# Chapter Five - Philadelphia University 5... · Chapter Five Stoichiometric Calculations: The Workhorse…

• Published on
01-Sep-2018

• View
212

• Download
0

Transcript

• Chapter Five

Stoichiometric Calculations:

The Workhorse of the Analyst

• BASICS: ATOMIC, MOLECULAR, AND FORMULA WEIGHTS

Gram-atomic weight for any element is the weight of Avogadro's number of atoms of that element, and that number is the same from one element to another.

Gram-atomic weight of any element contains exactly the same number of atoms of that element as there are carbon atoms in exactly 12 g of carbon- 12. This number is the Avogadro's number , 6.022 X 1023 atoms/g-at wt.

Since naturally occurring elements consist of mixtures of isotopes, the chemical atomic weights will be an average of the isotope weights of each element, taking into account their relative naturally occurring abundances.

Thus, none of the elements has an integral atomic weight

• 3

DALTON (ATOMIC MASS UNIT, amu)

Atomic and molecular weights are generally expressed in

terms of atomic mass units (amu),or dalton. The atomic mass unit, or dalton, is based upon a relative

scale in which the reference is the carbon-12 isotope which is assigned a mass of exactly 12 amu.

Thus, the amu, or Da, is defined as 1/12 of the mass of one neutral C-12 atom.

This definition makes 1 amu, or 1 Da, of carbon equal to

• Mole

To simplify calculations, chemists have developed the concept of the mole.

The mole is Avogadro's number (6.022 X 1023) of atoms, molecules, ions or other species.

The mole is the atomic, molecular, or formula weight of substance expressed in grams

# of moles = #grams

formula weight(g/mole)

• Millimole

# of milligrams = #millimoles X formula weight (mg/mmole)

Note: g/mole is the same as mg/mmole g/L is the same as mg/mL

# of millimoles = #milligrams

formula weight(mg/mmole)

• Exanmple

• Example

How many milligrams are in 0.250 mmol Fe2O3 (iron(III)oxide)

• Examples

1. Calculate the number of moles in 500 mg Na2S04 (sodium sulfate).

Formula weight of Na2S04 (23X2 + 32+64)= 160 mg/mmol

# Moles =

# of moles = #grams

formula weight(g/mole)mole12.3

160

500

mole12.3160

500

• Molarity Molarity is M= number of moles per liter (mole/L) or

millimoles per milliliter (mmole/mL)

M = #moles

volume (L) =

#mmoles

volume (mL)

#moles = M X volume (L)

# mmoles = M X volume (mL)

Expressing the concentrations of solutions

• Example A solution is prepared by dissolving 1.26 g AgN03 in a 250-

mL volumetric flask and diluting to volume. Calculate the molarity of the silver nitrate solution. How many millimoles AgNO3 were dissolved?

M = #moles/ volume in L= (mass/ formula mass)/ vol in L

• Example

How many grams per milliliter of NaCI are

contained in a 0.250 M solution?

#moles = M X volume (L)

# mmoles = M X volume (mL)

# mmoles = 1 mL X 0.250M = 0.250 mmol

Mass = #mmoles X formula mass of NaCl

• Example

How many grams Na2SO4 should be weighed out to prepare 500mL of a 0.100 M solution?

Mass = # mmoles X formula mass

# mmoles = M X Vol (mL) Mass = M X Vol (mL) X formula mass (mg)

Mass of Na2SO4 = 500 mL X 0.100 M X 142 mg/mmol= 7100 mg = 7.100 g

• Example

Calculate the concentration of potassium ion, K+ in grams per liter after mixing 100 mL of 0.250 M KCl and 200 mL of 0.100 M K2S04

Mmol K+ = 100 mL X 0.250 mmole/mL + 2 X 0.100 mmol/mL = 65.0 mmol /300 mL

mg of K =

• Normality

A one-normal solution, 1N, contains one equivalent of species per liter.

An equivalent 'represents the mass of material providing Avogadro's number of reacting units.

A reacting unit is a proton (H+) or an electron

The number of equivalents = # moles X # reacting units per molecule or atom

The equivalent weight is the formula weight divided by the number of reacting units.

• Normality

For acids and bases, the number of reacting units is based on the number of protons (i.e., hydrogen. ions) an acid will furnish or a base

will react with. H2S04, has two reacting units of

protons; that is, there are two equivalents of protons in each mole.

Therefore

• Normality

For oxidation-reduction reactions it is based 'on the number, of electrons an oxidizing or reducing agent will take on or supply.

• In normality calculations, the number of equivalents is the number of

moles times the number of reacting units per molecule or atom.

Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

• Normality and equivalents

# equivalents (eq) = wt (g)

eq wt (g/eq) = N (eq/L)X volume (L)

# mequivalents (eq) = wt (mg)

eq wt (mg/meq) = N (meq/mL)X volume (mL)

• Formality

Chemists sometimes use the term formality for solutions of ionic salts that do not exist as molecules in the solid or in solution.

