Trans:It Science

trans:it science trans:it science APPENDIX: PRACTICAL WORKChemistry, like other sciences, is a practical subject. To become successful chemists, students need to master a range of laboratory techniques. University chemistry courses provide plenty of chances to practice the required skills, but to get the most out of these opportunities you need to prepare well.Planning an experimentEven if you are provided with a fully comprehensive lab script, it is worth spending the time to carefully read it through before you start any practical work. This provides you with a chance to read up on any techniques you are unfamiliar with or have not done for a while, as well as a chance for you to understand the chemistry involved.Sometimes you may be required to do some pre-planning for an experiment, such as calculating amounts of reagents needed.Even if it is not required, it is often useful to draw up a plan for the experiment – this will help you use your time more effectively when in the laboratory, as well as help prevent you missing any vital steps! Careful planning allows you to spot where time can be saved e.g. when a reaction is being heated under reflux, things for the next step can sometimes be prepared.Try to make yourself familiar with common laboratory techniques such as methods of purification (recrystallisation, separation and chromatography) and standard apparatus setups (e.g. reflux and distillation).101 trans:it scienceCarrying out the experiment Record any molar calculations, observations and results in a lab book.  Remember it is better to record too much information than too little.  If unsure about anything, ask, especially if it relates to using something safely.  Make sure your glassware is clean before you start. It is much better to spend time ensuring everything is properly clean at the beginning, than finding out that your reaction has been contaminated by dirty glassware later on.Useful book:Dean, J. et al (2002). Practical Skills in Chemistry. Harlow: Pearson. A book explaining various practical techniques and methods including the running and interpretation of spectra, purification techniques, distillation and reflux.Useful unitsTo make it easier for scientists to communicate with each other and to simplify equations, a standard set of units, the SI units, was adopted by the scientific community around the middle of the 20th century. However, you will still find that non- standard units are used sometimes for convenience on a day-to-day basis. If you look at glassware in the laboratory you may find that it is marked in millilitres (ml) rather than in cm3 (the SI unit for volume is the cubic metre, m3). Thankfully, this is a very simple conversion, as 1 ml = 1 cm3, but some others are not quite so straightforward. Whenever you are writing up a report, or carrying out a calculation you will need to use SI units, so if you have a volume in cm3, you will need to convert it to m3 first. At the very least, you should have all your units the same (for example, you shouldn’t have a mixture of m3, dm3 and cm3 in the same equation). quantity SI unit Mass kg (kilogram) Length m (metre) Time s (second)102 trans:it scienceTemperature K (kelvin) Amount of substance mol (mole) Volume m3 (cubic metre) Energy or work J (joule) Force N (newton) Density kg m-3 (kilograms per cubic metre) Concentration mol m-3(moles per cubic metre)SI prefixes and multiplication factorsIt is not always sensible to use the appropriate SI unit, if the quantity you are dealing with is very large or very small it is convenient to use a prefix to denote multiples of the unit. Some of these prefixes may be familiar to you, others less so: multiplication factor prefix symbol 1 000 000 000 000 000 000 000 000 = 1024 yotta Y 1 000 000 000 000 000 000 000 = 1021 zetta Z 1 000 000 000 000 000 000 = 1018 exa E 1 000 000 000 000 000 = 1015 peta P 1 000 000 000 000 = 1012 tera T 1 000 000 000 = 109 giga G 1 000 000 = 106 mega M 1 000 = 103 kilo k 100 = 102 hecto h 10 = 101 deca da 0.1 = 10-1 deci d 0.01 = 10-2 centi c 0.001 = 10-3 milli m 0.000 001 = 10-6 micro  0.000 000 001 = 10-9 nano n 0.000 000 000 001 = 10-12 pico p 0.000 000 000 000 001 = 10-15 femto f 0.000 000 000 000 000 001 = 10-18 atto a 103 trans:it science0.000 000 000 000 