che121-lec8-sp120

Chemistry 121

Lecture 8

Outline

I. Solutions

- Molarity

- Stoichiometry

- Dilutions

Aluminum Lab

Several concepts will be defined in the next

week or two.

Oxidation state = charge

In the ionic compound,NaCl

Na oxidation state = +1

Cl oxidation state = -1

Molarity

Molarity (M)

A Conversion Factor between the amount of solute

and the volume of solution!!!

M = Molarity = moles of solute

L of solution

Molarity Problems

How many liters of 0.300M KBr solution can be

prepared from 100.g of KBr?

A. 5.60 B. 1.40 C. 2.80 D. 0.600

0.300 mols KBr

1 L soln

100.g KBr = L soln

119.0g KBr

1 mol KBr 1 L soln

0.300 mols KBr

Molarity in Chemical Reactions

In a chemical reaction,

The volume and molarity of a solution are used to

determine the moles of a reactant or product.

mol solute x volume (L) = moles solute

L

M

Molarity in Chemical Reactions

In a chemical reaction,

If molarity, M (mol/L), and moles are given, the volume (L) can be determined.

moles solute x 1 L = volume (L) solution

moles solute

1

M

Summary of Chemical Reactions

Copyright 2008 by Pearson Education, Inc.

Publishing as Benjamin Cummings

Basic gram to gram stoichiometry

Summary of Chemical Reactions

Copyright 2008 by Pearson Education, Inc.

Publishing as Benjamin Cummings

Stoichiometry using Molarity

Stoichiometric Equivalence - Titration

Reactant 1

Reactant 2 plus indicator (changes color when exactly

combined)

Acid/Base Reaction Lab

REDOX Reaction Lab

Using Molarity of Reactants

How many mL of 3.00 M HCl solution are needed to completely react with 4.85 g CaCO3?

2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)

B: 3.00 mols HCl

1L soln 4.85 g

Volume(mL)=?

Mols HCl

A:

0.0 g 0.0 g 0.0 g

0.0 g 0.0 g X Y Z

Using Molarity of Reactants

How many mL of 3.00 M HCl solution are needed to completely react with 4.85 g CaCO3?

2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)

3.00 mols HCl

1 L soln

1mol CaCO3

= mL soln 100.09g CaCO3

4.85g CaCO3

1mol CaCO3

2mol HCl

3.00mol HCl

1 L

1 L

1000mL

= 32.3 mL soln

Learning Check

How many mL of a 0.150 M Na2S solution are needed to

completely react 18.5 mL of 0.225 M NiCl2 solution?

NiCl2(aq) + Na2S(aq) NiS(s) + 2NaCl(aq)

BEFORE:

18.5 mL -> 0.0185L

0.225mols

1L

Mols NiCl2

0.150mols

1L

Volume(mL) = ?

? Mols Na2S

AFTER: 0.0 0.0

Learning Check

How many mL of a 0.150 M Na2S solution are needed to

completely react 18.5 mL of 0.225 M NiCl2 solution?

NiCl2(aq) + Na2S(aq) NiS(s) + 2NaCl(aq)

18.5 mL

1000 mL

1 L

0.0185L

= mL

NiCl2 solution Na2S solution

1 L

0.225 mols NiCl2 1molNa2S

1molNiCl2

1 L

.150 molsNa2S 1 L

1000mL

= 27.8mL Na2S

More on Molarity

Solution preparation from a concentrated

stock solution - Solution Dilution

A concentrated solution (high Molarity) is

converted to a dilute solution (lower Molarity)

by adding solvent to it.

More on Molarity - Dilution

Concentration = amount of solute

amount of solution

More on Molarity - Dilution

Concentration = amount of solute

amount of solution

Solute Amount Remains the Same!

More on Molarity

A concentrated solution (high Molarity) is

converted to a dilute solution (lower Molarity)

by adding solvent to it.

dilute concentrated

Mdil x Vdil = Mcon x V con

Moles of Solute = Moles of Solute

More on Molarity

Muriatic acid, an industrial grade of concentrated HCl, is

used to clean masonry and cement. Its concentration is

11.7M. How would you go about making 11.4L of 3.5M

acid for routine use?

