Ch06 Final

8/3/2019 Ch06 Final

1/46

John E. McMurry Robert C. Fay

Lecture NotesAlan D. EarhartSoutheast Community College Lincoln, NE

General Chemistry: Atoms First

Chapter 6

Mass Relationships in ChemicalReactions

Copyright 2010 Pearson Prentice Hall, Inc.

8/3/2019 Ch06 Final

2/46

Copyright 2010 Pearson Prentice Hall, Inc. Chapter 6/2

Balancing Chemical Equations

A balanced chemical equation shows that the law ofconservation of mass is adhered to.

In a balanced chemical equation, the numbers and

kinds of atoms on both sides of the reaction arrow areidentical.

2NaCl(s)2Na(s) + Cl2(g)

right side:

2 Na2 Cl

left side:

2 Na2 Cl

8/3/2019 Ch06 Final

3/46



HgI2(s) + 2KNO3(aq)Hg(NO3)2(aq) + 2KI(aq)

right side:

1 Hg2 I2 K2 N6 O

left side:

1 Hg2 N6 O2 K2 I

8/3/2019 Ch06 Final

4/46



2. Find suitable coefficientsthe numbers placedbefore formulas to indicate how many formulaunits of each substance are required to balancethe equation.

2H2O(l)2H2(g) + O2(g)

1. Write the unbalanced equation using the correctchemical formula for each reactant and product.

H2O(l)H2(g) + O2(g)

8/3/2019 Ch06 Final

5/46



3. Reduce the coefficients to their smallest whole-number values, if necessary, by dividing themall by a common divisor.

2H2O(l)2H2(g) + O2(g)

4H2O(l)4H2(g) + 2O2(g)

divide all by 2

8/3/2019 Ch06 Final

6/46



4. Check your answer by making sure that thenumbers and kinds of atoms are the same onboth sides of the equation.

2H2O(l)2H2(g) + O2(g)right side:

4 H

2 O

left side:

4 H

2 O

8/3/2019 Ch06 Final

7/46


Chemical Symbols on aDifferent Level

2H2O(l)2H2(g) + O2(g)

2 molecules of hydrogen gas react with1 molecule of oxygen gas to yield 2

molecules of liquid water.

microscopic:

8/3/2019 Ch06 Final

8/46


Chemical Symbols on aDifferent Level

2H2O(l)2H2(g) + O2(g)

0.56 kg of hydrogen gas react with4.44 kg of oxygen gas to yield 5.00 kg

of liquid water.

macroscopic:

2 molecules of hydrogen gas react with1 molecule of oxygen gas to yield 2

molecules of liquid water.

microscopic:

8/3/2019 Ch06 Final

9/46

Chapter 6/9

Chemical Arithmetic:Stoichiometry

8/3/2019 Ch06 Final

10/46


2(12.0 amu) + 4(1.0 amu) = 28.0 amuC2H4:


HCl: 1.0 amu + 35.5 amu = 36.5 amu

Molecular Mass: Sum of atomic masses of all atomsin a molecule.

Formula Mass: Sum of atomic masses of all atoms in

a formula unit of any compound, molecular or ionic.

8/3/2019 Ch06 Final

11/46


1 mole = 28.0 gC2H4:

6.022 x 1023 molecules = 28.0 g


One mole of any substance is equivalent to itsmolecular or formula mass.

1 mole = 36.5 g

6.022 x 1023 molecules = 36.5 g

HCl:

8/3/2019 Ch06 Final

12/46



How many moles of chlorine gas, Cl2, are in 25.0 g?

25.0 g Cl2

70.9 g Cl2

1 mol Cl2x = 0.353 mol Cl2

8/3/2019 Ch06 Final

13/46



How many grams of sodium hypochlorite, NaOCl, are in0.705 mol?

0.705 mol NaOCl

1 mol NaOCl

40.0 g NaOClx = 28.2 g NaOCl

8/3/2019 Ch06 Final

14/46



Stoichiometry: The relative proportions in whichelements form compounds or in which substancesreact.

aA + bB cC + dD

Moles ofA

Grams ofA

Moles ofB

Grams ofB

Mole RatioBetween A

and B(Coefficients)

Molar Massof B

Molar Massof A

8/3/2019 Ch06 Final

15/46



How many grams of NaOH are needed to react with 25.0 gCl2?

2NaOH(aq) + Cl2(g) NaOCl(aq) + NaCl(aq) + H2O(l)

Aqueous solutions of sodium hypochlorite (NaOCl), bestknown as household bleach, are prepared by reaction ofsodium hydroxide with chlorine gas:

Moles ofCl2

Grams ofCl2

Moles ofNaOH

Grams ofNaOH

Mole Ratio Molar MassMolar Mass

8/3/2019 Ch06 Final

16/46



How many grams of NaOH are needed to react with 25.0 gCl2?

