Stoichiometry: Basics and Calculations

• Stoichiometry involves fundamental concepts like atomic and molecular masses, moles, percentage composition, and empirical/molecular formulas.
• Knowledge of dimensional analysis of formulae and equations is required.

Significant Figures and Rounding

• The rules for significant figures and rounding are not strictly enforced in this unit, but consistency in calculations is encouraged.
• Using either 15.999 or 16.0 g/mol as the atomic mass of oxygen is acceptable; students are encouraged to practice proper significant figures.

Atomic Mass, Molecular Mass and Moles

• Atomic mass is the weighted average mass of naturally occurring isotopes of an element.
• Isotopes are atoms with the same number of protons but different numbers of neutrons.
• Molecular mass is the sum of the atomic masses of all elements in a compound, each multiplied by its stoichiometric index.
• Both atomic and molecular masses are also referred to as molar masses.
• Molar masses are expressed in grams per mole (g/mol).
• Each element has a specific, unique atomic mass, useful for element identification.

Calculating Molecular Masses

• To calculate the molecular mass of dioxygen (O2), multiply the atomic mass of oxygen (15.999 g/mol) by 2, resulting in 31.998 g/mol.
• To calculate the molecular mass of potassium sulfate (K2SO4), sum the atomic masses of its elements: (2 x 39.0983) + 32.065 + (4 x 15.999) = 174.258 g/mol.

Moles

• The mole concept is introduced, where the units of molar masses indicate the mass of one mole of a substance in grams.
• One mole of atomic oxygen has a mass of 15.999 g; one mole of potassium sulfate has a mass of 174.256 g.
• The number of moles (n) can be determined using the formula: n = m/W, where m is mass and W is molecular mass.
• A mole represents 6.022 x 10^23 particles (Avogadro's number). Avogadro's number is defined as 6.022 × 1023 mol-1

Calculating Moles

• To find the moles of aluminum in 700.0 mg: n = (700.0 x 10^-3 g) / (27.0 g/mol) = 2.60 x 10^-2 mol.
• To find the mass of 7.00 x 10^20 molecules of dichlorine (Cl2): n = (7.00 x 10^20) / (6.023 x 10^23 mol^-1) = 1.16 x 10^-3 mol, then m = (1.16 x 10^-3 mol) x (70.906 g/mol) = 8.23 x 10^-2 g.

Mass Percentage Composition

• Mass percentage composition is the ratio of the mass of an element in a compound to the total mass of the compound.
• The percentage composition of atoms in a compound is independent of the compound's mass.
• If the chemical formula and molecular mass are known, calculations can be made relative to 1 mole of the substance.
• Formula: %X = (mx / mtot) x 100, where %X is percentage composition of element X, mx is mass of element X, and mtot is total mass.
• The sum of the percentage compositions of all elements in a substance must equal 100%.

Calculating Percentage Composition

• For aluminum phosphate (AlPO4), molar mass is 122.0 g/mol. In 1 mol, there's 27.0 g of Al, 31.0 g of P, and 64.0 g of O.
  • %Al = (27.0 g / 122.0 g) x 100 = 22.1%
  • %P = (31.0 g / 122.0 g) x 100 = 25.4%
  • %O = (64.0 g / 122.0 g) x 100 = 52.5%
• Mass percentage compositions can refer to the percentage of each pure compound in a mixture.
• Simultaneous equations might be necessary to determine the percentage composition in a mixture.

Example of Simultaneous Equations

• Two main methods can solve simultaneous equations: substitution or subtraction.
• Isolate a variable in the first equation and substitute it into the second, or multiply the first equation by a constant so that the coefficients of one of the variables in both equations are the same, then subtract one equation from the other.

Mixture Composition Example

• A mixture of Cu2O and CuO with a total mass of 3.00 g reacts with excess hydrogen to form 2.55 g of metallic Cu.
• Two equations are needed, linked to Cu2O and CuO, relating to the mass of the mixture and the mass of Cu formed.
• If nCu2O and nCuO are the number of moles of Cu2O and CuO, respectively, the equations can be set up.
• The calculated number of moles of CuO in the mixture can be used to determine mass and % composition. The mass of CuO is divided by the total mass and multiplied by 100.

Empirical and Molecular Formulae

• The empirical formula lists elements in a compound with their stoichiometric ratios as the lowest integer numbers.
• It can be determined by combining molar masses and % composition of elements.

Determining Empirical Formulae

• For sodium sulfate (Na2SO4), the number of moles of Na is twice that of S, and the number of moles of O is four times that of S. Empirical formula: Na2SO4

Empirical Formula Example

• Unknown compound contains 32.81% chromium and 67.19% chlorine.
• Assume a 100 g sample: 32.81 g Cr and 67.19 g Cl.
• Determine moles of Cr (atomic mass 52.0 g/mol) and Cl (atomic mass 35.5 g/mol) in 100 g.
• Divide moles by the lowest amount to get the ratio.

Molecular Formula

• Discrete molecules connected by covalent bonds are represented with a molecular formula rather than an empirical formula.
• Glucose example: empirical formula is CH2O, but the molecular formula is C6H12O6.

Molecular Formula Determination

• It is necessary to know the empirical formula and the molecular mass of the compound.
• The combustion of 150.0 mg of fructose yields 220.0 mg of CO2 and 90.0 mg of H2O; the molecular mass of fructose is 180 g/mol.

Molecular Formula Example

• All C and H atoms in CO2 and H2O products originally belong to fructose.
• Number of moles of C equals moles of CO2 produced; moles of H equal twice the moles of H2O.
• The amount of O present in fructose is the difference between the total mass of the compound and the combined masses of C and H. Total mass can be determined with the calculated number of moles.

Determining Molecular Formula

• All C and H atoms in CO2 and H2O products originally belong to fructose.
• Establish the empirical formula of fructose as CH2O.
• Calculate molar mass corresponding to the empirical formula.
• Calculate the ratio between the molar mass of the compound and the molar mass corresponding to the empirical formula; multiply the stoichiometric ratios in the empirical formula by the value obtained. Multiplying by 6 the stoichiometric ratios 1:2:1 in the empirical formula of C, H and O, respectively, we obtain C6H12O6 (glucose and fructose have the same molecular formula, but different connectivity between atoms).

Dimensional Analysis

• It is essential to accompany the value with its associated units, unless the value is dimensionless.
• Dimensional analysis confirms that a calculation has been performed correctly and that the final value represents the actual quantity when associated to its unit.
• Units can be algebraically manipulated.
• Always ensure that the units simplify to the expected quantity's units.

Dimensional Analysis tips

• An example of dimensional analysis shows that multiplying the gas constant by the temperature results in an energy value.
• If R is in J mol−1 K−1 and T in K, R x T has units of J/mol (energy per mole).
• If R is in L atm mol−1 K−1, the multiplication Latm mol-1 K−1 × K = L atm mol-1.

Conversions

• Convert L atm into SI units (m³ and Pa) to show it represents energy.
• Pa is kg m−1 s−2, so L atm becomes m³ kg m−1 s−2 = kg m² s−2, which equals Joules (J).
• Multiplying a volume by a pressure yields an energy.