Mdil x Vdil = Mcon x V con

Concentrated

Dilute

(11.7mols HCl)

1L

(11.4L)(3.5 mols HCl)

1L

= V con

V con = 3.4L

To Make the Solution, Take 3.4L of concentrated acid and

add enough water (11.4L - 3.4L = 8.0L) to make 11.4L

Dilution Problem

How many mL of 18.0 M H2SO4 are required to prepare 650. mL of

6.00 M H2SO4 by dilution?

mmol(before dilution) = mmol(after dilution)

mL 18.0 M = 650. mL 6.00 M

mL = (650. mL 6.00 M) / 18.0 M

= 217 mL

Dilution Problem

How many mL of H2O must be added to 375 mL of a 3.50 M NaCl

doln to prepare a 1.25 M NaCl soln?

mmol(before dilution) = mmol(after dilution)

375 mL 3.50 M = mL 1.25 M

mL = (375 mL 3.50 M) / 1.25 M

= 1050 mL

H2O(added) = Vfinal Vinitial = 1050 ml - 375 mL

= 675 mL

Additional Molarity Problem

Assuming additive volumes, what is the molarity of a KBr soln

prepared by mixing 200. mL of a 0.053 M KBr soln with 500. mL of a

0.072 M KBr soln?

mol KBr soln 1 = 0.200 L 0.53 M = 0.011 mol KBr

mol KBr soln 2 = 0.500 L 0.072 M = 0.036 mol KBr

_______ _____________

0.700 L 0.047 mol KBr

Mfinal soln = 0.047 mol KBr / 0.700 L = 0.067 M

Safety Tip When working with Acids

Generally speaking - care must be taken

when adding water to an acid!!!

CHEM 121 - EXAM I REVIEW

Chapters 1, 2 & 3

EXAM I - REVIEW

Bring pencil & calculator (no cell phones)

Periodic Table & Scantron form will be provided.

Scantron (problem 5 - start with #1 on form)

Multiple Choice 1-19

You may leave (quietly) when finished.

Chapter 1 - Matter & Measurement

Matter -

Physical/Chemical

Elements/Compounds (Ch2)

Mixtures (Ch2)

Math in Chemistry

Measurement & Sig Figs

Dimensional Analysis

Density

Percent

Not On Exam: Energy

Scientific Method

Temperature

Speed as a Conversion Factor

The distance from the Earth to the moon is approximately

240,000 miles. The peregrine falcon has been measured as

traveling up to 350 km/hr in a dive. If this falcon could fly to the

moon at this speed, how many seconds would it take? (Useful conversion: 1 mile = 1.609km)

240,000 miles = seconds

1 mile

1.609km

350 km

1 hour 60 minutes

1 hour

60 secs

1 minute

A. 4.0 x 106 B. 3.97 x 106 C. 1.45 x 104 D. None of these

Distance Time Speed

2 sig figs 4 sig figs 2 sig figs exact exact

Density Learning Check

A 32.65 g sample of a solid is placed in a flask. Toluene (a liquid),

in which the solid is insoluble, is added to the flask so that the total

volume of solid and liquid together is 50.00mL. The solid and

toluene together weigh 58.58 g. The density of toluene at the

temperature of the experiment is 0.864g/mL. What is the density of

the solid?

Density = Mass Volume

A. 1.17g/mL B. 0.653g/mL C. 1.63g/mL D. None of these







the solid?

Find mass of Toluene, so that you can find the volume of Toluene:

58.58 g (toluene + solid) - 32.65 g (solid)

25.93g (toluene) = Volume Toluene

0.864g

1 mL = 30.0mL








the solid?

