IV. Stoichiometry and Solution

Key focus of this chapter: percent yield of chemical reactions

This chapter reviews percent yield, the mole, empirical and molecular formulas, balancing equations, solution concentrations, precipitation, acid–base neutralization, oxidation numbers, net ionic equations, and balancing redox reactions.

A. Density, D

  • Question: What is the density of a sample with a mass of 2 g and a volume of 5 mL, measured in a graduated cylinder?
Mass percent of an element equals its total atomic-mass contribution divided by the molecular or formula mass, times100 percent.

B. Percentage of composition

1. Molecular and formula mass

: the sum of atomic masses in one molecule (molecular mass) or one formula unit (formula mass), expressed in amu. The corresponding molar mass has the same numerical value in g/mol.

  • Example: Molecular mass of H2O → 1×2+16 = 18 amu; molar mass = 18 g/mol.
  • Example: Formula mass of Fe(ClO)3 → 56 + (35+16)×3 = 209 amu; molar mass = 209 g/mol (using the rounded atomic masses shown).

2. Percentage of mass composition

: the mass of a given element divided by the total mass of the compound, multiplied by 100%. Count all atoms of that element in the chemical formula.

  • Question: What is the mass percent of hydrogen in C2H4?
Mass percent of hydrogen in C2H4 = (4 × atomic mass of hydrogen)/(total atomic mass of C2H4) × 100 = 14.3%.

C. Mole

  • Mole: the SI unit of amount of substance; one mole contains exactly 6.02214076 × 10²³ specified entities, such as atoms, molecules, or ions.
  • Avogadro’s law: equal volumes of ideal gases at the same temperature and pressure contain equal numbers of particles.
  • Avogadro’s number: 6.02×1023 specified entities per mole (approximately); this count is independent of temperature and pressure.
  • Example: Amounts of the constituent elements in a compound

1 mol

4 mol

3 mol

4×3 = 12 mol

2 mol

4×2 = 8 mol

40×3+14×2

= 148 g

4(40×3+14×2)

= 592g

  • Example: Numbers of atoms and masses of elements in a compound

Compound

Mole of

Calcium

Nitrogen

Particles of

Calcium

Nitrogen

Grams  of

74g

Calcium

40g/mol × 1.5 mol Ca = 60g

40g/mol × 6.0 mol Ca = 240g

Nitrogen

14g/mol × 1.0 mol N = 14g

14g/mol × 4.0 mol N = 56g

D. Empirical and molecular formula

  • Empirical formula: the simplest whole-number ratio of atoms of each element in a compound.
  • Molecular formula: the actual number of atoms of each element in one molecule.

Unknown

compound

186g P, 240g O,

M.W= 284g/mol

85.8% C, 14.2% H,

M.W= 84g/mol

M. W. of each atom

P = 31g/mol

O = 16g/mol

C = 12g/mol

H = 1g/mol

Moles of each atom

For a100g sample with85.8percent C and14.2percent H, carbon85.8/12 equals7.15mol and hydrogen equals14.2mol.

Relative mol numbers

Intermediate relative mole ratio C7.15H14.2.

Empirical formula

Dividing by7.15 gives C:H approximately1:2 and empirical formula CH2.

M.W. of Empirical formula

Molecular formula

E. Balancing equations

Adjust coefficients so that each element has the same number of atoms on both sides. Do not change chemical subscripts. Balanced ionic equations must also conserve total charge.

  • Start with the most complex compound first.
  • Question: Write a balanced chemical equation for the unbalanced reaction.

Fe2O3(s) + C(s)  →   Fe(s) + CO2(g)

Balanced for Fe

Balanced for O

Balanced for C

Multiply by 2 to make the simplest ratio

F. Complete combustion

: complete burning of a hydrocarbon fuel in sufficient oxygen (CH4, C2H6, C3H8, or C4H10) with O2, producing CO2 and H2O

     Fuel(g) + O2(g)  →  CO2(g) + H2O(l)

  • Question: Write a balanced equation for the complete combustion of methane gas in air.

Solution: Combustion equation:  CH4(g) + O2(g)  →   CO2(g) + H2O(l)

Balanced equation:   CH4(g) + 2O2(g)  →  CO2(g) + 2H2O(l)

G. Yields of chemical reaction

1. Limiting reactant

: the reactant that is consumed first and determines the maximum amount of product that can form.    

  • Question: Identify the limiting reactant when the reaction is started with 1 mol N2 and 2 mol H2.

     N2 + 3H2  →  2 NH3

Solution: Compare the available moles of each reactant divided by its stoichiometric coefficient in the balanced equation.

1) Figure out moles of each reactant from the information given 1 mole N2 , 2 moles H2

2) Calculate n/coefficient: N2: 1/1 = 1 mol; H2: 2/3 mol. The smaller value identifies H2 as the limiting reactant.

Comparing n/coefficient identifies H2 as limiting for1molN2 and2molH2.

3) Determine which reactant is associated with the smaller n/coefficient value (in this case, H2).

This is the limiting reactant.

2. Percent yield

  • Actual yield: the amount of product obtained experimentally
  • Theoretical yield: the maximum amount of product predicted from the limiting reactant and the balanced equation

Question: When 640.0 g Fe2O3 reacts with 60.0 g C, the actual yield of Fe is 300.0g. Calculate the percent yield of Fe from the reaction.

