CPACS FROM ADVANCED INFORMATION
what are the solvents for proton NMR spectroscopy
The solvent shouldnt contain any hydrogen atoms - tetrachloromethane, CCl4 - solvent where hydrogen atoms are replaced by its isotope. eg. deuterium CDCl3 instead of CHCl3
Molecular formula = C4H8O2 Contains OH and C=O Ketone, 1o or 2o alcohol The high resolution proton nmr spectrum of X is shown below. The relative number of protons causing the peaks shown are: J = 1, K = 1, L = 3 and M = 3. Use the information above to determine the structural formula of X. In your answer, you should refer to the number of peaks, their relative sizes and their splitting patterns.
There are 4 peaks = 4 hydrogen environments Relative intensities of each peak indicates the number of hydrogen in each environment J = 1 K = 1 L = 3 M = 3 The splitting pattern indicates the number of hydrogen atoms on adjacent C J = quartet so is adjacent to CH3 M = doublet so adjacent to CH L and K = singlet so no H on adjacent C L is adjacent to C=O and K is due to H on O
what are sources of error in this experiment (CPAC 16) -> why is the yield higher -> why is the yield lower -> why is there a difference in melting temperature?
- impurities in the sample; the crystals may not be dry -> eg. unreacted 2-hydroxybenzoic acid - the reaction not to completion, solid lost in filtration - if impurities = larger range in melting point - if impurities = lower melting point
What are some errors caused in this practical (CPAC 11) and some improvements
- Stirring not sufficient to ensure that all the iron dissolves = warming the solution may help - Transfer of the solution and filtering = ensure beaker + filter paper are rinsed with water - Solution not mixed = invert the volumetric flask several times to ensure thorough mixing - Glassware measurements may not be read accurately = bottom or top of the meniscus (same titre obtained) - The end-point may not be clear = use a white tile
answer these further questions about CPAC 9: - explain why Ka = [H+] at half the volume of the equivalence point - A titration using NaOH and ethanoic acid had an equivalence point of pH=8.4. Using appropriate equations, explain why a pure solution of sodium ethanoate has an alkaline pH - ammonium ion can act as an acid: NH4+(aq) + H2O <-> NH3(aq) + H30+(aq). The pKa for this equilibrium is 9.3. use this information to calculate the pH of 0.025moldm3 of NH4Cl, stating any assumptions
- at equivalence point half the acid neutralised so we can assume that the same number of deprotonated A- ions as HA molecules. If [HA] = [A-], cancel to Ka = [H+] - CH3COO-Na+ is a salt of a strong acid & strong base. It dissociates in water to form CH3COO- + Na+. Na+ ions do not act as conjugate bases but ethanoate ions do. so: § CH3COO- + H20 <-> CH3COOH + OH- §. Also, Kw=[OH][H+] so conc of H+ ions drops to maintain Kw Ka = 5.6x10^-10 [H+] = √5.6x10^-10 x 0.025 = 3.74x10^-6 = pH = 5.4 -> we assume Cl- ion has negligible effect as conjugate base and so will not accept H+ ions to form HCl
how to use a pH meter why are they better than indicators
- calibrate the pH meter with deionised water and buffer solutions of known pH - wash with deionised water between readings to remove ions attached to the bulb - no ions are added or removed when used, reads to 2dp which is more accurate than interpreting colours
what are the errors caused in this experiment (CPAC 9)
- inaccuracy of burette readings -> read glassware from the bottom of the meniscus - difficulty identifying the exact equivalence point -> use a white tile so the colour change is clear -> smaller volumes alkali added, better approximation - not calibrating pH meter (lose accuracy on storage) -> calibrate using distilled water and pH buffer
What are the physical properties of aldehydes and ketones?
