CHEMICAL REACTIONS Chapter 4 · Chemical Reactions. ... Convert moles reactant to moles product ......

Jeffrey Mack

California State University,

Sacramento

Chapter 4

Stoichiometry:

Quantitative

Information about

Chemical Reactions

The study of mass relationships in chemical reactions is called

Stoichiometry.

Stoichiometry provides quantitative information about chemical

reactions.

Mass must be conserved in a chemical reaction.

Total mass of

reactants

Total mass of

products =

Chemical equations must therefore be balanced for mass.

Numbers of atoms

on the reactant side =

Numbers of atoms

on the product side

Stoichiometry

• The stoichiometric balancing coefficients are

the numbers in front of the chemical

formulas.

• They give the ratio of reactants and

products.

• The balancing coefficients allow us to convert

between numbers of reactants and products.

ratio Conversion

factors

Stoichiometry

The ratio of any two

species (reactants or products) in a balanced

chemical reaction.

Stoichiometric ratio:

The branch of chemistry that deals

with the mole proportions of chemical reactions.

Stoichiometry:

2A + 3B A2B3

2 A’s combine with 3B’s

CONVERSION FACTORS!!!

2A

3B

3B

2A

or

Stoichiometry

2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)

Given the following reaction

How many moles of water are produced when 3.0 mols of

oxygen react?

Stoichiometry

2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)



oxygen react?

2 2.6 mol H O23.0 mol O 2

2

6 mol H O×

7 mol O

mol O2 mol H2O

Stoichiometry

2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)



oxygen react?

2 2.6 mol H O23.0 mol O 2

2

6 mol H O×

7 mol O

mol O2 mol H2O

2 sig. figs. 2 sig. figs. exact

conversion

factor

Stoichiometry

2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)


How many moles of C2H6 must react to produce 1.75 mols

of CO2?

Stoichiometry

2C2H6(l) + 7 O2(g) 4CO2(g) + 6H2O(l)


How many moles of C2H6 must react to produce 1.75 mols

of CO2?

1.75 mol CO2 2 6

2

2 mol C H×

4 mol CO= 0.875 mol C2H6

Stoichiometry

• Mole/Mass relationships.

• It is not possible to relate masses in reactions

without going through moles.

Stoichiometry

Consider the following reaction:

If 454 g of solid ammonium nitrate decomposes to form

dinitrogen monoxide gas and liquid water, how many

grams of dinitrogen monoxide and water are formed?





Strategy Map for the solution to the problem:





Data Information:

Mass of reactant

Write the balanced

chemical equation Step 1:

Eq. gives mole ratios

(stoichiometry)

Convert mass to

moles Step 2:

amount of reactant in

moles

Convert moles reactant

to moles product Step 3:

amount of products

in moles

Convert moles of

products to mass Step 4:

mass of products in

grams





Step 1: Write the balanced chemical equation



dinitrogen monoxide gas and liquid water, how many grams of

dinitrogen monoxide and water are formed?


NH4NO3(s) + 2 H2O(l) N2O(g)







Step 2: Convert mass of reactant (454 g NH4NO3) to moles.







Step 2: Convert mass of reactant (454 g NH4NO3) to moles.

4 34 3

4 3

1mol NH NO454g NH NO ×

80.04g NH NO= 5.67 mol NH4NO3





Step 3: Using the balanced chemical equation, convert moles

of reactant (NH4NO3) to moles of products.

NH4NO3(s) + 2 H2O(g) N2O(g)








5.67 mol NH4NO3 2

4 3

1 mol N O×

1 mol NH NO= 5.67 mol N2O








5.67 mol NH4NO3 2

4 3

1 mol N O×


5.67 mol NH4NO3 2

4 3

2 mol H O×

1 mol NH NO= 11.3 mol H2O








5.67 mol NH4NO3 2

4 3

1 mol N O×


5.67 mol NH4NO3 2

4 3

2 mol H O×

1 mol NH NO= 11.3 mol H2O

Recall that the coefficients represent the molar ratios of reactants to

products and vice versa…





Step 4: Now convert the moles of products (N2O and H2O) to

grams using the molar mass of each.







