Chapter 3. Molecules, Moles, and Chemical Equations

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Chapter 3 Molecules, Moles, and Chemical Equations

Law of Conservation of Matter Matter is neither created nor destroyed. In chemistry the number of atoms going into a chemical reaction must be the same as the number of atoms coming our of the reaction. Reactions must be balanced.

Chemical Equation or Reaction: Statement in formulas depicting a chemical change. Reactants Products (initial) (final)

Example: What chemical reaction occurs when a flashbulb is used to take a photograph? MACROSCOPIC VIEW BALANCED EQUATION 2Mg(s) + O 2 (g)

MACROSCOPIC VIEW ATOMIC-SCALE VIEW Mg Mg BALANCED EQUATION 2Mg(s) + O 2 (g)

MACROSCOPIC VIEW ATOMIC-SCALE VIEW Mg Mg O 2 BALANCED EQUATION 2Mg(s) + O 2 (g)

MACROSCOPIC VIEW ATOMIC-SCALE VIEW Mg Mg electricity O 2 BALANCED EQUATION 2Mg(s) + O 2 (g) electricity

MACROSCOPIC VIEW ATOMIC-SCALE VIEW Mg Mg electricity O 2 BALANCED EQUATION 2Mg(s) + O 2 (g) electricity 2MgO(s)

MACROSCOPIC VIEW ATOMIC-SCALE VIEW Mg Mg electricity Mg 2+ O 2 O 2 Mg 2+ O 2 BALANCED EQUATION 2Mg(s) + O 2 (g) electricity 2MgO(s)

translate the statement balance the atoms adjust the coefficients check the atom balance specify states of matter

Write a balanced chemical equation for the following: If aqueous silver nitrate is mixed with aqueous sodium phosphate a chemical reaction occurs and solid silver phosphate and aqueous sodium nitrate are formed. Take out and complete Exercise 5 pg 26 of Lab Manual.

Interpreting Equations and the Mole

Common Terms that Represent Specific Quantities 1 pair = 2 1 dozen = 12 1 baker s dozen = 13 1 case = 24 1 gross = 144 1 mole = 6.022 X 10 23

mole(mol) - the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12. This amount is 6.022x10 23. The number is called Avogadro s number and is abbreviated as N. One mole (1 mol) contains 6.022x10 23 entities (to four significant figures)

One mole of common substances. Oxygen 32.00 g CaCO 3 100.09 g Water 18.02 g Copper 63.55 g

Summary of Mass Terminology Term Definition Unit Isotopic mass Mass of an isotope of an element amu Atomic mass (also called atomic weight) Molecular (or formula) mass (also called molecular weight) Molar mass (M) (also called grammolecular weight) Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units) amu amu g/mol

Let s Consider 1.00 mole of Si 1 Si atom = 28.09 amu 1 mole Si atoms = 28.09 g Here s the Proof! 1.00 mol Si X 6.022X10 23 Si X 28.09amu X 1.66054X10-24 g = 28.09 g Si 1 mol Si 1 Si atom 1 amu Now you have two unit factors depending upon whether you are talking about individual atoms or a group of atoms.

Information Contained in the Chemical Formula of Glucose C 6 H 12 O 6 ( M = 180.16 g/mol) Carbon (C) Hydrogen (H) Oxygen (O) Atoms per molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms per mole of compound 6 moles of atoms 12 moles of atoms 6 moles of atoms Atoms per mole of compound 6(6.022 x 10 23 ) atoms 12(6.022 x 10 23 ) atoms 6(6.022 x 10 23 ) atoms Mass of atoms per molecule of compound Mass of atoms per mole of compound 6(12.01 amu) =72.06 amu 12(1.008 amu) =12.10 amu 72.06 g 12.10 g 6(16.00 amu) =96.00 amu 96.00 g Molar Mass of Compound: 72.06g + 12.10g + 96.00g = 180.16 g/mol

Interconverting Moles, Mass, and Number of Chemical Entities Mass (g) = no. of moles x no. of grams 1 mol g No. of moles = mass (g) x 1 mol no. of grams n No. of entities = no. of moles x 6.022x1023 entities 1 mol No. of moles = no. of entities x 1 mol 6.022x10 23 entities

Another Example: What is the formula unit mass of K 2 S? What is the molar mass of K 2 S? What is the mass of 0.250 moles of K 2 S?

