LU2 Stoichiometry Calculation

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LU2 : Stoichiometric Calculation

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Learning Objectives

• At the end of this learning unit, students should be

able to:

• Explain and calculate molarity, molality, formality,

normality and other units of concentrations (ppm, ppt,

ppb etc).

• Explain dilution and solve problems related to dilution,

• Solve problems using stoichiometric approach.

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Concentration of Solutions

• Behaviour of solutions depend on compound itselfand on how much is present i.e., concentration.

• Two solutions can contain the same compounds but

behave quite different because the proportions ofthose compounds are different.

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S.I. unit for chemical quantity: The mole

The mole is also called Avogadro's number: 6.022 x

1023.

A mole is defined as the number of carbon atoms in

exactly 12.00 grams of carbon-12.

The definition is selected so that the formula weight (in

amu) and the molar mass (in grams/mol) have the

same numerical value.

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…continued

For example: consider water, H2O

Formula Weight:

2 atoms H(1.0 amu/atom) = 2.0 amu

1 atom O(16.0 amu/atom) = 16.0 amuFormula Weight = 18.0 amu/formula

Molar Mass = 18.0 grams/mol

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Units of Concentration

•

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….continued

Common units used by chemists include:

Molarity (M) = moles solute/liters of solution

• This unit refers to the molarity at equilibriumcondition.

• A 1.00 molar (1.00 M) solution contains 1.00 molsolute in every 1 liter of solution.

• Units of molarity are : mol/L = M

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….continued

Mole of solutesMolarity (M) = ----------------------------------

Volume of solution in liters

molMolarity (M) = ----------------

liters (L)

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Preparing a 1.0 Molar Solution

• One liter of a 1.00 M NaCl solution

• Need 1.00 mol of NaCl

•

Weigh out 58.5 g NaCl (1.00 mole) and

• Add water to make 1.00 liter (total volume) ofsolution

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Example 1

What is the molar NaCl concentration if you have 0.5 mol NaClin 1.00 L of solution?

0.5 mol NaCl/1.00 L = 0.5 mol/L = 0.5 M

What is the molar NaCl concentration if you have 0.5 mol NaClin 0.50 L of solution?

0.5 mol NaCl/0.50 L = 0.5/0.50 mol/L

= 1 mol/L = 1 M

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Example 2

What is the molar NaCl concentration if you have 10.0 g ofNaCl in 1.00 L of solution?

Answer:

Molar mass of NaCl – 58.5 g/molSo, 10.0 g x (1 mol/58.5 g) = 0.17 mol

Molar concentration = 0.17 mol/1.00 L = 0.17 M

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Example 3

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How many moles of HCl are present in 2.5 L of 0.10 M HCl?

Answer:

Given : 2.5 L of solution 0.10 M HClFind : mol HCl = 0.10 mol/1 L HCl

Use : mol = molarity x volume

Mol HCl = 0.10 M HCl x 2.5 L = 0.10 mol HCl x 2.5 L

1 L= 0.25 mol HCl


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Example 4

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What volume of a 0.10 M NaOH solution is needed to provide

0.50 mol of NaOH?Solution:

Given : 0.50 mol NaOH

0.10 M NaOH = 0.10 mol NaOH/1L

Find : volume of solutionUse : mol = molarity x volume

Volume of soln = 0.50 mol NaOH = 0.50 mol NaOH

0.10 M NaOH 0.10 mol NaOH

1L

= 0.50 mol NaOH x 1 L = 5 L

0.10 mol NaOH


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Example 5

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How many grams of CuSO4 are needed to prepare 250.0 mL of1.00 M CuSO4?

Solution:

Given : 250.0 mL solution of 1.0 M CuSO4Find : g CuSO4Use : mol CuSO4 = molarity x volume

Molarity = mol/1 L

Volume = 250.00 mL


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g CuSO4 = 250.0 mL sol x 1 L x 1.00 mol x 159.6 g CuSO41000 mL x 1 L soln x 1 mol

= 39.9 g CuSO4

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Example 5 continued


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Example 6

•

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Example 7


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Formality (F)

•

Formality (F) = formula units (regardless of chemical form)per liter of solution.