000 000 001 = 10-21 zepto z 0.000 000 000 000 000 000 000 001 = 10-24 yocto y104 trans:it scienceSome useful conversions and equationsTo convert a temperature in °C into K, simply add 273.15. For example, 22°C in kelvin = 22°C + 273.15 = 295.15 K mass of substance (g) Number of moles (mol)  mass of one mole (g mol-1) mass (kg) Density (kg m-3 )  volume (m3 ) molar amount (mol) Concentration (mol m-3 )  volume (m3 ) molar amount of product Percentage yield (%)  100 maximum molar yield from limiting reagentDetermining the yield of a reaction…Example One:For the preparation of copper(II) sulfate from copper(II) carbonate:CuCO3 + H2SO4 + 5H2O  CuSO4.5H2O + H2O + CO2(g)3 We react 5 g copper carbonate (assume this to be anhydrous CuCO3) with 90 cm of -3 1 mol dm H2SO4, and get 9.2 g of the product, CuSO4.5H2O. What is the reaction yield?To start, we can calculate the relative molecular masses and hence the molar amounts of the reagents:Relative atomic masses:Cu 63.546C 12.011105 trans:it scienceO 15.9994Relative molecular mass of CuCO3 = 63.546 + 12.011 + (3  15.9994) = 123.5552-1 So the molar mass of CuCO3 is 123.56 g mol5.00 g Number of moles of CuCO 3   0.04 mol 123.56 g mol-1As the volume of sulfuric acid is given in cm3, we need to convert to dm3 by dividing by 1000 (as there are 1000 cm3 in 1 dm3).90 cm3 Number of moles of H SO  1 mol dm-3   1 mol dm-3  0.09 dm3 = 0.09 mol 2 4 1000Relative Reagent Amount used Molar amount molecular massCuCO3 123.55 5.00 g 0.04 mol 3 3 H2SO4 90 cm = 0.09 dm 0.09 molFrom the equation: CuCO3 + H2SO4 + 5H2O  CuSO4.5H2O + H2O + CO2(g)… we can see that one mole of CuCO3 reacts with one mole of H2SO4. We are reacting 0.04 moles of CuCO3 with 0.09 moles of H2SO4, so we can see that the acid is in excess and the CuCO3 is the limiting reagent. The maximum yield of the reaction is therefore equal to the number of moles of CuCO3, i.e. 0.04 mol.Now we need to calculate the relative molecular mass (Mr) and the number of moles of product (CuSO4.5H2O) we have obtained:106 trans:it scienceRelative atomic masses:Cu 63.546S 32.06O 15.9994H 1.008Mr CuSO4.5H2O = 63.546 + 32.06 + (9 × 15.9994) + (10 × 1.008) = 249.6806-1 So the molar mass of CuSO4.5H2O is 249.68 g mol9.2 g Number of moles of CuSO .5H O   0.037 mol 4 2 249.68 g mol-1The percentage yield of the reaction is therefore: actual yield of product 0.037 mol Percentage yield  100  100  92.5% theoretical yield of reagent 0.040 mol107 trans:it scienceExample Two: Try this one for yourself then check your answers on pages 11 -12 of this Appendix.You react 1.65 cm3 benzaldehyde with 1.0 cm3 pyrrole and obtain 0.51 g of the desired product, tetraphenylporphyrin:H O NH H N N 4 4 N HN benzaldehyde pyrrole tetraphenylporphyrinTo calculate the molar amounts, you would need to look up the densities and the molar masses of the reactants, but we have done that for you:Reagent Volume used Density at 20C Molar mass benzaldehyde 1.65 cm3 1.044 g cm-3 106.12 g mol-1 pyrrole 1.00 cm3 0.969 g cm-3 67.09 g mol-11a: Using the volume and the density, calculate the mass of benzaldehyde used. 1b: Using the mass and the molar mass, calculate the number of moles of benzaldehyde used.2: Repeat the process for pyrrole, to work out how many moles of pyrrole were used.Now looking at the product:108 trans:it scienceProduct Mass produced Molar mass tetraphenylporphyrin 0.51 g 614.74 g mol-13: Calculate how many moles of tetraphenylporphyrin were produced.From the equation for the reaction, we can see that four molar equivalents of benzaldehyde are reacted with four of pyrrole to give one mole of product. 4: Looking at the results of your calculations, work out which of the two reagents is the limiting reagent. This is the one that should be used to calculate the yield of the reaction.5: What is the maximum theoretical yield of product for this reaction?6: What is the percentage yield of tetraphenylporphyrin?