The Mole, Stoichiometry , and Solution Chemistry

The Mole, Stoichiometry, and Solution Chemistry

H Chemistry IUnit 8

Objectives #1-4 The Mole and its Use in Calculations

I. Fundamentals1 mole of an element or compound

contains 6.02 X 1023 particles and has a mass = to its molar mass

“The Triad of the Mole”1 mole = 6.02 X 1023 particles =

molar mass*henceforth the number 6.02 X 1023 will

be known as the Avogadro’s number

*examples of mole quantities:1 mole of iron = 55.8 grams1 mole of copper =63.5 grams1 mole of water = 18.0 grams

II. Introduction to Mole Problems(Follow the procedures outlined in Unit

1 for dimensional analysis problems)

1. Calculate the number of atoms in .500 moles of iron.

*Determine known:.500 atoms Fe

*Determine unknown:atoms Fe*Use Triad of the Mole to determine

conversion factor:1 mole = 6.02 X 1023 atoms*Use conversion factor solve problem:.500 moles Fe X 6.02 X 1023 atoms/1

mole

Check answer for units, sig. figs,, and reasonableness

3.01 X 1023 atoms Fe2. Calculate the number of atoms

in .450 moles of zinc..450 moles Zn X 6.02 X 1023 atoms

Zn /1 mole = 2.71 X 1023 atoms Zn

3. Calculate the number of moles in 2.09 X 1025 atoms of sulfur.

2.09 X 1025 atoms S X 1mole S / 6.02 X 1023 atoms S = 34.7 moles S4. Calculate the number of moles in

3.06 X 1022 atoms of chlorine.3.06 X 1022 atoms Cl X 1 mole Cl / 6.02 X 1023 atoms Cl = .0508 moles Cl

5. Calculate the number of atoms in 35.7 g of silicon.

35.7 g Si X 6.02 X 1023 atoms Si / 28.1 g

= 7.65 X 1023 atoms Si

II. Molar Mass1. Calculate the molar mass of H3PO4:

3 H at 1.00 g = 3.00 g1 P at 31.0 g = 31.0 g4 O at 16.0 g = 64.0 gTotal = 98.0 g

2. Calculate the molar mass of Al(OH)3:

1 Al at 27.0 g = 27.0 g3 O at 16.0 g = 48.0 g3 H at 1.0 g = 3.0 gTotal = 78.0 g

3. Calculate the molar mass of BaCl2.2H2O:

1 Ba at 137.3 g = 137.3 g2 Cl at 35.5 g = 71.0 g2 H2O at 18.0 g = 36.0 g

Total = 244.2 g

III.Mole-Mass Problems1. How many grams are in 7.20 moles

of dinitrogen trioxide?*determine known:7.20 moles N2O3

*determine unknown:grams N2O3

*determine molar mass of compound if grams are involved:

76.0 g*use “Triad of the Mole” to determine

conversion factor:1 mole = 76.0 g*use conversion factor to solve problem:7.20 moles N2O3 X 76.0 g N2O3 / 1 mole =

547.2 g

*check answer for units, sig. figs., and reasonableness:

547 g N2O3

2. What is the mass in grams of 4.52 moles of barium chloride?

4.52 moles BaCl2 X 208.2 g / 1 mole = 941 g

3. Calculate the number of ammonium ions in 3.50 grams of ammonium phosphate.

3.50 g (NH4)3PO4 X

1 mole (NH4)3PO4 / 149. 0 g X

3 NH4+1 / 1 mole (NH4)3PO4 X

6.02 X 1023 ions / 1 mole NH4+1 ions =

4.24 X 1022 NH4+1 ions

4. Calculate the mass of carbon in 7.88 X 1026 molecules of C8H18.

7.88 X 1026 molecules C8H18 X

1 mole C8H18 / 6.02 X 1023 molecules C8H18

X 8 moles C / 1 mole C8H18 X

12.0 g C / 1 mole C = 1.26 X 105 g C

5. Calculate the number of molecules present in 2.50 moles of water.

2.50 moles H2O X 6.02 X 1023 molecules /

1 mole = 1.51 X 1024 molecules H2O

6. Calculate the mass in grams of 4.50 X 1025 molecules of C6H10.

4.50 X 1025 molecules C6H10 X

82.0 g / 6.02 X 1023 molecules =6130 g C6H10

7. Calculate the mass of 1 molecule of propane.

1 molecule C3H8 X

44.0 g / 6.02 X 1023 molecules = 7.31 X 10-23 g C3H8

Objective #5 Characteristics of Solutions

Characteristic

Components solute and solvent

Particle Size .01 – 1 nm (atoms, ions, and molecules)