Conversion Problem Solving 1. How many atoms of lead are present in 15 g of lead? 2. What is the mass of 1.36X 10 21 atoms of silver? 3. How many moles of Ba(NO 3 ) 2 are in 20.0 grams of Ba(NO 3 ) 2? 4. How many atoms of oxygen are present in 20.0 grams of Ba(NO 3 ) 2? 5. How many moles of nitrate ions are present in 20.0 grams of Ba(NO 3 ) 2?

EXERCISE 4 Work in Groups and complete this exercise.

Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the empirical formula.

Example Acetylene C 2 H 2 Benzene C 6 H 6 What is the empirical formula for each of the above compounds? Note: Two completely different compounds can have the same empirical formula.

Determination of Empirical and Molecular Formulas Example: A compound of nitrogen and oxygen is analyzed and a sample weighing 1.587 grams is found to contain 0.483 grams of nitrogen and 1.104 grams of oxygen. Determine the empirical formula of this compound.

Example: A compound contains 68.8% C, 5.0% H, and 26.2 % O by mass. The molecular mass of the compound is determined to be 122.13 amu. Determine the empirical and molecular formulas of the compound.

Example: A vitamin is analyzed and found to contain 40.9%C, 4.58%H, and 54.5% O. The molar mass of the compound is 176.13 g/mol. Determine the empirical and molecular formula.

Combustion Analysis: a procedure used to determine the chemical composition of organic compounds. General Equation for Combustion Analysis Reactions Compound + n O 2(g) n CO 2(g) + n H 2 O (g) + other elements

Example: A compound is combusted and the products are massed. Using the information provided below, determine the empirical and molecular formulas of the compound. Products: 6.21 mg CO 2 2.54 mg H 2 O Compound Mass: 4.24 mg Molar Mass of Compound: 60.6 g/mol

Some Compounds with Empirical Formula CH 2 O (Composition by Mass: 40.0% C, 6.71% H, 53.3% O) Name Molecular Formula Whole-Number Multiple (g/mol) Use or Function formaldehyde acetic acid lactic acid erythrose CH 2 O C 2 H 4 O 2 C 3 H 6 O 3 C 4 H 8 O 4 1 2 3 4 30.03 60.05 90.09 120.10 disinfectant; biological preservative acetate polymers; vinegar(5% soln) sour milk; forms in exercising muscle part of sugar metabolism ribose glucose C 5 H 10 O 5 C 6 H 12 O 6 5 6 150.13 180.16 component of nucleic acids and B 2 major energy source of the cell CH 2 O C 2 H 4 O 2 C 3 H 6 O 3 C 4 H 8 O 4 C 5 H 10 O 5 C 6 H 12 O 6

Solutions Solute + Solvent = Solution x x x x x x x x x x x x x x x = solute particles More Dilute More Concentrated

Molarity Molarity is a form of representing concentration of a solution. Molarity (M) = moles of Solute liters of Solution Units of Molarity: moles/l

Calculating Molarity Example: A sample of sodium nitrate weighing 0.38 grams is dissolved in 50.0 ml of solution. Determine the molarity of the solution.

Laboratory preparation of molar solutions. A Weigh the solid needed. Transfer the solid to a volumetric flask that contains about half the final volume of solvent. B Dissolve the solid thoroughly by swirling. C Add solvent until the solution reaches its final volume.

Problem Solving with Molarity 1. A student must prepare 250.0 ml of a 0.250 M sodium hydroxide solution. What mass of solute is required? 2. How many ml of 0.163 M sodium chloride solution are required to contain 0.0958 grams of solute?