5.67 mol N2O 2

2

44.01 g N O×

1 mol N O= 250. g N2O







5.67 mol N2O 2

2

44.01 g N O×

1 mol N O= 250. g N2O

11.3 mol H2O 218.02 g H O×

1 mol H O= 204 g H2O







5.67 mol N2O 2

2

44.01 g N O×

1 mol N O= 250. g N2O

11.3 mol H2O 218.02 g H O×

1 mol H O= 204 g H2O

…or 454g NH4NO3 250. g N2O = 204 g H2O





0 mol + 5.67 mol +11.3 mol

0 g 250. g 204 g

454 g 0 g 0 g

Compound NH4NO3 N2O H2O

Initial (g)

Initial (mol)

Change (mol)

Final (mol)

Final (g)

5.67 mol 0 mol 0 mol

Look what happens!

5.67 mol + 5.67 mol +11.3 mol

Mass is conserved but Moles are not!

grams moles moles grams

Using

molar

mass

Using

mole

ratios

Using

molar

mass

• In order to solve stoichiometry problems, one must

go through moles using molar masses and mole

ratios as conversion factors.

• One cannot do this without writing a balanced

chemical equation first.

Stoichiometry Problems

• In some cases involving two or more

reactants, there is insufficient amount of one

reactant to consume the other reactants

completely.

• The reactant that is in short supply therefore

LIMITS the quantity of product that can be

formed.

• The theoretical yield of products is limited by

this “Limiting Reactant”.

Reactions Involving a Limiting

Reactant

32

Limiting Reactant Analogy

If we have 10

sandwiches, 18

cookies, and 12

oranges …

… how many

packaged meals

can we make?

• The theoretical yield is the maximum product yield that

can be expected based on the masses of the reactants

and the reaction stoichiometry.

• The actual yield is the experimentally measured amount of

products that results upon completion of the reaction.

• The percent yield is a measure of the extent of the reaction

in terms of the actual vs. the theoretical yield.

% Yield Actual Yield (in grams or moles)

Theoretical Yield (in grams or moles)100

Reaction Yields

Consider the reaction of aluminum and oxygen:

• Which is the limiting reactant if we start with 50.0 g

Al and 50.0 g O2?

• What is the Theoretical Yield for the reaction?


• Which is the limiting reactant if we start with 50.0 g Al

and 50.0 g O2?


Strategy Map: Data Information: Masses of two

reactants. Recognize as a limiting

reactant problem.

Write the balanced


Balanced Eq. gives mole ratios

(stoichiometry).

Convert masses of each

reactant to moles Step 2:



and 50.0 g O2?


Strategy Map:

amount of each reactant in moles.

Decide which reactant limits by comparing

mole amounts to stoichiometric ratios. Step 3:

Limiting reactant is know.

Calculate the moles of product

based on limiting reactant. Step 4:



and 50.0 g O2?


Strategy Map: Amount of product based on the

L.R.

Calculate the mass of product

based on moles. Step 5:

This is the Theoretical Yield



and 50.0 g O2?


4Al(s) + 3O2(g) 2Al2O3(s)



and 50.0 g O2?


4Al(s) + 3O2(g) 2Al2O3(s)

g Al(initial) mols Al(initial)

g O2(initial) mols O2(initial)

1 mol Al50.0g Al 1.85 mols Al

26.98g Al

22 2

2

1 mol O50.0g O 1.56 mols O

32.00g O



and 50.0 g O2?


4Al(s) + 3O2(g) 2Al2O3(s)

Compare moles of Al to moles of O2

= =2 2 2 2

1.85 mols Al 1.19 mols Al 4 mols Al 1.33 mols Al vs.

1.56 mols O 1.00 mols O 3 mols O 1.00 mols O

Since 1.19 < 1.33, Aluminum is the Limiting Reactant



and 50.0 g O2?


4Al(s) + 3O2(g) 2Al2O3(s)

Calculate the moles and grams of product based on the

limiting reactant, Al.

2 3 2 32 3

2 3

2 mols Al O 101.96 g Al O1.85 moles Al 94.3 g Al O

4 mols Al 1 mol Al O

Consider the reaction of Al and O2 previously

discussed: From the calculations it was determined that 94.3g of Al2O3 could be

formed from the reaction of 50.0 g of Al and 50.0 g of O2. This is the

theoretical yield.




theoretical yield.

If 48.5 g of Al2O3 was collected at the completion of the

reaction (actual yield), what is the % yield for the reaction?




theoretical yield.