(Now check your answers on pages 11-12 of this Appendix).Report writing. Make sure you carefully read what is required. . If things have not gone according to plan, don’t panic! Having a bad day in the lab will occasionally happen to everyone, even the best of us! If things have gone wrong, can you work out in which part of the experiment this happened? Often you will be given some credit if you can objectively offer plausible explanations as to why things did not go right. . When answering questions on the experiment, it is often helpful to use textbooks to provide some background information. . If you have to do a full experimental or project report, taking time to plan it first will help you to write a focused and well structured piece of work. 109 trans:it scienceThe full report (stages):Abstract: A short summary of what was done and the main results.For example: Mono- and bis-hydride ruthenium(II) complexes containing triphenylphosphine were prepared in 58% and 42% yields respectively, and were characterised using 1H and 31P NMR and infrared spectroscopy. Introduction: Introduce background theory to put the science of the experiment into context.Experimental: This is where you set out exactly what you did, the techniques used, the amounts of reagents etc.Results and Analysis: Clearly present all your experimental data in an appropriate format, such as tables and properly labelled figures.Discussion: Relate your experimental results to theory; explain the science behind your results using academic paper references and text books. Where possible, compare your results to similar results that have already been published. Where appropriate, comment on how reliable your data is.Conclusions: Clearly restate the significant conclusions you have drawn from your experimental data including qualitative (e.g. state or colour changes) and quantitative results, such as reaction yields and other calculated or measured values.110 trans:it scienceReferences: You need to clearly state what sources you have used for any particular pieces of information. Sources can include textbooks, journals and sometimes websites (but be very careful about the reliability of information on websites).Useful books to help with your Chemistry degreeUsually you can find out which textbooks are recommended before or when you start your degree, but if not you might find the following helpful:1. Keynotes in Organic Chemistry (Andrew F. Parsons, Blackwell Publishing). This book is very useful when revising organic chemistry, but also to use in conjunction with a larger organic textbook.2. Chemistry3: introducing inorganic, organic and physical chemistry. (Andrew Burrows, John Holman, Andrew Parsons, Gwen Pilling and Gareth Price, Oxford University Press). A text book covering the principles from the three main areas of Chemistry that you will cover within the first year of a degree course.3. Maths for chemists (Martin C.R. Cockett, Graham Doggett, Royal Society of Chemistry). Volume one covers numbers, functions and calculus whilst volume two covers power series, complex numbers and linear algebra.111 trans:it science4. Beginning Mathematics for Chemists. (Stephen K. Scott, Oxford University Press). A workbook designed to help with mathematical skills by providing practice through working through problems.5. Oxford University Press has published an extensive series of Chemistry Primers; these are short textbooks on a wide range of chemistry areas such as aromatic chemistry, chemical bonding, and first-row transition metal chemistry. Details available: http://ukcatalogue.oup.com/nav/p/category/academic/series/chemistry/ocp.do ?sortby=bookTitleAscend&thumbby_crawl=10&thumbby=50 Useful equations (You will probably have come across some of these before) Gradient of a straight line: y = mx + c y = value on the y axis x = value on the x axis m = gradient c = the intercept on the y axis Ideal gas equation: pV = nRT p = pressure in Pa V = volume in dm3 n = molar amount in mol R = gas constant = 8.314 J K-1 mol-1 T = temperature in K Calculating the enthalpy change of a reaction: ΔH = Hproducts – Hreactants Calculating the entropy change of a reaction: ΔS = Sproducts – Sreactants112 trans:it science Relationship between free energy change, enthalpy change, entropy change and temperature: ΔH = ΔG -TΔSΔH = change in enthalpy ΔG = change in free energy T = temperature ΔS = change in entropy+  Relationship between pH and proton concentration: pH = -log10[H ][H+] = concentration of hydrogen ions (protons) Calculation of the acid dissociation constant, Ka, for a particular acid:[A- ][H ] K  a [HA]HA H+ + A- acid proton conjugate baseFor example:+ - CH3CO2H H + CH3CO2 acid proton conjugate base[A-] = the concentration of the conjugate base [H+] = proton concentration [HA] = concentration of acid  Relationship between pKa and Ka: pKa = -log10Ka113 trans:it science [A- ]     Relationship between pH and pKa: pH  pKa  log10    [HA] The laws of thermodynamics: 1. Energy can neither be created nor destroyed, only interconverted between forms. 