Chemical Reactions CH8 by Owen Borville October 16, 2025
Chemical reactions use chemical symbols to denote what occurs in a chemical reaction. Ex. NH3 + HCl => NH4Cl Ammonia and hydrogen chloride react to produce ammonium chloride.
Each chemical species that appears to the left of the arrow is called a reactant. Ex. NH3 + HCl => NH4Cl
Each species that appears to the right of the arrow is called a product. Ex. NH3 + HCl => NH4Cl
Labels are used to indicate the physical state of a chemical species, such as (g) for gas, (l) for liquid, (s) for solid, and (aq) for aqueous, dissolved in water.
Ex. NH3(g) + HCl (g) => NH4Cl (s)
Ex. SO3 (g) + H2O (l) => H2SO4 (aq)
Chemical equations must be balanced so that the law of conservation of mass is obeyed. Balancing is achieved by writing stoichiometric coefficients to the left of the chemical formulas. 2H2 (g) + O2 (g) => 2H2O (l).
Balancing chemical equations requires: (1) Change the coefficients of compounds before changing the coefficients of elements. (2) Treat polyatomic ions that appear on both sides of the equation as units. (3) Count atoms and/or polyatomic ions carefully, and track their numbers each time you change a coefficient.
Ex. Balance the equation (combustion of propane): C3H8 (g) + O2(g) => CO2 (g) + H2O(l)
Solution: C3H8 (g) + 5O2 (g) => 3CO2 (g) + 4H2O (l) The number of atoms of each element on each side must be balanced.
Ex. Barium hydroxide and perchloric acid => barium perchlorate and water Ba(OH)2(aq) + HClO4 (aq) => Ba (ClO4)2(aq) + H2O(l)
Solution: Ba(OH)2(aq) + 2HClO4(aq) => Ba(ClO4)2(aq) + 2H2O(l)
Ex. Metabolism of butanoic acid: C4H8O2(aq) + O2(g) => CO2(g) + H2O(l)
Solution: C4H8O2(aq) + 5O2(g) => 4CO2(g) + 4H2O(l)
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Chemical Reaction Types = The three most common chemical reactions: combination, decomposition, and combustion.
Combination > two or more reactants combine to form a single product. Ex. NH3 (g) + HCl(g) => NH4Cl(s)
Decomposition > two or more products form from a single reactant. Ex. CaCO3(s) => CaO(s) + CO2(g)
Combustion > a substance burns in the presence of oxygen. Combustion of a compound that contains C and H (or C, H, and O) produces carbon dioxide gas and liquid water.
CH2O(l) + O2(g) => CO2(g) + H2O(l)
Combustion analysis: The experimental determination of an empirical formula is carried out by combustion analysis.
Ex. In the combustion of 18.8 grams of glucose CxHyOz, it is possible to determine the mass of carbon and hydrogen in the original sample.
Mass of C = 27.6g CO2 x (1 mol CO2/44.01 g C) x (1 mol C/1 mol CO2) x (12.01 g C/1 mol C) = 7.53 g C
Mass of H = 11.3 g H2O x (1 mol H2O/18.01 g C) x (2 mol H/1 mol H2O) x (1.oo8 g H/1 mol H) = 1.26 g H
The remaining mass is oxygen: 18.8 g glucose - (7.53 g C + 1.26 g H) = 10.0 g O
Determine the number of moles of each element:
moles of C = 7.53 g C x (1 mol C/12.01 g C) = 0.627 moles C
moles of H = 1.26 g H x (1 mol H/1.008 g H) = 1.25 moles H
mols of O = 10.0 g ) x (1 mol O/16.00g O) = 0.626 moles O
Write the empirical formula and divide by the smallest subscript to find the whole number ratio: 1:2:1 => CH2O
The molecular formula may be determined from the empirical formula if the approximate molecular mass is known. To determine the molecular formula, divide the molar mass by the empirical formula mass:
Empirical formula = CH2O
Empirical formula mass [12.01 g/mol + 2(1.008g/mol) + 16.00g/mol] = 30 g/mol
Molecular mass = 180 g/mol
Molecular mass/Empirical mass = 180/30 = 6
Molecular formula = [CH2O] x 6 = C6H12O6
Balanced chemical equations are used to predict how much product will form from a given amount of reactant.
2CO(g) + O2(g) => 2CO2(g)
Two (2) moles of CO combine with one (1) mole of O2 to produce 2 moles of CO2. 2 moles of CO is stoichiometrically equivalent to 2 moles of CO2.
Consider the complete reaction of 3.82 moles of CO to form CO2. Calculate the number of moles of CO2 produced. 2CO(g) + O2(g) =>2CO2(g).
