Analytical instruments week 1
Upcoming SlideShare
Loading in...5
×
 

Analytical instruments week 1

on

  • 2,742 views

 

Statistics

Views

Total Views
2,742
Views on SlideShare
2,742
Embed Views
0

Actions

Likes
0
Downloads
31
Comments
0

0 Embeds 0

No embeds

Accessibility

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Analytical instruments week 1 Analytical instruments week 1 Presentation Transcript

    • Analytical chemistry
    • Qualitative Analysis What substances are present in the product?  Testing for the presence of pesticide and herbicide residues in milk  Actual oils used in a margarine  Ethanol in petrol  Arsenic in food
    • Quantitative Analysis How much of a particular substance is present in a product?  Determination of the actual concentration of pesticide and herbicide residues in milk  Percentage of canola oil used in margarine Analysis uses modern instrumentation or the old fashioned reactions
    • Analysis Instruments Chemistry spectroscop y Gravimetric Volumetricchromatography HPL AA UV Acid base RedoxTLC GLC IR C
    • Balance equations Na2CO3(aq) + HCl(aq) aq) + HNO3(aq) NaOH(aq) + HCN(aq) HCOOH(l) + NaOH(aq)
    • Ionic EquationsFull Equation NaOH(aq) + HCN(aq) NaCN(aq) + H2O(l)Ionic Equation Na+(aq) + OH-(aq) + H+(aq) + CN-(aq) Na+(aq) + CN-(aq) + H2O(l)Partial Ionic equation Just shows the substances that undergo change OH-(aq) + H+(aq) H2O(l)
    • Units Convert  125oC to K  23mL to L  2.4atm to Pa  740mmHg to Pa
    • Relative atomic mass Mass of an atom or compound relative to 12C having a mass of 12  MR of Mg =  MR of MgO =  MR of Al2O3 =
    • Examples Calculate the volume of  3.5 mole of helium at STP  12g of hydrogen at SLC  12g of hydrogen at 110oC and 200000Pa
    • Percentage Composition % mass = MR of element x number present x 100 MR of compound Examples  % S in SO2  % Al in Al2O3
    • Empirical Formulae A calculation frequently asked for is the determination of an empirical formula given the percentage composition of the atoms or just a mass ratio
    • Example Determine the empirical formula for cholesterol given that the percentage composition of cholesterol is 83.938% carbon, 11.917% hydrogen and 4.145% oxygen. The empirical formula is therefore C27H46O
    • Example When 0.864g of nitrogen burns, it forms 2.839g of oxide. Find the empirical formula of the oxide. If the molar mass of the oxide is 92g, determine its molecular formula as well.
    • Example When a 0.995g sample of an organic molecule containing carbon, hydrogen and oxygen is burnt in air, the only products are 1.468g of carbon dioxide and 0.602g of water. What is the empirical formula?
    • Example 10.848g hydrated copper(II) sulfate is dried until there is no further change in mass. After drying, the anhydrous salt has a mass of 6.935g. Determine the degree of hydration of the salt.
    • The value of MR The value of molar mass can sometimes be provided in a round about way. A 51g sample of a compound occupied the same volume as 16g of oxygen at the same temperature and pressure.  n(O2) = 16 / 32 = 0.5 mol  Therefore the number of mole of the unknown compound must also be 0.5 mol.  If 0.5 mol has a mass of 51, then the molar mass must be 102g
    • Mole 1 mole is the amount of substance that contains the same number of particles as there are in 12g of 12C The number of particles in 1 mole = 6.02 x 1023.  This is Avagadros number Examples  1 mole of aluminium = 26.9g  1 mole of copper =  1 mole of Al2O3 =
    • Mole calculations m n = MR Example  Calculate the number of mole in  200g of aluminium  0.34g of Al2O3
    • Significant figures Your answer must only have the same number of significant figures as the least acurate data given The zeros before and immediately after a decimal point are not counted as significant
    • Masses of solids Take special care with questions that give mass in kilograms or milligrams. Moles are calculated using grams so you must convert kg and mg to grams before you start.
    • Stoichiometry Calculate the number of mole of each substance for the following reactions  2FeCl3 2FeCl2 + Cl2  3 mole _____ ___  ______ 0.45mole ____  4Al + 3O2 2Al2O3  10mole ___ ______  _____ 0.45mole ______
    • Excess 20g of magnesium is added to 200mL of 2M hydrochloric acid. Which chemical is in excess? Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) The amount of product will be determined by the reactant which is scarce, as it all reacts. The volume of H2(g) must be calculated from the limiting reactant.
    • Gravimetric Analysis Is a method of analysis that involves accurate measurement of masses in a precipitation reaction An ion in solution is caused to precipitate out of solution. The precipitate is filtered, then dried. Its mass is related to the concentration of the original ion. Gravimetric analysis relies on stoichiometry Gravimetric analysis will work if either of the ions present can be precipitated.
    • Simplified solubility table All nitrates, acetates (ethanoates), group 1 and ammonium compounds are soluble All chlorides, bromides and iodides are soluble except Ag+ and Pb2+ compounds All sulfates are soluble except BaSO4 and PbSO4 All carbonate compounds are of low solubility except group 1 carbonates and (NH4)2CO3 All hydroxide compounds are of low solubility except group 1 hydroxides and NH4OH, Sr(OH)2 and Ba(OH)2
