125

9. CHEMICAL ANALYSIS Chemical analysis is a necessary requirement of many industrial processes, particularly in satisfying government regulations. It is necessary for chemists to be able to identify the composition of raw materials, manufactured products and substances which are discharged as wastes. An integral component of a typical chemist’s role is in Quality Control. 9.1 Qualitative Vs Quantitative Chemical Analysis When determining the composition of a substance we need to find out: What chemicals are present? This identification process is called qualitative analysis.

Sometimes this is all that is required, but often we also need to know: how much of each chemical is present. This measurement process is called

quantitative analysis. In the mercury chloride laboratory task you determined whether your unknown sample contained mercury (I) or mercury (II) chloride. The methods you used involved quantitative analysis since you measured masses of reactants and products in order to analyse your sample. Consider the following examples: A jar of peanut butter has listed as ingredients: peanuts, vegetable oils, sugar salt. This represents a qualitative analysis only. Normally ingredients are listed in order of decreasing mass but no actual quantities have been given. On a packet of pain relief capsules the following ingredients are listed: Paracetamol: 500 mg ; Codeine phosphate: 8 mg This represents a quantitative analysis. Note that there may be other ingredients also which are unlisted, since the manufacturers are only obliged to list active ingredients, not fillers, binders etc. Written Exercise / Research 1. Examine the labels of other containers for examples of qualitative and quantitative analysis. List three further examples of each. 2. Care must be taken with some ingredients to make sure that their concentrations stay within set limits. For example, the allowable pesticide residue limits in foods is carefully controlled. Find another example and find out how this product is analysed. What methods can we use in chemical analysis? Chemists employ many methods in order to analyse the millions of different substances recognised today. We will look at some examples from which you may be able to choose for your major investigation in this module - some involve simple laboratory apparatus

126

while others will involve sophisticated instruments. You will need to adapt any procedure given to make use of the resources available to you. Consult a chemistry teacher once you have made your choice. You should practice as many techniques as possible. Consult references to find out about other techniques.

9.2 Gravimetric Analysis Gravimetric analysis involves the separation and weighing of a substance obtained from a sample. This form of analysis separates the substance of interest from the sample based on its physical or chemical properties. The concentration of separated substance is calculated and reported as a percentage by mass. Often there are three main steps to this procedure:

1. Obtain and weigh sample mass

2. Employ means of physical or chemical separation of the analyte

3. Reweigh sample (or separated analyte component) Gravimetric analysis in its most simple form could be the analysis of the moisture content of say, a food substance like cereal. A suitable sample container needs to be chosen as it must not change mass during the analysis. An accurately weighed mass of cereal is placed into the container and heated in air oven at around 120 oC for a minimum time period to evaporate the entire moisture content. The container is removed, cooled and reweighed. The difference in mass (confidently assuming it became lighter !) is calculated and expressed as a percentage of the sample mass. All calculations in moisture determinations are purely mass based, but there are further examples of gravimetric analysis involving mole based calculations . . . If we were to determine the salt content of a food sample we must first dissolve the food in water. Any undissolved substance will need to be removed using a separation method such as filtration. The food solution can be analysed for salt, as NaCl by adding an excess of AgNO3 solution. This precipitates out the entire chloride ion, Cl- (aq) content as AgCl (s) according to the following chemical equation: and then filtering and weighing the precipitate which forms: Ag+

(aq) + Cl-(aq) AgCl(s) The precipitate is separated by filtration. After the precipitate is dried, it is then weighed and its mass recorded. The number of moles of AgCl formed can be calculated. From the equation, this is the same as the number of moles of Cl- in the food solution.

127

This is therefore equal to the number of moles of salt, NaCl in the food sample analysed. We can convert this number of moles to a mass and express it as a percentage of the mass of the original sample. Example: A 25.0 mL sample of soup was analysed for its salt content, as NaCl. The sample was filtered to remove any suspended solids. An excess of Silver nitrate solution was added to the filtrate to precipitate out the available salt component. After standing for a few minutes a further few drops of silver nitrate were added to make sure that precipitation was complete. The precipitate was collected by filtration, then dried and weighed. A mass of 0.23g of precipitate was obtained. Find the percentage (g / 100 mL) of salt in the soup. No of moles of AgCl produced = mass of AgCl formula weight = 0.23 143.4 = 1.60 x 10-3 Chemical equation: Ag+

(aq) + Cl-(aq) AgCl(s) From the equation above, moles Cl- = moles of AgCl = 1.60 x 10-3 We are assuming that all the chloride ions present came from dissolved salt - NaCl. This is a reasonable assumption in most food samples. Therefore, moles of NaCl in the soup sample = 1.60 x 10-3 mass of NaCl in the soup sample = moles x FW = 1.60 x 10-3 x 58.45 = 0.0938 % of NaCl in soup sample = 0.0938 x 100 25.0 = 0.38 % w/v You should note the following important points: • solid samples need to be dissolved so that the substance to be detected is in solution.

Any undissolved material should be removed by filtration. • the precipitate formed needs to be a highly insoluble substance, so that the proportion

of soluble ions remaining in solution for analysis is negligible. Check the table of solubility rules, under the chapter “properties of compounds” – your teacher will assist you with this!

