chapter 1 introduction · introduction analytical chemistry • qualitative analysis (chem 101-103)...

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Chapter 1

Introduction

Analytical Chemistry

• qualitative analysis (Chem 101-103)

• quantitative analysis

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Quantitative Analysis

• gravimetric

• volumetric

• electroanalytical

• spectroscopic

• chromatographic (if time allows)

Analytical Terminology

• heterogeneous

• homogeneous

• analyte

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Steps in a Chemical Analysis

• Sampling: representative

• Sample Preparation

• Analyzing the Sample

• Interpreting the results

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Grind to small pieces

Extraction

Decant, quantitative transfer, supernatant liquid

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Analyzing the Sample

Step 1. Obtain a representative bulk sample.

Step 2. Extract from the bulk sample asmaller, homogeneous laboratory sample.

Step 3. Convert the laboratory sample into aform suitable for analysis, a process thatusually involves dissolving the sample.

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Analyzing the Sample

Step 4. Remove or mask species that willinterfere with the chemical analysis.

Step 5. Measure the concentration of theanalyte in several aliquots.

Step 6. Interpret your results and drawconclusions.

SI Prefixes

especially useful in this course

mega M 106

kilo k 103

centi c 10-2

milli m 10-3

micro 10-6

nano n 10-9

pico p 10-12

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Classification of Analytical MethodsAccording to Size of Sample

<1Ultramicro

<501-10Micro

50-10010-100Semimicro

>100>100Meso

Sample Volume( L)

Sample Weight(mg)

Method

Constituents

<0.1%trace

0.1-1%minor

>1%major

ConstituentsMethod

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Solution Terminology

• solute

• solvent

• aqueous solution

• liter

• atomic weight

• molecular weight

Molarity

# moles AMolarity => M = -------------------

# liters solutionor

# millimoles AMolarity => M = -------------------------

# milliliters solution

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Useful Algebraic Relationships

wt A (g)# mol A = -----------------

fw A (g/mol)

# mol A = V (L) x M (mol A/L soln)

or wt A (mg)

# mmol A = ----------------- fw A (g/mol)

# mmol A = V (mL) x M (mmol A/mL soln)

Types of Solutions

• strong electrolyte

• weak electrolyte

• non-electrolyte

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Useful Algebraic Relationships

n = M ×V

n = W (g)FM(g /mol)

M = nV

(mol /L) Molarity

Formal Concentration

• used for systems which separate (ionize) insolution

• same form for equation as molarity,substitute formula weight for molecularweight for those substances which do notform molecules

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Molality => m# moles A

molality => m = ------------------------- # kilograms solvent

• this concentration unit is temperatureindependent as the mass does not changewith temperature whereas volume does

• used in freezing point depression/boilingpoint elevation

• not commonly used.

p-Functions

pX = - log10[X]

examples:

pH

pOH

pCl

pAg

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Percent Composition

wt of a solutew - w% = -------------------- 102

wt of solutionvol of a solute

v - v% = -------------------- 102

vol of solutionwt of a solute

w - v% = --------------------- 102

vol of solution

Parts per Million

wt of a solutecppm = ------------------- 106

wt of solution

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Parts per Billion

wt of a solutecppb = ------------------- 109

wt of solution

Preparing Solutions

EXAMPLE: Describe the preparation of 1.00L of 0.100 M NaOH solution (f.w.40.00) from reagent grade solid.

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EXAMPLE: Describe the preparation of 1.00L of 0.100 M NaOH solution (f.w. 40.00)from reagent grade solid.

(1.00 L soln)# g NaOH = ----------------


(1.00 L soln)(0.100 mol NaOH)# g NaOH = ---------------------------------------

(1 L soln)

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(1.00 L soln)(0.100 mol NaOH)# g NaOH = ---------------------------------------

(1 L soln)


(1.00)(0.100 mol NaOH)# g NaOH = ------------------------------

(1)

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(1.00)(0.100 mol)(40.00g NaOH)# g NaOH = ----------------------------------------

(1) (1 mol)


(1.00)(0.100 mol)(40.00g NaOH)# g NaOH = ----------------------------------------

(1) (1 mol)

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(1.00)(0.100)(40.00g NaOH)# g NaOH = ----------------------------------------

(1) (1)


(1.00)(0.100)(40.00g NaOH)# g NaOH = ----------------------------------------

(1) (1)

= 4.00 g NaOH

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(1.00)(0.100)(40.00g NaOH)# g NaOH = ----------------------------------------

(1) (1)

= 4.00 g NaOH

Weigh 4.00 g of NaOH, transfer to a 1.00 Lvolumetric flask, and dilute to the line.

Dilution

#moles solute in conc. soln

equals

#moles solut in dil. soln

therefore

Mconc Vconc = Mdil Vdil

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Examples

• 2. How can one prepare 500 mL 1.0 M HClsolution from concentrated HCl solution(12M)?

Examples• 3. What is the molarity of 1:6 (v:v) HNO3

solution?

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Examples

• 3. Calculate the molarity of concentratedsulfuric acid (Strength = 95.5-96.5%,Density = 1.84)?

• AAcetic Acid, Glacial, 100% 117 MolarAmmonia, 29% 115 MolarHydrochloric Acid, 37%, 12 MolarNitric Acid, 70%, 16 MolarPhosphoric Acid, 85%, 15 MolarPerchloric Acid, 71%, 11 MolarSodium Hydroxide, 50%, 19 MolarSulfuric Acid, 96%, 18 Molar (36 Normal)

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Comparison of Different AnalyticalMethods

chapter 1 introduction · introduction analytical chemistry • qualitative analysis (chem 101-103)...

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