If 48.5 g of Al2O3 was collected at the completion of the

reaction (actual yield), what is the % yield for the reaction?

=% Yield Actual Yield

Theoretical Yield ´ =100

2 3

2 3

48.5 g Al O

94.3 g Al O´ 100

= 51.4 %


discussed:

From the calculations it was determined that 94.3g of Al2O3 could be


theoretical yield.

How many grams of oxygen remain?


discussed:



theoretical yield.


By conservation of mass:

mass of reactants = mass of products

mass Al + O2 (initial) = mass of Al2O3 + O2 (left over)


discussed:



theoretical yield.


2

2 3

50.0 g Al

50.0 g O

94.3 g Al O

+

-

5.7g O2 remains

48

Here's a nice limiting reagent problem we will practice

Consider the reaction:

2 Al + 3 I2 ------> 2 AlI3

Determine the limiting reagent if one starts with:

a) 1.20 mol Al and 2.40 mol iodine.

b) 1.20 g Al and 2.40 g iodine

c) How many grams of Al are left over in part b?

ANSWES a) Al b) Iodine c) 1.03g Al

Now Try This

Chemical Analysis

• A 0.123 g sample of the mineral thenardite

contains sodium sulfate along with several inert

compounds.

• The sodium sulfate in the sample is converted to

insoluble barium sulfate by adding aqueous

barium chloride in excess to the dissolved

thenardite sample.

• The mass of recovered barium sulfate is 0.177 g

• What is the mass percent of sodium sulfate in

the mineral?

Chemical Analysis

• Recognize that all of the sulfate in sodium sulfate

ends up as barium sulfate.

Chemical Analysis

Chemical Analysis



Na2SO4(aq) + BaCl2(aq) BaSO4(aq) + 2NaCl(aq)




Chemical Analysis

Chemical Analysis




• The moles and mass of sodium sulfate can

therefore can be found from the mass of barium

sulfate






sulfate

g BaSO4 mols BaSO4 mols Na2SO4 g Na2SO4 % Na2SO4

Chemical Analysis

Chemical Analysis






sulfate

• This yields the % of sodium sulfate in the sample.

g BaSO4 mols BaSO4 mols Na2SO4 g Na2SO4 % Na2SO4

Na2SO4(aq) + BaCl2(aq) BaSO4(s) + 2 NaCl(aq)

= ´2 42 4

mass Na SOMass % Na SO 100

mass Sample

4 2 4 2 44 2 4

4 4 2 4

1 mol BaSO 1 mol Na SO 142.0 g Na SO0.177 g BaSO 0.108 g Na SO

233.4 g BaSO 1 mol BaSO 1 mol Na SO´ ´ ´ =

molar

mass

BaSO4

mole

ratio

molar

mass

Na2SO4

´ = 2 4

0.108g100 61.0% Na SO

0.177g

Chemical Analysis

• Combustion involves the addition oxygen to another

element.

• When hydrocarbon molecules burn completely, the

products are always carbon dioxide gas and water.

Determining the Formula of a Hydrocarbon by Combustion

• A streaming flow of pure O2 is passed over a weighed mass

of the unknown sample in an apparatus like the one above.

• The heat from the furnace converts the carbon in the

compound to CO2 and the hydrogen into H2O.

• The CO2 and H2O are collected in pre-weighed traps.

A hydrocarbon is a

compound that contains

predominately carbon

and hydrogen and

possibly smaller amount

of elements such as O, N

and S.

Combustion Analysis

When combustion is completed, the traps are weighed and the

masses of CO2 and H2O are determined.

mass trap + H2O

mass trap

= mass H2O

mass trap + CO2

mass trap

= mass CO2

Combustion Analysis

• All of the carbon in the sample ends up in the CO2.

• All of the hydrogen in the sample ends up in the water.

• Using stoichiometry, one can relate the carbon in CO2 and

the hydrogen in H2O back to the carbon and hydrogen in the

original sample.

• Any oxygen in the sample can be deduced via conservation

of mass.

Combustion Analysis

Combustion Analysis Problem:

• 1.516 g of a compound containing carbon, hydrogen and oxygen

(CXHYOZ) is subjected to combustion analysis.

• The results show that 2.082 g of CO2 and 1.705 g of H2O were

produced.

• What is the empirical formula for the compound?