• the filtered precipitate needs to be washed carefully with small volumes of water to remove other soluble substances present in the solution.

• the precipitate should be dried in an oven at 110oC until constant mass is obtained and allowed to cool before weighing.

128

Experiment – Sulfate Content of Plant Fertilizer Background: Sulfate is an important nutrient for plant growth. It is added to many fertilizer materials, present in soluble forms, most notably as Ammonium Sulfate. This method aims to dissolve the sulfate component of a fertilizer sample into a slightly acidified aqueous solution. The available Sulfate, SO4

2- (aq) will be precipitated, filtered and weighed as BaSO4 (s) by addition of a soluble Barium Chloride solution, BaCl2 (aq) Method:

1. Obtain a 600 mL beaker, place onto a 3 decimal place balance and tare. 2. Accurately weigh approximately 0.75g of fertiliser and record this mass into the

results table below. 3. Dissolve the weighed material in around 100 mL of distilled water. 4. Add 2M HCl dropwise until just acid to litmus paper. 5. If any solid residue remains, filter solution through #4 (fast) filter paper. Ensure

quantitative transfer and rinse final residue in filter paper with distilled water. 6. Place solution onto a hotplate and heat until just boiling. 7. Remove from heat and add 25 mL of 5% BaCl2. 8. Cover beaker with a watchglass and transfer solution onto a hotplate (! low heat). 9. Add a further 2 mL of the BaCl2 solution after 10 min. 10. Continue to digest your solution on low heat for a further 20 min. 11. Obtain and record the mass of a sintered glass crucible in the results table. 12. Remove from hotplate and cool in an ice bath for 10 min. 13. Set up vacuum filtration apparatus and filter your solution through the pre-weighed

sintered glass crucible. Wash remaining residue with a small volume of distilled water.

14. Place into an air oven to dry and leave until next class to weigh back.

15. Remove from oven. Allow 10min to cool and reweigh. Results:

Mass (g): Fertiliser

Mass (g): Sintered Glass crucible

Mass (g): Sintered Glass Crucible + BaSO4 (s)

Mass (g) : BaSO4 (s)

129

Calculations:

• Moles BaSO4 (s)

• Balanced Chemical Equation (precipitation process):

• Reaction Ratio BaSO4 : SO4 2- = _________ : __________

• Moles SO4 2-

• Mass SO4 2-

• % SO4 2-

Questions: 1 a) Why is it important to initially filter and remove any undissolved fertilizer in step 5 ? 1 b) Why would this undesirable solid residue be washed with distilled water? 2) Explain the purpose of cooling the precipitate prior to filtering in step 12. 3) Explain the importance of washing the solid BaSO4 compound with distilled water prior to drying in step 13 4) Why was the measured volume of distilled water (ie 100 mL) instructed as approximate and not accurate?

% SO4 2- = Mass SO4

2- x 100 Sample Mass

130

9.3 Volumetric Analysis Unknown concentrations of soluble chemicals in aqueous solution can be determined by a technique referred to as Volumetric Analysis, or Titration. A carefully measured volume of the unknown solution is transferred to a conical flask by pipette. This volume is referred to as an aliquot. A second chemical solution is chosen that is known to chemically react with the unknown solution in a predictable manner, via a balanced chemical reaction. This solution is must have a known concentration as it is referred to as a standard solution. A burette is filled with the standard solution where it is then slowly delivered into the aliquot of unknown solution, allowing the chemical reaction to progressively “consume” the chemical analyte within the unknown solution aliquot. This method of analysis continues to deliver standard solution into the unknown solution aliquot (in the conical flask) until all of the chemical analyte in the unknown solution is consumed. At this point the chemical analysis is complete, and addition of further solution is ceased. The volume of standard solution delivered to the unknown solution by burette is carefully measured and recorded. This is referred to as an endpoint volume. Diagram of basic Titration apparatus: The fundamental basis of all volumetric analyses is the chemical reaction. It is essential that a balanced chemical equation be expressed as part of the calculation procedure. Generic Equation: HX(aq) + YOH(aq) XY (aq) + H2O (l) 1 : 1 In this example the reaction ratio, and hence mole ratio is 1:1, however different chemical reactants may have mole ratios like 1:2, 1:3, 3:2 and so on. This ratio is an integral part of the calculation procedure in determining the unknown solution concentration.

Bulb Pipette Apparatus

131

Volumetric analysis (titration) is commonly employed to determine the concentrations of dilute acids and bases. As covered previously, acids and bases react together predictably in what is referred to as a neutralisation reaction, ie; Acid + Base Salt + Water The general steps involved in a titration, in summary are: • a measured pipette volume (aliquot) of the unknown solution is added to a conical flask. • a few drops of a pH sensitive indicator are normally added to the conical flask to

produce a noticeable colour change at endpoint. • a burette is filled with the standard solution and volume adjusted to the top volume

reading of 0 mL. • the standard solution is carefully delivered to the flask. As the colour change becomes

more apparent, the delivery rate from the burette is slowed from a steady stream to smaller “squirts” and ultimately dropwise until a permanent colour change is observed.