- no hydrogen bonding with each other only london forces and permenant dipole forces - soluble (HB bt O on carbonyl and H on water) - low boiling point (increases with size) - smell: aldehydes fruity/sweet, ketones = pungent
What are errors that arise from the poor technique of the titration
- not rinsing all the solid from the weighing boat - not rinsing stirrer, funnel into the volumentric flask - not shaking the volumetric flask thouroughly - not ensuring there is no air below the tap of burette - getting air bubbles in the stem of pipette - running solution from burette too quickly= overshoot - not swirling the flask after each addition of solution
answer these questions of this practical? (CPAC 11) - the flask with the iron tablet was left un-stopped overnight, when the experiment was repeated the next day with the same sample, the amount of iron calculated dropped by 15% why? - explain why the potassium manganate(VII) both act as oxidant and indicator - Titrate solution until mixture just turns pink (colourless to pink) Pink colour disapears due to secondary reactionz - one of the filling agents used in iron tablets can be start or cellulose. can u suggest why after the end of the titration the end point colour may fade slowly after 5 mins (hint: reaction must be acidic)
- oxygen would have oxidised with Fe2+ to make Fe3+ thus titre would be reduced - Manganate(VII) ions are reduced to Mn2+ ions, colour change from purple to colourless. colour change happens as long as there is Fe2+ ions present in solution to oxidise. as soon as Fe2+ run out, manganate(VII) ions stay oxidised and colour is intense - Titrate solution until mixture turns pink (colourless to pink). pink colour disapears due to secondary reaction - gradual fading of pink suggests a further redox reaction, in acidic condition, start is hydrolysed to glucose molecules -> aldehyde -> carboxylic acid
Answer these questions about CPAC 13a - what does sulfuric acid do - what is the role of sodium hydrogen carbonate - what solutions are in excess and why - why is starch added - effect of doubling concentration of species in rate equation - what happens if u used 0.5moldm3 of propanone (not 1moldm3)
- reaction only occurs in acidic conditions, H+ = catalyst - quench the reaction by reacting with H+ (HCO3- + H+ -> CO2 + H20), H+ ions cannot catalyse reaction so slows - concentration remains constant during reaction so the influence of iodine on the rate of reaction measured - acts as an indicator, deep dark blue when iodine and starch reaction, when iodine reacts with thiosulphate, less iodine so from blue-black - colourless - doubling the rate of reaction, doubling both = x4 - rate is 1/2, propanone is 1st order, if halves, rate halves
how to reduce errors in a titration
- white tile to make the colour change more noticeable - allow titrant to drain down burette before reading - swirl conical flask properly - dilute solution for smaller percentage error - larger-scale = less percentage error - using a balance that weighs to more d.p
what are the errors with making a standard solution
-> could lose solid when transferring from the weighing bottle to the beaker -> weigh by difference and then calculate the mass of a solid in a beaker: mass of (weighing bottle + beaker) - (mass of the bottle after emptying solid)
describe how to make a standard solution
1. Weigh an empty test tube and add 2.5g sulfamic acid in it, reweigh test tube (weight by difference) 2. transfer solid into a clean beaker 3. Add 50 cm3 of distilled water stir with glass rod 4. transfer solution and washings to graduated volumetric flask and make up to the 250 cm3 mark -> read from below the level of the meniscus 5. stopper the flask, shake by inverting + shaking flask
interpret this proton NMR spectrum
0.94, 3 hydrogens, triplet (t) = CH2 = HA 1.57, 2 hydrogens, sextet (sex) =CH3, CH3 = HB 2.85, 1 hydrogen, singlet (s) = HD 3.58, 2 hydrogens, triplet (t) CH2 = HC
what is the procedure of this practical? (CPAC 11)
1. Crush 5 iron supplement tablets using mortar + pestle 2. transfer into weighing boat and weigh to least 2dp 3. add 100cm3 of 1.5moldm3 sulfuric acid to beaker and stir to dissolve as much of solid as possible 4. transfer the dissolved solution into a 250cm3 volumetric flask (make sure no solid remains in beaker) 5. make up volumentric flask to volume of 250cm3 using distilled water, invert flask and mix 6. transfer 50cm3 of solution from flask to beaker and pipette 25cm3 of this solution into conical flask 7. fill burette with standardised 0.005moldm3 KMnO4 8. titrate contents of the conical flask with KMnO4, swirl contents of the flask as you add standardised solution 9. end point = permanent faint pink colour remains 10. repeat steps 7-10 until concordant titres achieved, ensuring all titres are recorded to nearest 0.05cm3.
Answer these questions about CPAC 16 1. Which functional group of the 2-hydroxybenzoic acid reacts with the ethanoic anhydride? 2a. an alternative method uses ethanoyl chloride instead of ethanoic anhydride. write a balanced equation for this + atom economy 2b. what other considerations need to be taken into account when choosing either of the synthetic routes for individual scale synthesis 3. old asprin left in moist environment can small like vinegar. why? 4. there are three possible isomers for hydroxybenzoic acid. draw all three and suggest which instrumental technique to distinguish between them.