• If the molecular weight of the compound is 160.2 g/mol,

what is the molecular formula of the compound?





produced.

Data Information: Mass of

CO2 & H2O, MM of the

compound.

calculate moles of CO2 and

H2O using molar masses Step 1:

Amounts of CO2 & H2O in moles

Convert mols CO2 & H2O into mols

C and H from the sample. Step 2:

Strategy Map





produced.

Strategy Map Moles of Carbon and hydrogen in

the sample.

Convert the moles of C

and H to grams Step 3:

Grams of Carbon and hydrogen in

the sample.

Subtract the mass of C &

H from the sample mass Step 4:





produced.

Strategy Map This gives the mass of Oxygen in the

sample. If the mass is zero then there is

no oxygen in the sample.

Convert grams of oxygen

to moles of oxygen Step 5:

moles of O, C and H are now known. This

becomes and empirical formula problem.

Divide each mole quantity by the smallest mole

value to get the elemental mole ratios. Step 6:





produced.

Strategy Map This gives x, y and z in

CXHYOZ, the empirical

formula.

Determine the ratio of the experimental

molar mass to the empirical molar mass. Step 6:

This yields the molecular formula..

Solution to the problem:

22.082 g CO 2

2

1 mol CO 44.01g CO 2

1 mol C 1 mol CO

g CO2 mol CO2 mol C

0.04731 mols C


22.082 g CO 2

2


1 mol C 1 mol CO

0.04731 mols C

g CO2 mol CO2 mol C

21.705 g H O 2

2

1 mol H 0 18.02 g H 0 2

2 mol H 1 mol H O

0.1892 mol H

g H2O mol H2O mol H


22.082 g CO 2

2


1 mol C 1 mol CO

0.04731 mols C

g CO2 mol CO2 mol C

21.705 g H O 2

2

1 mol H 0 18.02 g H 0 2

2 mol H 1 mol H O

0.1892 mol H

g H2O mol H2O mol H

Don’t forget the mole ratio…

But what about the oxygen?

To find the moles of oxygen in

the sample you could use the

grams of CO2 or H2O…

Right?

Survey says….

NO!

To find the moles of oxygen in

the sample you could use the

grams of CO2 or H2O…

You can’t determine the oxygen in the sample from

either CO2 or H2O…

Think about what is happening to the sample:

CxHyOz + O2(g) CO2(g) + H2O (l)

The oxygen in the CO2 and H2O came from both the sample

and the O2 in the furnace!

Since there is no way to know how much of that oxygen

came from the sample and the O2, you cannot determine the

oxygen in the sample directly!

Combustion Analysis Problem

Mass of O in the sample:

1.516 g 0.5682 g 0.1907 g 0.757 g O

mols of O: 0.757 g O1 mol O

16.00 g O

0.0473 mols O

original sample mass

mass of C and H in sample


0.04731 mols C

0.1892 mol H

0.0473 mols O

Collecting the results:

Both carbon and oxygen have the smallest numbers of

mols: Divide each set of mols by 0.04731.

CXHYOZ

0.0473 1

0.0X

473

0.1892 4

0.0473Y

0.0473 1

0.0Z

473

yields

The empirical formula is: CH4O


Determining the Molecular Formula

For some compounds, the molecular formula is a multiple of the

empirical formula:

empirical formula molecular formula

Since the molecular formula is a multiple is scaled by a factor “n”,

the molecular and empirical molar masses must also scale by

the same ratio.

n x CXHYOZ = CnXHnYOnZ n = 2,3,4…

gmolar formula mass mol

gempirical formula mass mol

n


From the combustion analysis it was found that the empirical

formula for the unknown compound was CH4O.

This yields an empirical molar mass or empirical weight of:

[12.01 + 4x(1.0079) 16.00] g/mol =

The ratio of the molar mass to the empirical mass is:

g160.2 mol

g32.05.

4mol

000

4Therefore, the molecular formula is 5 CH O

5 20 5C H O

32.04 g/mol


78

Combust 11.5 g ethanol

Collect 22.0 g CO2 and 13.5 g H2O

ANSWER: Empirical formula C2H6O

Now Try This for Combustion Analysis

• Most chemical studies require quantitative

measurements.

• Experiments involving aqueous solutions are

measured in volumes of solutions rather than

masses of solids, liquids, or gases.