• Formality of a solution may be defined as the number of gram

formula masses of the ionic solute dissolved per liter of the

solution.

• This unit describes how to make a solution, not what exists at

equilibrium.

• Analytical molarity is another term for Formality.

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Formality (F)

• The difference between formality and molarity is that

you can express the molarity of the different ions

individually and the formality is the entire compound

irrespective of ionization.

• For example 1.0 F solution of Ca(NO3)2 is 2.0 M NO3- or

1.0 M Ca(NO3)2.

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Normality (N)

• Definition : The number of equivalents of solute per liter of

solution.

• Normality (N) = #equivalents solute/#liters of solution.

• Equivalents depend on the reaction taking place in the solution.

a. Acid-base reactions, 1 equivalent = 1 mole of hydrogen ions (or1 mole of hydroxide ion) donated.

b. Oxidation-reduction reactions, 1 equivalent = 1 mole ofelectrons.

c. For determining electrolyte concentration, 1 equivalent = 1mole of charge.

Normality must be specified with respect to a definite reaction. 20

Normality is equivalents / liter


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Normality (N)

• Normality, N, is similar to molarity, moles of soluteper liter of solution. However, instead of the entire

solute, the normality is based on the number of

moles of the active part of the solute, called a

chemical equivalent.

• For an acid, the chemical equivalent is the number

of moles of H+ ion. For a base, the chemical

equivalent is the number of moles of OH- ions.

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Normality (N)

• The normality of hydrochloric acid, HCl, is the same as the

molarity of hydrochloric acid, because there is one mole of H+1

ions for every one mole of HCl. The normality of sulfuric acid,

H2SO4, is twice the molarity because there are two moles of

H+1 ions per mole of sulfuric acid.

• The advantage to using normality is that it gives an effective

concentration (3M sulfuric acid is twice as acidic as 3M

hydrochloric acid—this is clear if they are labeled 6N and 3N,

respectively).

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Molality (m)

Molality (m) = moles of solute/kilograms of solvent

m = mol solute

kg solvent

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Molality (m)

What is the molarity of 6.0 moles of x dissolved into200g of methanol?

Molality = 6 moles = 30 m

0.2 kg of solution

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Parts per million (ppm)

ppm = weight of substance x 106

weight of solution

ppm = mg = µg

liter ml

Micrograms of solute per gram of solution; for aqueous

solutions the units expressed as milligrams of solute per

liter of solution.

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Parts per million (ppm)

Cppm = weight of substance x106 ppm

weight of solution

Note that for dilute, aqueous solutions, 1 ppm = 1 mg/L

because 1 L approximately equals1 kilogram of solution.

Cppm = weight of substance(mg) x106 ppm

volume of solution(L)

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Parts per billion (ppb)

ppb = weight of substance x 109weight of solution

ppb = ug = ng

liter ml

Nanograms of solute per gram of solution; for aqueous

solutions the units expressed as micrograms of solute

per liter of solution.

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Example 8

A sample weighing 1.3535 g contain 0.4701 g of Fe. Calculate

the percentage of Fe in the sample. Calculate the content ofFe in ppt and ppm

% Fe = weight of solute (g) x 100%

weight of sample (g)

= 0.4701 g x 100%

1.3535 g

ppt Fe = 0.4701 g x 103

1.3535 g

= 347.3 ppt

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ppm Fe = 0.4701 g x 106

1.3535 g

= 3.473 x 105 ppm


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Percent Concentration

•

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Example 9

• What is the w/w % of an aqueous ammonia (NH3)

solution at 14.3 M, with density = 0.900 g/mL (900g/L)?