• cease delivering standard solution from burette. Record burette volume. This is the volume of standard solution that was required to consume the last of the analyte in the standard solution.

There are 4 measurable quantities recorded in a single titration analysis. 3 of them are known quantities and 1 is unknown initially. Place a tick in the boxes below which correspond with known quantities. Place a cross in the box corresponding with the value to be calculated (unknown). Burette Standard solution:

Volume

Concentration

Conical flask Unknown solution

aliquot

Volume

Concentration

The titration technique requires considerable practice in order to obtain reliable results. Volumetric analysis is commonly performed in triplicate. It is desirable that your 3 titration volume results are within 0.2 mL 0.3 mL. This can be difficult to achieve initially!

132

Titrations are used in many situations such as when we want to find the proportion of particular pollutants present in industrial waste water, or the proportion of vitamin C in commercially produced fruit juices. There is a variety of other chemical reaction types that can be used for the technique of volumetric analysis including: Reduction – Oxidation (Redox: Covered in TPC Chemistry B) Complexometric (TAFE: Cert IV / Diploma in Laboratory Techniques)

Precipitation The following examples use acid / base reactions although many other types of chemical reactions can be used in volumetric analysis. Example 1 A 20.0 mL aliquot of hydrochloric acid, HCl was titrated to endpoint with 15.5 mL of a standardised 0.112 M solution of sodium carbonate, Na2CO3. Determine the concentration of the HCl. moles of Na2CO3 titrated = Concentration x Volume (L) = 0.112 x 0.0155 = 1.74 x 10-3 Equation: 2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2O(l) + CO2(g) Mole ratio: 2 : 1 The mole ratio above indicates there will be twice as many moles of HCl than there will be of Na2CO3. Therefore, moles of HCl reacted = 2 x 1.74 x 10-3 = 3.47 x 10-3 Concentration of HCl = moles HCl Volume (L) = 3.47 x 10-3 0.02 L = 0.174 M (ie; moles / L)

133

Example 2 How much ammonia (a base) is in ‘Cloudy Ammonia’ ? A 25.0 mL aliquot of ‘Cloudy Ammonia’ was diluted to 250.0 mL in a volumetric flask. A 25.0 mL aliquot of this diluted solution was transferred by pipette into a conical flask. A few drops of methyl orange indicator were added (a yellow solution results). A standardised 0.099 M HCl solution was delivered by burette with constant swirling of the flask. The titration was completed when the colour of the solution just changed to a permanent tinge of apricot-pink. The volume of HCl added was noted. The procedure was repeated three times until consistent results were obtained. Results:

Titration No Volume HCl added (mL)

1 28.90

2 27.60

3 27.60

4 27.65 Titration 1 result is ignored as it is not consistent with the following titrations. The results from titrations 2, 3 and 4 are used to find an average titre. Average titre = 27.60 + 27.60 + 27.65 3 = 27.62 mL moles of HCl titrated = concentration x volume (L) = 0.0991 x 0.02762 L = 2.74 x 10-3 Equation: HCl(aq) + NH3(aq) NH4Cl(aq) Mole ratio: 1 : 1 Therefore: moles of NH3 = 2.74 x 10-3 (since reaction ratio is 1:1) These 2.74 x 10-3 moles of NH3 are contained in the 25.0 mL titrated. Concentration of NH3 in the diluted solution = moles NH3 Volume (L)

= 2.74 x 10-3 0.025 L = 0.110 M ** ** The original sample had been diluted by 10x in the 250 mL volumetric flask. Therefore, the concentration of NH3 in the ‘Cloudy Ammonia’

= 0.110 M x 10 = 1.10 M

134

TITRATION CALCULATION STEPS Remember - a titration is always a chemical reaction between 2 reacting solutions: A Solution of KNOWN concentration (Standard Reagent) or quantity (Primary Standard) is fully titrated with your solution of UNKNOWN concentration. REMEMBER MOLE EQUATIONS: Equation 1. Moles = Mass (g)

FW Equation 2. Molarity = Moles

Volume (L) 1. You must firstly read the scenario carefully and identify your KNOWN reagent and

UNKNOWN reagent. 2. Calculate moles of Known Reagent 3. Write down a BALANCED CHEMICAL EQUATION between the 2 reagents:

Acid + Base Salt + H2O Acid + Carbonate Salt + CO2 + H2O

4. Determine and write down the REACTION RATIO between your Known and Unknown

reagent. 5. Calculate MOLES of your Unknown Reagent:

Multiply or Divide your Moles of KNOWN reagent (step 1) by the Reaction Ratio (step 3).