• Solution concentration, expressed as

molarity, relates the volume of solution in

liters to the amount of substance in moles.

Measuring Concentrations of

Compounds in a Solution

Solution = solute + solvent

That which is

dissolved (lesser

amount)

That which is

dissolves

(greater amount)

Terminology

per liter of solution. Molarity: Moles of solute

Molarity of X (cX) = moles of solute

L of Solution {units: mol/L}

Molarity is a conversion factor that transforms units of volume

to mole and vise–versa.

cNaCl = [NaCl] mol NaCl

= 1.00 1L

= 1.00 M

Solutions and Concentration

Molarity

moles

Liters

Molarity relates

mols and

volume (L)

knowing 2

corners, you can

find the 3rd

If you know moles & L, you know Molarity (M) and so on.

Moles, Liters, & Molarity

Molarity Volume = moles

mols

L

moles = Volume

mol L

mol

L

If you know Molarity and volume, you know moles!

If you know mols and molarity, you know volume!

L = moles

Moles, Liters, & Molarity

Preparing Solution of Known

Concentration

PROBLEM:

Dissolve 5.00 g of NiCl2 • 6 H2O in enough

water to make 250. mL of solution. Calculate

molarity.

Step 1: Calculate moles of NiCl2 • 6H2O

2 2 2 2

1 mol5.00g NiCl 6H O 0.0210 NiCl 6H O

237.7 g

Step 2: Calculate molarity

3

2 2

1 10 mL0.0210 NiCl 6 H O 0.0841 M

250. mL 1 L

[CuCl2] = 0.30 M = [Cu2+] = 0.30 M

[Cl-] = 2 0.30 M = 0.60 M

CuCl2(aq)

Cu2+(aq) + 2Cl (aq)

For every one copper

(II) ion in solution,

there are two chloride

ions.

Ion Concentrations in a Solution

Concentration Problem:

Determine the nitrate ion concentration (in mols/L) of a

solution that is made from 50.1451 g of iron (III) nitrate

diluted to a final volume of 250.0 mL.





Solution:

3 3g Fe(NO ) 3 3mol Fe(NO ) 3mole NO 1

Vol. (L)

molar mass

mole ratio

3mole NO

1 L





Solution:


Vol. (L)

molar mass

mole ratio

3mole NO

1 L

Fe(NO3)3 (aq) Fe3+(aq) + 3NO3– (aq)





Solution:


Vol. (L)

molar mass

mole ratio

3mole NO

1 L


3 350.1451g Fe(NO ) 3 3

3 3

1 mol Fe(NO ) 241.88 g Fe(NO )

3

3 3

3 mol NO 1 mol Fe(NO )

1 250.0 ml

1000 ml×

L= 2.488 M NO3

–





Solution:


Vol. (L)

molar mass

mole ratio

3mole NO

1 L


3 350.1451g Fe(NO ) 3 3

3 3

1 mol Fe(NO ) 241.88 g Fe(NO )

3

3 3

3 mol NO 1 mol Fe(NO )

1 250.0 ml

1000 ml×

L= 2.488 M NO3

–

from the balanced equation!

33NO

C = [NO ] = 2.488 M


How many grams of sodium phosphate are in

35.0 mL of a 1.51 M Na3PO4 solution?

mL solution L mols Na3PO4 g Na3PO4

use M as a

conversion

factor

Solution:

molar mass


How many grams of sodium phosphate are in

35.0 mL of a 1.51 M Na3PO4 solution?

mL solution L mols Na3PO4 g Na3PO4

use M as a

conversion

factor

Solution:

molar mass

35.0 mL3

L×

10 mL3 41.51 mol Na PO

× L

3 4

163.94 g×

1 mol Na PO

3 4= 8.66 g Na PO

• Weigh out the solid solute

and carefully transfer it to the

volumetric flask.

• Dissolve in a volume of

solvent less than the flask

volume.

• Fill the volumetric flask to the

calibration mark with solvent.

• Invert the flask many times to

ensure homogeneous mixing.

Preparing Solutions

Preparing a Solution by Dilution

When more solvent is added to a solution, the concentration

of solute decreases:

Why? molcon

es centra

Solutetion (

L SoluM) =

tion

as solvent is added, the

denominator increases in

magnitude…

since the moles of

solute remains

constant…

The magnitude of the concentration must decrease!!!