Tyndall Effect Result negative

Ability to be separated by filtration

no

Homogeneous or heterogeneous

homogeneous

Examples in different phases solid – brassliquid – saltwatergas - air


II. The Solution Process*dissociation and hydration of solutes:solute is split apart by solvent (dissociation)solute particles are surrounded by solvent

particles (hydration or solvation)*the rate of solution formation can be

increased by:stirring, raising temperature, and powdering


*behavior of ionic, polar, and nonpolar solutes in water:water is polar and will dissolve many ionic and polar solutes

NaCl → Na+1 + Cl-1

HCl → H+1 + Cl-1

C6H12O6 - C6H12O6

nonelectrolytes vs. electrolytes

*”like dissolves like”: materials that have similar bonds will dissolve each other

NaCl (ionic) H2O (polar)

HCl (polar) H2O (polar)

I2 (nonpolar) CCl4 (nonpolar)


*miscible vs. immiscible liquids:miscible liquids dissolve in each other and form one

phase

immiscible liquids don’t dissolve in each other and form two phases

*unsaturated vs. saturated vs. supersaturated solutions:

Unsaturated (less solute dissolved than possible)Saturated (limit of solute dissolving)Supersaturated (beyond limit of solute dissolving)


*effect of pressure on solubility:gases will be more soluble in a liquid if the

atmospheric pressure is increased*effect of temperature on solubility:for solids in liquids – increasing temp. usually

increases solubilityfor gases in liquids – increasing temp.

decreases solubility (interpreting solubility graphs)

Interpreting Solubility Graphs

Example Problems:1. Which chemical is the most soluble at 20oC? _________2. Which chemical is the least soluble at 40oC? ________3. What is the solubility of KCl at 25oC?________4. At 90oC you dissolved 10 g of KCl in 100 g of water. Is

this solution saturated or unsaturated? _________5. Amass of 80 g of KNO3 is dissolved in 100 g of water

at 50oC. The solution is heated to 70oC. How many more grams of potassium nitrate must be added to make the solution saturated. _________

Objective #6 Molarity

*formula: Molarity = moles of solute / liter of

solution*examples:1. Calculate the molarity of a solution

prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution.

11.5 g NaOH X 1 mole / 40.0 g = .288 molesM = .288 moles / 1.50 L = .192 M

2. A chemist requires 1.00 L of .200 M potassium dichromate solution. How many grams of solid potassium dichromate will be required?

Moles = M X L = (.200 M) (1.00 L) = .200 moles K2Cr2O7

.200 moles K2Cr2O7 X 294.2 g / 1 mole =

58.8 g K2Cr2O7

.200 moles K2Cr2O7 X 294.2 g / 1 mole =

59 g K2Cr2O7

3. What is the molarity of each ion in the following solutions (assuming all are strong electrolytes)

.15 M calcium chloride:CaCl21 Ca to 2 Cl.15 M Ca+2

2(.15 M) = .30 M Cl-1

.22 M calcium perchlorateCa(ClO4)2

1 Ca to 2 ClO4

.22 M Ca+2

2 (.22 M) = .44 M ClO4-1

.18 M sodium chloride

.18 M Na and .18 M Cl

4. What is the molarity of an HCl solution made by diluting 3.50 L of a .200 M solution to a volume of 5.00 L?

moles before dilution = moles after dilution

recall moles = MVM1V1 = M2V2

M1V1 = M2V2

#1 = concentrated#2 = dilutedM2 = M1V1 / V2

= (3.50 L) (.200 M) / 5.00 L = .140 M

5. An experiment calls for 2.00 L of a .400 M HCl, which must be prepared from 2.00 M HCl. What volume in milliliters of the concentrated acid is needed to be diluted to form the .400 M solution?