The concentration is given as formal (F).

Operationally, formality is identical to molarity.

For convenience, we shall use molarity exclusively, a common practice.

• Molality, m

A one-molal solution contains one mole of species per 1000 g of solvent.

Molal concentrations are not temperature dependent as molar and normal concentrations are (since the solvent volume in molar and normal concentrations is temperature dependent).

Molality = #moles

mass of solvent(kg)

• Density Calculations How do we convert to Molarity

Density = mass solute /unit volume

Specific Gravity = Dsolute/DH20 DH2O = 1.00000 g/mL @ 4

oC

DH2O = 0.99821 g/mL @ 20oC

• Example

How many milliliters of concentrated sulfuric acid, 94.0% (g/100 g .solution), density 1.831 are required to prepare 1 liter of a 0.100 M solution?

• Analytical and equilibrium concentrations

The analytical concentration represents the concentration of total dissolved substance, i.e., the sum of all species of the substance in

solution = Cx

An equilibrium concentration is that of a given dissolved form of the substance = [X].

For ions of strong electrolytes, NaCl

CNa+= CCl- = CNaCl ; [Na

+] = [Cl-] = [NaCl]

For ions of weak electrolytes, HOAc

C H+ = COAc- CHOAc (It depends upon the degree of dissociation)

• Dilutions

Stock standard solution are used to prepare a series of more dilute standards.

The millimoles of stock solution taken for dilution will be identical to the millimoles in the final diluted solution .

# mmoles used for dilution = # mmoles after dilution

M stock(initial) X Vstock(initial) = M diluted(final) X Vdiluted(final)

• Example (dilution)

You have a stock 0.100 M solution of KMn04 and a series of I00-mL volumetric flasks. What volumes of the stock solution will you have to pipet into the flasks to prepare standards of 1.00, 2.00, 5.00, and 10.0 X 10-3 M KMn04 solutions?

• Example

What volume of 0.40 M Ba(OH)2 must be added to 50 mL of 0.30 M NaOH to give a solution 0.50 M in OH-?

• Other concentration units

Percentage

Parts per thousands, ppt

Parts per million, ppm

Parts per billion

%(w/w) = wt sloute (g)

wt sample (g)X100 =

wt solute (g)

wt solution (g)X100

ppt (w/w) = wt sloute (g)

wt sample (g)X103 =

wt solute (g)

wt solution (g)X103

ppm (w/w) = wt sloute (g)

wt sample (g)X106 =

wt solute (g)

wt solution (g)X106

ppb (w/w) = wt sloute (g)

wt sample (g)X109 =

wt solute (g)

wt solution (g)X109

• Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

• Example

A 2.6-g sample of plant tissue was analyzed and found to contain 3.6 g zinc. What is the concentration of zinc in the plant in ppm? In ppb?

Concentration of Zn in ppm =

Concentration of Zn in ppb = ppbXg

gX138010

6.2

106.3 96

ppmXg

gX380.110

6.2

106.3 66

• Other units (wt/vol)

%(w/v) = wt sloute (g)

vol sample (mL)X100 =

wt solute (g)

vol solution (mL)X100

ppm (w/v) = wt sloute (g)

vol sample (mL)X106 =

wt solute (g)

vol solution (mL)X106

ppb (w/v) = wt sloute (g)

vol sample (mL)X109 =

wt solute (g)

vol solution (mL)X109

• Example

A 25.0-L serum sample was analyzed for glucose content and found to contain 26.7 g. Calculate the concentration of glucose in ppm and in mg/dL.

• Example

(a) Calculate the molar concentrations of 1.00 ppm (w/v) solutions each of Li+ and Pb2+.

(b) What weight of Pb(N03)2 will have to be dissolved in 1 liter of water to prepare a 100 ppm Pb2+ solution?

• Example

The concentration of zinc ion, Zn2+, in blood serum is about 1 ppm. Express this as meq/L.

• http://www.chem.wits.ac.za/chem212-213-280/0%20Introduction%20-%20Lecture.ppt

• Stoichiometric calculations in volumetric analysis Volumetric titration

A solution of accurately known concentration (Standard

solution) is gradually added to another solution of unknown

concentration until the chemical reaction between the two

solutions is complete.

Equivalence point the point at which the reaction is complete

Indicator substance that changes color at (or near) the equivalence point

The titrant is add

Slowly until

The indicator

changes color

(pink)

Endpoint the point at which the color of indicator changes

• Requirements for titration

The reaction must be stoichiometric. That is, there must be a well-defined and known reaction between the analyte and the titrant.

NaOH + HCl NaCl + H2O The reaction should be rapid. Most ionic

reactions are very rapid.