2. The entropy within a closed system increases i.e. becomes more disordered. 3. Zero entropy can only be achieved in a perfect crystal at absolute zero i.e. when the temperature is 0 K.Answers to Example 2Determining the yield of a reactionExample Two: The answersYou react 1.65 cm3 benzaldehyde with 1.0 cm3 pyrrole and obtain 0.51 g of the desired product, tetraphenylporphyrin:H O NH H N N 4 4 N HN benzaldehyde pyrrole tetraphenylporphyrinTo calculate the molar amounts, you would need to look up the densities and the molar masses of the reactants, but we have done that for you:Reagent Volume used Density at 20C Molar mass benzaldehyde 1.65 cm3 1.044 g cm-3 106.12 g mol-1 pyrrole 1.00 cm3 0.969 g cm-3 67.09 g mol-1114 trans:it science1a: Using the volume and the density, calculate the mass of benzaldehyde used. Mass of benzaldehyde used = 1.65 cm3  1.044 g cm-3 = 1.72 gNote that the answer is given to 2 decimal places, in line with the mass that was measured. We should not imply a greater level of accuracy than this.1b: Using the mass and the molar mass, calculate the number of moles of benzaldehyde used.1.72 g Number of moles benzaldehyde used = = 0.016 mol 106.12 g mol-1Here the answer is given to 2 significant figures.115 trans:it science2: Repeat the process for pyrrole, to work out how many moles of pyrrole were used.Mass of pyrrole used = 1.00 cm3  0.969 g cm-3 = 0.97 g0.97 g Number of moles of pyrrole used = = 0.014 mol 67.09 g mol-1Now looking at the product:Product Mass produced Molar mass tetraphenylporphyrin 0.51 g 614.74 g mol-13: Calculate how many moles of tetraphenylporphyrin were produced.0.51 g Number of moles of product obtained = = 0.00083 mol 614.74 g mol-1Answer given to 2 significant figures.From the equation for the reaction, we can see that four molar equivalents of benzaldehyde are reacted with four of pyrrole to give one mole of product. 4: Looking at the results of your calculations, work out which of the two reagents is the limiting reagent. This is the one that should be used to calculate the yield of the reaction.From the molar calculations, we can see that the benzaldehyde (at 0.016 mol) is in slight excess compared to the pyrrole (0.014 mol), so pyrrole is the limiting reagent for the reaction and therefore will be used to calculate the yield.116 trans:it science5: What is the maximum theoretical yield of product for this reaction? no. of moles pyrrole used Maximum theoretical yield of product = 40.014 = = 0.0035 mol 46: What is the percentage yield of tetraphenylporphyrin? actual yield of product 0.00083 mol Percentage yield  100  100  23.7% theoretical yield of product 0.0035 molCREATIVE COMMONS LICENSING: You are free:  to Share — to copy, distribute and transmit the work  to Remix — to adapt the work Under the following conditions:  Attribution — You must attribute the work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work).  Noncommercial — You may not use this work for commercial purposes.  Share Alike — If you alter, transform, or build upon this work, you may distribute the resulting work only under the same or similar license to this one. With the understanding that:  Waiver — Any of the above conditions can be waived if you get permission from the copyright holder.  Public Domain — Where the work or any of its elements is in the public domain under applicable law, that status is in no way affected by the license.  Other Rights — In no way are any of the following rights affected by the license: o Your fair dealing or fair use rights, or other applicable copyright exceptions and limitations; o The author's moral rights; o Rights other persons may have either in the work itself or in how the work is used, such as publicity or privacy rights. 117 Notice — For any reuse or distribution, you must make clear to others the license terms of this work

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