1) MW of NH3 = 17.0 g/mol

2) Mole of NH3 = 14.3 (mol/L) x 1.00L = 14.3 mol

3) Weight of NH3 = mole of NH3 x MW of NH3= 14.3 mol x 17.0 (g/mol) = 243 g

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Example 9 continued

•

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Example 10

What is the molar concentration of an aqueousammonia (NH3) solution with density = 0.900 g/mL (900

g/L) and 27.0% (w/w)?

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1) MW of NH3 = 17.0 g/mol2) Weight of NH3 = moles of NH3 x MW of NH3

= CV(NH3) x MW of NH3

3) Weight of 1.00L solution = vol x density= 1.00L x 900(g/L) = 900 g


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Example 10 continued

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Example 11

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continued

Usage of percent concentration varies.

For example, commercial aqueous reagents are usually sold

in a weight % (w/w):

•37% HCl means 37 g HCl/100 g solution.

• To calculate the molarity of the HCl requires knowledge of

the density of the solution.

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Confusion Over Concentration Units

•

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Stoichiometric Calculations and the Factor-Label

Method

A. Steps in the method:

1. Write a balanced equation for the reaction.2. Compute all the molar masses.3. Write down the given and unknown quantities.

4. Setup conversion factors using the stoichiometriccoefficients in the balanced equation.

5. For example, suppose 4.0 g of hydrogen gas reacts withexcess oxygen gas to produce water. How many grams ofoxygen are required, and how many grams of water can

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continued •

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continued

7. Note the rationale followed for all stoichiometric

calculations:

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Preparing solution:

Steps involved in preparing solutions frompure solids

Calculate the amount of solid required

Weigh out the solid

Place in an appropriate volumetric flask

Fill flask about half full with water and mix

Fill to the mark with water and invert to mix.


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Dilutions

Many laboratory chemicals such as acids are purchased as

concentrated solutions (stock solutions)

Examples : 12 M HCl; 12 M H2SO4

More dilute solutions are prepared by taking a certain

quantity of the stock solution and diluting it with water

Stock solution – A solution of known

concentration from which other solutions are

prepared.


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Dilutions

• A stock solution of 1.00 M NaCl is available. How many mL are

needed to make 100.0 mL of 0.750 M ?

• (0.750 mol/L) (100.0 mL) = (1.00 mol/L) (x)

• x = 75.0 mL

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Dilutions

Although the number of moles of solutes does not

change, the volume of solution does change.

The concentration of the solution will change since

Molarity = Moles solute

Volume of solution

Moles solute before dilution = Moles solute after dilution


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Dilution Calculation

When a solution is diluted, the concentration of thenew solution can be found using:

Mc x Vc = Md x Vd

Where Mc = initial concentration (mol/L)

Vc = initial volume of more conc. Solution

Md = final concentration (mol/L) in dilution

Vd = final volume of diluted solution.


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Example 12

You have 50.0 mL of 3.0 M NaOH and you want 0.50 M

NaOH. What do you do?


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Solution to Example 12

Minitial x Vinitial = Mfinal x Vfinal

(3.0 mol/L) (0.050 L) = (0.50 mol/L)Vfinal

Vfinal = 0.3 L or 300 mL.

Add 250 mL of water to 50.0 mL of 3.0 M NaOH to make

300 mL of 0.50 M NaOH


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Solution to Example 12

Add 250 mL of water to

50.0 mL of 3.0 M NaOH

to make 300 mL of 0.50

M NaOH.


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Dilution

Make a diluted solution once you know Vc and Vd;

Use a pipet to deliver a volume of the concentrated

solution to a new volumetric flask;

Add solvent to the line on the neck of the new flask;

Mix well


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Example 13

How many mL of 5.0 M K2Cr2O7 solution must be diluted

to prepare 250 mL of 0.10 M solution?

Vc = ? ; Mc = 5.0 M; Vd = 250 mL ; Md = 0.10 M

If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to

250 mL, what is the concentration of the resultingcondition?