2NaOH(aq) + Cl2(g) NaOCl(aq) + NaCl(aq) + H2O(l)

Aqueous solutions of sodium hypochlorite (NaOCl), bestknown as household bleach, are prepared by reaction ofsodium hydroxide with chlorine gas:

1 mol NaOH

40.0 g NaOH25.0 g Cl2

70.9 g Cl2

1 mol Cl2

1 mol Cl2

2 mol NaOH

= 28.2 g NaOH

xx x

8/3/2019 Ch06 Final

17/46


Yields of Chemical Reactions

The amount actually formed in a reaction.

The amount predicted by calculations.

Actual Yield:

Theoretical Yield:

Actual yield of product

Theoretical yield of productx 100%Percent Yield =

8/3/2019 Ch06 Final

18/46


Reactions with LimitingAmounts of Reactants

Limiting Reactant: The reactant that is present inlimiting amount. The extent to which a chemicalreaction takes place depends on the limiting reactant.

Excess Reactant: Any of the other reactants stillpresent after determination of the limiting reactant.

8/3/2019 Ch06 Final

19/46



Because water is so cheap and abundant, it is used inexcess when compared to ethylene oxide. This ensuresthat all of the relatively expensive ethylene oxide is entirelyconsumed.

At a high temperature, ethylene oxide reacts with water toform ethylene glycol, which is an automobile antifreeze anda starting material in the preparation of polyester polymers:

C2H4O(aq) + H2O(l) C2H6O2(l)

8/3/2019 Ch06 Final

20/46




If 3 moles of ethylene oxide react with 5 moles of water,which reactant is limiting and which reactant is present in

excess?


8/3/2019 Ch06 Final

21/46

Chapter 6/21




8/3/2019 Ch06 Final

22/46



Li2O(s) + H2O(g) 2LiOH(s)

Lithium oxide is used aboard the space shuttle to removewater from the air supply according to the equation:

If 80.0 g of water are to be removed and 65.0 g of Li2O areavailable, which reactant is limiting? How many grams ofexcess reactant remain? How many grams of LiOH areproduced?

8/3/2019 Ch06 Final

23/46




Which reactant is limiting?

65.0 g Li2O

1 mol Li2O

1 mol H2O

29.9 g Li2O

1 mol Li2O

= 4.44 moles H2O80.0 g H2O

18.0 g H2O

1 mol H2O

= 2.17 moles H2O

Amount of H2

O given:

Amount of H2O that will react with 65.0 g Li2O:

Li2

O is limiting

x

x

x

8/3/2019 Ch06 Final

24/46




80.0 g H2O - 39.1 g H2O = 40.9 g H2O

2.17 mol H2O

1 mol H2O

18.0 g H2O

= 39.1 g H2O (consumed)

How many grams of excess H2O remain?

remaininginitial consumed

x

8/3/2019 Ch06 Final

25/46




2.17 mol H2O

1 mol LiOH

23.9 g LiOH= 104 g LiOH

How many grams of LiOH are produced?

1 mol H2O

2 mol LiOHxx

8/3/2019 Ch06 Final

26/46


Concentrations of Reactants inSolution: Molarity

Molarity (M): The number of moles of a substancedissolved in each liter of solution. In practice, asolution of known molarity is prepared by weighing anappropriate amount of solute, placing it in a container

called a volumetric flask, and adding enough solventuntil an accurately calibrated final volume is reached.

Solution: A homogeneous mixture.

Solute: The dissolved substance in a solution.

Solvent: The major component in a solution.

8/3/2019 Ch06 Final

27/46

Chapter 6/27

8/3/2019 Ch06 Final

28/46



Molarity converts between mole of solute and liters ofsolution:

Molarity =Moles of solute

Liters of solution

Lmol or 1.00 M

1.00 L1.00 mol = 1.00

1.00 mol of sodium chloride placed in enough water tomake 1.00 L of solution would have a concentrationequal to:

8/3/2019 Ch06 Final

29/46



Molar mass C6H12O6 = 180.0 g/mol

How many grams of solute would you use to prepare1.50 L of 0.250 M glucose, C6H12O6?

1 mol

0.275 mol 180.0 g= 49.5 g

1 L

1.50 L 0.250 mol= 0.275 mol

x

x

8/3/2019 Ch06 Final

30/46

Chapter 6/30

C

8/3/2019 Ch06 Final

31/46


Diluting ConcentratedSolutions

dilute solutionconcentrated solution + solvent

Since the number of moles of solute remains constant,all that changes is the volume of solution by addingmore solvent.

Mi

x Vi

= Mf

x Vf

finalinitial

Dil i C d

8/3/2019 Ch06 Final

32/46


Diluting ConcentratedSolutions

Add 6.94 mL 18.0 M sulfuric acid to enough water tomake 250.0 mL of 0.500 M solution.