6. Calculate the MOLARITY of your Unknown Reagent:

MOLARITY = Moles Unknown Reagent Volume (L) Unknown Reagent

135

Experiment 1 – PRACTICE TITRATION Indicator Colour Screened Methyl

Orange Methyl Orange Phenolphthalein

Acid (HCl)

Base (NaOH)

Endpoint

Practice Titration Part A

Obtain triplicate 25 mL volumes of HCl (0.1M) by pipette and place into 3 conical flasks. Add phenolphthalein indicator. Fill your burette with with NaOH (0.1M) to the 0mL mark. (don’t forget sacrificial rinsings where required….) Titrate your solutions to endpoint placing emphasis on the precision of your results. Aim to titrate each set of 3 analyses within 0.2 mL 0.3 mL. Repeat practice titration above using (2) screened MO and if time, (3) MO. RESULTS

NaOH Titration Volume (mL)

HCl Aliquot Volume (mL)

1. Phenolphthalein

2. Screened MO

3. MO

AVERAGE

136

Part B Obtain triplicate 25 mL volumes of NaOH (0.1M) by pipette and place into 3 conical flasks. Add phenolphthalein indicator. Fill your burette with with HCl (0.1M) to the 0mL mark. (don’t forget sacrificial rinsings where required….) Titrate your solutions to endpoint placing emphasis on the precision of your results. Aim to titrate each set of 3 analyses within 0.2 mL 0.3 mL. Repeat practice titration above using (2) screened MO and if time, (3) MO. RESULTS

HCl Titration Volume (mL)

NaOH Aliquot Volume (mL)

1. Phenolphthalein

2. Screened MO

3. MO

AVERAGE

QUESTIONS

1. Summarise the rinsing procedure(s) for the following pieces of laboratory equipment:

• Burette

• Conical Flask

• Pipette

2. How is endpoint different to equivalence point when performing volumetric analysis (ie titration)?

137

Experiment 2 – HCl ANALYSIS Aim: To determine the concentration of an unknown solution of hydrochloric acid, HCl with a known concentration of the base sodium hydroxide, NaOH solution. Method: HCl Analysis ! Recall and carry out appropriate rinsing techniques for all glassware.

• Pour around 100 mL of an approximately 0.1M HCl into a beaker. • Using a pipette, transfer triplicate 25 mL aliquots of the HCl into 3 conical flasks. • Add about 5 drops of an appropriate indicator…… • Fill your burette with the standardised NaOH solution. • Titrate your triplicate HCl solutions with the standard NaOH until endpoint is reached. • Record your individual titration volumes in the table below and calculate the

concentration of the HCl. Results:

NaOH concentration: ______________ M Indicator choice: ___________________

Titration Results

HCl aliquot volume NaOH Titration Volume (mL)

Average Titration Volume, NaOH = ________________ mL Calculations:

1. Moles NaOH Titrated: = Molarity x Volume (L)

2. Chemical Equation: HCl + NaOH _______________ + _______________

3. Reaction Ratio: ____ : ____

4. Moles HCl =

5. Molarity HCl: = Moles Volume(L)

138

Experiment 3 – HCl STANDARDISATION Aim: To standardise a solution of hydrochloric acid using the primary standard Na2CO3. Prework: Calculate the mass of Sodium Carbonate, Na2CO3 required to react with 25 mL of an approximately 0.1M HCl solution. Method: HCl Standardisation

• Weigh accurately, approximately the mass of Na2CO3 calculated above into 3 conical flasks.

• Dissolve in a small volume of water. Add 5 drops of the methyl orange indicator. • Fill your burette with the HCl solution to be standardised. • Titrate your standard Na2CO3 solution with the HCl until endpoint is reached. • Record the volume of HCl and calculate the concentration of the HCl. Results:

Mass Na2CO3 (g) HCl Volume (mL)

Calculations: Perform separately for each titration result.

i) Calculate moles of sodium carbonate ii) Write a balanced chemical reaction, and hence determine the reaction ratio. iii) Calculate moles of HCl iv) Calculate the molarity of HCl

Questions:

1. Why are titration volumes between 15 – 30 mL the most desirable for titrations?

2. Why is it important to give a burette a “sacrificial rinse” with the known solution?

3. Why is it OK to rinse a conical flask with distilled water before addition of reagent, but NOT with the burette?

4. Why is it OK to average the titration volumes performed on triplicate analyses of solution aliquots, but NOT OK for 3 separate masses of Sodium Carbonate carefully weighed and dissolved within 3 separate conical flasks?

139

Experiment 4 – VINEGAR ANALYSIS Aim: To determine the concentration of Ethanoic Acid, CH3COOH in Vinegar. Method:

1. Perform a 10x dilution on a vinegar sample by transferring a 25 mL aliquot of vinegar sample into a 250.0 mL volumetric flask and making up to volume.

2. Transfer 25.0 mL aliquots of the diluted vinegar in triplicate by pipette into 3 conical flasks.

3. Add 3 – 5 drops of phenolphthalein indicator to each flask and titrate with standardised sodium hydroxide, NaOH reagent.

Results Table: NaOH = ____________________ M

Aliquot volume (mL) Volume NaOH (mL)

25

25

25

Volume NaOH (Average)

Calculate the concentration of the vinegar in %w/v…. 1. Moles NaOH titrated = 2. Chemical Reaction (and hence reaction ratio): 3. Moles CH3COOH present in 25 mL (diluted) aliquot = 4. Molarity CH3COOH (diluted) = 5. Dilution Factor = 6. Molarity CH3COOH (undiluted solution) = 7. % w/v, (ie g/100 mL vinegar) = ……………..

i) Moles / L = ii) g / L = iii) g /100 mL =

140

Experiment 5 – ACID / BASE TITRATION: RESULTS PROFORMA 1. Sample ID:

_________________________________

4. Total sample volume prepared:

________________________________

2. Analyte ID:

_________________________________

5. Titrant ID:

________________________________

3. Sample amount taken:

_________________________________

6. Titrant molarity:

________________________________

Aliquot (mL) Titrant Volume (mL)

1.

