. Molar mass of Cu₂S = 159.2 g.

  3. Percentage Composition Calculations:

    • Cu: 127.1 g Cu159.2 g Cu₂S×100=79.85%\frac{127.1 \text{ g Cu}}{159.2 \text{ g Cu₂S}} \times 100 = 79.85\% Cu

    • S: 32.07 g S159.2 g Cu₂S×100=20.15%\frac{32.07 \text{ g S}}{159.2 \text{ g Cu₂S}} \times 100 = 20.15\% S

Focusing on formulas, molar mass calculations, and percentage composition will significantly aid in grasping essential concepts for your exam. Practice these calculations thoroughly to build confidence in your understanding!