Dilutions: Volume by Volume

• A Student adds a 50.0 mL aliquot of a 0.515 M HCl

solution to a 500.0 mL volumetric flask.

• She then filled the volumetric flask to the calibration mark

with water.

• What is the new molarity of the HCl solution?




with water.


Solution

50.0 mL HCl3

L 10 ml

0.515 mol HCl

1 L




with water.


Solution

50.0 mL HCl3

L 10 ml

0.515 mol HCl

1 L

at this point I

have mols HCl

• A Student adds a 50.0 mL aliquot of a 0.515 M HCl solution

to a 500.0 mL volumetric flask.


with water.


Solution

50.0 mL HCl3

L 10 ml

0.515 mol HCl

1 L

at this point I

have mols HCl

1 500.0 mL




with water.


Solution

50.0 mL HCl3

L 10 ml

0.515 mol HCl

1 L

at this point I

have mols HCl

1 500.0 mL

dividing by the “new

volume” results in a new

M




with water.


Solution

50.0 mL HCl3

L 10 ml

0.515 mol HCl

1 L

at this point I

have mols HCl

1 500.0 mL

dividing by the “new

volume” results in a new

M

310 ml

L

converting

to L gives:

1/10th the

original M =HClC 0.0515 M

What we call pH is actually a mathematical function, "p"

(the negative logarithm,

based 10 )

p is a shorthand notation for " log10"

Quick Review of logs… (see Appendix A, p. A2 of your text)

log x = n where x = 10n

log 1000 = log(103) = 3

log 10 = log(101) = 1

log 0.001 = log(10 3) = 3

The pH Scale

Example: A student is given a solution that is labeled pH = 4.72,

what is the molarity of H+ in this solution?

+ -= pH[H ] 10plugging in 10 4.72 into you calculators yields:

1.90546 10 5

but wait… how many sig. figs. are allowed?

- - -= = ´4.72 (0.28 5) 0.28 510 10 10 10

100.28 = 1.9 2 sig. figs.!

therefore the concentration

should be reported as: - +´ 51.9 10 M [H ]

The pH of a solution provides a way to express the acidity, or

the concentration of H+ in solution:

low pH = high [H+]

acidic solution

high pH = low [H+]

basic solution

a pH of 7 indicates that the solution is neutral

The pH Scale: 0 to 14

Prior to now, we have discussed reactions in solution

from a qualitative aspect.

Reactants Products

With the addition of molarity to our tools of chemistry,

we now can perform quantitative calculations for

reactions in aqueous solutions.

volume moles moles grams

grams moles moles volume

Solution Stoichiometry

Example:

How many grams of calcium carbonate can be

consumed by 35.5 mL of 0.125 M H2SO4 (aq) ?

Example:



H2SO4 (aq) + CaCO3 (s) H2O (l) + CO2(g) + CaSO4(s)

Solution: We know that acids react with carbonate salts to produce CO2(g)

Example:





mL H2SO4 mol H2SO4 mol CaCO3 g CaCO3

convert volume to grams

Example:





mL H2SO4 mol H2SO4 mol CaCO3 g CaCO3

convert volume to grams

35.5 mL 3

L

10 mL2 40.125 mol H SO

L3

2 4

1 mol CaCO

1 mol H SO3

3

100.1 g CaCO

1 mol CaCO

= 30.444g CaCO

• Some solutions cannot be accurately made by

weight and dilution methods if the solute is

impure or unstable.

• When this is the case, one can make up a

solution of approximate concentration then

“Standardize” the solution against a “Standard”

compound that reacts with the solute in solution.

• A standard compound is one that is very stable

with a known molar mass that yields a

necessary number of sig. figs. (4 or more)

Standard Solutions

• To standardize a solution, one performs a

“Titration” where a measured volume of the

solution is added to a known amount of the

standard.

• An “Indicator” is added to signal the point

where the moles of standard = moles of

solute reacted (Endpoint), knowing moles

and volume, one can compute the

concentration of the solution.

Standard Solutions

Setup for Titrating an Acid with a

Base

1. Add solution from the buret.

2. Reagent (base) reacts with

compound (acid) in solution in

the flask.

3. Indicator shows when exact

stoichiometric reaction has

occurred.