2.

3.

Average Titrant Volume

1. Calculation of Analyte Concentration (M) i) Moles of titrant:

ii) Balanced chemical reaction and reaction ratio:

iii) Moles of analyte in sample solution:

Calculations continued over page

141

iv) Molarity of sample solution:

v) Dilution factor (if applicable)

2. Report your analyte concentration as a percentage, ie; Liquid samples as % w/v, (ie g/100 mL) i) Moles / L =

ii) g / L =

iii) g /100 mL = _______________ OR Solid samples % w/w, (ie g/100 g) i) Moles / L =

ii) g / L =

iii) g / total sample volume (volumetric flask)

iv) g / sample amount taken

v) % w/w

142

9.4 Chromatography Chromatography is a procedure for separating mixtures containing several dissolved substances. Each dissolved substance has a difference in attraction for the solvent, referred to as the mobile phase, versus the blotting paper, referred to as the stationary phase. By comparing the distance moved by one substance in the mixture with that of a known substance, we can identify the presence of that substance in the mixture. You can separate some food colourings or inks using this method. Concentrated dots of colour are placed along one edge of a piece of blotting or other absorbent paper as shown. In this example, three coloured inks A, B and C and a known substance S are spotted. The paper is suspended in a suitable solvent in a covered glass container for several hours. At the beginning ` cover covered glass container blotting paper spots are small and concentrated pencil line marking solvent level is below the origin spots on the paper A B C S Known reference substance (S) can be spotted next to the substances to be tested. After several hours ---------------- solvent front distance solvent front known reference substance has moved (S) A B C S

143

Substance A contains substance S and at least one other substance (since this solvent may not have separated all the substances present). B and C do not contain S because there are no spots at this height on the chromatogram. The components of the inks can be identified by the distance they travel through the paper compared to the distance that the solvent front has travelled. This is expressed as an Rf value for each component. Rf = distance component spot has moved from the origin distance solvent front has moved from the origin solvent front component P component Q Distance moved by Q = 5.2 cm Distance moved by P = 11.8 cm Distance moved by solvent = 15.0 cm Rf for component P = 11.8 = 0.79 15.0

Rf for component Q = 5.2 = 0.35 The solvent that is best for a particular separation is found by experimentation. You might try various mixtures of water, alcohol (colourless eg vodka, white rum), white vinegar or ammonia solutions. Other absorbent materials such cotton or even a stick of chalk can be used for the stationary phase of the chromatogram. Thin layer chromatography makes use of a fine layer of silica or aluminium oxide on a glass plate. The absorbent material is packed in to a glass column for column chromatography and a solvent is dripped through after the sample is placed on top.

Practical Activity / research: Chromatography 1. Use chromatography to separate some pigments of your choice. In your report

comment on the effectiveness of your procedure and suggest improvements. 2. Research instrumental methods such as high performance liquid chromatography and gas-liquid chromatography. Observe demonstrations of these methods if possible. 3. Chromatography is used extensively in industry. Use some modern sources to

investigate one application. Ref: CC1 pp172-174 C2 pp13-14 CC2 pp25-29; 58-61

144

9.5 Electrophoresis Anything which conducts an electric current must contain mobile electric charge carriers. Conduction can arise in two different ways: by movement of electrons or by movement of ions. In metal wires, only electrons move through the rigid crystal lattice but this is not the case in solutions where ions move. Solutions which contain ions and therefore conduct electricity are called electrolytic solutions while any substance which produces ions in a solution is an electrolyte. Electrolytic solutions conduct because they contain charge-carriers (ions) which are absent from solutions of non-electrolytes and virtually absent from water itself. The electrically charged ions in the compound sodium chloride, Na+ & Cl- will conduct an electric current if they are free to move. The ions can move if the compound is molten (liquid) or dissolved in water. Electrophoresis uses this movement of ions to separate them in a process which is, in some ways is similar to chromatography. The positive ions are attracted to the wire and metal clip attached to the negative terminal of the electricity source. The negative ions are attracted to the oppositely charged positive electrode. Smaller ions can move faster than larger ones and so a separation and identification can take place. If the ions are not coloured they can be made visible by spraying with a suitable dye or other means.

Diagram

145

9.6 Colorimetry The intensity of a coloured solution is directly related to its concentration. The more intense the colour, the more concentrated the solution. A copper(II) sulfate solution of unknown concentration is mid-blue. We can estimate its concentration by comparing its colour intensity with other copper sulfate solutions of known concentrations. This analysis technique is called colorimetry. The colour intensity of solutions can be more reliably and accurately compared using an instrument called a colorimeter than using our eyes. The instrument measures the intensity of light which passes through the solution of unknown concentration. A series of measurements is first carried out on the standard solutions and a calibration graph is drawn. The concentration of the unknown solution can then be read from the graph. Colourless substances need to be converted first to coloured compounds before simple colorimetry can be used. Phosphates are converted to a blue complex by reacting them with ammonium molybdate. In this way colorimetry can be used to calculate the amount of phosphates in detergent samples. Example: The concentration of a solution of dye orange-X needed to be determined. Standard solutions of orange-X were made up using distilled water to the concentrations shown in the table. A blank (plain distilled water) was measured for absorbance in a colorimeter as was each of the standards.

