Chapter 6

Chemical Reactions:
Classification and Mass Relationships

Chapter 6 suggested problems: 38, 40, 46, 48, 50, 53, 54, 58, 62, 66, 70, 72, 75, 82, 84, 90, 94, 98

  1. Empirical formulas and molecular formulas

    1. Empirical formula: the formula of a compound with the simplest whole number ratio of elements involved in the compound

    2. Molecular formula: the types and actual number of atoms in a compound

      1. Glucose: C6H12O6 vs. CH2O

      2. Ethene: C2H4 vs. CH2

      3. Imaginary compound: C25H75O15N5S10 vs. ???

    3. Empirical formulas are often determined by elemental analysis, molecular formulas can be determined using a variety of techniques

  2. Molecular formulas and molecular weight (formula mass)

    1. Given the molecular formula of a compound, we find the molecular weight by finding the sum of the masses of all of the atoms in the compound

    2. The mass of individual atoms is found on the Periodic Table and is given in amu

    3. Examples

      1. Sodium chloride: 22.99 amu + 35.45 amu = 58.44 amu

      2. Methane: 12.01 amu + (4)(1.008 amu) = 16.04 amu

      3. Strychnine (C21H22N2O2): 334 amu

      4. TNT (C7H5N3O6): 227 amu

    4. Molecular weight vs. formula weight (formula mass): many ionic compounds do not exist as discrete molecules but as interconnected lattices of ions

      1. The formula weight is the weight of one formula unit of an ionic compound

  3. The mole and Avogadro's number

    1. Avogadro's number

      1. The mass of a single C atom is 12.01 amu (from the atomic weight on the Periodic Table). How many C atoms are there in 12.01 g of C?
        (12.01 g C) x (1 kg / 1000 g) x (1 amu / 1.6605402 x 10-27 kg) x (1 C atom / 12.01 amu) = 6.022 x 1023 C atoms

      2. The mass of a single Au atom is 196.97 amu (from the atomic weight on the Periodic Table). How many Au atoms are there in 196.97 g of Au?
        (196.97 g Au) x (1 kg / 1000 g) x (1 amu / 1.6605402 x 10-27 kg) x (1 Au atom / 196.97 amu) = 6.022 x 1023 Au atoms

      3. The mass of a single glucose molecule is 180 amu (from the sum of the atomic weights of the elements from the Periodic Table). How many glucose molecules are there in 180 g of glucose?
        (180 g glucose) x (1 kg / 1000 g) x (1 amu / 1.6605402 x 10-27 kg) x (1 glucose molecule / 180 amu) = 6.022 x 1023 glucose molecules

      4. This holds true for any atom or molecule - the mass in grams of an atom or molecule equal to its weight in amu contains 6.022 x 1023 of that atom or molecule

      5. Avogadro's number = 6.022 x 1023

      6. 1 mole (x) = Avogadro's number of (x)

    2. Note the relationship: the atomic weight on the Periodic Table is both the mass of a single atom in amu and the mass of a mole of the substance in grams

    3. Molar mass - the sum of the weights of the moles of atoms in one mole of the compound (units of MW are g/mole) note: molar mass and molecular weight, while technically different, are used more ore less synonymously

      1. H2O
        1. The mass of a single molecule: 18.02 amu

        2. The mass of a mole of molecules: 18.02 g

      2. CO2

        1. The mass of a single molecule: 44.01 amu

        2. The mass of a mole of molecules: 44.01 g

      3. C6H12O6

        1. The mass of a single molecule: 180 amu

        2. The mass of a mole of molecules: 180 g

      4. Diazinon is a pesticide with the molecular formula C12H21N2O3PS

        1. The mass of a single molecule: 304.34 amu

        2. The mass of a mole of molecules: 304.34 g

    4. The linking relationship between the microscopic and the macroscopic is the mole. If we know how many moles of a substance we have, we also know how many atom/molecules of the substance we have. Conversely, if we know how many atom/molecules of the substance we have, we know how many moles of the substance we have.