2Fe2O3(s) + 3C(s)  →  4Fe(s)  + 3CO2(g)

Solution:

M.W. of each compound

Fe2O3: 56×2+16×3 = 160 g/mol

C: 12 g/mol

Fe: 56 g/mol

Moles of each reactant

Limiting reactant

Comparing4molFe2O3 divided by2 with5molC divided by3 identifies carbon as limiting.

Moles of product

Theoretical yield of

Fe

56 g/mol Fe × (20/3) mol Fe ≈ 373.3 g

Percent yield of Fe

H. Concentration

1. Solute, solvent, and solution

Category


Features

Solute

• Dissolved in solvent

• Usually present in a smaller amount than the solvent

• Can be solid, liquid, or gas

Solvent

• Can dissolve solute

• Usually the component present in the greatest amount

• Can be solid, liquid, or gas

Solution

• Solute + solvent

2. Concentration

Category


Features

Mass percent

(mass %)

Molarity (M)

Molality (m)

Normality (N)

(equivalents per liter)

(reaction dependent)


Normality is reaction dependent: N equals equivalent factor times molarity;3M sulfuric acid is6N for complete neutralization.

Dilution and neutralization

(use the appropriate

stoichiometric relation)


NV = NV

(neutralization)

At the specified neutralization endpoint, acid and base equivalents are equal: NaVa=NbVb and faMaVa=fbMbVb.
  • Question: Calculate the mass percent of NaCl when 30 g of NaCl is dissolved in 70 g of water.
  • Question: Calculate the molarity (M) of a solution containing 800 g of NaOH in a total volume of 4 L.
  • Question: Calculate the molality (m) when 800 g of NaOH is dissolved in 5 kg of water.
  • Question: For complete neutralization of all three acidic protons, calculate the normality when 10 mol H3PO4 is present in 5 L of solution.
Ten moles of phosphoric acid in5L is2M and6N for complete neutralization of all three acidic protons.
  • Question: What volume of 4 M HCl is required to neutralize 30 mL of 8 M KOH?
  • Question: What volume of 8 M HBr is required to neutralize 40 mL of 3 M Ba(OH)2?

This is an idealized stoichiometry exercise: assume the stated Ba(OH)2 is fully dissolved and dissociated. The given concentration is used for the calculation and does not establish its solubility at room temperature.

I. Aqueous reactions

The hydroxide neutralization examples below produce water. Acid–base reactions more generally involve proton transfer and do not always form water. Electrolyte strength describes the extent of ionization or dissociation in solution; it is distinct from solubility.

Category


Features

Precipitation

Balanced silver-bromide precipitation and common solubility guidelines with separate halide and sulfate exceptions.

Acid–base reaction


Electrolyte classes, dissociation, and non-electrolytes including CH3OH methanol. Strength is distinct from solubility.

Redox reaction

Oxidation loses electrons and increases oxidation number; reduction gains electrons and decreases oxidation number.
Chlorine oxidizes bromide; Cl changes0 to minus1 and Br minus1 to0 per atom, with balanced half-reactions.

J. Oxidation numbers

: a formal bookkeeping value assigned by an ionic approximation of bonding. It tracks oxidation and reduction and is not the same as an atom’s actual partial charge. The sum of oxidation numbers equals the total charge of the molecule or ion.

In common compounds, fluorine is −1 and oxygen is usually −2. Exceptions include peroxides (oxygen −1) and OF2 (oxygen +2). Hydrogen is usually +1 with nonmetals and −1 in metal hydrides. Halogens other than fluorine can have positive oxidation numbers when bonded to oxygen or a more electronegative halogen.

Classifications

Oxidation number


Elemental state atoms

Ionic state atoms

Atoms in

polyatomic

ion or molecule

H

O

Halogens

Cl, Br, and I can have positive oxidation numbers when bonded to oxygen. HOI: I(+1); KClO3: Cl(+5).

Sum of oxidation numbers


Neutral compound


Polyatomic ion


K. Net ionic equation

1. Precipitation

  • Question: Write the net ionic equation for the following reaction.

Pb(NO3)2(aq) + 2NaI(aq)  →

Processing

Equation

Dissociation

Reaction equation

Completed ionic

 equation

Net ionic equation

2. Acid–base neutralization

  • Question: Write the net ionic equation for the following reaction.

HNO3(aq) + KOH(aq)  →

Processing

Equation

Dissociation

Reaction equation

Complete Ionic

 equation

Net Ionic equation

3. Redox reaction

  • Question: Write the net ionic equation for the following reaction.

2HCl(aq) + Zn(s)  →

Processing

Equation

Dissociation

Reaction equation

Complete ionic

 equation

Net ionic equation

L. Balancing redox reactions (half-reaction method)

  • Question: Balance the following redox reaction in acidic aqueous solution.

NO2–(aq)  +  I–(aq)  →  I2(s)  +  NO(g)

Processing

Equation

Separation into oxidation and reduction

Balance atoms except H, O

Balance O by adding H2O

Balance H by adding H+

Balance charges by adding electrons

Multiply by coefficient to cancel electrons

Cancel electrons

    and balance redox reaction

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