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Analytical ChemistryWeek 4 ENVI 15 - Analytical Chemistry Lecture 2 Last Week We Discussed Measurement Uncertainty -Systematic and random error -Accuracy vs precision -Significant figure

Analytical Chemistry

Week 4

ENVI 15 - Analytical Chemistry Lecture 2

Last Week We Discussed Measurement Uncertainty

-Systematic and random error

-Accuracy vs precision -Significant figures -Average and standard deviations

Quality Assurance/Quality Control

(QA/QC)

Review Harris Chapter 5

Quality Assurance

„ A program to produce high quality data with known uncertainty.

ENVI 15 - Analytical Chemistry Lecture 2

Quality Assurance

„ Elements of a QA program are:

„ Quality Control

„ Quality Assessment

„ An accredited laboratory has a

well-developed QA/QC program

„ ISO 17025 – Industry Standard

ENVI 15 - Analytical Chemistry Lecture 2

Quality Assurance

„ Quality Control

„ Establish Method Detection and Reporting Limits

„ Reagent Blanks

„ Calibration standards

„ Analysis of standard reference materials

„ Analysis of duplicates

„ Recovery of known additions (spikes)

„ Control Charts

„ Analyst proficiency checks

ENVI 15 - Analytical Chemistry Lecture 2

Quality Assurance

„ Quality Assessment

„ Performance Evaluation Samples

„ Laboratory Round Robins

„ Performance Audits

Analytical Chemistry

Concentration Concepts

Harris Chapter 1

ENVI 15 - Analytical Chemistry Lecture 2

Atomic Mass

„ Atomic mass of element = mass of one

atom of that element

„ (a really, really small number!!)

ENVI 15 - Analytical Chemistry Lecture 2

Molecular Mass

„ Molecular mass of a molecule = mass of one molecule

„(which is the sum of the atomic masses that make up the molecule)

„(still a really, really small number!!)

The MOLE

„ A Mole is Avogadro’s Number (6.022 x

10 23 ) of atoms or molecules

The MOLE Concept

NaCl NaCl NaCl NaCl NaCl NaCl NaCl NaCl NaCl

NaCl NaCl NaCl NaCl

ENVI 15 - Analytical Chemistry Lecture 2

The MOLE

„ Atomic mass X 1 mole of atoms = Atomic

mass in grams

„ e.g Na (22.98 amu) X 1 mole = 22.98 g

ENVI 15 - Analytical Chemistry Lecture 2

The MOLE

„ Molecular mass X 1 mole of molecules=

Molecular mass in grams

„ Na(22.99) + Cl (35.45) x 1 mol = 58.44 g

ENVI 15 - Analytical Chemistry Lecture 2

Find the Molecular Mass

„ BaCl 2

„ H 2 SO 4

„ Hg 2 (NO 3 ) 2

Ba = 137.327 g/mol

H = 1.008 g/mol

Cl = 35.453 g/mol

S = 32.062 g/mol

O = 15.999 g/mol

Hg = 200.59 g/mol

N = 14.007 g/mol

ENVI 15 - Analytical Chemistry Lecture 2

Solutions

„ What happens when

we put a small amount of NaCl in water?

„ Solute and solvent

ENVI 15 - Analytical Chemistry Lecture 2

Concentration

„ Refers to how much solute is in a given

volume or mass of solvent.

„ Standard concentration measure in

chemistry is: # of moles / L

„ We call this molarity (M for short)

„ [Compound] = concentration of compound

ENVI 15 - Analytical Chemistry Lecture 2

Molarity of Sea Salts

„ What is the molarity of NaCl in seawater if it contains 2.7

g NaCl per 100 mL?

„ The molecular mass of NaCl is 22.99 (Na) + 35.45 (Cl) = 58.44 g/mol

„ # of moles of NaCl = 2.7 g = 0.046 mol

58.44 g /mol

Molarity of Sea Salts

„ What is the molarity of NaCl in seawater if it contains 2.7 g NaCl per

100 mL?

The molecular mass of NaCl is 22.99 (Na) + 35.45 (Cl) = 58.44 g/mol

# of moles of NaCl = 2.7 g = 0.046 mol

58.44 g /mol

100 mL = 0.100 L

[NaCl] = mol NaCl = 0.046 mol = 0.46 mol/L or M

L 0 100 L

ENVI 15 - Analytical Chemistry Lecture 2

Molarity of Sea Salts

[NaCl] = mol NaCl = 0.046 mol = 0.46 mol/L

L 0 100 L

How many millimoles of NaCl in 1 L of seawater?

How many micromoles?

ENVI 15 - Analytical Chemistry Lecture 2

ENVI 15 - Analytical Chemistry Lecture 2

Concentration

„ Other measures of concentration are:

„ g/L, mg/L, µg/L, ng/L etc

„ g/kg

„ Parts per … ppm, ppb, ppt etc

ENVI 15 - Analytical Chemistry Lecture 2

Mass Concentration

„ What is the NaCl concentration (in mg/L) in seawater if it contains 2.7 g of NaCl per 100 mL?

„ [NaCl] = 2.7 g we want answer in mg/L

NaCl (mg/L) = 2.7 g x 1000 mg x 1000 mL

100 mL

100 mL 1 g 1 L

= 2700 mg/L

ENVI 15 - Analytical Chemistry Lecture 2

Percent Concentration

„ Weight percent (w/w) = weight solute/weight solution x 100%

„ Volume percent (v/v) = volume solute/volume solution x 100%

„ Weight/volume (w/v) = weight solute, g/volume solution , mL x

100%

ENVI 15 - Analytical Chemistry Lecture 2

Percent Concentration

„ For %w/w and %vol/vol, the units must be the same

„ A 10% (vol/vol) solution is prepared by dissolving 10 mL of a pure substance in the amount of solvent it takes to produce a 100

mL solution

„ Best done with a volumetric flask

Parts per Million and Parts per Billion

„ C ppm = mass solute/mass solution x 10 6

„ C ppb = mass solute/mass solution x 10 9

Note units in numerator and denominator must agree

Parts per Million and Parts per Billion

„ C ppm = mass solute/mass solution x 10 6

„ C ppb = mass solute/mass solution x 10 9

„ What is parts per trillion?

ENVI 15 - Analytical Chemistry Lecture 2

Parts per Million and Parts per Billion

„ C ppm = mass solute/mass solution x 10 6

„ For aqueous solutions, 1 ppm = 1 mg/L

„ C ppb = mass solute/mass solution x 10 9

„ For aqueous solutions, 1 ppb = ?

ENVI 15 - Analytical Chemistry Lecture 2

The Dilution Equation

V 1 x C 1 = V 2 X C 2

„ Where:

„ V 1 is the initial volume

„ C 1 is the initial concentration

„ V 2 is the final volume

„ C 2 is the final concentration

Lab Volumetric Glassware

Calibrating Burets and Pipets

ENVI 15 - Analytical Chemistry Lecture 2

Buret

Lecture 2 Lecture 2

Reading the Buret

Estimate each reading

to ±0.01 mL For example 13.63 mL

ENVI 15 - Analytical Chemistry

Lecture 2

ENVI 15 - Analytical Chemistry Lecture 2

Pipets

TD or TC marks

ENVI 15 - Analytical Chemistry

Lecture 2 ENVI 15 - Analytical Chemistry Lecture 2

Lecture 2