1. OH group 2a. total mass of reactants = 216.5 (180/216.5 x100 = 83%) 2b. second method has higher atom economy but HCl is a strong acid so disposal/env + corrodes equipment 3. ester functional group slowly hydrolyse to 2-hydroxybenzoic acid/ethanoic acid = vinegar 4. H1 NMR would enable to distinguish focusing on type and number of aromatic hydrogen env + chemical shift
describe the alternative method of finding pH using mass find the Ka of the solution when 0.49g benzoic acid is added to distilled water and transferred to a 250cm3 volumetric flask. When pH was recorded after mixing, it was 3.0
1. accurately weigh between 0.4g and 0.5g benzoic acid and dissolve in 50cm3 distilled water 2. transfer to 250cm3 volumentric flask with washings 3. put stopper in flask and turn upside down to mix 4. withdraw sample and place in small beaker 5. measure pH using calibrated pH meter C6H5COOH <-> C6H5COO- + H+ [H+] = 1x10^-3 nC6H5COOH = 0.49/122 = 4.02x10^-3 [C6H5COOH] = 4.02x10^-3/(250/1000) = 0.016 Ka = [C6H5COO-][H+]/[C6H5COOH] Ka = [1x10^-3]^2/0.016 Ka = 6.22 x 10^-5
how does nuclear magnetic resonance (NMR) spectrometry work
1. atoms with odd number of protons OR neutrons spin will behave as a magnetic dipole (magnetic field) -> NMR detects how field is affected by large field 2. nuclei spins randomly but when an external magnetic field is applied the magnetic field of this dipole aligned: -> parallel (same direction) to applied magnetic field -> opposed (op direction) to the applied magnetic field 3. nuclei that spin parallel with the external magnetic field have lower energy than ones opposite 4. NMR sends out radio waves 5. at specific frequency nuclei aligned with magnetic field absorb radio waves + move from lower (parallel) to higher energy (antiparallel) state = resonance condition 6. nuclei will jump back down to base state (lower state) which releases energy (proportional to absorption) 7. NMR measures amount of energy absorbed -> absorption depends on chemical environment -> amount that it differs is called the chemical shift, δ
how to do a titration to find the unknown concentration of a solution
1. burette is rinsed out with distilled water (to remove impurities) and with some solution A (so not dilute) 2. pipette is rinsed out with distilled water + solution B 3. conical flask is rinsed out with water 4. burette is filled with solution A, tap is opened to release any air bubbles, burette read to 0.05cm3 5. pipette is filled with solution B using pipette filler 6. transfer to conical flask, drop of indicator added 7. solution from burette added with constant swirling until indicator changes colour, burette is read and the difference between starting and final volume is calculated = titre (rough as endpoint overshot) 8. repeat with solution B added until 1cm3/2cm3 short of approx titre, then added drop by drop 9. repeat until at least 2 concordant titres (0.2cm3) 10. average titre calculated with concordant values
what is the method for finding the Ka of a weak acid (CPAC 9)
1. calibrate pH meter is calibrated using a 3.0 buffer 2. Pipette 25cm3 of 0.1moldm3 acid into conical flask and add 2-3 drops of phenolphthalein indicator 3. Record the pipette with 0.1moldm3 to 1 or 2 dp 4. Fill a burette with 0.1moldm3 NaOH solution 5. Add 5cm3 of NaOH to flask stir w meter + record pH 6. Add further 5cm3 of NaOH, record pH 7. repeat step 6 until total volume of 20cm3 NaOH 8. Repeat step 6 but make 1cm3 additions of NaOH 9. Once pH above 7, recalibrate meter with pH = 10 10. add 1cm3 NaOH additions until a total of 30cm3 11. 5cm3 additions of NaOH until a total of 50cm3 NaOH added (total volume 75cm3)
What is enthalpically + entropically happening in these reactions: O dissolving ammonium nitrate crystals in water O reacting ethanoic acid with ammonium carbonate O burning magnesium ribbon in air O mixing solid barium hydroxide, Ba(OH)2.8H2O, with solid ammonium chloride.
