Stoichiometry

• Stoichiometry means there are ratios between your reactants and your products.

• It is an area of study that examines the quantities of substances consumed and produced in chemical reactions.

Chemical Equations

• Chemical equations are how chemists represent chemical reactions on paper.

• Arrows separate the starting materials (on the left), called reactants, from the ending materials (on the right), called products.

• Reactants → Products

Balancing Equations

• Ensure there are the same amount of elements on each side.

• Do not change the coefficients.

• The visual above shows there are 4 Hydrogens and 2 Oxygens on each side.

Other Symbols in Chemical Equations

• The states of matter for the reactants and products are often written in parentheses to the right of each formula or symbol.

  • gas (g)

  • liquid (l)

  • solid (s)

  • aqueous (aq)

• Example: CH₄(g) + 2O₂ (g) → CO₂ (g) +2H₂O (g)

Patterns of Chemical Reactivity

Combustion Reactions

• In a combination reaction, two or more substances react to form one product.

• A + B → C

• Example of a Combination Reaction Formula: C(s) + O₂(g) → CO₂(g)

Decomposition Reactions

• In a decomposition reaction one substance breaks down into two or more substances.

• C → A + B

• Example of a Decomposition Reaction: 2KClO₃(s) → 2KCl(s) + 3O₂(g)

Combustion Reactions

• In a combustion reactions there are rapid reactions that produce a flame, they often involve oxygen in the air as a reactant.

• When burning compounds with C and H in them, the products are always H₂O and CO₂.

• Example of a Decomposition Reaction: C₃H₈ (g) + 5O₂ (g) —> 3CO₂ (g) + 4H₂O (g)

  • Carbon dioxide

  • Water vapor

Formula Weight (FW)

• A formula weight is the sum of the atomic weights (AW) for the atoms in a chemical formula.

• For an element like sodium (Na), the formula weight is the atomic weight (23.0 amu).

• The formula weight for compounds and polyatomics requires finding the atomic weight for each individual element. Multiply the element’s atomic weight by the coefficient (if there is one).

• Formula Weight equation: #(AW) + #(AW) = FW

  • AW = Atomic Weight

  • FW = Formula Weight

• Example: Find the formula weight of sulfuric acid H₂SO₄:

  • There are 2 Hydrogens, 1 Sulfur, and 4 Oxygens.

  • 2(AW of H) + 1 (AW of S) + 4(AW of O) = FW

  • 2(1.0 amu) + 32.1 amu + 4(16.0 amu)

  • FW of H₂SO₄ = 98.1 amu

Percent Composition

• One can find the percentage of the mass of a compound that comes from each of the elements in the compound.

• Percent Composition equation:

• Example: Determine the percentage of carbon in glucose (C₆H₁₂O₆)

Avogadro’s Number

• Avogadro’s number: 6.02 × 10²³ atoms or molecules is the number of particles in one mole (mol).

Molar Mass

• A molar mass is the mass of 1 mol of a substance.

• The molar mass of an element is the atomic weight for the element from the periodic table. If it is diatomic, it is twice that atomic weight.

• The formula weight (in amu) will be the same number as the molar mass (in g/mol).

  • H = 1.01 g/mol

Converting Amounts

• Moles provide a bridge from the molecular scale to the real-world scale.

• Using equalities, we can convert from mass to atoms or from atoms to mass.

• Example: Convert 3 grams of copper (Cu) into atoms

Determining Empirical Formulas

• One can determine the empirical formula from the percent composition by following these three steps:

Steps for determining empirical formulas

• Assume the sample is 100 grams.

• Convert percentages to grams.

• Covert to grams to mols.

• Calculate the mole ratio by dividing by the smallest number of moles.

• Use results as subscripts in empirical formula.

Example

The compound para-aminobenzoic acid is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Assuming it is a 100 g sample, find the empirical formula.

Step 1: Convert percentage to grams (assume is is a 100 gram sample).

Step 2: Convert grams to mols and determine the smallest value.

Step 3: Divide mols by smallest value.

Step 4: Use results from previous step to create the empirical formula.

Final answer: C₇H₇N₁O₂

Determining a Molecular Formula

• The number of atoms in a molecular formula is a multiple of the number of atoms in an empirical.

• If we find the empirical formula and know a molar mass (molecular weight) for the compound, we can find the molecular formula.

• Example: If CH has a molar mass of 78 g/mol what is the molecular formula?

  • CH mass = 13 g/mol

    • C = 12.01 g/mol

    • H = 1.01 g/mol

    • Added together: 13 g/mol

  • Equation: $\dfrac{molar\;mass\;of\;molecular\;formula}{molar\;mass\;of\;empirical\;formula}$

  • Ratio: $\dfrac{78}{13}$ = 6

  • (CH) x 6 = C₆H₆

    • Benzene

Quantitative Information from a Balanced Equation

• The coefficients in the balanced equation show:

  • Relative numbers of molecules of reactants and products.

  • Relative numbers of moles of reactants and products, which can be converted to mass.

Percent Yield

• The theoretical yield is the maximum amount of product that can be made.

  • It is the amount of product possible as calculated through the stoichiometry problem.

• The actual yield is the amount one actually produces and measures.

• The percent yield can be found by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield)

• Percent yield equation:

  

Limiting Reactants and Excess Reagents

• The limiting reactant is the reactant present in the smallest stoichiometric amount.

  • It is the reactant you will run out of the first.

  • It is used in all stoichiometry calculations to determine amounts of products that are produced and amounts of any other reactant(s) that are used in a reaction.

• The excess reagent is what is left over.