Chapter 5 -A

Stoichiometric Calculations:

The Workhorse of the Analyst

BASICS: ATOMIC, MOLECULAR, AND FORMULA WEIGHTS

• Gram-atomic weight for any element: It is the weight of Avogadro's number of atoms of that element • 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.

• None of the elements has an integral atomic weight; why? Since naturally occurring elements consist of mixtures of isotopes, the atomic weight of an element will be an average of the isotope weights of this element, taking into account their relative naturally occurring abundances.

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

g X10 1.06605 atoms 12-C 10 X 6.0221

g 12.00

amu 12

atom 12-C 1 X1amu ? 24-

23 Xg

amu 12 atom 12-C 1 of Mass

atoms 12-C10 X 6.0221

12.0000g ;

amu 12

atom 12-C 1

23

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.

• 1mole of atoms, or molecules or formulas =

= Avogadro’s number of atoms or molecules or formulas

= mass (grams)of Avogadro’s number of atoms or molecules or formulas

= molar mass of atoms or molecules or formulas

• Formula weight (g/mole) = molar mass (mass of 1 mole of atoms or molecules or

ions expressed in grams)

• Mass (g) = # moles X formula mass (g/mol)

ionsor moleculesor atoms of (g/mole)weight formula

ionsor moleculesor atoms of grams# = ionsor moleculesor atomes of moles of #

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)

Example

Mass = #moles X molar mass

# g Fe2O3 = 0.250 mmol X X X

Example: How many milligrams are in 0.250 mmol Fe2O3 ?

mmol 1000

mol 1

32

32

OFe 1mol

OFe 159.7g

g 1

mg 1000

= 39.9 mg

Example: How many milligrams are in 0.250 mmol Fe2O3 ?

3232 OFe mmol 0.250in OFe g?m

3232 OFe mg 159.7 OFe 1mmol

32

32

32

32

OFe mmol 1

OFe mg 7.159or

OFe mg 7.159

OFe 1mmol:factor Conversion

mg 39.9 OFe mmol 1

OFe mg 7.159 X OFe mmol 0.250 OFe #

32

323232 mg

Examples

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

Formula weight of Na2S04 (23X2 + 32+64)= 142 mg/mmol

mol 0.00352 mmol 10

1mol X mmol 12.3

52.3/142

500

3

mmolmmolmg

mg

/mmole) weight(mgformula

mg# = mmoles of #

4242 SONa mg 500in SONa mol ?

42

42

42

42

SONa 1mmol

SONa mg 142or

SONa mg 142

SONa 1mmol:factor Conversion

mol 0.00352 SONa mmol 1

SONa mol 10

SONa mg 142

SONa 1mmol X SONa mg 500 SONa mol ?

42

42

3-

42

424242 X

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

mass (molar) formula

mass # moles

mol 0.007416 AgNO /1molAgNO g 169.9

AgNO 1.260g AgNO of #

33

33 moles

mol/L 0.0297

ml 1000

L 1 X ml 250

AgNO mol 0.007416 AgNO 3

3 Molarity

(ml) volume

mmoles #

(L) volume

moles # Molarity

(ml) V X M AgNO mmoles # 3

mmol 7.42 ml 250 X ml

mmol 0.0297 AgNO mmoles # 3

Example

How many grams per milliliter of NaCI are

contained in a 0.250 M solution?

(mL) volumeX M = mmoles #

(L) volumeX M = moles#

# 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 (mg) = # mmoles X formula (molar) mass (mg/mmol)

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

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

mg

g

1000

1

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+ in 300 ml =100 mL X 0.250 mmol/mL+2 X 0.100mmol/ml

= 65.0 mmol

# mmol K+ = # mmol in KCl + 2 X # mmol of K2SO4

# mg of K+ in 300 ml = # mmmol K+ X 39.1 mg/mmol =

= 65.0 mmol K+ X 39.1 mg/mmol = 2541.5 mg

Conversion factor = ml 300

K 5.2541 mg

# mg K+ = 1L X 1L

1000ml ml 300

K mg 2541.5

X = 8471.7 mg/L

# K+ / L = 8471.7 mg/L X mg

g

1000

1= 8.472 g /L

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

= # moles X (# protons or electrons involved in the reaction)

• The equivalent weight is the formula weight (molar mass) divided by the number of reacting units.

involved) electronsor protons (# units reacting #

mass)(molar weight formula # weight Equivalent

Normality in acid base reactions

• 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

tg/eqivalen 49.04 2

SOH g 08.98

2

SOH =SOH weight Equivalent 4242

42 wtFormula

Normality of H2S04 = M X number of protons = M (H2S04) X 2

Normality of base = M X number of OH- reacting with protons

Normality of Ca (OH)2 =M (base) X 2

Normality of acid = M X number of protons

Normality in oxidation reduction reactions

• For oxidation-reduction reactions normality calculations is based 'on the number, of electrons involved in an oxidizing or reducing agent (number of electrons gained or released)

Fe2+ + Ce 4+ → Fe 3+ + Ce 2+

• Normality of Fe2+ solution = M of Fe2+ X 1

• Normality of Ce4+ solution = M of Ce4+ X 2

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)

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

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). Explain?