M1V1 = M2V2

V1 = M2V2 / M1

= (.400 M) (2.00 L) / (2.00 M) = .400 L = 400. mlHow would one prepare solution?

Objs. #7-10 Acids, Bases, and pH/pOH

I. Strong vs. Weak Acids and Bases*strong acids and bases completely ionize

in waterHCl(aq) ---- H+1

(aq) + Cl-1(aq)

NaOH(aq) ---- Na+1(aq) + OH-1

(aq)

*common examples:HCl, H2SO4, HNO3, HClO4, NaOH, KOH,

Ca(OH)2

*weak acids and bases don’t completely ionize completely in water

HCN(aq) + H2O(aq) < > CN-1(aq) + H3O+1

(aq)

NH3(aq) + H2O(aq) < > NH4(aq)+1 + OH-1

(aq)

*common examples:HC2H3O2, NH3

II. Self-Ionization of Water*water has the ability to act as both a Bronsted-

Lowery acid and baseH2O(l) + H2O(l) < > H3O(aq)

+1 + OH(aq)-1

Kw = {H3O+1} {OH-1}

where at 25oC the Kw = 1.00 X 10-14 so

1.00 X 10-14 = {H3O+1} {OH-1}

1.00 X 10-14 = x2

1.00 X 10-7 = x

*these values for the H3O+1 and OH-1 concentrations are valid for room temperature conditions; K does vary with temperature however

III. pH and pOH*pH and pOH are convenient methods for

expressing the relative H3O+1 or OH-1 of a solution

*pH is defined as the –log{H3O+1}

*pOH is defined as the –log{OH-1}*at 25oC pH +pOH = 14

*pH and pOH scales compared:1-6 7 8-14

pH acidic / pOH basic

Neutral pH basic / pOH acidic

{H+1} > 1 X10-7 M {H+1}={OH-1}=1X10-7M

{H+1} < 1 X 10-7

{OH-1} < 1 X10-7M {OH-1} > 1 X10-7 M

pH < 7.0 pH = pOH = 7.0 pH >7.0

pOH > 7.0 pOH < 7.0

*examples:Calculate the pH for each of the

following solutions:1. 1.0 M H+1

2. 1.0 X 10-9 M H+1

3. 2.5 X 10-4M H+1

4. Calculate the hydronium ion concentration of a solution having a pH of 4.5.

5. Calculate the hydronium ion concentration of a solution having a pH of 7.52

6. Calculate the H+1 and OH-1 for a 1.0 X 10-3 M NaOH solution. Calculate

the pH and pOH.

IV.Buffers*buffers are solutions that resist

changes in pH when small amounts of acid or base are added

*buffer solutions can be produced from a combination of a weak acid and a salt containing a common ion or a combination of a weak base and a salt containing a common ion

*example:adding acid to a mixture of sodium acetate and

acetic acid:CH3COO-1 + H+1 ---- CH3COOH + H2O

The hydrogen ion of the additional acid will be “captured” by the acetate ion

adding base to a mixture of sodium acetate and acetic acid:

CH3COOH + OH-1 --- CH3COO-1 + H2O

The hydroxide ion of the base will be “neutralized” by the acid

*Examples:Which of the following mixtures would

form buffers?1. HCl and NaCl2. HNO2 and NaNO2

3. HNO2 and NaCl

Objective #11 Percentage Composition

I. Formula% element in compound = mass of element in sample of

compound /mass of sample of compound

*examples:1. Calculate the percentage composition

potassium phosphate (K3PO4)

*determine molar mass of compound:212.4 g2. Determine total mass of each component:3 K at 39.1 g each = 117.3 g1 P at 31.0 g each = 31.0 g4 O at 16.0 g each = 64.0 g

3. Divide total mass of each component by

molar mass of compound:% K = 117.3 g / 212.4 g = .55% P = 31.0 g / 212.4 g = .15% O = 64.0 g / 212.4 g = .30