Md = ? ; Vc = 10.0 mL; Mc = 10 mL ; Vd = 250 mL


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Solution Stoichiometry

Remember : reactions occur on a mole to molebasis.

For pure reactants, we measure reactants using

mass.

For reactants that are added to a reaction as

aqueous solutions, we measure the reactants

using volume of solution.


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Solution Stoichiometry


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Example 14

If 25.0 mL of 2.5 M NaOH are needed to

neutralize (i.e react completely with) a solution

of H3PO4, how many moles of H3PO4 were

present in the solution?

3NaOH (aq) + H3

PO4

(aq) Na3

PO4

(aq) + 3H2

O (l)

Balanced eqn : 3 mol NaOH/1 mol H3PO4

Find : moles of H3PO4


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Approach


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Approach

Next …. Complete the calculation by multiplying by

stoichiometric ratio …

No of mole H3PO4 = 0.0625 mol NaOH x 1 mol H3PO43 mol NaOH


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Titrimetry

• Titrimetry refers to that group of analytical techniques which

takes advantage of titers or concentrations of solutions.

• The word "titer" is also used to denote “equivalence" or that

amount of a solution required to complete a chemical

reaction.

• Titrimetry often refers to the use of some volume of a solution

of known concentration to determine the quantity of an

analyte.

• But there are some variations on the use of the term.


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Types of Titrimetry

Volumetric titrimetry

• establishes a quantity of analyte using volumes of reagents ofknown concentrations and the knowledge of the stoichiometry

of the reactions between the reagents and the analytes.

Gravimetric titrimetry

• determines the quantity of analyte by a measure of the mass

of a solution of known concentration.

Coulometric titrimetry

• arrives at the amount of analyte by measuring the number of

coulombs or total charge required to complete a reaction with

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Definitions

Equivalence point:

•

is the point where sufficient titrant has been added to bestoichiometrically equivalent to the amount of analyte.

Chloride determination

•50 mL of a 0.1M AgNO3 solution would be required to

completely react with 0.005 moles of NaCl.

AgNO3 (aq) + NaCl (aq) AgCl (s)↓+ NaNO3 (aq)

1 mol 1 mol

0.050 L x 0.1 MC x V = 0.050 L x 0.1 M = 0.005 moles 59


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continued

End point is the point at which some detection technique tells

you that chemical equivalence has been reached.

Ideally the end point and the equivalence point should

coincide.

But this rarely happens due to the methods used to detect

the end point.The difference between the end point and the equivalence

point is called a titration error (typically an over-titration).

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continued

Primary Standard is a high purity compound that serves as a

reference material in all volumetric and mass titrimetric

methods.

Ideally the titrant solution would be made from a

primary standard.

Titrant solutions must be of known concentration.

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Primary standard requirements

1. High purity;

2. Stability in presence of air;

3. Absence of any water of hydration which might vary

with changing humidity and temperature;

4. Dissolves readily to produce stable solutions in

solvent of choice;

5. A relatively large molar mass to minimize weighing

errors;

6. Reacts rapidly and stoichiometrically with analyte;

7. Cheap (no really necessary).

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Examples of primary standards

Primary standards for standardization of NaOH -Potassium acid phthalate (KHP)

Primary standards for standardization of HCl - Sodium

carbonate (pure)

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Secondary standards

•Few materials meet all of the primarystandard requirements;

• Instead a secondary standard that is

standardized with a primary standard.

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Standardization

is a process in which the concentration of a secondary standard isdetermined to a high level of accuracy by titration with a primary

standard.

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Example 15

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Example 15

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x mole = Mole of BaCl2 (mole) = (0.500/2)= 0.250 (mole)

MW of BaCl2 • 2H2O = 244.2 g/mol

W of BaCl2 • 2H2O = Mole of BaCl2 (mole) x MW of BaCl2.2H2O

= 0.250 mol x 244.2 g/mol= 61.1 g

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LU2 Stoichiometry Calculation