Converting a concentrated solution to a dilute solution. M 1 V 1 = M 2 V 2 Or M d V d = M c V c

Dilution Problem Solving Explain how you would prepare 100.0 ml of a 0.100 M ammonia solution from 14.8 M ammonia.

The Role of Water as a Solvent The Universal Solvent: H 2 O Let s consider the (1) shape of the water molecule. (2) bond angle. (3) unequal sharing of electrons. (4) partial charges. (5) dipole moment.

Electron distribution in molecules of H 2 and H 2 O.

The dissolution of an ionic compound in water. The separation of an ionic compound caused by water molecules results in a solution containing solvated ions.

+ + + + + + +

+ + + + + + +

+ + + + + + +

+ + + + + + + +

The electrical conductivity of ionic solutions.

Solvation of Covalent Compounds by H 2 O Water does not dissociate some covalent compounds (nonelectrolytes). Examples: sucrose, ethanol, acetone, etc. Water can slightly dissociate some covalent compounds (weak electrolytes). Examples: HC 2 H 3 O 2 HC 7 H 5 O 2 etc. Note: Why is there one H in the front of the chemical formula?

Water can completely dissociate some covalent compounds (strong electrolytes). HCl HNO 3 HI etc. Water dissociates soluble ionic compounds completely into ions (strong electrolytes).

Useful Tables Page 17 of lab manual: Summary of Electrolyte Rules Page 16 of lab manual: Solubility Rules Do not memorize. However, be prepared to use similar tables on the exam.

Problem Solving When one mole of the following compounds dissociates in H 2 O, how many moles of each ion are present? Aluminum chloride Copper (II) nitrate Sodium phosphate

Problem Solving If 150.00 ml of 0.200 M iron (III) sulfate are present in a beaker, how many moles of each ion are present?

Acid-Base Reactions (Neutralization) Acids Release H + Turn blue litmus paper pink ph < 7 Electron pair acceptors React with some metals to produce H 2(g) Taste Sour

Bases Release OH - Turn pink litmus paper blue ph > 7 Electron pair donors React with many metal ions to produce precipitates (ppt s) Taste Bitter

Dissociation of Strong Acids in H 2 O HCl(g) + H 2 O (l) H 3 O + (aq) + Cl - (aq) (hydronium ion)

The hydrated proton.

Dissociation of Weak Acids in H 2 O HF(g) + H 2 O (l) H 3 O + (aq) + F - (aq) F - = Anion of a weak acid

GENERAL Neutralization Reaction Acid + Base H 2 O (l) + Salt HCl (aq) + KOH (aq) H 2 O (l) + KCl (aq)

An aqueous strong acid-strong base reaction on the atomic scale.

Determining the Molarity of H + Ions in Aqueous Solutions of Acids Nitric acid is a major chemical in the fertilizer and explosives industries. Nitric acid is a strong acid/strong electrolyte. What is the molarity of H + (aq) in 0.900M nitric acid?

Writing Equations for Aqueous Ionic Reactions The molecular equation: shows all of the reactants and products as intact, undissociated compounds. The total ionic equation: shows all of the soluble ionic substances dissociated into ions. The net ionic equation: eliminates the spectator ions and shows the actual chemical change taking place.

How to Begin Writing Net Ionic Equations Write the molecular equation and balance. Determine the solubility of each compound. Write the total ionic equation. Dissociate ionic compounds that are soluble. Dissociate strong acids and bases. (Solubility Rules) Do not dissociate weak acids, weak bases, solids, or gases. (Electrolyte Rules) Determine the spectator ions. Write the balanced net ionic equation (w/o spectators).

A precipitation reaction and steps to writing its net ionic equation.

Problem Solving Aqueous barium chloride reacts with aqueous sodium phosphate to produce aqueous sodium chloride and barium phosphate. Write the net ionic equation for this reaction.

Another Example: The reaction of Pb(NO 3 ) 2 and NaI. double displacement reaction (metathesis)

Another Example: An acid-base reaction that forms a gaseous product. Sodium carbonate reacts with acetic acid

Practice Writing Net Ionic Equations Exercise 7 of Lab Manual, page 30.