4. Multiply each result by 100 % and round appropriately.

.55 X 100 = 55 % K

.15 X 100 = 15 % P

.30 X 100 = 30 % O

2. Calculate the percentage composition of sodium carbonate (Na2CO3)

molar mass = 106.0 g% Na = (46.0 g / 106.0 g) X 100 =

43%% C = (12.0 g / 106.0 g) X 100 = 11%% O = (48.0 g / 106.0 g ) X 100 = 45%

Objective #12 Determining Empirical Formulas

*empirical vs. molecular formulas:H2O2 --› HO

H2O --› H2O

*examples:1. Determine the empirical formula for

a compound that when analyzed contained 70.9% potassium and 29.1% sulfur by mass.

*If given percentages, assume 100 gram sample and convert percents to grams:

70.9% K --› 70.9 g K29.1% S --› 29.1 g S

*convert masses to moles:70.9 g K X 1 mole K / 39.1 g = 1.81

moles K29.1 g S X 1 mole S / 32.1 g =.907

moles S*divide all mole values by the smallest

mole value in the set:.907 / .907 = 1 S1.81 / .907 =1.995 K

*round resulting values to the nearest whole number or multiply by factor:

1 S2 K*use final values as subscripts and write

formula with the elements in order of increasing electronegativity; if compound is organic, list carbon first, then hydrogen, and the remainder by electronegativity:

K2S

2. A compound of iron and oxygen when analyzed showed 70.0% iron and 30.0% oxygen by mass. Determine the empirical formula.

70. 0 g Fe X 1 mole/55.8 g = 1.25 moles Fe30.0 g O X 1 mole/16.0 g = 1.88 moles O1.25 / 1.25 = 11.88 / 1.25 = 1.51 Fe X 2 = 2 Fe 1.5 O X 2 = 3 OFe2O3

3. Determine the empirical formula for a compound that contains 10.88 g of calcium and 19.08 g of chlorine.

10.88 g Ca X 1 mole / 40.1 g = .271 mole

19.08 g Cl X 1 mole / 35.5 g = .537 mole

.271 / .271 = 1 Ca

.537 / .271 = 2 ClCaCl2

Objective #13 Determining Molecular Formulas

*examples:1. Analysis of a compound showed it to

consist of 80.0% carbon and 20.0% hydrogen by mass. The gram molecular mass is 30.0 g. Determine the molecular formula.

*determine empirical formula if needed:

80.0 g C X 1 mole / 12.0 g = 6.67 moles C20.0 g H X 1 mole / 1.0 g = 20.0 moles H6.67 / 6.67 = 1 C20.0 / 6.67 = 3 HCH3

*determine empirical formula mass (efm):15 g

*if efm matches gram molecular mass (gmm), then empirical formula is the same as molecular

15 g ≠ 30 g*if masses don’t match, divide gmm by efm

to determine factor:30 g / 15 g = 2*multiply subscripts of empirical formula by

factor to obtain molecular formula:C2H6

2. If the empirical formula for a compound is C2HCl, determine the molecular formula if the gram molecular mass is 181.5 g.

emp. mass 60.5 g181.5 g / 60.5 g = 3C6H3Cl3

3. A compound contains 58.5% carbon, 9.8% hydrogen, 31.4% oxygen and the gram molecular mass is 102 g. Determine the molecular formula.

58.5 g C X 1 mole / 12.0 g = 4.88 mole C9.8 g H X 1 mole / 1.0 g = 9.8 mole H31.4 g O X 1 mole / 16.0 g = 1.96 mole O4.88 / 1.96 = 2.5 C9.8 / 1.96 = 5 H1.96 / 1.96 = 1 O

2.5 C X 2 = 5 C5 H X 2 = 10 H1 O X 2 = 2 Oemp. formula = C5H10O2

emp. mass = 102 gmolecular formula = C5H10O2

Objectives #14-15 Introduction to Stochiometry

*Stoichiometry is a method of calculating amounts in a chemical reaction

I. Interpreting Chemical Equations Quantitatively

*example: 4Al + 3O2 --› 2Al2O3

*the following information can be determined from this reaction:

1. Number of particles2. Number of moles3. Mass

1. Particles:4Al + 3O2 --› 2Al2O3

atom molecule formula unit4 atoms Al3 molecules O2

2 formula units Al2O3

2. Moles4Al + 3O2 --› 2Al2O3

4 moles Al3 moles O2

2 moles Al2O3

3. Mass:4Al + 3O2 --› 2Al2O3

108 g Al96 g O2

204 g Al2O3

What is conserved in a chemical reaction?