Solution concentration (mg / L) Absorption

0 (blank) 0.00

0.10 0.16

0.20 0.30

0.30 0.45

0.40 0.61

0.50 0.76

Unknown 0.37

A calibration graph has been plotted. Your task is to interpolate from the graph a corresponding for the unknown solution.

See over page

146

From the calibration graph the concentration of the orange-X solution was

determined to be ................ mg / L.

Ref: CC2 pp38-39 C2 pp49-50

Orange - X Analysis by Colorimetry

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 0.1 0.2 0.3 0.4 0.5 0.6

Concentration of Orange - X (mg / L)

Abs

orba

nce

147

Experiment – COLORIMETRY Aim: To analyse the relationship between colour intensity and concentration of an analyte in aqueous solution.

! Graph Paper required as part of this exercise Method:

1. Devise a simple method for the preparation of a series of standard copper sulfate, CuSO4 solutions from a 1.0 M stock solution:

2. Analyse standard solutions through the colorimeter. Record each standard absorbance value in the results table.

3. Record the absorbance of your unknown(s). 4. Make a calibration graph from your CuSO4 standards 5. Determine the concentration of your unknown solutions.

Results:

Standard Solution Method of Preparation Absorbance

0 M

0.2 M

0.4 M

0.6 M

0.8 M

Unknown Solution

Unknown Solution Concentration: _________________________ Questions: 1. Consider the accuracy of this method. Suggest ways in which it could be improved.

2. Discuss the use of a calibration graph as a mathematical model in determining the concentration of your unknown.

148

9.7 Spectroscopy Chemicals are capable of emitting or absorbing energy in the form of visible light, ultraviolet and infrared radiation. By measuring the amount and types of energy absorbed or emitted we have a method of identifying the chemical (qualitative analysis) and in some cases determining the amount present (quantitative analysis).

Visible and U.V. spectra

An electron cannot move in an orbit of just any radius - there are in fact very few orbits whose radius values are “legal”, and an electron must be in one or another of these. If an electron in a “legal” orbit near the nucleus absorbs enough energy it can move to an orbit which is further away. The electron is now said to be in an “excited” state and will usually fall back fairly promptly into an empty “legal” orbit of lower energy closer to the nucleus. When this happens the energy the electron no longer carries is released as a pulse of radiant energy such as light. This process can only be studied properly in a collection of isolated atoms, for in polyatomic molecules or in condensed phases (solids or liquids) adjacent atoms interfere with the free movement of outer electrons. Ways of obtaining isolated atoms include: 1) vaporising a substance at high temperatures (eg in a bunsen flame); 2) subjecting a gas at low pressure to an electric discharge (eg in a fluorescent tube). Both processes result in a collection of isolated gaseous atoms in excited states which will release light or other radiant energies as they revert to less energetic states. Common applications of the principle are the red neon advertising signs and both mercury (blue) and sodium (orange-yellow) streetlights. Each element produces energies of a characteristic pattern which makes this phenomenon a powerful tool in chemical analysis. The instrument for examining the wavelengths of these energies is a spectroscope. An incandescent liquid (eg molten iron) or solid (eg the tungsten filament of an electric light globe) when viewed through a spectroscope shows a “continuous” spectrum, ie; a rainbow of colours with no gaps, ranging from red to violet. Incandescent gaseous atoms however produce light at only certain sharply-defined positions in the spectrum (only at specific wavelengths), with darkness in between. Through the spectroscope discrete bright coloured lines appear across a dark background. It is the number and positions (colours) of these lines which constitute the unique “fingerprint” of each element called its emission spectrum. Continuous spectrum

149

Sodium spectrum Hydrogen spectrum It is also possible to analyse which frequencies of light from the continuous spectrum are absorbed by a particular element - an absorption spectrum. The spectrum appears as the continuous spectrum with black lines across it corresponding to those frequencies which would be emitted by the same element (ie. a reverse of the emission spectrum). Flame Emission Spectrophotometry Many metal compounds at the high temperature of a bunsen flame or similar produce excited metal atoms whose emission spectra are analytically useful. The technique is very simple; the only apparatus required is: · a burner which can be adjusted to give an almost non-luminous flame · a platinum wire holder or · a plastic squeeze bottle with a very fine nozzle from which a fine spray of a solution of

the compound may be delivered. Either method will result in a coloured flame characteristic of a particular element. Not all metals emit energy in the visible light region of the electromagnetic spectrum. Magnesium, for example, does not. However for those that do, the colour may be used to identify the metal either directly by observation or by spectroscopic analysis. Non-metals do not emit in this way. Each metal gives off its own unique spectrum even if other metals are present in the specimen as well. The spectrum produced will contain all the lines from each individual metal’s spectrum. Potassium alone Sodium alone If the colour of one metal masks that of another because of overwhelming brightness, spectroscopic examination will still reveal the presence of the fainter one. Sodium, for example, generally outshines all others, and a mixed sodium-potassium specimen will give a flame colour which looks yellow like the sodium alone to the naked eye.