Molality = #moles

mass of solvent(kg)

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 before dilution = # mmoles after dilution

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

M 1 X V1 = M 2 X V2

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?

M1 X V1 = M2 X V2

0.100 X V1 = 1.00X 10-3 X 100 ml (first flask)

ml 00 1. 0.100M

ml 100 X 101

3

1 MX

V

Similarly, the other solutions will need 2.00 and 5.00 and

10.0 ml of the stock solution

Density Calculations How do we convert to Molarity

• Density = mass solute /unit volume

• Specific Gravity = Dsolute/DH20

• DH2O = 1.00000 g/mL @ 4oC

• DH2O = 0.99821 g/mL @ 20oC

Example

How many milliliters of concentrated sulfuric 94% (g sulfuric acid/ 100g solution), density 1.831 g/ cm3 are required to prepare 1 liter of 0.100M solution? First calculate the molarity of the concentrated H2SO4 solution: To calculate molarity we have to find: # moles H2SO4 /liter solution, that is # g H2SO4 /liter solution

From density

1 ml solution weighs 1.831 g solution

1 liter solution weighs ml

g

1

831.1X

L

ml

1

1000 = 1831 g

The concentration of H2SO4 solution is 94% (g sulfuric acid/100g solution)

So, the amount of H2SO4 in 1831 g sulfuric acid (1 liter) solution

= 1831 g solution X solution 100

SOH g 94 42

g= 1721 g H2SO4

Molarity of sulfuric acid solution= (L)

#

volume

moles =

solution acid sulfuric of volume

SOH mass /SOH 4242 molarmass=

L 1

g/mol /98g 1721 = 17.56 M

From dilution: M1 X V1 = M2 X V2

Mdil H2SO4 X Vdil H2SO4 = Mconc H2SO4 X Vcon H2SO4

0.1 M X 1 L = 17.56 mol/L X Vconc

Volume of concentrate sulfuric acid = M 17.56

L 1 X M 1.0 = 0.0057L

= 5.7 ml

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)

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

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

wt sample (g)X100 =

wt solute (g)

wt solution (g)X100

(kg) samplewt

(g) sloutewt = 10

(kg) samplewt

(kg) sloutewt = (w/w)ppt 3X

(g) samplewt

(mg) sloutewt = 10

(g) samplewt

(g) sloutewt = (w/w)ppt 3X

Parts per million, ppm

(kg) samplewt

g)( solutewt = 10

(kg) samplewt

(kg) sloutewt = (w/w) ppm 6 m

X

Since the solution is very dilute, that is the density of solution is still considered as 1g/1ml Consequently, 1kg of solution is equivalent to 1 liter of solution. So:

(ml) sample vol

g)( solutewt = 10

(ml) sample vol

(g) sloutewt = (w/vol) ppm 6

X

(L) sample vol

g)( solutewt = 10

(L) sample vol

(kg) sloutewt = (w/vol) ppm 6 m

X

(g) samplewt

g)( solutewt = 10

(g) samplewt

(g) sloutewt = (w/w) ppm 6

X

Parts per billion, ppb

(kg) samplewt

g)( sloutewt = 10

(kg) samplewt

(kg) sloutewt = (w/w) ppb 9

X

Since the solution is very dilute, that is the density of solution is still considered as 1g/1ml Consequently, 1kg of solution is equivalent to 1 liter of solution. So:

(ml) samplewt

(ng) sloutewt = 10

(ml) sample vol

(g) sloutewt = (w/vol) ppb 9X

(L) samplewt

g)( sloutewt = 10

(L) sample vol

(kg) sloutewt = (w/vol) ppb 9

X

©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?

Concentration of Zn in ppm =

Concentration of Zn in ppb = ppbXg

gX(ng/g) 138010

6.2

106.3 96

ppmXg

gX g/g)( 380.110

6.2

106.3 66

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.

mg/dl 107 dl 0.01

1ml

1g

mg 1000 X

ml 10 2.50X

g 10 X 2.67 dl /

2-

-5

Xmg

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.