4Al + 3O2 --› 2Al2O3

*playing with the mole ratios:4 moles Al --› ? Al2O3

4 moles Al --› 2 moles Al2O3

8 moles Al --› ? Al2O3

8 moles Al --› 4 mole Al2O3

2 moles Al --› ? Al2O3

2 moles Al --› 1 mole Al2O3

4Al + 3O2 --› 2Al2O3

.200945 moles Al --› ? Al2O3

.200945 moles Al X 2 moles Al2O3 / 4 mole Al =

.100473 mole Al

Objective #16 Stoichiometric Calculations

*examples:1. If 20.0 g of magnesium react with

excess hydrochloric acid, how many grams of magnesium chloride are produced?

*write balanced chemical equation if not provided:

Mg + 2HCl --› MgCl2 + H2

Mg + 2HCl -- H2 + MgCl2

*determine known:20.0 g Mg*determine unknown:g MgCl2*convert given to moles:20.0 g Mg X 1 mole Mg / 24.3 g *select and utilize appropriate mole-mole to

convert to moles of unknown:X 1 mole MgCl2 / 1 mole Mg

*if unknown is to be measured in moles skip to step 7. If unknown is to be measured in grams, multiply by unknown’s molar mass. If unknown is to be measured in particles, multiply by Avogadro’s number:

X 95.3 g MgCl2 / 1 mole MgCl2

*perform math:78.4 g MgCl2*check answer for units, sig figs, and

reasonableness

2. How many moles of water can be formed from 7.5 moles of ethyne gas (C2H2) reacting with excess oxygen gas?

2C2H2 + 5O2 --› 4CO2 + 2H2O

7.5 moles C2H2 X

2 moles H2O / 2 moles C2H2 =

7.5 moles H2O

3. If 50.00 g of Rb reacts with excess S8 how many formula units of Rb2S can be formed?

16 Rb + S8 --› 8Rb2S

50.00 g Rb X 1 mole Rb / 85.5 g Rb X8 mole Rb2S / 16 mole Rb X

6.02 X 1023 f. units Rb2S / 1 mole Rb2S

= 1.760 X 1023 f. units Rb2S

Obj. #17 Solution Stoichiometry

*stoichiometry that involves using molarityExample I:HCl + NaHCO3 --- NaCl + H2O + CO2

How many grams of NaHCO3 solution are needed to neutralize 18.0 ml of .100 M

HCl?

Example II:H2SO4 + 2NaOH --- Na2SO4 + 2H2O

How many grams of sodium sulfate is produced from 25.0 ml of .150 M NaOH solution?

Obj. #18-19 Titration

*description of titration:

*end point: the point in a titration when the

indicator changes colorfor example, phenolphthalein is clear

in a pH below 8 and is pink in a pH above 8

*equivalence point: the point in a titration when the moles of acid just neutralize the moles of base

*Example I:H2SO4 + 2NaOH --- Na2SO4 + 2H2O

What is the molarity of a 25.0 ml sample of sulfuric acid that required 37.5 ml of .15 M sodium hydroxide?

*Example II:HCl + NaHCO3 -- NaCl + H2O +

CO2

Objective #20 Limiting Reagents

*General Terms:excess reagent reactant that is leftoverlimiting reagent reactant that is used up*Examples:1. How many grams of carbon dioxide can be

obtained by the action of 50.0 g of sulfuric acid on 100.0 g of calcium carbonate?