1. endo reaction + less ordered: s -> aq - NH4NO3(s) -> NH4+(aq) + NO3-(aq) 2. exo + less ordered - gas produced - 2CH3COOH(aq) + (NH4)2CO3(s) -> 2CH3COONH4(aq) + CO2(g) + H20(l) 3. exo + less ordered (g->s) + 2 moles -> 1 - 2Mg(s) + O2(g)-> 2MgO(s) 4. endo less order - gas + liquid produced - Ba(OH)2.8H2O(s) + 2NH4Cl(s) -> BaCl2(s) + 2NH3(g) + 10H20(l)
Answer these questions about CPAC 2/3 1. a student left the top off NaOH for a period of several hours. suggest how this might affect the concentration 2. when investigating the concentration of HCl a set of class results indicate that the conc HCl is lower than true value. is this systematic or random error, explain why and suggest a reason for this error. 3. 5 mins after the end point has been reached the pink colour in the conical flask has disappeared. with aid of equations suggest why
1. evaporation of water would increase concentration or NaOH would react with CO2 and reduce concentration -> 2NaOH + CO2 -> Na2CO3 + H20 2. systematic, preparation of an inaccurate standard -> as repeating the experiment will give random error 3. CO2 in the air neutralises excess of NaOH at end point (2NaOH + CO2 -> Na2CO3 + H20)
what properties should the acid have before making the standard solution
1. high molar mass -> larger Mr = lower percentage error (weighing error) 2. pure -> stops sulfuric acid from being used 3. not be air-sensitive or react with air components (e.g. by absorbing CO2 or H2O) -> stops sulfuric acid from being used -> stops sodium hydroxide being used 4. not lose water of crystallisation to the air -> stops sodium carbonate from being used 5. high solubility in solvent (usually water)
what is the procedure of this practical? (CPAC 13a)
1. large excess aq propanone w sulfuric acid in beaker -> sulfuric acid is acting as a catalyst 2. measure I2 using measuring cylinder 3. add iodine solution to beaker and start stopwatch -> CH3COCH3 + I2 + H+ -> CH3COCH2I + HI + H+ -> Propanone + I2 -> iodopropanone + hydrogen iodide 4. after 5 mins withdraw 10cm3 of reaction mixture using graduated pipette to conical flask 5. record exact time added sodium hydrogencarbonate 6. titrate iodine present with sodium thiosulfate solution, adding starch when colour faded to a straw colour -> 2S2O3 + I2 -> S4O6 + 2I- 7. stop when the blue-black colour vanished 8. withdraw aliquots every 5mins until no solution left
method in how to make, purify + identify a carbonyl compound
1. react carbonyl compound with 2,4-DNPH 2. filter off precipitate 3. recrystallise the precipitate using hot ethanol 4. dry the purified product + measure melting temp 5. use data booklet to compare melting temp w those of 2,4-DNPH derivatives of A/K
how to identify a carbonyl compound
2,4-dinitrophenylhydrazine (2,4-DNPH) dissolved in sulfuric acid + methanol = nucleophilic substitution Forms an orange precipitate
25cm3 of hydrogen peroxide was made up to 250cm3 with water, from this solution 25cm3 was titrated against 23.45cm3 of 0.0200moldm3 KMnO4. calculate the concentration of the original solution of H2O2 in gdm3
2MnO4 + 5H202 + 6H+ -> 2Mn2+ + 8h20 + 5O2 nMnO4- = 23.45/1000 x 0.0200 = 0.000469 nH2O2 in 25 = 0.000469/2 x 5 = 0.001173 nH2O2 in 250 = 0.001173 x 10 = 0.01173 [H2O2] = 0.001173/(25/1000) = 0.04692 Gdm3 = 0.04692 x 34 = 1.59528 = 1.59gdm-3
what is a back titration what is the method of a back titration
A titration used when the substance being investigated is either insoluble so cannot be titrated directly 1. weigh sample of substance being investigated 2. add excess standard solution of acid or base into the beaker with the substance being invesigated 3. use funnel to pour solution made into 250cm3 volumetric flask with washings, shake by inverting flask 4. titrate 25cm3 portions of diluted solution against solution of known concentration (either acid or base) 5. repeat until concordant titres
how to prepare an aldehyde how to prepare a ketone
Aldehyde: oxidising a primary alcohol using acidified potassium dichromate and distilling off aldehyde as it is produced, orange -> green -> eg. CH3CH2OH + [O] -> CH3CHO + H20 -> eg. Cr2O7 + 14H+ + 6e- -> 2Cr3+ + 7H20 Ketone: heating a secondary alcohol under reflux with acidified potassium dichromate, orange -> green -> CH3CH(OH)CH3 + [O] -> CH3COCH3 + H2O
what is fehlings solution results with aldehydes and ketones what is tollens solution results with aldehydes and ketones what is potassium dichromate solution results with aldehydes and ketones?
