Chapter 3 Molecules, Moles, and Chemical...

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CHEM 101-GENERAL CHEMISTRY CHAPTER 3

MOLECULES, MOLES & CHEMICAL EQUATIONS

INSTR : FİLİZ ALSHANABLEH

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CHAPTER 3 MOLECULES, MOLES & CHEMICAL EQUATIONS

• Chemical Formulas and Equations

. Writing Chemical Equations

. Balancing Chemical Equations

• Aqueous Solution and Net Ionic Equations

. Solutions, Solvents and Solutes

. Chemical Equations for Aqueous Solutions

• Interpreting Equations and The Mole

. Avogadro’s Number and the Mole

. Determinig Molar Mass

• Calculations Using Moles and Molar Masses

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Explosions

1. Explosions release a large amount of energy when a fairly complex molecule decomposes into smaller, simpler compounds.

2. Explosions occur very quickly.

3. Modern explosives are generally solids.

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Explosions • Dynamite is an explosive made from liquid nitroglycerin and an inert binder to form a solid material.

Solids are easier to handle than liquids The destructive force of explosions is due in part to expansion of gases, which produces a shockwave.

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Chemical Formulas and Equations

• Chemical formulas provide a concise way to represent chemical compounds. • Nitroglycerin, shown earlier, becomes C3H5N3O9

• A chemical equation builds upon chemical formulas to concisely represent

a chemical reaction.

• Chemical equations represent the transformation of one or more chemical species into new substances.

• Chemical formulas represent reactants and products.

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Writing Chemical Equations

• Phase labels follow each formula.

• solid = (s) • liquid = (l) • gas = (g) • aqueous (substance dissolved in water) = (aq)

• Some reactions require an additional symbol placed over

the reaction arrow to specify reaction conditions. • Thermal reactions: heat (∆) • Photochemical reactions: light (hν)

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Writing Chemical Equations

• Different representations for the reaction between hydrogen and oxygen to produce water.

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Balancing Chemical Equations

• The law of conservation of matter: matter is neither created nor destroyed. • Chemical reactions must obey the law of conservation of

matter. • The same number of atoms for each element must

occur on both sides of the chemical equation.

• A chemical reaction simply rearranges the atoms into new compounds.

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• Balanced chemical equation for the combustion of methane.

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• Chemical equations may be balanced via inspection, which really means by trial and error. • Numbers used to balance chemical equations are called

stoichiometric coefficients. • The stoichiometric coefficient multiplies the number of

atoms of each element in the formula unit of the compound that it precedes.

• Stoichiometry is the various quantitative relationships

between reactants and products.

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Exercise

Which of the following correctly balances the chemical equation given below? There may be more than one correct balanced equation. If a balanced equation is incorrect, explain what is incorrect about it.

CaO + C → CaC2 + CO2

I. CaO2 + 3C → CaC2 + CO2 II. 2CaO + 5C → 2CaC2 + CO2 III. CaO + (2.5)C → CaC2 + (0.5)CO2

IV. 4CaO + 10C → 4CaC2 + 2CO2

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Aqueous Solutions and Net Ionic Equations

• Reactions that occur in water are said to take place in aqueous solution. • Solution: homogeneous mixture of two or more

substances.

• Solvent: solution component present in greatest amount.

• Solute: solution component present in lesser amount.

• The preparation of a solution is a common way to

enable two solids to make contact with one another.

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Solutions, Solvents, and Solutes

• For solutions, the concentration is a key piece of information. • Concentration: relative amounts of solute and solvent.

• Concentrated: many solute particles are present.

• Dilute: few solute particles are present.

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• Compounds can be characterized by their solubility. • Soluble compounds dissolve readily in water.

• Insoluble compounds do not readily dissolve in water.

• Solubility can be predicted using solubility guidelines.

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• Solubility guidelines for soluble salts

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Exercise

• Which of the following compounds would you predict are soluble in water at room temperature? a) Mg(NO3)2

b) CaCO3

c) BaSO4

d) KMnO4

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• Electrolytes are soluble compounds that conduct current when dissolved in water.

• Weak electrolytes dissociate partially into ions in solution.

• Strong electrolytes dissociate completely into ions in

solution.

• Nonelectrolytes do not dissociate into ions in solution.

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a) Sugar, a nonelectrolyte, does not conduct current when dissolved in water. b) Acetic acid, a weak electrolyte, weakly conducts current when dissolved in

water. c) Potassium chromate, a strong electrolyte, strongly conducts current when

dissolved in water.

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Chemical Equations for Aqueous Reactions

• When a covalently bonded material dissolves in water and the molecules remain intact, they do not conduct current. These compounds are nonelectrolytes.

• When an ionic solid dissolves in water, it breaks into its constituent ions. This is called a dissociation reaction. These compounds conduct electric current and are electrolytes.

C6H12O6 (s) → C6H12O6 (aq)

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Chemical Equations for Aqueous Reactions

• Aqueous chemical reactions can be written as a molecular equation. The complete formula for each compound is shown.

• Dissociation of reactants and products is emphasized by writing a total ionic equation.

• Spectator ions are ions uninvolved in the chemical reaction. When spectator ions are removed, the result is the net ionic equation.

