The Mole and Stoichiometry

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The Mole and Stoichiometry
H Chemistry I Unit 8

Objectives #1-4 The Mole and its Use in Calculations
Fundamentals 1 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
*examples of mole quantities: 1 mole of iron = 55.8 grams 1 mole of copper = 63.5 grams 1 mole of water = 18.0 grams

Introduction to Mole Problems
(Follow the procedures outlined in Unit 1 for dimensional analysis problems) Calculate the number of atoms in .500 moles of iron. *Determine known: .500 atoms Fe

Determine unknown: 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 Fe 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

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

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

Molar Mass Calculate the molar mass of H3PO4: 3 H at 1.00 g = 3.00 g 1 P at 31.0 g = 31.0 g 4 O at 16.0 g = 64.0 g Total = 98.0 g

Calculate the molar mass of Al(OH)3:
1 Al at 27.0 g = 27.0 g 3 O at 16.0 g = 48.0 g 3 H at 1.0 g = 3.0 g Total = 78.0 g

Calculate the molar mass of BaCl2.2H2O:
1 Ba at g = g 2 Cl at 35.5 g = 71.0 g 2 H2O at 18.0 g = 36.0 g Total = g

Mole-Mass Problems 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
*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 = g

*check answer for units, sig. figs., and reasonableness: 547 g N2O3

What is the mass in grams of 4.52 g of barium chloride?
4.52 g BaCl2 X 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 / g X 3 NH4+1 / 1 mole X 6.02 X 1023 ions / 1 mole NH4+1 = 4.24 X 1022 NH4+1 ions

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

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

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

Calculate the mass of 1 molecule of propane.
1 molecule C3H8 X 44.0 g / 6.02 X 1023 molecules = 7.31 X g C3H8

Objective #5 Characteristics of Solutions
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 – brass liquid – saltwater gas - air

Objective #5 Characteristics of Solutions
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

Objective #5 Characteristics of Solutions
*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 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)

Objective #5 Characteristics of Solutions
*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)

Objective #5 Characteristics of Solutions
*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 solubility for gases in liquids – increasing temp. decreases solubility (interpreting solubility graphs)

Interpreting Solubility Graphs

Molarity = moles of solute / liter of solution *examples:
Objective #6 Molarity *formula: Molarity = moles of solute / liter of solution *examples: 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 moles M = moles / 1.50 L = .192 M

A chemist requires 1. 00 L of. 200 M potassium dichromate solution
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 g / 1 mole = 59 g K2Cr2O7 What is the molarity of each ion in the following solutions (assuming all are strong electrolytes) .15 M calcium chloride: CaCl2 1 Ca to 2 Cl .15 M Ca+2 2(.15 M) = .30 M Cl-1

22 M calcium perchlorate Ca(ClO4)2 1 Ca to 2 ClO4. 22 M Ca+2 2 (
.22 M calcium perchlorate Ca(ClO4)2 1 Ca to 2 ClO4 .22 M Ca+2 2 (.22 M) = .44 M ClO4-1

What is the molarity of an HCl solution made by diluting 3. 50 L of a
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 = MV M1V1 = M2V2

M1V1 = M2V2 #1 = concentrated #2 = diluted M2 = M1V1 / V2 = (3.50 L) (.200 M) / 5.00 L = .140 M

An experiment calls for 2. 00 L of a
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. ml How would one prepare solution?

Objective #7 Percentage Composition
Formula % element in compound = mass of element in sample of compound / mass of sample of compound

*examples: Calculate the percentage composition phosphate (K3PO4) *determine molar mass of compound: 212.4 g Determine total mass of each component: 3 K at 39.1 g each = g 1 P at 31.0 g each = 31.0 g 4 O at 16.0 g each = 64.0 g

Divide total mass of each component by
molar mass of compound: % K = g / g = .55 % P = 31.0 g / g = .15 % O = 64.0 g / g = .30

Multiply each result by 100 % and round appropriately.
.55 X 100 = 55 % K .15 X 100 = 15 % P .30 X 100 = 30 % O

Calculate the percentage composition of sodium carbonate (Na2CO3)
molar mass = g % Na = (46.0 g / g) X 100 = 43% % C = (12.0 g / g) X 100 = 11% % O = (48.0 g / g ) X 100 = 45%

Objective #8 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 K 29.1% S --› 29.1 g S

convert masses to moles: 70. 9 g K X 1 mole K / 39. 1 g = 1
*convert masses to moles: 70.9 g K X 1 mole K / 39.1 g = 1.81 moles K 29.1 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 K 1.81 / .907 =1.995 S

*round resulting values to the nearest whole number or multiply by factor: 1 K 2 S *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

A compound of iron and oxygen when analyzed showed 70. 0% iron and 30
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 Fe 30.0 g O X 1 mole/16.0 g = 1.88 moles O 1.25 / 1.25 = 1 1.88 / 1.25 = 1.5 1 Fe X 2 = 2 Fe O X 2 = 3 O Fe2O3