      1. How many moles of carbon are in one mole of diazinon? how many carbon atoms?
        12; (12) x (6.02 x 1023 ) = 7.22 x 1024 carbon atoms

      2. How many molecules are there in 1.4 moles of ethanol?
        (1.4 moles) x (6.02 x 1023 / 1 mole) = 8.43 x 1023 molecules

      3. A sample contains 7.75 x 1015 molecules of ethanol, how many moles is this?
        (7.75 x 1015 molecules) x (1 mole / 6.02 x 1023) = 1.29 x 10-8 mole

      4. A sample of magnesium phosphate weighs 2.50 g, how many molecules is this?
        (2.50 g) x (1 mole / 262.87 g) x (6.02 x 1023 / 1 mole) = 5.73 x 1021 molecules

      5. A chemical assay based on the detection of phosphorus can detect 5.65 pg of magnesium phosphate, how many P atoms are present?
        (5.65 pg) x (1 g / 1012 pg) x (1 mole / 262.87 g) x (6.02 x 1023 / 1 mole) = 1.29 x 1010 molecules x (2 P atoms / molecule) = 2.58 x 1010 P atoms

      6. How many mg will 6.1 x 1020 butane molecules weigh?
        (6.1 x 1020 molecules) x (1 mole / 6.02 x 1023) x (58.12 g / 1 mole) x (1000 mg / 1 g) = 58.9 mg butane

  4. Chemical equations

    1. Chemical equations - shorthand representations of chemical reactions

      1. The reaction of aqueous silver (I) nitrate and aqueous ammonium chloride results in the formation of solid silver (I) chloride and aqueous ammonium nitrate

      2. AgNO3 (aq) + NH4Cl(aq) => AgCl(s) + NH4NO3 (aq)

        1. Reactants and products

        2. Indications of state

        3. Coefficients vs. subscripts

          1. 3 Ba(NO3)2 (aq) + 2 Na3PO4 (aq) => Ba3(PO4)2 (s) + 6 NaNO3 (aq)

    2. Balancing chemical equations

      1. Remember mass balance!

      2. Use the Periodic Table and a knowledge of polyatomic formulas and charges to decide the ratios of ions and how they go together

      3. There is a difference between the subscripts in molecular formulas and the coefficients of balanced chemical equations

      4. Common reaction types

        1. Double displacement reactions

          1. 2 KCl(aq) + Pb(NO3)2 (aq) => PbCl2 (s) + 2 KNO3 (aq)

        2. Combustion reactions

          1. C5H12 (l) + 8 O2 (g) => 5 CO2 (g) + 6 H2O(g)

      5. Examples - don't worry about states (yet)

        1. NaBr + Fe(NO3)3 =>

        2. AlCl3 + BaSO4 =>

        3. Ammonium phosphate and calcium sulfite

        4. Sodium cyanide and gold (III) nitrate

        5. CH4 + O2 =>

        6. The combustion of propanol

  5. Stoichiometry

    1. Definitions

      1. As defined by Ebbing "calculation of the quantities of reactants and products involved in a chemical reaction"

      2. My definition: calculation of the quantities of reactants and/or products based on their relationships in a balanced chemical equation

      3. Note: a balanced chemical equation is essential to stoichiometry; a knowledge of molar masses is often also necessary

    2. Mole relationships and conversion factors: in any balanced chemical equation equivalencies exist between all of the reactants and all of the products

      1. C2H5OH(l) + 3 O2 (g) => 2 CO2 (g) + 3 H2O(g)
        1 mol C2H5OH(l) = 3 mol O2 (g)
        3 mol O2 (g) = 2 mol CO2 (g)
        2 mol CO2 (g) = 3 mol H2O(g)
        1 mol C2H5OH(l) = 2 mol CO2 (g)
        3 mol O2 (g) = 3 mol H2O(g)
         
        1 mol C2H5OH(l) = 3 mol H2O(g)
         
         

    3. These equivalencies are useful when calculating amounts of reactants needed and/or amounts or product formed, both in terms of number of moles and mass