Mi

= 18.0 M Mf

= 0.500 M

Vi = ? mL Vf = 250.0 mL

= 6.94 mL18.0 M

250.0 mL

Vi =

Vf 0.500 M

=

Sulfuric acid is normally purchased at a concentration of18.0 M. How would you prepare 250.0 mL of 0.500 Maqueous H2SO4?

xMi

Mf

x

S l i S i hi

8/3/2019 Ch06 Final

33/46


Solution Stoichiometry

aA + bB cC + dD

Moles ofA

Volume ofSolution of A

Moles ofB

Volume ofSolution of B

Mole RatioBetween A

and B

(Coefficients)

Molar Massof B

Molarity ofA

S l i S i hi

8/3/2019 Ch06 Final

34/46



H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)

What volume of 0.250 M H2SO4 is needed to react with50.0 mL of 0.100 M NaOH?

Moles ofH2SO4

Volume ofSolution of H2SO4

Moles ofNaOH

Volume ofSolution of NaOH

Mole RatioBetween H2SO4

and NaOH

Molarity ofNaOH

Molarity ofH2SO4

S l ti St i hi t

8/3/2019 Ch06 Final

35/46



H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)

2 mol NaOH1 mol H2SO4

0.250 mol H2SO41 L solution

1 L

0.100 mol

1 L1000 mL

= 0.00500 mol NaOH

Volume of H2SO4 needed:

1000 mL

1 L

10.0 mL solution (0.250 M H2SO4)

0.00500 mol NaOH

50.0 mL NaOH

Moles of NaOH available:

x

x

x

x x

Tit ti

8/3/2019 Ch06 Final

36/46


Titration

How can you tell when the reaction is complete?

HCl(aq) + NaOH(aq) NaCl(aq) + 2H2O(l)

Titration: A procedure for determining the concentrationof a solution by allowing a carefully measured volume toreact with a solution of another substance (the standardsolution) whose concentration is known.

Once the reaction is complete you can calculate theconcentration of the unknown solution.

Tit ti

8/3/2019 Ch06 Final

37/46

Chapter 6/37

Titration

unknown concentration solutionErlenmeyer

flask

buretstandard solution

(known concentration)

An indicator is added which changescolor once the reaction is complete

8/3/2019 Ch06 Final

38/46

Tit ti

8/3/2019 Ch06 Final

39/46


Titration


48.6 mL of a 0.100 M NaOH solution is needed to reactwith 20.0 mL of an unknown HCl concentration. What isthe concentration of the HCl solution?

Moles ofNaOH

Volume ofSolution of NaOH

Moles ofHCl

Volume ofSolution of HCl

Mole RatioBetween NaOH

and HCl

Molarity ofHCl

Molarity ofNaOH

Tit ti

8/3/2019 Ch06 Final

40/46


Titration


20.0 mL solution

0.00486 mol HCl= 0.243 M HCl

Concentration of HCl solution:

Moles of NaOH available:

1 L

0.100 mol

= 0.00486 mol NaOH

48.6 mL NaOH

1000 mL

1 L

Moles of HCl reacted:

1 mol NaOH

1 mol HCl= 0.00486 mol HCl

0.00486 mol NaOH

1 L

1000 mL

x

x

x

x

P t C iti d

8/3/2019 Ch06 Final

41/46


Percent Composition andEmpirical Formulas

Percent Composition: Expressed by identifying theelements present and giving the mass percent of each.

Empirical Formula: It tells the smallest whole-number

ratios of atoms in a compound.

Molecular Formula: It tells the actual numbers of atomsin a compound. It can be either the empirical formula or amultiple of it.

Multiple =Molecular mass

Empirical formula mass

Percent Composition and

8/3/2019 Ch06 Final

42/46



A colorless liquid has a composition of 84.1 % carbonand 15.9 % hydrogen by mass. Determine the empiricalformula. Also, assuming the molar mass of thiscompound is 114.2 g/mol, determine the molecular

formula of this compound.

MolesMass percents Mole ratios Subscripts

Relativemole ratios

Molarmasses


8/3/2019 Ch06 Final

43/46



Mole of carbon:

Assume 100.0 g of the substance:

12.0 g C

1 mol C84.1 g C

= 7.01 mol C

1.0 g H

1 mol H

= 15.9 mol H

Mole of hydrogen:

15.9 g H

x

x


8/3/2019 Ch06 Final

44/46



Molecular formula:

7.01

C1x4H2.27x4 = C4H9

7.01

= C1H2.27

need whole numbers

C1H2.27

C7.01H15.9

smallest value for the ratio

C7.01H15.9

Empirical formula:

= 257.0

114.2multiple = C4x2H9x2 = C8H18

Determining Empirical

8/3/2019 Ch06 Final

45/46


Combustion Analysis: A compound of unknowncomposition (containing a combination of carbon,hydrogen, and possibly oxygen) is burned with oxygen toproduce the volatile combustion products CO2 and H2O,

which are separated and weighed by an automatedinstrument called a gas chromatograph.

Determining EmpiricalFormulas: Elemental Analysis

hydrocarbon + O2(g) xCO2(g) + yH2O(g)

carbonhydrogen

8/3/2019 Ch06 Final

46/46

Date post:	07-Apr-2018
Category:	Documents
Upload:	ngoctung4him
View:	256 times
Download:	0 times

Ch06 Final

Documents