• Net ionic equation:

HNO3(aq) + NH3(g) → NH4NO3(aq)

H+(aq) + NO3− (aq) + NH3(g) → NH4

+(aq) + NO3− (aq)

H+(aq) + NH3(g) → NH4+(aq)

Spectator ion = NO3−

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Exercise

Write the molecular equation, total ionic equation, and net ionic equation for the reaction between cobalt(II) chloride and sodium hydroxide. Molecular Equation: CoCl2(aq) + 2NaOH(aq) → Co(OH)2(s) + 2NaCl(aq) Total Ionic Equation:

Co2+(aq) + 2Cl−(aq) + 2Na+(aq) + 2OH−(aq) →

Co(OH)2(s) + 2Na+(aq) + 2Cl−(aq)

Net Ionic Equation: Co2+(aq) + 2Cl−(aq) → Co(OH)2(s)

• Na+ and Cl− are spectator ions.

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Acid-Base Reactions

• Acids are substances that dissolve in water to produce H+ (or H3O+) ions. • Examples: HCl, HNO3, H3PO4, HCN

• Bases are substances that dissolve in water to produce OH–

ions. • Examples: NaOH, Ca(OH)2, NH3

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Acid-Base Reactions

• Strong acids and bases completely dissociate in water.

• Weak acids and bases partially dissociate in water.

• Notice the two-way arrows, which emphasize that the reaction does not proceed completely from left to right.

HCl(g) + H2O(l ) → H3O+(aq) + Cl− (aq)

NaOH(s) → Na+(aq) + OH− (aq)

CH3COOH(aq) + H2O(l ) →← H3O+(aq) + CH3COO− (aq)

NH3(aq) + H2O(l ) →← NH4+(aq) + OH− (aq)

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Acid-Base Reactions

• Mixing an acid and a base leads to a reaction known as neutralization, in which the resulting solution is neither acidic nor basic. • Net ionic equation for neutralization of strong acid and

strong base.

Moleculer Equation: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Ionic Equation:

Na+(aq) + OH−(aq) + H+(aq) + Cl−(aq) → Na+(aq) + Cl−(aq) + H2O(l)

Net Ionic Equation:

OH−(aq) + H+(aq) → H2O(l)

H3O+(aq) + OH− (aq) → 2H2O(l )

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Precipitation Reactions

• A precipitation reaction is an aqueous reaction that produces a solid, called a precipitate.

• EXAMPLE: When aqueous silver nitrate and sodium chloride are combined, the solution becomes cloudy white with solid silver chloride. Let’s write molecular, total ionic and net ionic equations:

• Molecular Equation: • AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

• Total Ionic Equation:

Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq) → AgCl(s) + Na+(aq) + NO3

−(aq)

• Net İonic Equation: Ag+(aq) + Cl−(aq) → AgCl(s) • Na+ and NO3

− are spectator ions.

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Avogadro’s Number and the Mole • A mole is a means of counting the large number of particles in

samples. • One mole is the number of atoms in exactly 12 grams of

12C (carbon-12). • 1 mole contains Avogadro’s number (6.022 x 1023

particles/mole) of particles.

• The mass of 6.022 x 1023 atoms of any element is the molar mass of that element.

Avogadro’s Number and the Mole

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• One mole samples of various elements. All have the same number of particles.

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• 1 mole C = 6.022 x 1023 C atoms = 12.01 g C • 1 mole H = 6.022 x 1023 H atoms = 1.008 g H • 1 mole O = 6.022 x 1023 O atoms = 16.00 g O • , etc.

• Balanced chemical reactions also provide mole ratios

between reactants and products.

• 2 moles H2 : 1 mole O2 : 2 moles H2O

2H2 (g) + O2 (g) → 2H2O(g)

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Determining Molar Mass

• The molar mass of a compound is the sum of the molar masses of all the atoms in a compound.

• Mass in grams of one mole of the substance: Molar Mass of N = 14.01 g/mol Molar Mass of H2O = 18.02 g/mol (2 × 1.008 g) + 16.00 g Molar Mass of Ba(NO3)2 = 261.35 g/mol 137.33 g + (2 × 14.01 g) + (6 × 16.00 g)

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Calculations Using Moles and Molar Mass

• Molar mass allows conversion from mass to number of moles, much like a unit conversion. • 1 mol C7H5N3O6 = 227.133 g C7H5N3O6

300.0 g C 7 H5N3O6 ×1 mol C 7 H5N3O6

227.133 g C 7 H5N3O6

= 1.320 mol C 7 H5N3O6

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Calculations Using Moles and Molar Mass

• Avogadro’s number functions much like a unit conversion

between moles to number of particles. • 1 mol C7H5N3O6 = 6.022 × 1023 C7H5N3O6 molecules

• How many molecules are in 1.320 moles of

nitroglycerin?