      1. How many moles of ethanol (EtOH) must be burned to produce 16.7 moles of carbon dioxide?
        (16.7 mol CO2) x (1 mol EtOH / 2 mol CO2) = 8.35 mol EtOH

      2. The combustion of 2.78 moles of ethanol will produce how many moles of water vapor?
        (2.78 mol EtOH) x (3 mol H2O / 1 mol EtOH) = 8.34 mol H2O

      3. How many moles of oxygen are required for the complete combustion of 33.6 moles of ethanol?
        (33.6 mol EtOH) x (3 mol O2 / 1 mol EtOH) = 100.8 mol O2

      4. The combustion of ethanol produces 16.62 moles of carbon dioxide. How many moles of water vapor are also produced?
        (16.62 mol CO2) x (3 mol H2O / 2 mol CO2) = 24.93 mol H2O

      5. Mass conversions - these must always go through mole/mole conversions first; cannot do direct mass - mass conversions

        1. How many grams of ethanol must be burned to produce 125 grams of carbon dioxide?
          (125 g CO2) x (1 mol CO2 / 44.01 g CO2) x (1 mol EtOH / 2 mol CO2) x (46.07 g EtOH / 1 mol EtOH) = 65.4 g EtOH

        2. How many grams of oxygen are required for the complete combustion of 1500 grams of ethanol?
          (1500 g EtOH) x (1 mol EtOH / 46.07 g EtOH) x (3 mol O2 / 1 mol EtOH) x (32.0 g O2 / 1 mol O2) = 3125.7 grams of oxygen

        3. The combustion of a certain amount of ethanol produces 6.5 milligrams of carbon dioxide. How many milligrams of water vapor are also formed?
          (6.5 mg CO2) x (1 g CO2 / 1000 mg CO2) x (1 mol CO2 / 44.01 g CO2) x (3 mol H2O / 2 mol CO2) x (18.02 g H2O / 1 mol H2O) x (1000 mg H2O / 1 g H2O) = 3.99 mg H2O

    4. Theoretical yield and percent yield

      1. Theoretical yield: the calculated value

      2. Actual yield: the measured value, seldom the same as the theoretical yield

      3. Percent yield: (actual yield / theoretical yield) x 100

      4. Examples

        1. In V.C.5.a above we calculated that 65.4 g of EtOH must be burned to produce 125 g of CO2. If only 88.6 g of CO2 are produced, what is the percent yield of the reaction?

        2. In V.C.5.b above we calculated that 3125.7 g of O2 are required for the complete combustion of 1500 g of EtOH. If the percent yield of CO2 in this reaction is only 65.0%, how many grams of CO2 are produced?

  6. General properties of aqueous solutions

    1. When anything dissolves in water it is said to be in aqueous solution

      1. Solution: a homogenous mixture consisting of a solvent and one or more solutes

      2. Solvent: "the stuff that does the dissolving" and/or the material present in greatest amount

      3. Solute: "the stuff that gets dissolved" and/or the material(s) present in lesser amount

    2. When ionic compounds dissolve in water they dissociate i.e. they break apart into ions

      1. Not all ionic compounds will dissolve in water

      2. Ions can serve as charge carriers, i.e., aqueous solutions of ions will conduct electricity

    3. Electrolytes and nonelectrolytes

      1. Substances that dissolve and create electrically conductive aqueous solutions are electrolytes

        1. If an ionic compound will dissolve in water it will almost certainly ionize and behave as an electrolyte

      2. Most - but not all - molecular compounds do not conduct electricity when they dissolve; substances that do not create conducting solutions when they dissolved are called nonelectrolytes

      3. Examples

        1. NaCl (aq) => Na+(aq) + Cl-(aq)

        2. C6H12O6 (aq) => C6H12O6 (aq)

    4. Strong, weak, and nonelectrolytes

      1. Strong electrolytes - dissociate (ionize) ~100%

        1. This includes virtually all ionic compounds and molecular compounds like HCl and other mineral acids

      2. Weak electrolytes - dissociate less than 100%, usually 1-10% or less

        1. The most common weak electrolytes are the organic acids

      3. Nonelectrolytes - do not dissociate

        1. This includes virtually all covalent compounds except organic acids and bases

  7. Chemical reactions: an overview

    1. Note that many of the reactions we discuss take place in aqueous solution

    2. Chemical equations are a shorthand way of describing a chemical reaction

    3. There are different types of chemical equations

      1. Molecular equations - all reactants and products are written as complete molecules even though they may exist as ions in solution