150

Spectroscopically however the flame appears like this: Mixture (sodium-potassium) Flame spectroscopy is the most convenient way of detecting the various “alkali metals” (Group I) in their compounds. They have very similar properties and often occur naturally together.

Written exercises: Visible spectra 1. Why does salty water boiling over in a saucepan turn a gas flame yellow? 2. How can hydrogen with only one electron produce multiple lines in its spectrum?

Why are some lines brighter than others? 3. The following spectra are provided so that you can answer the questions below.

(a) Identify the metallic elements in the unknown specimen. (b) Which alkali metal is most abundant in the atmosphere of the sun?

Experiment 1 – FLAME TESTS

The colour of light emitted when different metal ions are sprayed into a flame are characteristic for each metal element, as it appears on the Periodic Table. PART A - METAL ION ID

Metal Ion Colour

151

PART B - UNKNOWN METAL ION IDENTIFICATION Try matching an element to the colour of the unknown metal ion upon being aspirated into the flame. Unknown Metal Ion ID Colour Metal Ion

PART C - COLOUR INTENSITY AND METAL ION CONCENTRATION

1. Prepare a solution containing an approximately 10% concentration of Sodium Chloride.

2. Aspirate this through the flame. 3. Dilute 10 mL of this solution up to 100 mL with distilled water. 4. Aspirate the diluted solution through the flame. 5. Repeat steps 1 – 4 another 5 times.

• Summarise your observations of this task below.

Ref: CC1 p27 (flame tests) C2 pp10-13 CC2 pp29; 383

152

Analysis of Sodium and Potassium by Flame Emission Spectrophotometry

Potassium (K+) and Sodium (Na+) are soluble metal ions commonly found in natural water bodies. The ocean is a salt water environment. This salt is composed predominantly of table salt, known chemically as Sodium Chloride (NaCl). Sodium ions (Na+) exist in concentrations of around 1.2%, or 1.2 g/100 mL Potassium (K+) ions exist at lower concentration - around 0.4%, or 0.4 g/100mL We have seen in previous lessons how various metal ions release different colours of light when placed into a hot gas flame. This is referred to as Light Emission. A relationship exists between the concentration of soluble metal ion and colour intensity. As concentration increases, flame colour emitted becomes brighter, or more INTENSE. Flame colour intensity therefore relates directly to metal ion concentration. A machine called a FLAME EMISSION SPECTROPHOTOMETER is designed to aspirate (or spray) small volumes of solution into a gas flame. The machine has a detector, designed to measure the intensity of coloured light being emitted by the metal ion. If we can prepare a series of different known concentrations of a given metal ion (eg Sodium or Potassium), we will see that the colour intensity of light emitted changes according to the relationship described above. If we make a graph of metal ion concentration versus colour intensity (vertical axis), we will see on graph paper that colour intensity increases as soluble metal ion concentration increases (see graph). An unknown concentration of the same metal ion will give a particular flame intensity when aspirated through the flame photometer. We can use our graph to convert this intensity to a concentration. Flame Emission Spectrophotometers are designed to analyse for the elements Sodium (Na) and Potassium (K). The machine will accurately analyse concentrations of these elements between 1mg/L and 100 mg/L. Outside of these concentrations the light emitted is too weak or too strong to analyse accurately. Samples which are too concentrated must be accurately diluted until the Sodium / or Potassium concentration is lowered back within this range.

153

Written Exercises: 1. What is the chemical name of the salt found predominantly in sea water? 2. Name 2 soluble metal ions commonly found in sea water. 3. What do the following symbols stand for: Na+ ______________________

K+ ______________________

g: ______________________

mg: ______________________

L: ______________________

mL: __________________________

g/100mL: ___________________________

g/L: __________________________

mg/L: __________________________

4. What is the name of the instrument used to analyse for potassium and sodium ion

concentration?

0

20

40

60

80

100

120

0 20 40 60 80 100

Col

our I

nten

sity

Concentration (mg / L)

Graph of Metal Ion Concentration versus Colour Intensity

154

5. A small volume of solution is aspirated into the photometer flame to measure analyte concentration. What is meant by the term "aspirate"?

6. When metal ions release different colours of light in a gas flame, this referred to as

Fl_________________ E___________________.

7. What is the purpose of the detector within the photometer machine? 8. According to the Flame Emission principle, what is colour intensity dependant on? 9. What concentration range of Na+ and K+ is most suited for analysis by flame

emission? 10. How might we be able to still analyse a sample for both Na+ and K+ when their

concentrations are too high? 11. Convert the expected concentrations of Sodium and Potassium into different units of

concentration according to the table below: Concentration

Sodium (Na+) 1.2 g/100 mL ....…………….. g / L ……………… mg / L

Potassium (K+) 0.4 g/100 mL ……………….. g / L ……………… mg / L

155

Experiment 2 – INTRODUCTION TO FLAME EMISSION Background:

We can see the relationship between metal ion concentration and colour intensity through the method of Flame Emission. A series of known concentrations (or standards) of the sodium metal ion will be aspirated through a Flame Emission Spectrophotometer for this purpose. This experiment aims to develop the following skills:

1. Preparation of standard solutions through serial dilution. 2. Tabulating and graphing results. 3. Determining the concentration of an unknown solution. Method:

- Complete the information below summarising how you will prepare your standard solutions.

