Chemistry Chapter 4: Moles and Mass

The Mole and Stoichiometry of Formulas and Equations

• The mole (mol) is the amount of substance containing the same number of entities as the number of atoms in exactly 12 g of carbon-12
• 'Entities' refer to atoms, ions, molecules, formula units or electrons—any particle type
• One mole (1 mol) contains 6.022x10²³ entities (to four significant figures)
• This number is called Avogadro's number (N_A)
• 1 mol of carbon-12 contains 6.022x10²³ carbon-12 atoms and has a mass of 12 g
• Mass in atomic mass units (amu) of one atom/molecule/formula unit is numerically the same as the mass in grams of 1 mole of atoms/compound

Photosynthesis

• Photosynthesis is a process where plants use light energy to convert carbon dioxide and water into sugar
• Chemical formula for photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

Stoichiometry of Formulas and Equations (Page 3)

• The mole
• Formula of compounds
• Writing and balancing chemical equations
• Calculating quantities of reactants and products
• Reversibility of reactions and equilibrium state

Molar Mass

• The molar mass (M) of a substance is the mass per mole of its entities (atoms, molecules, or formula units).
• Unit is g/mol
• For monatomic elements, the molar mass is the same as its atomic mass in grams per mole. Atomic mass is from the Periodic Table.
• The molar mass of Neon (Ne) = 20.18 g/mol
• To find molar mass of diatomic molecules and compounds, add molar masses of atoms in the formula

Example: Chemical Formula of Glucose

• Subscripts in a formula refer to both the atoms and moles of atoms
• Glucose: C₆H₁₂O₆
• 6 atoms of Carbon, 12 atoms of Hydrogen, 6 atoms of Oxygen
• Moles of atoms per mole of compound (glucose): 6 mol C, 12 mol H, 6 mol O
• Atoms/mole of compound: 6x(6.022x10²³) atoms C, 12x(6.022x10²³) atoms H, 6x(6.022x10²³) atoms O
• Mass/molecule of compound (glucose): 72.06 amu C, 12.10 amu H, 96.00 amu O ; Total = 180.16 amu
• Mass/mole of compound (glucose): 72.06 g C, 12.10 g H, 96.00 g O; Total = 180.16 g

Moles, Mass, and Number of Chemical Entities

• Mass (g) = no. of moles x molar mass (g/mol)
• No. of moles = mass (g) / molar mass (g/mol)
• No. of entities = no. of moles x Avogadro's number (6.022 x 10²³ entities/mol)
• No. of moles = no. of entities / Avogadro's number

Amount-Mass-Number Relationship

• Relationship between mass (g), amount (mol), and number of atoms/molecules of elements
• Mass (g) of element is related to molar mass (g/mol) and Amount (mol)
• Amount (mol) of element is related to Avogardro's number and number of atoms/molecules

Sample Problems (Page 11, 12, 13, 14, and 17)

• Students will find examples or problems calculating between mass and amount of an element or compound, relating amount/number of entities to amount of an element, converting between number of entities and mass of compound

Mass Percent

• Mass % of an element = (atoms of X in formula x atomic mass of X (amu)) / molecular mass of compound (amu) x 100%
• or, Mass % of element X = (moles of X in formula x molar mass of element X (g/mol)) / mass (g) of 1 mol of compound x 100%

Mass of an Element

• Mass of any element in sample = mass of compound x (mass of element in 1 mol of compound / mass of 1 mol of compound)

Sample Problem 4

• Determining the mass of an element in a compound (Calculating the mass of nitrogen in ammonium nitrate)

Empirical and Molecular Formulas

• Empirical formula is the simplest formula for a compound that agrees with elemental analysis (shows lowest whole number ratio of atoms)
• Molecular formula is the actual number of atoms of each element in the compound (actual # of atoms per molecule)
• Structural formula shows the relative placement and connections of atoms in the molecule

Determining the Empirical Formula

• Finding the empirical formula of a compound
• Finding mass (g) of each component
• Convert each mass (g) to amount (mol) to write a preliminary formula
• Convert the amount (mol) into integer subscripts to get the formula

Determining the Molecular Formula

• The molecular formula is a whole-number multiple of the empirical formula.
• (Molar mass (g/mol)) / (empirical formula mass (g/mol)) = whole number multiple.