Moles CO2 produced = 3.82 mol CO x (2 mol CO2/2 mol CO) = 3.82 mol CO2
Moles O2 needed = 3.82 mol CO x (1 mol O2/2 mol CO) = 1.91 mol O2.
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Limiting Reactants: The reactant used up first in a chemical reaction is called the limiting reactant. Excess reactants are those present in a chemical reaction in quantities greater than necessary to react with the quantity of the limiting reactant.
Ex. Determine the limiting reactant: CO(g) + 2H2(g) => CH3OH(l)
How many moles of H2 are necessary in order for all the CO to react? Moles of H2 = 5 mol CO x (2 mol H2/1 mol CO) = 10 mol H2
How many moles of CO are necessary in order for all of the H2 to react? Moles of CO = 8 mol H2 x (1 mol CO/2 mol H2) = 4 mol CO
Therefore, 10 moles of H2 are required. 8 moles of H2 are available=limiting reactant
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Reaction Yield: The theoretical yield is the amount of product that forms when all the limiting reactant reacts to form the desired product. The actual yield is the amount of product actually obtained from a reaction. The percent yield tells what percentage the actual yield is of the theoretical yield.
Percent % yield = (actual yield/theoretical yield) x 100%
Ex. Asprin (C9H8O4) is produced from (C7H6O3) and (C4H6O3)
C7H6O3 (104.8 g)+ C4H6O3 (110.9 g)=> C9H8O4 (105.6 g) + HC2H3O2
104.8 g C7H6O3 x (1 mol C7H6O3/138.12 g C7H6O3) = 0.7588 mol C7H6O3
110.9 g C4H6O3 x (1 mol C4H6O3/102.09 g ) = 1.086 mol C4H6O3
0.7588 mol C9H8O4 x (180.15 g C9H8O4/1 mol C9H8O4) = 136.7 g C9H8O4 = Theoretical Yield
Actual Yield = (105.6 g /136.7g ) x 100% = 77.25% Percent yield can never be greater than 100 %
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Periodic Trends in Reactivity of the Main Group Elements:
Ionization energy and electron affinity enable us to understand types of reactions that elements undergo and the types of compounds formed.
General Trends in Reactivity:
Chemical behavior can be similar in a group due to similar valence electron configuration. The differences can be attributed to the small size of the first element in each group.
A trend in chemical reactivity of main group elements is the diagonal relationship (or similarities between pairs of elements in different groups and periods). Li, Be, and B exhibit chemical properties similar to Mg, Al, and Si, located diagonally below them. They also show similar patterns in reactivity.
6 Li(s) + N2(g) => 2 Li3N(s)
3 Mg(s) + N2(g) => Mg3N2(s)
Hydrogen (1s1): Grouped by itself. Forms a cation with a +1 charge (H+). Forms an anion with a -1 charge (H-). Hydrides react with water to produce hydrogen gas and a base.
CaH2 (s) + H2O (l) => Ca(OH)2(aq)+H2(g)
Reactions of the Active Metals: Group 1A Elements (ns1, n>=2)= Low IE; Never found in nature in pure elemental state; Reacts with oxygen to form metal oxides. Alkali metals react violently with water.
Reactions of Other Main Group Elements: Group 3A elements (ns2, np1, n>=2) Metalloid (B) and metals (all others). Al forms Al2O3 with oxygen. Al forms +3 ions in acid. Others form +1 and +3. Finely-divided aluminum sprinkled into a flame to form Al2O3.
Group 4A elements (ns2, np2, n>=2) Nonmetal (C); metalloids (Si, Ge) and other metals. Forms +2 and +4 oxidation states. Sn, Pb react with acid to produce H2.
Group 5A elements (ns2np3, n>=2) Nonmetals (N2, P), metalloids (As, Sb), and metal (Bi). Nitrogen N2 forms a variety of oxides. Phosphorus P4; As, Sb, Bi (crystalline). HNO3 and H3PO4 are important industrially.
Group 6A elements (ns2, np4, n>=2) Nonmetals (O, S, Se) Oxygen O2, Sulfur S8, Selenium Se8. Metalloids (Te, Po) (crystalline). SO2, SO3, H2S, H2SO4. Nonmetal oxides added to water produce an acid.
Group 7A elements (ns2, np5, n>=2) All diatomic and do not exist in elemental form in nature. Form ionic salts. Form molecular compounds with each other. React with hydrogen to form hydrogen halides.
Group 8A elements (ns2, np6, n>=2) All monatomic. Filled valence shells. Considered inert until 1963 when Xe and Kr were used to form compounds. No major commercial use.
Comparison of Group 1A and Group 1B Elements: Have a single valence electron. Properties differ. Group 1B much less reactive than 1A. High IE of 1B because of incomplete shielding of nucleus by inner "d" outer "s" electron of 1B strongly attracted to nucleus. 1B metals often are found elemental in nature (coinage metals).