Aldehyde: red precipitate forms in alkaline conditions -> CH3CHO + [O] + OH- -> CH3COO- + H2O -> 2Cu2+ + 2OH- + 2e -> Cu2O + H20 (copper(I) oxide) Ketone: no effect, remains blue Aldehyde: warmed + silver ''mirror'' formed -> CH3CHO + [O] + OH- -> CH3COO- + H2O -> Ag(NH3)2+ +e- -> Ag + 2NH3 Ketone: do not form a silver mirror Aldehyde: orange -> green (oxidised to RCOOH) -> CH3CHO + [O] -> CH3COOH Ketone: remains orange, no further oxidation
What are errors that arise from the apparatus in a titration
Burette: 0.1cm3 every division, error = 0.1/2 = 0.05cm3 -> eg. (0.05x2/23.40)x100 = 0.43% Pipette: single measurement = 0.05cm3 -> eg. 0.05 x (25/1000) = 0.20% Mass balance: reads to 0.01g, error = 0.01/2 = 0.005g -> eg. 0.005x2/1.34g = 0.75% Beaker: very inaccurate for measuring volumes Titrations below 20-30cm3 cause large error
describe the reaction between group 2 metals and oxygen describe the reaction between group 2 metals and chlorine describe the reaction between group 2 metals and water(l) describe the reaction between group 2 metals and water(g)
Burn in the air to form white ionic oxides, MO. -> 2X(s) + O2(g) -> 2XO(s) - metal is oxidised (Be: reluctant, Mg: bright flame, Ca: brick red flame, Sr: crimson flame, pale apple green flame) heating with chlorine forms white ionic chlorides, MCl2 -> M(s) + Cl2(g) --> MCl2(s) (eg. Ca + Cl2 -> CaCl2) -> dissolve in water to form [M(H20)6]2+ form alkaline suspension of a metal hydroxide, X(OH)2 & hydrogen gas. Increasing in vigour down the group magnesium heated with steam forms magnesium oxide and hydrogen - Mg(s) + H20(g) -> MgO(s) + H2(g)
what is the data analysis of this practical? (CPAC 16) Example data: - mass 2-hydroxybenzoic acid (2.03g) - mass pure product (2.06g) - melting point (130-132) - Draw the structural formulae for the reactants and product involved in the formation of aspirin from 2-hydroxybenzoic acid. - Calculate atom economy - calculate the percentage yield - ethanoic anhydride has a density of 1.08gcm3. Show that the amount of ethanoic acid used in an experiment is in excess
C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2 - 180/240 = 75% - (2.03/138 x 180 = 2.65g) so 2.06/2.65 x100 = 78% - mass = density x volume = 1.08 x 5 = 5.4g -> 5.4/102 = 0.053 (well in excess of 0.0147 of 2-hydroxybenzoic acid)
deduce the structure of compound from proton and carbon NMR -> C6H10O3
CH3CH2C(O)CH2C(O)OCH3
describe the ionisation energy of group 2 metals? describe the reactivity of group 2 metals?
Down the group decreases - higher proton number outweighed by higher shielding - more electron repulsion, easier to remove increases down the group as IE decreases making it easier to remove the two outer electrons (2+ ion)
would ∆G be feasible if: - ∆S= +ve ∆H = +ve - ∆S= -ve ∆H = -ve - ∆S= +ve ∆H = -ve - ∆S= -ve ∆H = +ve
EQUATION: ∆G = ∆H - T∆S 1. ∆G only neg if T∆S more pos than ∆H 2. ∆G = only neg if T∆S is less neg than ∆H 3. negative (feasible) 4. ∆G = positive (not feasible)
would ∆Stotal be feasible if: - ∆S= +ve ∆H = +ve - ∆S= +ve ∆H = -ve - ∆S= -ve ∆H = -ve - ∆S= -ve ∆H = +ve
EQUATION: ∆Stotal = ∆Ssystem - ∆H/T 1. only if ∆S outweighs ∆Ssur (at high temps) 2. never 3. only if ∆Ssurr outweighs ∆S (at low temps) 4. always
1.41g of a sample of chalk which is mostly calcium carbonate with some inert impurities was placed in a beaker and 50cm3 of 1.00moldm3 hydrochloric acid was slowly added. When the fizzing had ceased the contents of the beaker were washed into a standard 250cm3 volumentric flask made up to the mark and shaked. A pipette was used to transfer a 25cm3 sample into a conical flask. the sample was titrated against 0.10cm3 sodium hydroxide solution. The mean titre was 23.6cm3 Calculate the mass of calcium carbonate and the percentage purity