Sample Problem 5

• Determining the empirical and molecular formulas of lactic acid (Example of compound and mass percentages of constituent elements) (mass % C, mass % H, mass % O)

Chemical Formulas and Molecular Structures

• Empirical formula does not show molecular structure
• Molecular formula does not show molecular structure
• Isomers are compounds with the same molecular formula but different properties

Compounds with Empirical Formula CH₂O

• List of compounds with the same empirical formula, their molecular formulas, whole number multiples, and uses

Chemical Equations

• A chemical equation represents identities and quantities in a physical or chemical change

Balancing a Chemical Equation

• Balancing equations is needed to follow the Law of Conservation of Mass
• Steps to balance equations:
  • Step 1: Write a skeleton equation
  • Step 2: Balance the atoms (match the numbers of each type of atom on left and right sides of reaction)
  • Step 3:Adjust coefficients to balance
  • Step 4: Check the balance
  • Step 5: Specify states of substances (solid, liquid, gas, or aqueous)

A Three-level View of a Reaction

• Shows macroscopic level reaction, atomic level reaction (with diagrams of atoms rearranging), and the balanced chemical equation

Balancing Process

• Chemical formulas cannot be altered, only coefficients can be altered
• Balancing equations implies maintaining same number of atoms on each side of reaction
• Coefficients can be modified in order to achieve a balanced equation

Sample Problem 6

• Balancing a chemical equation related to combustion of octane in a car's engine

Visualizing a Reaction with a Molecular Scene

• Visual representation of reactants and products

Stoichiometric Calculations

• Stoichiometry is the chemical arithmetic related to mole-mass conversion
• In a balanced equation, amounts in moles of substances are stoichiometrically equivalent to one another.
• Quantitative relationships are expressed as molar ratios

Information Contained in a Balanced Equation

• Reactants and products showing quantities in terms of molecules and amounts in moles and grams

Amount-Mass-Number Relationships

• Relationship between mass, amount, and number of entities through balanced equations

Sample Problem 7

• Calculating quantities using amounts in moles and relating them to mass in grams of a specific reactant or product

Reactions in Sequence

• Multiple overall reaction steps in which one substance forms as a reactant is reacted or produced.
• Steps to write overall equation:
  • Step 1: Write the separate balanced equations for each reaction
  • Step 2: Adjust the equations arithmetically to cancel common substances (from 1)
  • Step 3: Add the adjusted equations together

Sample Problem 8

• Example of calculating overall equation for a specific reaction sequence

Limiting Reactants

• Defined as the substance in a reaction that is present in limiting amount and completely used up before the other reactants
• Determines the maximum amount of product that can be formed in a chemical reaction
• Excess Reactants are any other reactants that are still present after the limiting reactant is used up (and determines how much reactant is left)

Limiting Reactants - Example

• Example problem of finding a limiting reactant and/or excess reactant; including calculation from a balanced chemical equation

Reaction Table

• Illustrates the initial amount, change (of moles), and final amount (of moles) of Reactants and Products in a balanced reaction

Sample Problem 9

• Example calculations based on amount to amount calculations in a limiting reactant reaction

Stoichiometric Relationships

• Shows the relationship among various quantities in chemical reactions: mass, amount (mol), and number of atoms/molecules (of each element, from both reactants and products)

Reaction Yields

• Shows the relationship between theoretical yield (calculated yield of product using the balanced reaction and given mass) vs. the actual yield (actual amount produced by performing an experiment)
• Theoretical yield vs. Actual yield

Percent Reaction Yield

• The percent yield of product = (actual yield/theoretical yield) x 100%

Sample Problem 10

• Example of calculating the percent yield. (Example problem about determining percent yield in processing SiC from reacting sand with carbon)

The Equilibrium State

• Equilibrium reaction in open and closed container
• Balanced reaction (Reactants vs Products)
• Equilibrium reactions in closed vs open containers