There should be no side reactions, and the reaction should be specific.

• There should be a marked change in some property of the solution when the reaction is complete. This may be a change in color of the solution.

A color change is usually brought about by addition of an indicator, whose color is dependent on the properties of the solution, for example, the pH.

• The point at which an equivalent or stoichiometric amount of titrant is added is called the equivalence point.

The point at which the reaction is be complete is called the end point

The reaction should be quantitative. That is, the equilibrium of the reaction should be far to the right so that a sufficiently sharp change will occur at the end point to obtain the desired accuracy. If the equilibrium does not lie far to the right, then there will be gradual change in the property marking the end point (e.g., pH) and this will be difficult to detect precisely.

• Standard Solutions A solution that contains a known concentration of an analyte

It is prepared by dissolving accurately weighed quantity of highly pure material called Primary Standard material.

The primary standard material should fulfill the following requirements:

• Examples of Primary Standards

A primary standard is a reagent that is extremely pure, stable, has no waters of hydration, and has a high molecular weight.

Some primary standards for titration of acids: sodium carbonate: Na2CO3, mol wt. = 105.99 g/mol

tris-(hydroxymethyl)aminomethane (TRIS or THAM): (CH2OH)3CNH2, mol wt. = 121.14 g/mol

Some primary standards for titration of bases: potassium hydrogen phthalate (KHP): KHC8H4O4, mol wt. = 204.23

g/mol

potassium hydrogen iodate: KH(IO3)2, mol wt. = 389.92 g/mol

Some primary standards for redox titrations: potassium dichromate: K2Cr2O7, mol wt. = 294.19 g/mol

• Desirable Properties of Standard Solutions

Standard solutions should

be sufficiently stable to establish its concentration only once

react rapidly with analyte so that the time required between addition of reagent is minimized

react quantitatively to accurately determine end points

undergo a selective reaction with the analyte that can be represented by a single balanced equation

• Preparation of standard solutions Standard solutions are prepared in two ways:

1. Direct method (Primary standard solution) a primary standard compound is carefully weighed and dissolved in an

exactly known volume of solution. The direct method is the best method to be utilized.

2. Indirect method-Standardization (Secondary standard solution) - the prepared (approximately) standard solution is standardized by titrating it against :

a. A carefully weighed primary standard compound. b. A carefully weighed secondary standard compound. c. A carefully measured volume of another standard solution.

• Classification of Volumetric Methods

Acid-Base

Precipitation

Complex Formation

Oxidation-Reduction

• Volumetric calculation

• Things to know for molarity calculations

Many substances do not react on the basis of 1:1 mile ratio

Consider the general reaction:

• The percent analyte

• Example

• Example

• Standardization and titration calculations- (they are the reverse of one another)

• Example

• Example

• Precipitation and complex formation reactions

• Example

• Example

• Example

• Back titration

• Example

• Example

• Calculations based on normality

formula weight

• Example

• Equivalent weight, Equivalents and Normality

• Example

• Conversion between mole and equivalents; normality and molarity

Stoichiometry factor, n, (units if eq/mol) can be used to convert between moles and equivalents; noramlity and molarity

• Calculations in Gravimetric Analysis

In gravimetric analysis the analyte is converted into an insoluble form.

The precipitate is dried and weighed

From the weight of the precipitate formed and the weight relationship between the analyte and the precipitate, the weight of the analyte is calculated.

-

--

Cl mole 1

Cl weight X

AgCl mole 1

Cl mole 1 X

AgCl wt formula g

AgCl mole 1 X AgCl g Cl of

atomicgg

• Cl- (CaCl2) 2 AgCl

2

2

2

22

Cl mol 1

Cl wt f g

CaCl mol 1

Cl mol 1

AgCl mol 2

CaCl mol 1

AgCl wt f

AgCl mol 1X AgCl g Cl XXXg

• Gravimetric factor, GF

Gravimetric factor is the weight of analyte per unit weight of precipitate. OR

The ratio of the formula weight of the substance sought to that of the substance weighed.

Wt of substance sought = wt of precipitate (substance weighed) X GF

• Example

Ba 5.16Ba 1mole

g 208.2X

BaCl 1mole

Ba 1moleX

(137.3)BaClfwt

BaCl 1mole X BaCl g 25.0 Ba ?

22

22 gg

Cl 51.8Cl 1mole

g 35.5X

BaCl 1mole

Cl mole 2X

(137.3)BaClfwt

BaCl 1mole X BaCl g 25.0 Cl ?

22

22 gg

• Example

Wt of substance sought = wt of precipitate (substance weighed) X GF

3232 OAl mole 1

Al mol 2X

OAlfwt

Alfwt GF

Al g 0.1262 1

2

101.96

26.982X 0.2385 Al XWt