Find volume of the solid, and then compute the density:

- 30.0mL (toluene)

50.00mL (solid + toluene)

20.0 mL (solid)

Density (solid) = 32.65g

20.0mL

= 1.63g/mL


Calculate the following:

[(5.031-4.96)(2.38)] 3.91 = _____

A. 0.04

B. 0.043

C. 0.0432

D. None of these

5.031 -4.96

0.07

1 sig fig

(0.07)x(2.38)

3.91

Chapter 2 - Matter

Matter

Laws that lead to Atomic View of Matter

History of Atom

Atomic Symbols & Atomic Mass

Periodic Table

Ionic Compounds

Predicting formulas

Nomenclature

Polyatomic Ions

Hydrates

Covalent Compounds & Elements

Nomenclature

Learning Check

Which one of the following is not a physical property of iron pentacarbonyl [Fe(CO)5]

a. color

b. density

c. melting point

d. flammability

e. boiling point

Learning Check

Oxygen, O2, is an example of

a. an atom

b. a molecular compound

c. an ionic compound

d. a molecular element

Learning Check

Which species has 54 electrons?

a. 132 Xe+

b. 123 Te2-

c. 118 Sn2+

d. 112 Cd2+

54

52

50

49

36

Learning Check

Name each compound.

A. N2O4

B. MnCl2

C. (NH4)3PO4

D. Cu2CO3

Dinitrogen tetroxide (Covalent - Greek prefixes)

(Ionic - Variable charge on metal) Manganese (II) Chloride

(Ionic - Polyatomics) Ammonium Phosphate (Ionic - variable charge on metal & polyatomic)

Copper (I) Carbonate

Chapter 3 - Stoichiometry

The Mole

Empirical/Molecular

Formulas (combustion)

Chemical Equations

Writing from scratch

balancing

Stoichiometry

Basic

Limiting Reactant & %

yield

Learning Check

The combustion of 12.62 grams of a compound which contains only C,H and Br yields 12.09 grams of CO2 and

4.95 grams of H2O. What is the empirical formula of the

compound?

a. C4H12Br2 b. C5H10Br2 c. C5H10Br5 d. C7H10Br4

e. C5H8Br2

12.09g CO2

g C + g H + g Br

= ? g C

44.01g CO2

1 mol CO2

1 mol CO2

1 mol C

1 mol C

12.011g C

= 3.300g C

- 3.300g C 12.62 g

Learning Check



compound?


e. C5H8Br2

4.95g H2O

- 3.300g C 12.62 g

g C + g H + g Br

= g H

18.02g H2O

1 mol H2O

1 mol H2O

2 mol H

1 mol H

1.0079 g H

= 0.554g H

- 0.554g H

8.77 g Br

Learning Check



compound?


e. C5H8Br2

- 3.300g C 12.62 g

g C + g H + g Br

0.554g H

- 0.554g H

8.77 g Br

3.300g C

8.77 g Br

12.011g C

1 mol C

1.0079g H

1 mol H

79.90 g Br

1 mol Br

= 0.2747 mol C

= 0.550 mol H

= 0.110 mol Br

0.110

0.110

0.110

= 2.5 C

= 5 H

= 1 Br

Learning Check



compound?


e. C5H8Br2

- 3.300g C 12.62 g

g C + g H + g Br

0.554g H

- 0.554g H

8.77 g Br

3.300g C

8.77 g Br

12.011g C

1 mol C

1.0079g H

1 mol H

79.90 g Br

1 mol Br

= 0.2747 mol C

= 0.550 mol H

= 0.110 mol Br

0.110

0.110

0.110

= 2.5 C

= 5 H

= 1 Br

X 2

X 2

X 2

Learning Check

Analysis of a metal chloride, XCl3, shows that

it contains 67.2% Cl by mass. Calculate the

atomic mass of X .

