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Chapter 3: Chemical Reactions and Reaction Stoichiometry

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).

  • The formula for compounds can be found using the equation #(AW) + #(AW) = FW

  • 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 by using the following 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.

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: molar  mass  of  molecular  formulamolar  mass  of  empirical  formula\dfrac{molar\;mass\;of\;molecular\;formula}{molar\;mass\;of\;empirical\;formula}

    • Ratio: 7813\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):

Limiting Reactants

  • 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.

S

Chapter 3: Chemical Reactions and Reaction Stoichiometry

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).

  • The formula for compounds can be found using the equation #(AW) + #(AW) = FW

  • 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 by using the following 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.

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: molar  mass  of  molecular  formulamolar  mass  of  empirical  formula\dfrac{molar\;mass\;of\;molecular\;formula}{molar\;mass\;of\;empirical\;formula}

    • Ratio: 7813\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):

Limiting Reactants

  • 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.