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

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

3 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

#grams # of moles = formula weight(g/mole) Millimole #milligrams # of millimoles = formula weight(mg/mmole)

# 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 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). #grams 500 # of moles =  3.12mole formu160la weight(g/mole)

Formula weight of Na2S04 (23X2 + 32+64)= 160 mg/mmol 500  3.12mole # Moles = 160 Expressing the concentrations of solutions Molarity • Molarity is M= number of moles per liter (mole/L) or millimoles per milliliter (mmole/mL)

#moles #mmoles M = = volume (L) volume (mL)

#moles = M X volume (L) # mmoles = M X volume (mL)

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

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

wt (mg) # mequivalents (eq) = = N (meq/mL)X volume (mL) eq wt (mg/meq) 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. # moles Molality = mass of solvent(kg) • Molal concentrations are not temperature dependent as molar and normal concentrations are (since the solvent volume in molar and normal concentrations is temperature dependent).

Density Calculations How do we convert to Molarity • Density = mass solute /unit volume

• Specific Gravity = Dsolute/DH20 o • DH2O = 1.00000 g/mL @ 4 C o • DH2O = 0.99821 g/mL @ 20 C 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 wt sloute (g) wt solute (g) %(w/w) = X100 = X100 wt sample (g) wt solution (g)

• Parts per thousands, ppt wt sloute (g) wt solute (g) ppt (w/w) = X103 = X103 wt sample (g) wt solution (g) • Parts per million, ppm wt sloute (g) wt solute (g) ppm (w/w) = X106 = X106 wt sample (g) wt solution (g) • Parts per billion wt sloute (g) wt solute (g) ppb (w/w) = X109 = X109 wt sample (g) wt solution (g)

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

Example

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

3.6X10 6 g Concentration of Zn in ppm = X10 6 1.380 ppm 2.6g 3.6X10 6 g Concentration of Zn in ppb = X10 9 1380 ppb 2.6g Other units (wt/vol)

wt sloute (g) wt solute (g) %(w/v) = X100 = X100 vol sample (mL) vol solution (mL)

wt sloute (g) wt solute (g) ppm (w/v) = X106 = X106 vol sample (mL) vol solution (mL)

wt sloute (g) wt solute (g) ppb (w/v) = X109 = X109 vol sample (mL) vol solution (mL) Example • A 25.0-μL serum sample was analyzed for glucose content and found to contain 26.7 μg. 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 Endpoint –the point at which the color of indicator changes

The titrant is add Slowly until The indicator changes color (pink) 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.

1 mole AgCl 1 mole Cl - g atomic weight Cl g of Cl -  g AgCl X X X g formula wt AgCl 1 mole AgCl 1 mole Cl - - Cl (CaCl2) 2 AgCl

1 mol AgCl 1 mol CaCl 1 mol Cl g f wt Cl g Cl  g AgCl X X 2 X 2 X 2 f wt AgCl 2 mol AgCl 1 mol CaCl 2 1 mol Cl 2 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

1mole BaCl 2 1mole Ba 208.2 g ?g Ba  25.0 g BaCl 2 X X X 16.5g Ba fwt BaCl 2 (137.3) 1mole BaCl 2 1mole Ba

1mole BaCl 2 2 mole Cl 35.5 g ?g Cl  25.0 g BaCl 2 X X X  8.51g Cl fwt BaCl 2 (137.3) 1mole BaCl 2 1mole Cl Example

Wt of substance sought = wt of precipitate (substance weighed) X GF fwt Al 2 mol Al GF  X fwt Al2O3 1 mole Al2O3 26.982 2 Wt Al  0.2385 X X  0.1262 g Al 101.96 1