Equation 1: CaCO3 + 2HCl -> CaCl2 + CO2 + H20 Equation 2: HCl + NaOH -> NaCl + H20 nNaOH = 0.10 x (23.6/1000) = 2.36x10^-3 nHCl in 25cm = 2.36x10^-3 x 1 = 2.36x10^-3 nHCl in 250cm3 = 2.36x10^-3 x 10 = 0.0236 nHCl in original = (50/1000) x 1.0 = 0.05 nHCl reacted = 0.05 - 0.0236 = 0.0264 nCaCO3 = 0.0264/2 = 0.0132 mass CaCO3 = 0.0132 x 100 = 1.32g Percentage purity = 1.32/1.41 x 100 = 93.6%
calculate the conc of sodium hydroxide using the following data: - mass of weighing boat = 28.20g - mass of weighing boat + ethanedioic acid = 29.34g - volume of ethandioic acid = 250cm3 - volume of sodium hydroxide in each titration = 25cm3 - mean titre = 23.45cm3
Equation: H2C2O4 + 2NaOH -> Na2C2O4 + 2H20 nH2C2O4 = 1.14g nH2C2O4 = 1.14/90 = 0.0127 [H2C2O4] = 0.0127/(250/1000) = 0.0507moldm3 nH2C2O4 in mean titre = 0.0507 x (23.45/1000) = 0.00118 nNaOH = 0.00118 x 2 = 0.002376 [NaOH] = 0.002376/(25/1000) = 0.0951moldm3
what is the data analysis of this practical? (CPAC 9) - Plot a graph of pH on y axis and sodium hydroxide on x axis - use graph to calculate volume NaOH added at equivalence point - Use graph to find pH and then the Ka - A databook value for this same acid is 1.7x10^-5moldm3. suggest reasons for any difference between the sample data value and data book value
Equivilence point (middle of vertical region) = 24cm3 Half neutralisation (1/2V) = 12cm3 So pH = 4.5 at 1/2V, pH = pKa so log [H+] = Ka -> [H+] = 10^-4.5 = Ka = 3.16x10^-5 (pH -> [H+] -> Ka) ○ Small changes in pH causes large changes in log scale. We have to estimate the midpoint of the vertical part of the curve and extrapolate
how to test for group 1 and 2 cations
Flame test (as soluble, cannot be detected w ppt) 1) take a nichrome/platinum (inert) wire + nickel rod 2) dip wire in concentrated hydrochloric acid on a watch glass + in bunsen -> remove impurities 3) dip wire in HCl + solid compound under test 4) put in hottest part of flame + observe colour
describe the trend in thermal stability of group 1 carbonates describe the trend in thermal stability of group 2 carbonates
Group 1: Only Li2CO3 decomposes when heated. -> Li2CO3 -> Li2O + CO2 -> other group 1 cations = large radii = polarising power can't cause decomposition of anhydrous carbonates Group 2: decompose (less readily down group) to form metal oxide and carbon dioxide: CaCO3 -> CaO + CO2
what are the colours of the flame test explain the flame test
Heat from flame = absorbed + promotes an electron into an excited state. This is not stable, drops back down releasing energy in form of visible light. Gaps between energy levels = different in different cations = specific Colourless as wavelength not on visible spectrum
describe reduction of aldehydes and ketones
In dry ether with lithium tetrahydride aluminium (LiAlH4): - aldehyde forms a primary alcohol -> CH3CHO + 2[H] -> CH3CH2OH - ketone forms a secondary alcohol -> CH3COCH3 + 2[H] -> CH3CH(OH)CH3
what is the data analysis of this practical? (CPAC 11) 1.86g of FeSO4 was made up in a 250cm3 solution, 25cm3 of this was titrated against 0.005moldm3 potassium manganate (vii). the titre needed to oxidise Fe2+ was 19.75cm3. calculate the mean mass of iron per tablet Find the percentage error for each measurment made with balance, pipette and burette, then find the total percentage error and expected range for mass of iron per tablet
Ionic: 8H+ + MnO4- + 5Fe2+ -> 5Fe3+ + Mn2+ + 4H20 n in mean titre: 0.005 x (19.75/1000) = 9.875 x10^-5 mean mass of iron per tablet: 5:1 ratio of Fe2+:MnO4- nFe2+ = (9.875 x10^-5 x 5) = 4.938x10^-4 in 25cm3 nFe2+ 250cm3 = 4.938x10^-4 x 10 = 4.938x10^-3 Mass = n x mr 4.938x10^-3 x 55.8 = 0.276g of 5 tablets 0.276/5 = 0.055g (55mg) per tablet - balance = +0.01 (0.01/1.86)x100= 0.54%, pipette = +0.05 (0.05/25)x100 = 0.2%, burette = +0.05 (0.05x2/19.75)x100 = 0.51% = total possible error = (0.54 + 0.2 + 0.51) = 1.25 -> 1.25/100 x 55 = 55+-0.69 (54.3-55.7g)
what is the aim of CPAC 9 (acid base)
Ka of weak acid measured by titrating a known volume of acid against NaOH, phenolphthalein as an indicator.