Chemical reactions use chemical symbols to denote what occurs in a chemical reaction. Ex. NH3 + HCl => NH4Cl Ammonia and hydrogen chloride react to produce ammonium chloride.
Each chemical species that appears to the left of the arrow is called a reactant. Ex. NH3 + HCl => NH4Cl
Each species that appears to the right of the arrow is called a product. Ex. NH3 + HCl => NH4Cl
Labels are used to indicate the physical state of a chemical species, such as (g) for gas, (l) for liquid, (s) for solid, and (aq) for aqueous, dissolved in water.
Ex. NH3(g) + HCl (g) => NH4Cl (s)
Ex. SO3 (g) + H2O (l) => H2SO4 (aq)
Chemical equations must be balanced so that the law of conservation of mass is obeyed. Balancing is achieved by writing stoichiometric coefficients to the left of the chemical formulas. 2H2 (g) + O2 (g) => 2H2O (l).
Balancing chemical equations requires: (1) Change the coefficients of compounds before changing the coefficients of elements. (2) Treat polyatomic ions that appear on both sides of the equation as units. (3) Count atoms and/or polyatomic ions carefully, and track their numbers each time you change a coefficient.
Ex. Balance the equation (combustion of propane): C3H8 (g) + O2(g) => CO2 (g) + H2O(l)
Solution: C3H8 (g) + 5O2 (g) => 3CO2 (g) + 4H2O (l) The number of atoms of each element on each side must be balanced.
Ex. Barium hydroxide and perchloric acid => barium perchlorate and water Ba(OH)2(aq) + HClO4 (aq) => Ba (ClO4)2(aq) + H2O(l)
Solution: Ba(OH)2(aq) + 2HClO4(aq) => Ba(ClO4)2(aq) + 2H2O(l)
Ex. Metabolism of butanoic acid: C4H8O2(aq) + O2(g) => CO2(g) + H2O(l)
Solution: C4H8O2(aq) + 5O2(g) => 4CO2(g) + 4H2O(l)
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Chemical Reaction Types = The three most common chemical reactions: combination, decomposition, and combustion.
Combination > two or more reactants combine to form a single product. Ex. NH3 (g) + HCl(g) => NH4Cl(s)
Decomposition > two or more products form from a single reactant. Ex. CaCO3(s) => CaO(s) + CO2(g)
Combustion > a substance burns in the presence of oxygen. Combustion of a compound that contains C and H (or C, H, and O) produces carbon dioxide gas and liquid water.
CH2O(l) + O2(g) => CO2(g) + H2O(l)
Combustion analysis: The experimental determination of an empirical formula is carried out by combustion analysis.
Ex. In the combustion of 18.8 grams of glucose CxHyOz, it is possible to determine the mass of carbon and hydrogen in the original sample.
Mass of C = 27.6g CO2 x (1 mol CO2/44.01 g C) x (1 mol C/1 mol CO2) x (12.01 g C/1 mol C) = 7.53 g C
Mass of H = 11.3 g H2O x (1 mol H2O/18.01 g C) x (2 mol H/1 mol H2O) x (1.oo8 g H/1 mol H) = 1.26 g H
The remaining mass is oxygen: 18.8 g glucose - (7.53 g C + 1.26 g H) = 10.0 g O
Determine the number of moles of each element:
moles of C = 7.53 g C x (1 mol C/12.01 g C) = 0.627 moles C
moles of H = 1.26 g H x (1 mol H/1.008 g H) = 1.25 moles H
mols of O = 10.0 g ) x (1 mol O/16.00g O) = 0.626 moles O
Write the empirical formula and divide by the smallest subscript to find the whole number ratio: 1:2:1 => CH2O
The molecular formula may be determined from the empirical formula if the approximate molecular mass is known. To determine the molecular formula, divide the molar mass by the empirical formula mass:
Empirical formula = CH2O
Empirical formula mass [12.01 g/mol + 2(1.008g/mol) + 16.00g/mol] = 30 g/mol
Molecular mass = 180 g/mol
Molecular mass/Empirical mass = 180/30 = 6
Molecular formula = [CH2O] x 6 = C6H12O6
Balanced chemical equations are used to predict how much product will form from a given amount of reactant.
2CO(g) + O2(g) => 2CO2(g)
Two (2) moles of CO combine with one (1) mole of O2 to produce 2 moles of CO2. 2 moles of CO is stoichiometrically equivalent to 2 moles of CO2.
Consider the complete reaction of 3.82 moles of CO to form CO2. Calculate the number of moles of CO2 produced. 2CO(g) + O2(g) =>2CO2(g).
Moles CO2 produced = 3.82 mol CO x (2 mol CO2/2 mol CO) = 3.82 mol CO2
Moles O2 needed = 3.82 mol CO x (1 mol O2/2 mol CO) = 1.91 mol O2.