- Run your prepared standards and unknown sample through the spectrophotometer. - Record the intensity readings of each solution in the table at right of page.

Results:

• Sodium (Na+) ion Stock Solution: ___________________ mg/L

• Intermediate solution ___________________ mg/L

Preparation: __________ mL (Pipette) __________ mL (Volumetric flask)

STANDARD SOLUTION PREPARATION

ANALYSIS: FLAME EMISSION

SPECTROPHOTOMETER

Intermediate (mL) - Pipette

Total Volume - Volumetric

Flask

Sodium (Na+) Intensity

Standard 1 0 mg/L

Standard 2 20 mg/L

Standard 3 40 mg/L

Standard 4 60 mg/L

Unknown

156

Analysis:

1. Construct a calibration graph of Concentration (axis → ) versus Colour Intensity (axis ↑ ) for your Sodium (Na+) standard solutions.

PASTE CALIBRATION GRAPH HERE 2. From your graph, describe how the colour intensity changes with sodium concentration. 3. Use your graph to calculate the concentration of Na+ in the unknown solution.

157

Experiment 3 – SODIUM AND POTASSIUM IN BEER Background:

We have seen there is a relationship between metal ion concentration and colour intensity through the method of Flame Emission. The concentration of Sodium (Na+) and Potassium (K+) ions in beer can be determined via a similar method. Method:

- Complete the information below summarising how you will prepare your standard Na+ and K+ solutions.

- De-gas a beer sample by pouring from one beaker to another several times. - Prepare and aspirate your prepared standards through the spectrophotometer and

record your results in the table below. - Quantitatively dilute beer sample until Na+ / K+ intensities fall within the range of

standard concentrations. - Aspirate diluted beer sample and record this diluted Na+ / K+ intensity. - Draw a calibration graph for both Na+ and K+ - Calculate the K+ and Na+ in your original beer sample. Results:

PART A: Sodium

Na+ Stock Solution: ___________________ mg/L

STANDARD SOLUTION PREPARATION Na+

ANALYSIS: FLAME EMISSION

SPECTROPHOTOMETER

Stock Na+ (mL) - Pipette

Total Volume - Volumetric

Flask

Sodium (Na+) Intensity

Standard 1 0 mg/L

Standard 2 10 mg/L

Standard 3 20 mg/L

Standard 4 30 mg/L

DILUTED BEER

158

Dilution Factor - Beer:

Beer Volume PIPETTE Total Volume VOL FLASK DILUTION FACTOR

= ____________ mL = ____________ mL x __________

PART B: Potassium • K+ ion Stock Solution: ___________________ mg/L

STANDARD SOLUTION PREPARATION K+

ANALYSIS: FLAME EMISSION

SPECTROPHOTOMETER

Stock K+ (mL) - Pipette

Total Volume - Volumetric

Flask

Potassium (K+) Intensity

Standard 1 0 mg/L

Standard 2 20 mg/L

Standard 3 40 mg/L

Standard 4 60 mg/L

DILUTED BEER

Dilution Factor - Beer:

Beer Volume PIPETTE Total Volume VOL FLASK DILUTION FACTOR

= ____________ mL = ____________ mL x __________

Analysis - Over Page

159

Analysis:

PART A - Sodium • Construct a calibration graph of Concentration (axis → ) versus

Colour Intensity (axis ↑ ) for your Sodium (Na+) standard solutions.

PASTE CALIBRATION GRAPH HERE Calculations: Sodium Ion Concentration: 1. Na+ intensity (diluted beer) = _________________ 2. Na+ concentration (diluted beer) = ________________ mg/L 3. Dilution Factor (beer) = x ______________

Na+ concentration (beer sample) ___________________ mg/L

160

Analysis: PART B - Potassium • Construct a calibration graph of Concentration (axis → ) versus

Colour Intensity (axis ↑ ) for your Potassium (K+) standard solutions.

PASTE CALIBRATION GRAPH HERE Calculations: Potassium Ion Concentration: 1. K+ intensity (diluted beer) = _________________ 2. K+ concentration (diluted beer) = ________________ mg/L 3. Dilution Factor (beer) = x ______________

K+ concentration (beer sample) ___________________ mg/L

161

Infra-Red (I.R.) Spectroscopy Other radiations such as infra-red can also be absorbed or emitted by chemicals in a spectrum which allows them to be identified and even measured. You may be able to observe infra-red spectroscopy being used. It is particularly useful in identifying covalently bonded molecules and it is the type of bond which absorbs the energy enabling it to be identified. X-rays are another radiation which is often used in this identification process. Figure 1: An infrared spectrum of the compound butanoic acid Figure 2: A general infrared spectrum. Fingerprint region is unique to each compound