describe the reaction with aldehydes and ketones and hydrogen cyanide
Nucleophilic addition reaction to form hydroxynitriles Conditions: HCN in pH8 is added to carbonyl -> too low: too few CN- catalyst -> too high: not enough HCN Aldehyde: ethanal + HCN -> 2-hydroxypropanenitrile -> RCHO + HCN -> RCH(OH)CN + CN- Ketone: propanone + HCN -> 2-hydroxy-2-methylpropanenitrile -> RCOR' + HCN -> RCR'(OH)CN + CN-
describe the features of a proton NMR spectrum
Peak: represent the number of hydrogen environments Peak integral: ratio of area under each peak which equal to number of hydrogen atoms in the environment -> divide by smallest, multiply by common denominator Peak at 0: protons in reference of tetramethylsilane Spin coupling pattern: number of adjacent protons Chemical shift: different external magnetic fields to bring them to resonance at particular radio frequency -> allows us to see what environment hydrogen is in
what are the colour changes for indicators - Phenolphthalein - Methyl orange
Phenolphthalein - pink in basic solution, colourless in acidic and neutral solutions. Methyl orange - yellow in basic solution, orange in neutral solution, and red in acidic solution
How to find the temperature at which ∆G is feasible?
use equation: T = ∆H/∆Ssystem -> Endothermic reaction will BECOME feasible -> Exothermic reaction will stop being feasible
what is the procedure of this practical? (CPAC 16)
Preparation of Asprin: 1. weigh 2g of 2-hydroxybenzoic acid and transfer 2-hydroxybenzoic acid into a pear-shaped flask 2. clamp flask so it suspends in a beaker of cold water 3. add 5cm3 ethanoic anhydride to pear-shaped flask followed by 5 drops of conc sulfuric acid (catalyst) 4. fix a condenser on flask (for reflux) 5. use bunsen to heat the beaker in a water bath + swirl to dissolve solid 6. remove the pear-shaped flask and add ice/water to break down XS ethanoic anhydride (form CH3COOH) -> (CH3CO)O2 + H20 -> 2CH3COOH 7. place pear flask in beaker of water until precipitation 8. filter off solid product using Buchner funnel 9. wash crystals product with min amount ice water Recrystallisation of Asprin: 10. crystals into BT tube (1:3 ratio ethanol:water) solvent -> min amount solvent = saturated solution 11. place tube into water bath and dissolve crystals 12. filter through hot funnel, separate insoluble impurity 13. cool the solution until crystals reappear 14. filter crystals under suction, wash with cold water 15. pat dry crystals bt filter papers then into a desiccator 16. measure mass of product to nearest 2 dp Measure of melting point: 17. take the capillary tube, heat to seal one end 18. place crystals at other end and tap to sealed end 19. into apparatus until white solid -> colourless liquid 20. compare melting temperature to literature value
What is the aim of CPAC 2 / CPAC 3
Prepare a standard solution from a solid acid salt and find the concentration of NaOH by titration using this standard solution
describe the iodoform reaction of carbonyl groups
Test for ethanal + methyl ketones, ethanol + methyl secondary alcohols (methyl group on carbonyl) Add excess iodine + alkali (OH), warm and observe -> very pale yellow precipitate of triiodomethane (CHI3) + disinfectant smell formed if carbonyl present Equation: CH3COR + 3I2 + 4NaOH -> CHI3 + RCOONa + 3NaI + 3H2O Alternative version: KI and sodium chlorate -> ClO- + 2I- + 2H+ -> I2 + H2O + Cl- -> ClO- + H2O -> HOCl + OH-
what is the data analysis for this experiment (CPAC 13a)
[I2] proportional to volume of sodium thiosulfate -> concentration [I2] /volume S2O3 against time graph Graph = straight line = 0 order w.r.t. iodine -> it is not involved in the rate-determining step of the reaction and therefore has no effect on the rate Propanone = 1st order, sulfuric acid = 1st order -> Rate = k[CH3COCH3][H+] zero for [I-]
What is used as a reference point in proton NMR and why
absorption of substance tetramethylsilane (TMS) -> (CH3)4Si = 1ppm 1. 12 hydrogen atoms with same environment 2. single, strong peak due to many hydrogen atoms 3. C-H bonds closer to hydrogens = hydrogen nuclei shielded from the external magnetic field 4. it is non-toxic and inert 5. it has a low boiling point (26oC) = easily removed
describe the reaction mechanism for the reaction in CPAC 13a
iodine = 0 order (not in RDS) propanone and sulfuric = 1st order (RDS) This mechanism forms an intermediate propen-2-ol molecule then to the formation of iodopropanone
describe the reaction between group 2 metal oxides and dilute acid describe the reaction between group 2 metal oxides and water describe the reaction between group 2 metal hydroxides and dilute acid
acts as a base to form salt and water (neutrilisation) -> MO(s) + 2H+(aq) -> M2+(aq) + H20(l) React to form metal hydroxides (> exo down group) -> XO + H20 -> X(OH)2) -> increase in solubility down the group react to form salts and water - neutralisation -> M(OH)2(s) + 2H+(aq) -> M2+(aq) + 2H20(l)
describe the trend in thermal stability of group 1 nitrates
all (except lithium) group 1 nitrates: MNO2 + O2 -> 2NaNO3 -> 2NaNO2 + O2 Lithium: small radius with high charge density, polarises more to form metal oxide, nitrogen gas and oxygen -> 4LiNO3 -> 2Li2O + 4NO2 + O2 (NO2 = brown gas)
what is gibbs free energy how to calculate it
combines enthalpy and entropy into a single value. -> reaction is feasible if ∆G is negative or < zero ∆G = ∆H - T∆System (units: kjmol^-1)
what is the aim of CPAC 11 (redox ii)
titration used to quantify the amount of iron (Fe2+) per tablet. no indicator needed as potassium manganate (VII) solution acts both as oxidising agent + indicator
what is the aim of CPAC 13a (kinetics ii)
continuous monitoring measurement see how changing concentration of different reactants (eg. iodine) affects the rate of reaction, allowing us to find the rate equation and reaction mechanism
what is a suitable acid to use when making a standard solution and why
ethanedioic acid (H2C2O4) -> can be obtained as a pure solid -> does not interact with the air
What is entropy (S) a measure of?