    • Gravimetric Analysis A chemist determined that the salt (NaCl) content of food by precipitating chloride ions as silver chloride. A 8.45g sample of food yielded 0.636g of precipitate.
    • An example of the steps involved in gravimetric analysis The equation needs to be written down first Find the molar mass of the precipitate and salt Find the number of moles of the precipitate and equate this to the number moles of the salt Find the mass in grams of the salt from n = m/MR Determine the percentage of salt
    • What can go wrong in gravimetric analysis? Although precipitates used are of low solubility, a very small amount will remain in solution Other insoluble compounds may also be precipitated The precipitate must be washed to prevent any other chemicals such as solute particles from crystallising out during the drying process. The washing must be limited, however, so as not to re-dissolve any precipitate The balance has an uncertainty that may be tiny for very precise instruments or quite large for simpler models. For this reason the mass of precipitate should be reasonably large.
    • Test yourself The formation of an insoluble salt in solution is known as __________________. Insoluble salts can be removed from solution by ________________. When two soluble salts, sodium chloride and silver nitrate, are reacted together they form a white precipitate of silver chloride. The overall equation is: NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq) The precipitate is ______________ The ionic equation is: Ag+(aq) + Cl-(aq) AgCl(s) The spectator ions are: _______________ + ________________
    • Test yourself An analytical procedure in which the masses of solids are measured in order to determine the concentration of a particular substance in a mixture is called _________________ Steps involved in gravimetric analysis are: (i) add an _____________ of a reactant to form a precipitate, (ii) ______________ the precipitate and wash it, (iii) dry the precipitate to ____________ mass.
    • Chicken soup worksheet
    • Example A 5.40g sample of potato chips is crushed and mixed in water to dissolve the potassium chloride that is used instead of sodium chloride. To analyse the KCl, excess silver nitrate solution is added. The precipitate is filtered, dried and weighed. Its mass is 0.192g. Calculate the % mass of potassium chloride in the chips (assume no NaCl is used.)
    • Tips Set out each question carefully Write the equation out, with each piece of data placed under the chemical it refers to Include the units Where to start will often be evident from what you have written
    • Example A 50.0mL solution containing iron(III) nitrate, Fe(NO3)3, has excess sodium hydroxide solution, NaOH, added to it to precipitate the iron as Fe(OH)3. After heating, the hydroxide decomposes to iron oxide, Fe2O3. The mass of precipitate obtained was 0.533g. Calculate the iron concentration in the original solution.
    • Fertiliser (Exam question) A soluble fertiliser contains phosphorus in the form of phosphate ions (PO43-) content by gravimetric analysis, 5.97g of the fertiliser powder was completely dissolved in water to make a volume of 250.0mL. A 20.00mL volume of this solution was pipetted into a conical flask and the PO43- ions in the solution were precipitated as MgNH4PO4. The precipitate was filtered, washed with water and then converted by heating into Mg2P2O7. The mass of Mg2P2O7 was 0.0352g.
    • Questions Calculate the amount, in mole, of Mg2P2O7. Calculate the amount, in mole, of phosphorus in the 20.00mL volume of solution Calculate the amount, in mole, of phosphorus in 5.97g of fertiliser Calculate the percentage of phosphate ions by mass in the fertiliser. Ensure you express your answer to an appropriate number of significant figures.
    • Continued Several actions which could occur during this analytical procedure are listed below. For each action indicate the likely effect on the calculated percentage of phosphate ions in the fertiliser.  The MgNH4PO4 precipitate was not washed with water  The conical flask had been previously washed with water but not dried  A 25.00mL pipette was unknowingly used instead of a 20.00mL pipette  The mass of the fertiliser was recorded incorrectly. The recorded mass was 0.2g higher than the actual mass
    • Method again Step 1: A known mass of the sample is dissolved in a suitable solvent  Polar substances are dissolved in  Non polar substances are dissolved in  Alloys are dissolved in Step 2: the substance to be analysed is precipitated by the addition of an appropriate chemical species. The chosen solution is one that exclusively precipitates the ion of interest. If the precipitating solution co-precipitates other ions, the weighed mass will be and therefore, the calculated percentage by mass will be
    •  Step 3: the precipitate is collected by filtration and thoroughly washed to remove substances that would otherwise contribute to the mass of the precipitate. Washing is usually performed using deionised water. If the washing step is omitted, the weighed mass will be and the calculated percentage by mass of precipitate will be Step 4: the precipitate is then dried in an oven at 110oC. If the drying step is incomplete, water will contribute to the mass; the weighed mass will be and the calculated percentage by mass of the precipitate will be