        1. NaCl(aq) + AgNO3 (aq) => AgCl(s) + NaNO3 (aq)

      2. Complete ionic equations - strong electrolytes are written as ions if they are in aqueous solution

        1. Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) => AgCl(s) + Na+(aq) + NO3-(aq)

      3. Net ionic equations - spectator ions are canceled and the actual reaction that takes place is left

        1. Cl-(aq) + Ag+(aq) => AgCl(s)

    4. Examples

      1. Write the molecular, complete ionic, and net ionic equations for the double-displacement reaction of barium nitrate and ammonium sulfide

        1. Ba(NO3)2 (aq) + (NH4)2S(aq) => BaS(s) + 2 NH4NO3 (aq)

        2. Ba2+(aq) + 2 NO3-(aq) + 2 NH4+(aq) + S2-(aq) => BaS(s) + 2 NH4+(aq) + 2 NO3-(aq)

        3. Ba2+(aq) + S2-(aq) => BaS(s)

      2. Write the molecular, complete ionic, and net ionic equations for the double-displacement reaction of lithium sulfate and lead acetate

        1. Li2SO4 (aq) + Pb(C2H3O2)2 (aq) => PbSO4 (s) + 2 LiC2H3O2 (aq)

        2. 2 Li+(aq) + SO42-(aq) + Pb2+(aq) + 2 C2H3O2-(aq) => PbSO4 (s) + 2 Li+(aq) + 2 C2H3O2-(aq)

        3. SO42-(aq) + Pb2+(aq) => PbSO4 (s)

    5. For chemical reactions to occur there has to be more than a mixing of chemicals. There has to be a "driving force" that makes the reaction happen.

      1. Formation of a solid (precipitate)

      2. Formation of a pure liquid

      3. Formation of a gas

      4. Electron transfer

  8. Precipitation reactions and solubility guidelines

    1. Precipitation reactions - the mixing of two aqueous solutions of ionic compounds may result in the formation of a new insoluble (will not dissolve) ionic compound called a precipitate

    2. Solubility rules (p. 147) make it possible to predict whether or not an ionic compound is soluble or insoluble

      1. Rule 1: Group 1 cations and ammonium ion are always soluble

      2. Rule 2: acetates and nitrates are always soluble

      3. Rule 3: halogens are always soluble unless the cation is Ag+, Hg22+, Hg2+, or Pb2+

      4. Rule 4: sulfates are always soluble unless the cation is Ag+, Hg22+, Hg2+, Pb2+, Ca2+, Sr2+, or Ba2+

      5. Rule 5: carbonates, phosphates, sulfides, and hydroxides are always insoluble

    3. Examples. Will a reaction occur if the aqueous solutions of the following ionic compounds are mixed? Assume that if a reaction occurs it is a double-displacement reaction:

      1. Ammonium phosphate and iron (III) nitrate

      2. Sodium iodide and mercury (II) acetate

      3. Ammonium sulfide and potassium hydroxide

      4. Silver (I) acetate and cesium sulfate

      5. Rubidium carbonate and chromium (III) chloride

      6. Gold (III) carbonate and nickel (IV) sulfide

  9. Acids, bases, and neutralization reactions

    1. Arrhenius definition: acid - H+ donor, base - OH- donor

    2. Examples of Arrhenius acids and bases, strong and weak acids and bases

    3. Neutralization reactions of Arrhenius acids: double displacement, the hydrogen ion and hydroxide ion combine and form water