100% - 67.2%(Cl) = 32.8%(X)

100 g sample

67.2g Cl

32.8g X

35.453g Cl

1 mol Cl = 1.90 mol Cl

1 mol X

Molar Mass X

= ? mol X

Smallest 1:3 ratio of atoms

? mol X

= 3 Cl

= 0.633 mol X

Molar Mass of X = 51.8g/mol

Learning Check

Magnesium metal and oxygen react to form magnesium

oxide, which is a white solid. If 36.5g of Mg and 35.0g of

O2 actually produce in the lab 52.3 grams of magnesium

oxide

What is the percent yield of Magnesium oxide as

determined by the limiting reactant problem?

2Mg + O2 ----> 2MgO BEFORE: 36.5g 35.0g 0.0g

Actual Yield

AFTER: Theory Yield

% Yield = Actual x 100%

Theory

Learning Check


Actual Yield = 52.3g MgO

AFTER: Theory Yield

36.5g Mg = g MgO

24.30g Mg

1 mol Mg

2 mol Mg

2 mol MgO

1 mol MgO

40.30g MgO

= 60.5g MgO

35.0g O2 = g MgO

32.00g O2

1 mol O2

1 mol O2

2 mol MgO

1 mol MgO

40.30g MgO

= 88.2g MgO

0.0g 60.5g

Learning Check


Actual Yield = 52.3g MgO

AFTER:

36.5g Mg = g MgO

24.30g Mg

1 mol Mg

2 mol Mg

2 mol MgO

1 mol MgO

40.30g MgO

= 60.5g MgO

0.0g

52.3g

60.5g X 100% = 86.4% % Yield=

60.5g

LIMITING REAGENT PROBLRM

A 30.8 g sample of PCl5 (MW = 208.22 g / mol) was reacted with

8.50 g H2O (MW = 18.02 g / mol) according to the following rxn:

PCl5 + H2O H3PO4 + HCl

What is the theoretical yield of HCl? Assuming a student isolated

13.0 g HCl, what is the %-yield?

PCl5 + 4 H2O H3PO4 + 5 HCl

mol PCl5 = 30.8 g 1 mol PCl5 / 208.22 g = 0.148 mol PCl5

0.148 mol / 1 = 0.148

Mol H2O = 8.50 g 1 mol H2O / 18.02 g = 0.472 mol H2O

0.472 mol / 4 = 0.118 LR

Note: The smaller of the two values 0.148 and 0.118 is the LR.

g HCl = 8.50 g H2O 1 mol H2O / 18.02 g H2O

5 mol HCl / 4 mol H2O 36.45 g HCl / mol HCl

= 21.5 g

%-yield = (actual yield / theortical yield) 100

= 13.0 g HCl / 21.5 g HCl (100) = 60.5 %

Mult-step Stoichiometry Problem

How many g of KClO3 would be needed to supply the proper amount

of oxygen to burn 33.2 g of methane (CH4) according to the following

reactions?

KClO3 KCl + O2

CH4 + O2 CO2 + H2O

First balance each reaction.

2 KClO3 2 KCl + 3 O2 CH4 + 2 O2 CO2 + 2 H2O

g KClO4 = 33.2 g CH4 1 mol CH4 / 1604 g CH4 2 mol O2 / mol CH4

2 mol KClO3 / 3 mol O2 122.55 g KClO3 / mol KClO3

= 338 g KClO3

Learning Check

If 22.8 mL of 0.100 M MgCl2 is needed to completely react with 15.0 mL

of AgNO3 solution, what is the molarity of the AgNO3 solution?

MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq)

A. 0.304M B. 1.20M C. 0.152M D. 0.405M

BEFORE: 22.8 mL

0.100 M

15.0 mL

Moles MgCl2 ? Moles AgNO3

? M

Learning Check




A. 0.304M B. 1.20M C. 0.152M D. 0.405M

0.0228 L 0.100 mols MgCl2

MgCl2

= moles AgNO3

1 L

2 mol AgNO3

1 mol MgCl2

= 0.00456 moles AgNO3

Learning Check




A. 0.304M B. 1.20M C. 0.152M D. 0.405M

Molarity = 0.00456 moles AgNO3

0.0150 L

che121-lec8-sp120

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