measure of disorder in a system (ways energy can be spread out between particles). units: J mol-1 K-1 affected by state, number of moles, temperature -> gaseous = more disorder, more movement -> more moles = more way of spreading energy -> higher temperature = higher disorder, more KE
0.126g of vanadium was made up to 50cm3 with an unknown acid. The resulting solution contained VO3- ions, this yellow solution was reduced and a violet solution containing Vn+ ions was left. 10cm3 of the violet solution was oxidised (back to vo3-) by 13.2cm3 of 0.0225 moldm3 aqueous potassium manganate(vii). Calculate the charge of the V^n+ ions.
nV = 0.126/51 = 0.00248 nKmnO4 = 13.2/1000 x 0.0225 = 0.000297 in 10cm3 0.000297 x 5 = 0.001485 in 50cm3 Mn04:Vn+ 1.48x10^-5:0.00248 1:1.67 3:5 3 moles MnO4- react with 5 moles V 3MnO4- + 5V^n+ + 6H+ -> 3Mn2+ + 5VO3- + 3H20 Electrons: -3 + 5x : +7 -3 + 5x = +7 5x = 10 X = 2
describe the trend in thermal stability of group 2 nitrates
nitrate ion (NO2-) polarised by 2+ charge on the cation so decompose to form metal oxide, nitrogen gas and oxygen (eg. 2Mg(NO3)2 -> 2MgO + 4NO2 + O2)
why might a reaction with a negative ΔG value may not occur in practice?
some reactions that are thermodynamically feasible may be inhibited by kinetic factors -> activation may be too high, too few molecules possess energy on collision and so the reaction does not take place at that temp. (catalyst can be used)
how to find the concentration of an unknown solution using calculations how to find the volume needed of an unknown solution using calculations
standard solution - solution with known concentration 1. calculate moles of reagent with known concentration -> moles = concentration x volume 2. calculate moles of other reagent using stochiometry 3. calculate the concentration of other solution -> concentration = moles/mean titre 1. calculate moles of reagent with known concentration -> moles = concentration x volume 2. calculate moles of other reagent using stochiometry 3. calculate the volume of other solution -> volume = moles/concentration
what is the aim of CPAC 16 (organic synthesis)
synthesising Asprin using ethanoic anhydride (acid anhydride) & 2-hydroxybenzoic acid (carboxylic acid) formed in the presence of a catalyst (sulfuric acid)
what is the relationship between K, ∆G and ∆Stotal?
∆G = -RT lnK ∆G is feasible, negative, K = large (>1) ∆G is not feasible, positive, K = small (<1)
how does temperature have an effect on gibbs free energy (and therefore the feasibility of a reaction)
∆H = negative, ∆S=positive, ∆G=negative (any temp) ∆H = positive, ∆S=negative, ∆G=positive(any temp) ∆H = negative, ∆S=negative, ∆G=neg (lower temp) -> eg. freezing ice (becomes less disordered -> s) ∆H = positive, ∆S=positive, ∆G=neg (higher temp) -> eg. decomp of NaHCO3 (more disordered -> g)
how to calculate the entropy of the system how to calculate the entropy of surroundings how to calculate the total entropy change
∆Ssystem = ∑S (products) - ∑S (reactants) units: JK^-1mol^-1 (or Jmol-1K-1) ∆Ssurr = -∆H/T units - JK^-1mol^-1 (or Jmol-1K-1) ∆Stotal = ∆Ssystem + ∆Ssurroundings units - JK^-1mol^-1 (or Jmol-1K-1) If positive = feasible