    •  Step 5: the sample is cooled in a dessicator and weighed. The dessicator removes moisture from the atmosphere and minimised the amount of moisture absorbed by the sample during cooling Step 6: steps 4 and 5 are repeated until there is no significant change in mass. This ensures that all the water has been evaporated.
    • The precipitate formed must exhibit the following properties: The precipitate must have a low solubility so that it does not go back into solution If the precipitate has a relatively high solubility, the weighed mass will be and the calculated percentage by mass of precipitate will be The precipitate must have a molar mass that does not vary, so that stoichiometric calculations can be accurately preformed If a sample absorbs substances from the atmosphere, the weighed mass will be and the calculated percentage by mass of precipitate will be If a sample gives off substances to the atmosphere, the weighed mass will be and the claculated percentage by mass of precipitate will be
    •  The precipitate must be stable when heated and dessicated The precipitate must be pure The precipitate must be easy to recover by filtration The molar mass should be high so that weighing errors are minimised
    • Effects of incorrect techniques on calculations ERROR Mass of collected Mass of sample % precipitate component compositionCo-precipitation ofionsHigh solubility ofprecipitateLoss of precipitateduring filtrationIncomplete dryingIncomplete washing
    • Question 1 A solution containing 10.0g of sodium chloride is mixed with a solution of silver nitrate. What mass of precipitate will be formed?
    • Question 2a A 1.595 g sample of silver alloy is dissolved in 50.00 ml (an excess) of nitric acid. A 10.00 ml sample was then treated with excess sodium chloride solution to produce a precipitate of silver chloride. The precipitate was filtered, dried and weighed. If the mass of silver chloride precipitated is 0.25g, find the percentage of silver in the alloy.
    • Question 2b,c State the assumptions that were necessary to determine the percentage of silver in the alloy. What are the principal sources of experimental error in a gravimetric analysis?
    • Question 2d, e What conditions could be arranged to precipitate as much of the substance to be analysed as possible? Why was excess sodium chloride added to the reaction mixture?
    • Question 2 f Consider the following reaction: BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + NaCl(aq) Discuss why you would choose filtration techniques in preference to evaporation when collecting the precipitate of this reaction.
    • Question 3 a A sample of contaminated hydrated copper sulphate (CuSO4.5H2O) was tested for purity. A 15.0 g sample was dissolved in water and filtered to remove insoluble impurities. The sulphate ions were precipitated by the addition of excess barium chloride and the resulting precipitate was collected, dries and weighed. If the final mass is 4.95 g: Find the percentage purity of the sample.
    • Question 3 b, c If the precipitate was not completely dry when weighed, what effect would this have on the calculated percentage purity of the sample? What would be the effect on the calculated percentage if barium nitrate was used instead of barium chloride?
    • Question 3 d, e What would be the effect on the calculated percentage if silver ions were present in the sample? Find the mass of contaminated hydrated copper sulphate that would be required to produce 100.00 ml of a 0.250 M copper sulphate solution.
    • Question 4 A 0.500 g sample of sodium sulphate and a 0.500 g of aluminium sulphate were dissolved in a volume of water, and excess barium chloride added to precipitate barium sulphate. What was the mass of barium sulphate produced?
    • Question 5 0.6238 g of copper(II) sulphate crystals with formula CuSO4.xH2O was dissolved in water, and the black copper (II) oxide was precipitated by treatment with boiling NaOH solution. The precipitate was collected by filtration, washed, dried and weighed. If the precipitate weighs 0.1988 g, calculate the value of x in the formula CuSO4.xH2O.
    • Question 6 In order to determine the molecular formula of a compound known to contain only carbon, hydrogen and oxygen, the following experiments were carried out.1. A 0.60 g sample of the compound was burnt in excess oxygen. When the gases evolved were passed through anhydrous CaCl2, its mass increased by 0.36g due to the absorption of H2O. The remaining gas(es) when bubbled through a NaOH solution, increased its mass by 0.88 g.
    • Question 6 a, b2. A 1.21 g sample of the compound was vaporised. The vapour occupied 0.403L at 150oC and 1.17 x 105 Pa Determine the mass of the gaseous products Determine the mass of carbon in the sample
    • Question 6 c, d Determine the mass of hydrogen in the sample Determine the mass of oxygen in the sample
    • Question 6 e, f Determine the empirical formula of CxHyOz Determine the relative molecular mass of CxHyOz
    • Question 6 g Determine the molecular formula of CxHyOz