Moles

Chemical Quantities

Measuring Quantities

  • Mass can be measured in grams.

  • Volume can be measured in liters.

  • We count pieces in MOLES.

    • A mole is the unit of quantity for atoms, molecules, and compounds.

Understanding Moles

  • 1 mole = 6.02 x 10^23 particles.

  • This is known as Avogadro's number.

    • It is analogous to how 12 represents 1 dozen and a pair represents 2.

  • Different types of atoms have different masses.

    • Example:

    • Sodium (Na) = 22.98 g/mol

    • Chlorine (Cl) = 35.45 g/mol

Counting Atoms in Compounds

  • Example of counting oxygen atoms in the following compounds:

    • CaCO3 (Calcium Carbonate): 3 oxygen atoms

    • Al2(SO4)3 (Aluminum Sulfate): 12 oxygen atoms

    • NO2 (Nitrogen Dioxide): 2 oxygen atoms

Measuring Moles and Atomic Mass

  • AMU (Atomic Mass Unit) is defined as the mass of atoms.

  • The decimal number on the periodic table indicates the mass of 1 mole of those atoms in grams.

Gram Atomic Mass

  • The mass of 1 mole of an element is expressed in grams:

    • Example:

    • 12.01 grams of carbon contains the same number of atoms as 1.008 grams of hydrogen and 55.85 grams of iron.

    • Mathematical representation:

    • 12.01extgC=1extmole12.01 ext{ g C} = 1 ext{ mole}

  • By knowing the mass of a substance, we can find the number of moles present.

Calculating Molar Mass

  • Example to find the mass of one mole of CH4 (Methane):

    • 1 mole of C = 12.00 g

    • 4 moles of H x 1.00 g = 4.00 g

    • Therefore, 1 mole of CH4 = 12.00 + 4.00 = 16.00 g

    • The molar mass of CH4 is 16.00 g.

Molar Mass of Ionic Compounds

  • Molar mass can be calculated the same way for ionic compounds.

  • Example calculation for Fe2O3 (Iron(III) oxide):

    • 2 moles of Fe x 55.85 g = 111.70 g

    • 3 moles of O x 16.00 g = 48.00 g

    • Total Molar Mass = 111.70 g + 48.00 g = 159.70 g

Representative Particles

  • The smallest pieces of a substance can vary:

    • For a molecular compound, it is a molecule.

    • For an ionic compound, it is a formula unit.

    • For an element, it is an atom.

Percent Composition

  • Percent Composition formula is analogous to all percentages:

    • extPercent=racextPartextWholeimes100ext{Percent} = rac{ ext{Part}}{ ext{Whole}} imes 100

  • Find the mass of each component and divide it by the total mass.

Example Calculation

  • For a compound with 29.0 g of Ag (Silver) and 4.30 g of S (Sulfur):

    • Total mass = 29.0 g + 4.30 g = 33.30 g

    • Percent Ag = rac{29.0 ext{ g}}{33.30 ext{ g}} imes 100 = 87.8 ext{%}

    • Therefore, percent Sulfur = 100% - 87.8% = 12.2%

Using Molar Mass

  • Molar mass indicates the weight in grams for 1 mole of atoms, ions, or molecules.

  • Conversion factors can be derived from molar mass to convert grams of a compound to moles of that compound.

Calculating Moles from Mass

  • Example: How many moles are in 5.69 g of NaOH?

    • To solve: Convert grams to moles using molar mass.

  • Molar mass of NaOH is calculated as follows:

    • Sodium (Na) = 22.99 g, Oxygen (O) = 16.00 g, Hydrogen (H) = 1.01 g

    • Total for NaOH = 22.99 + 16.00 + 1.01 = 40.00 g

  • Therefore:

    • extMolesofNaOH=rac5.69extg40.00extg/mol=0.14225extmolesext{Moles of NaOH} = rac{5.69 ext{ g}}{40.00 ext{ g/mol}} = 0.14225 ext{ moles}

Gases and Measurement

  • Gases are often challenging to weigh, hence we need to determine moles based on volume.

  • Two factors that affect the volume of gas:

    • Temperature

    • Pressure

    • It is essential to compare gas volumes at the same temperature and pressure for consistency.

Standard Temperature and Pressure (STP)

  • STP is defined as:

    • 0ºC (273.15 K) and 1 atm pressure

  • At STP, 1 mole of gas occupies 22.4 L, known as the molar volume.

  • Avogadro's Hypothesis states that at the same temperature and pressure, equal volumes of gas contain the same number of molecules.

Empirical and Molecular Formulas

  • Empirical Formula indicates the lowest whole number ratio of elements in a compound.

  • Molecular Formula indicates the actual number of atoms in the compound.

  • Example:

    • Molecular formula C2H4 corresponds to empirical formula CH2.

    • Molecular formula C3H6 corresponds to the same empirical formula CH2.

  • H2O exhibits both molecular and empirical formulas equally.

Calculating Empirical Formulas

  • The approach to calculating the empirical formula includes:

    • Finding the lowest whole number ratio of elements.

    • Using molecular ratios of moles to determine the ratios of atoms.

  • Example:

    • In 1 mole of CO2, there is 1 mole of carbon and 2 moles of oxygen.

    • In one molecule of CO2, there is 1 atom of carbon and 2 atoms of oxygen.

From Percent Composition to Empirical Formula

  • You can derive the empirical formula from percentage composition:

    • Assume you have 100 g of the compound to convert percentages directly to grams.

    • Convert grams to moles.

    • Divide by the smallest number of moles to find the simplest whole number ratio.

Example Calculation of Empirical Formula

  • For a compound composed of 38.67% C, 16.22% H, and 45.11% N:

    • Assume 100 g leading to:

    • 38.67 g C x rac1extmolC12.01extgC=3.220extmolCrac{1 ext{ mol C}}{12.01 ext{ g C}} = 3.220 ext{ mol C}

    • 16.22 g H x rac1extmolH1.00extgH=16.22extmolHrac{1 ext{ mol H}}{1.00 ext{ g H}} = 16.22 ext{ mol H}

    • 45.11 g N x rac1extmolN14.00extgN=3.219extmolNrac{1 ext{ mol N}}{14.00 ext{ g N}} = 3.219 ext{ mol N}

  • The ratios are:

    • 3.220extmolC:3.219extmolN=1:13.220 ext{ mol C}: 3.219 ext{ mol N} = 1: 1

    • 16.22extmolH:3.219extmolN=5:116.22 ext{ mol H}: 3.219 ext{ mol N} = 5: 1

  • Thus, the empirical formula is:

    • C1H5N1 (The 1 is understood and not typically written as a subscript).

    • Therefore, final empirical formula: CH5N