Determine the empirical formula for a compound that contains 10
Determine the empirical formula for a compound that contains g of calcium and 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 Cl CaCl2

Objective #9 Determining Molecular Formulas
*examples: 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 C 20. 0 g H X 1 mole / 1
80.0 g C X 1 mole / 12.0 g = 6.67 moles C 20.0 g H X 1 mole / 1.0 g = 20.0 moles H 6.67 / 6.67 = 1 C 20.0 / 6.67 = 3 H CH3 *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

If the empirical formula for a compound is C2HCl, determine the molecular formula if the gram molecular mass is g. emp. mass 60.5 g 181.5 g / 60.5 g = 2 C4H2Cl2

A compound contains 58. 5% carbon, 9. 8% hydrogen, 31
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 C 9.8 g H X 1 mole / 1.0 g = 9.8 mole H 31.4 g O X 1 mole / 16.0 g = 1.96 mole O 4.88 / 1.96 = 2.5 C 9.8 / 1.96 = 5 H 1.96 / 1.96 = 1 O

2.5 C X 2 = 5 C 5 H X 2 = 10 H 1 O X 2 = 2 O emp. formula = C5H10O2 emp. mass = 102 g molecular formula = C5H10O2

Objectives #10-11 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: Number of particles Number of moles 3. Mass

Particles: 4Al O › Al2O3 atom molecule formula unit 4 atoms Al 3 molecules O2 2 formula units Al2O3

Moles 4Al O › Al2O3 4 moles Al 3 moles O2 2 moles Al2O3

Mass: 4Al O › Al2O3 108 g Al 96 g O2 204 g Al2O3

4Al + 3O2 --› 2Al2O3. playing with the mole ratios: 4 moles Al --›
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 --› 1 mole Al2O3 2 moles Al --› ? Al2O3 2 moles Al --› 1 mole Al2O3

4Al + 3O2 --› 2Al2O3. 200945 moles Al --›
4Al + 3O2 --› 2Al2O moles Al --› ? moles Al X 2 moles Al2O3 / 4 mole Al = mole Al

Objective #12 Stoichiometric Calculations
*examples: 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 HCl --› MgCl H2

determine known: 20. 0 g. determine unknown: g MgCl2
*determine known: 20.0 g *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 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

How many moles of water can be formed from 7
How many moles of water can be formed from 7.5 moles of ethyne gas (C2H2) reacting with excess oxygen gas? 2C2H O › 4CO H2O 7.5 moles C2H2 X 2 moles H2O / 2 moles C2H2 = 7.5 moles H2O

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 X 8 mole Rb2S / 16 mole Rb X 6.02 X 1023 f. units Rb2S / 1 mole Rb2S = X 1023 f. units Rb2S

Objective #13 Limiting Reagents
*General Terms: excess reagent reactant that is leftover limiting reagent reactants that is used up *Examples: How many grams of carbon dioxide can be obtained by the action of 50.0 g of sulfuric acid on 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
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 CO g CaCO3 X 1 mole CaCO3 / g X 1 mole CO2 / 1 mole CaCO3 X 44.0 g CO2 / 1 mole CO2 = 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

How many grams of hydrogen gas can be produced by the reaction of 100
How many grams of hydrogen gas can be produced by the reaction of 100. g of phosphoric acid on 25.0 g of aluminum? 2H3PO Al --› 2AlPO H2 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 X 3 mole H2 / 1 mole H2 X 2
25.0 g Al X 1 mole Al / 27.0 g Al X 3 mole H2 / 1 mole H2 X 2.0 g H2 / 1 mole= 2.78 g H g H2 (smaller value)

Determine how much of the excess reagent is left over
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 X 2 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

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? Cl NaBr --› Br NaCl

15. 0 g Cl2 X 1 mole Cl2 / 71. 0 g Cl2 X 2 mole NaCl / 1 mole Cl2 X 58
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 g NaBr X 1 mole NaBr / g NaBr X 2 mole NaCl / 2 mole NaBr X 58.5 g NaCl / 1 mole NaCl = 8.53 g NaCl

15.0 g NaBr X 1 mole NaBr / g NaBr X 1 mole Cl2 / 2 mole NaBr X 71.0 g Cl2 / 1 mole Cl2 = 5.17 g Cl2 (used) g – 5.17 g = 9.8 g Cl2 leftover

Objective #15 Percentage Yield
*General Formula: Percent Yield = (actual yield / theoretical yield) X 100% *examples: 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
*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 X 2 mole AlPO4 / 2 mole Al X 122
9.00 g Al X 1 mole Al / 27.0 g Al X 2 mole AlPO4 / 2 mole Al X 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%

Calculate the percentage yield if 6
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 / 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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