1.320 mol C 7 H5N3O6 ×6.022 × 1023 molecules C 7 H5N3O6

1 mol C 7 H5N3O6

= 7.949 × 1023 molecules C 7 H5N3O6

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Exercise

Calculate the number of iron atoms in a 4.48 mole sample of iron. (Fe: 55.85) 1 mole Fe = 6.022 x 1023 Fe atoms = 55.85 g Fe 1 mole Fe 6.022 x 1023 Fe atoms 4.48 mole Fe X= 2.70×1024 Fe atoms

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Exercise

Which of the following is closest to the average mass of one atom of copper? (Atomic Mass of Copper= 63.55 amu) a) 63.55 g b) 52.00 g c) 58.93 g d) 65.38 g e) 1.055 x 10-22 g

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Exercises

Prob1. Calculate the number of copper atoms in a 63.55 g sample of copper. 6.022×1023 Cu atoms Prob1. Calculate the number of oxygen atoms in a 44.01 g sample of CO2 ( C:12.01, O:16.00) 1 mole of CO2 = 44.01 g = 6.022 × 1023 CO2 molecules = 2x 6.022 × 1023 O atoms

= 12.04 × 1023 O atoms

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Exercises

Prob3. Which of the following 100.0 g samples contains the greatest number of atoms? (Mg: 24.31 , Zn: 65.38 , Ag: 107.9) a) Magnesium b) Zinc c) Silver

Prob4. Rank the following 100.0 g samples (each) from greatest to least number of oxygen atoms. (C:12.0 , H:1.01, O:16.0 )

a) C3H6O2 b) H2O, c) CO2,

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Elemental Analysis: Determining Empirical and Molecular Formulas

• Empirical formulas can be determined from an elemental analysis. • An elemental analysis measures the mass percentage of

each element in a compound.

• The formula describes the composition in terms of the number of atoms of each element.

• The molar masses of the elements provide the connection between the elemental analysis and the formula.

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Percent Composition by Mass

• Mass percent of an element:

• For iron in iron(III) oxide, (Fe2O3):

mass of element in compoundmass % = × 100%mass of compound

2( 55.85 g) 111.70 gmass % Fe = = × 100% = 69.94%2( 55.85 g)+ 3( 16.00 g) 159.70 g

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Exercise

Consider separate 100.0 gram samples of each of the following: H2O, N2O, C3H6O2, CO2

Rank them from highest to lowest percent oxygen by mass.

Answer: H2O, CO2, C3H6O2, N2O

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Determining Empirical and Molecular Formulas

• Assume a 100 gram sample size

• Percentage element × sample size = mass element in compound. (e.g., 16% carbon = 16 g carbon)

• Convert mass of each element to moles using the molar mass.

• Divide by smallest number of moles to get mole to mole ratio for empirical formula.

• When division by smallest number of moles results in small rational fractions, multiply all ratios by an appropriate integer to give whole numbers. • 2.5 × 2 = 5, 1.33 × 3 = 4, etc.

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• A molecular formula is a whole number multiple of the empirical formula. • Molar mass for the molecular formula is a whole number

multiple of the molar mass for the empirical formula.

• If the empirical formula of a compound is CH2 and its molar mass is 42 g/mol, what is its molecular formula?

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• Empirical formula = CH Simplest whole-number ratio

• Molecular formula = (empirical formula)n

[n = integer] • Molecular formula = C6H6 = (CH)6 Actual formula of the compound

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Exercise

The composition of adipic acid is 49.3% C, 6.90% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. Find empirical and molecular Formula of apidic acid. (C:12.0 , H:1.01, O:16.0 ) Consider 100 g of CxHyOz 49.3 g C = 4.11 mol of C 6.90 g H = 6.83 mol of H 43.8 g O = 2.73 mol of O The empirical formula? C4.11H6.83O2.73 = C1.5H2.5O = C3H5O2

C3H5O2 : 73.1 g/mol

The molecular formula: 146 = n( 73.1) n= 2 n(C3H5O2 ) = 2(C3H5O2 )= C6H10O4

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Molarity

• Molarity, or molar concentration, M, is the number of moles of solute per liter of solution. • Provides relationship among molarity, moles solute, and

liters solution.

• n = M x V

Molarity (M ) = moles of soluteliter of solution

6 moles of HCl3 HCl = 2 liters of solutionM

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Exercise

A 500.0-g sample of potassium phosphate (K3PO4 ) is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution? (K3PO4 : 212.4 g/mol) Moles of Solute = n = 500.0 g / (212.4 g/mol) = 2.36 mol Volume of Solution = 1.50 L

M = 2.36 / 1.5 = 1.57 M

Molarity (M ) = moles of soluteliter of solution

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Dilution

• Dilution is the process in which solvent is added to a solution to decrease the concentration of the solution. • The number of moles of solute is the same before and

after dilution.

• Since the number of moles of solute equals the product of molarity and volume (M × V), we can write the following equation, where the subscripts denote initial and final values.

M i × Vi = M f × Vf

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Concept Check

A 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution?

a) Add water to the solution. b) Pour some of the solution down the sink drain. c) Add more sodium chloride to the solution. d) Let the solution sit out in the open air for a couple of

days.

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Exercise

What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution? M1V1 = M2V2

(2.00 M)(V1) = (0.800 M)(150.0 mL) V1 = 60.0 mL

Chapter 3 Molecules, Moles, and Chemical...

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