(follow general steps for stoichiometry, just do it twice)

H2SO4 + CaCO3 --› CaSO4 + CO2 +H2O

50.0 g H2SO4 X 1 mole H2SO4 / 98.1 g X

1 mole CO2 / 1 mole H2SO4 X

44.0 g CO2 / 1 mole CO2 = 22.4 g CO2

------------------------------------------100.0 g CaCO3 X 1 mole CaCO3 / 100.1 g X

1 mole CO2 / 1 mole CaCO3 X

44.0 g CO2 / 1 mole CO2 = 43.96 g CO2

(examine two answers produced; the smaller of the two is the right answer of product produced)

*the reactant quantity that provides the smaller correct answer is called the limiting reagent

*the reactant quantity that provides the larger incorrect answer is called the excess reagent

2. How many grams of hydrogen gas can be produced by the reaction of 100. g of phosphoric acid on 25.0 g of aluminum?

2H3PO4 + 2Al --› 2AlPO4 + 3H2

100. g H3PO4 X 1 mole H3PO4 / 98.0 g X

3 mole H2 / 2 mole H3PO4 X

2.0 g H2 / 1 mole H2 = 3.06 g H2

25.0 g Al X 1 mole Al / 27.0 g Al X3 mole H2 / 1 mole H2 X 2.0 g H2 / 1

mole=2.78 g H2

2.78 g H2 (smaller value)

Determine how much of the excess reagent is left over.

*determine limiting reagent:Al*use stochiometry to convert from mass of

limiting reagent to find mass of excess reagent actually used:

25.0 g Al X 1 mole Al / 27.0 g Al X2 mole H3PO4 / 2 mole Al X

98.0 g H3PO4 / 1mole = 90.7 g H3PO4 used

*Subtract amount of excess reagent used from original amount of excess reactant given in problem to determine mass of excess reactant left over:

100. g – 90.7 g = 9 g left over

3. How many grams of sodium chloride can be produced from then reaction of 15.0 g of chlorine gas and 15.0 g of sodium bromide in the above reaction? Determine how much of the excess reagent is left over?

Cl2 + 2NaBr --› Br2 + 2NaCl

15.0 g Cl2 X 1 mole Cl2 / 71.0 g Cl2 X

2 mole NaCl / 1 mole Cl2 X

58.5 g NaCl / 1 mole NaCl = 24.7 g NaCl

--------------------------------------------15.0 g NaBr X 1 mole NaBr / 102.9 g

NaBrX 2 mole NaCl / 2 mole NaBr X58.5 g NaCl / 1 mole NaCl = 8.53 g

NaCl

15.0 g NaBr X 1 mole NaBr / 102.9 g NaBr

X 1 mole Cl2 / 2 mole NaBr X

71.0 g Cl2 / 1 mole Cl2 = 5.17 g Cl2 (used)

--------------------------------------------15.0 g – 5.17 g = 9.8 g Cl2 leftover

Objective #21 Percentage Yield

*General Formula:Percent Yield = (actual yield / theoretical yield) X 100%*examples:1. When 9.00 g of aluminum react with an

excess of phosphoric acid, 30.0 g of aluminum phosphate are produced. What is the percentage yield of this reaction?

*write balanced chemical equation:2Al + 2 H3PO4 --› 3H2 + 2AlPO4

*determine actual yield of a product in reaction given in the problem:

30.0 g AlPO4

*use given reactant quantity and general stochiometry procedures to determine mass of appropriate product:

9.00 g Al X 1 mole Al / 27.0 g Al X2 mole AlPO4 / 2 mole Al X

122.0 g / 1 mole = 40.7 g AlPO4

*divide actual yield from problem by the theoretical yield calculated to find percentage yield:

%Y = (30.0 g / 40.7 g) X 100 = 74%

2. Calculate the percentage yield if 6.00 g of aluminum are produced from the decomposition of 25.0 g of aluminum oxide.

2Al2O3 --› 4Al + 3O2

25.0 g Al2O3 X 1 mole Al2O3 / 102.0 g X

4 mole Al / 2 mole Al2O3 X

27.0 g Al / 1 mole Al = 13.2 g Al%Y= (6.00 g / 13.2 g) X 100% = 45%

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The Mole, Stoichiometry , and Solution Chemistry