4. Net ionic equation

H3O+(aq) + OH-(aq)

2 H2O (l)

5. At equivalence point

moles H3O+ = moles OH-

Titrations

A common standard used in base (OH–) standardizations in

the mono–protic acid, KHP.

C

C

C

C

C

C

C

O

OH

C

OK

O

H

H

H

H

KHP is an acronym for:

potassium hydrogen phthalate (not potassium hydrogen phosphorous)

The H-atom in the upper right of the

compound dissociates to yield H+(aq)

1 mol of KHP yields 1 mol of H+ when dissolved in solution

KHP (aq) H+(aq) + KP–(aq)

The molar mass of KHP is 204.22 g/mol (C8H5O4K)

KPH

Problem: A student prepares a solution of approximately 0.5 M NaOH(aq).

He performs a standardization of the solution using KHP as the

standard to determine the exact concentration.

Determine the [OH–] from the data collected:





mass KHP + flask:

mass empty flask:

95.3641 g

95.0422 g

final buret reading:

initial buret reading:

30.12 mL

1.56 mL





mass KHP + flask:

mass empty flask:

95.3641 g

95.0422 g



30.12 mL

1.56 mL

mass KHP: 0.3219 g vol. NaOH: 28.56 mL





mass KHP + flask:

mass empty flask:

95.3641 g

95.0422 g



30.12 mL

1.56 mL

mass KHP: 0.3219 g vol. NaOH: 28.56 mL

- =[OH ]-mols OH

L of solution=

mol KHP

L titrated

Remember, [OH–] = [NaOH] Strong Electrolyte!!!





0.3219 g KHP1 mol KHP

204.22 g

1 mol OH 1 mol KHP

1 28.56 mL

310 mL

L

0.05519 M OH

[NaOH] = 0.05519 M

from the stoichiometry, one mole of

OH reacts with 1 mol of KHP

Problem: • Apples contain a compound know as malic acid, C4H6O5.

• Malic acid is a “diprotic” acid, meaning that one mole of the acid

reacts with two moles of base, OH (aq).

• If 76.80 g of apple requires 34.56 mL of 0.664 M NaOH for

titration, what is weight % of malic acid in the apples?


• Malic acid is a “diprotic” acid, meaning that one mole of the acid

reacts with two moles of base, OH (aq).



Data Information:

Apple contains acid

Write the balanced


Eq. gives mole ratios

(stoichiometry)

Convert volume to

moles Step 2:

Moles of OH used

to titrate

Convert moles reactant

to moles product Step 3:

amount of malic acid

in moles

Convert moles of

products to mass Step 4:

mass of malic acid in

grams / wt. % in apples


• Malic acid is a “diprotic” acid, meaning that one mole of the

acid reacts with two moles of base, OH (aq).



The equation for the reaction is:

C4H6O5(aq) + 2NaOH(aq) Na2C4H4O5(aq) + 2H2O(liq)

Mass % of Malic Acid:

4 6 54 6 53

4 6 5

1 mol C H O1 L 0.664 mol OH 134.09 g34.56 mL 154 g C H O

10 1 L 2 mol OH 1 mol C H O

4 6 54 6 5

154g C H O100 2.01 C H O

76.80g Sample´ =

Spectrophotometry is the interaction between light

and matter that provides quantitative about solution

concentration.

Spectrophotometry

Light

Source

Wavelength

Selection

Sample

Container Spectrum

An Absorption Spectrophotometer

• Amount of light absorbed by a sample depends on

path length and solute concentration.

Different concentration of Cu2+ Same concentration, different

pathlengths

Spectrophotometry

BEER-LAMBERT LAW relates amount of light

absorbed and the path length and solute

concentration.

A = absorbance l = path length

= molar absorptivity c = concentration

• There is a linear relation between A and c for a

given path length and compound.

• Once the instrument is calibrated, knowing

absorbance one can determine concentration.

Spectrophotometry

To use the Beer-Lambert law you must first calibrate the

instrument.

The calibration plot can be

used to find the unknown

concentration of a solution

from a measured

Absorbance.

Spectrophotometry

CHEMICAL REACTIONS Chapter 4 · Chemical Reactions. ... Convert moles reactant to moles product ......

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Transcript of CHEMICAL REACTIONS Chapter 4 · Chemical Reactions. ... Convert moles reactant to moles product ......