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Limiting Reactants: The reactant used up first in a chemical reaction is called the limiting reactant. Excess reactants are those present in a chemical reaction in quantities greater than necessary to react with the quantity of the limiting reactant.
Ex. Determine the limiting reactant: CO(g) + 2H2(g) => CH3OH(l)
How many moles of H2 are necessary in order for all the CO to react? Moles of H2 = 5 mol CO x (2 mol H2/1 mol CO) = 10 mol H2
How many moles of CO are necessary in order for all of the H2 to react? Moles of CO = 8 mol H2 x (1 mol CO/2 mol H2) = 4 mol CO
Therefore, 10 moles of H2 are required. 8 moles of H2 are available=limiting reactant
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Reaction Yield: The theoretical yield is the amount of product that forms when all the limiting reactant reacts to form the desired product. The actual yield is the amount of product actually obtained from a reaction. The percent yield tells what percentage the actual yield is of the theoretical yield.
Percent % yield = (actual yield/theoretical yield) x 100%
Ex. Asprin (C9H8O4) is produced from (C7H6O3) and (C4H6O3)
C7H6O3 (104.8 g)+ C4H6O3 (110.9 g)=> C9H8O4 (105.6 g) + HC2H3O2
104.8 g C7H6O3 x (1 mol C7H6O3/138.12 g C7H6O3) = 0.7588 mol C7H6O3
110.9 g C4H6O3 x (1 mol C4H6O3/102.09 g ) = 1.086 mol C4H6O3
0.7588 mol C9H8O4 x (180.15 g C9H8O4/1 mol C9H8O4) = 136.7 g C9H8O4 = Theoretical Yield
Actual Yield = (105.6 g /136.7g ) x 100% = 77.25% Percent yield can never be greater than 100 %
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Periodic Trends in Reactivity of the Main Group Elements:
Ionization energy and electron affinity enable us to understand types of reactions that elements undergo and the types of compounds formed.
General Trends in Reactivity:
Chemical behavior can be similar in a group due to similar valence electron configuration. The differences can be attributed to the small size of the first element in each group.
A trend in chemical reactivity of main group elements is the diagonal relationship (or similarities between pairs of elements in different groups and periods). Li, Be, and B exhibit chemical properties similar to Mg, Al, and Si, located diagonally below them. They also show similar patterns in reactivity.
6 Li(s) + N2(g) => 2 Li3N(s)
3 Mg(s) + N2(g) => Mg3N2(s)
Hydrogen (1s1): Grouped by itself. Forms a cation with a +1 charge (H+). Forms an anion with a -1 charge (H-). Hydrides react with water to produce hydrogen gas and a base.
CaH2 (s) + H2O (l) => Ca(OH)2(aq)+H2(g)
Reactions of the Active Metals: Group 1A Elements (ns1, n>=2)= Low IE; Never found in nature in pure elemental state; Reacts with oxygen to form metal oxides. Alkali metals react violently with water.
Reactions of Other Main Group Elements: Group 3A elements (ns2, np1, n>=2) Metalloid (B) and metals (all others). Al forms Al2O3 with oxygen. Al forms +3 ions in acid. Others form +1 and +3. Finely-divided aluminum sprinkled into a flame to form Al2O3.
Group 4A elements (ns2, np2, n>=2) Nonmetal (C); metalloids (Si, Ge) and other metals. Forms +2 and +4 oxidation states. Sn, Pb react with acid to produce H2.
Group 5A elements (ns2np3, n>=2) Nonmetals (N2, P), metalloids (As, Sb), and metal (Bi). Nitrogen N2 forms a variety of oxides. Phosphorus P4; As, Sb, Bi (crystalline). HNO3 and H3PO4 are important industrially.
Group 6A elements (ns2, np4, n>=2) Nonmetals (O, S, Se) Oxygen O2, Sulfur S8, Selenium Se8. Metalloids (Te, Po) (crystalline). SO2, SO3, H2S, H2SO4. Nonmetal oxides added to water produce an acid.
Group 7A elements (ns2, np5, n>=2) All diatomic and do not exist in elemental form in nature. Form ionic salts. Form molecular compounds with each other. React with hydrogen to form hydrogen halides.
Group 8A elements (ns2, np6, n>=2) All monatomic. Filled valence shells. Considered inert until 1963 when Xe and Kr were used to form compounds. No major commercial use.
Comparison of Group 1A and Group 1B Elements: Have a single valence electron. Properties differ. Group 1B much less reactive than 1A. High IE of 1B because of incomplete shielding of nucleus by inner "d" outer "s" electron of 1B strongly attracted to nucleus. 1B metals often are found elemental in nature (coinage metals).