    4. Examples

      1. HCl(aq) + NaOH(aq) => H2O(l) + NaCl(aq)

      2. Salt - the ionic compound formed during a neutralization reaction

      3. Polyprotic acids and neutralization reactions: H3PO4 (aq) + 3 NaOH(aq) => 3 H2O(l) + Na3PO4 (aq)

      4. Bases with multiple hydroxide groups and neutralization reactions: 2 HCl(aq) + Ba(OH)2 (aq) => 2 H2O(l) + BaCl2 (aq)

  10. Gas formation

    1. Gas forming reactions: only take place under a very limited set of circumstances

    2. One of the reactants must be an acid, and the anion of the other reactant must be either carbonate, sulfite, or sulfide

      1. 2 H+(aq) + CO32-(aq) => H2CO3 (aq) => H2O(l) + CO2 (g)

      2. 2 H+(aq) + SO32-(aq) => H2SO3 (aq) => H2O(l) + SO2 (g)

      3. 2 H+(aq) + S2-(aq) => H2S(g)

    3. Examples

      1. 2 HCl(aq) + Na2CO3 (aq) => 2 NaCl(aq) + H2O(l) + CO2 (g)

      2. 2 HNO3 (aq) + BaSO3 (aq) => Na(NO3)2 (aq) + H2O(l) + SO2 (g)

      3. 2 HBr(aq) + (NH4)2S(aq) => 2 NH4Br(aq) + H2S(g)

  11. Redox reactions

    1. Terminology

      1. Oxidation-reduction reaction (redox reaction): a reaction in which electrons are transferred between spp. and/or in which atoms involved in the reaction change oxidation number

      2. Oxidation number: a concept devised as way of keeping track of electrons in reactions: the actual charge on a monatomic ion, or the hypothetical charge assigned to an uncharged atom using a set of rules

      3. Oxidation (oxidized): the loss of one or more electrons

      4. Reduction (reduced): the gain of one or more electrons

      5. "OIL RIG:" oxidation is loss, reduction is gain

      6. Oxidizing agent: a chemical that oxidizes something else and reduces itself

      7. Reducing agent: a chemical that reduces something else and oxidizes itself

    2. Rules for determining oxidation numbers

      1. Rule 1: the oxidation number of atoms in their elemental state is zero

      2. Rule 2: the oxidation number of a monatomic ion is equal to its charge

      3. Rule 3: the oxidation number of oxygen is always equal to -2 unless in a peroxide (then -1)

      4. Rule 4: the oxidation number of hydrogen is always +1 unless in a hydride (then -1)

      5. Rule 5: Fluorine always has an oxidation number of -1. The other halogens always have an oxidation of -1 as anions in binary compounds. Halogens listed as the first member of a binary molecular compound or involved in oxyanions have positive oxidation numbers.

      6. Rule 6: for either a neutral compound or for any polyatomic ion, the sum of the oxidation numbers of the atoms in the molecule is equal to the net charge on the specie

    3. Examples of determining oxidation numbers:

      1. NO3-

      2. H2SO4

      3. Fe(OH)2

      4. Li3PO4

      5. HClO3

      6. W2(SO3)3

    4. Balancing simple redox equations: rules

      1. Assign oxidation numbers to all atoms in all reactants and products

      2. Break the reaction down into oxidation and reduction half-reactions

      3. Multiply if necessary to get the number of electrons in each half-reaction to be equal

      4. Add the half-reactions together

    5. Examples of balancing simple redox equations (single displacement reactions)

      1. 2 Ca(s) + O2 (g) => 2 CaO(s)

      2. Fe2O3(s) + 2 Al(s) => 2 Fe(s) + Al2O3 (s)

      3. Mg(s) + 2 HCl(aq) => MgCl2 (aq) + H2 (g)

    6. Common types of redox reactions

      1. Combination reactions - two substances combine to form a third substance (the formation of table salt from sodium and chlorine)

      2. Decomposition reactions - a single compound reacts to form two or more new substances (the formation of hydrogen and oxygen from water)

      3. Displacement reactions (single displacement reactions) - an element reacts with a compound and replaces an element in the compound

      4. Combustion reactions - a substance reacts with oxygen with the rapidly release of heat and light energy - oxygen is oxidized in combustion reactions

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