FC 26: Practical Chemistry Syllabus: 1.1.1 (a) experimental design, including to solve problems set in a practical context; (b) identification of variables that must be controlled, where appropriate; (c) evaluation that an experimental method is appropriate to meet the expected outcomes 1.1.2 (a) how to use a wide range of practical apparatus and techniques correctly; (b) appropriate units for measurements ; (c) presenting observations and data in an appropriate format 1.1.3 (a) processing, analysing and interpreting qualitative and quantitative experimental results; (b) use of appropriate mathematical skills for analysis of quantitative data; (c) appropriate use of significant figures; (d) plotting and interpreting suitable graphs from experimental results, including: (i) selection and labelling of axes with appropriate scales, quantities and units; (ii) measuring gradients. 1.1.4 (a) how to evaluate results and draw conclusions; (b) identification of anomalies in experimental measurements; (c) the limitations in experimental procedures; (d) precision and accuracy of measurements and data, including margins of error, percentage errors and uncertainties in apparatus; (e) refining experimental design by suggestion of improvements to the procedures and apparatus 4.2.3 (a) the techniques and procedures for: (i) use of Quickfit apparatus including for distillation and heating under reflux; (b) preparation and purification of an organic liquid including: use of a separating funnel to remove an organic layer from an aqueous layer; drying with an anhydrous salt (e.g. MgSO4, CaCl2); redistillation. Titrations A standard solution is a solution whose concentration is accurately known. You can prepare a standard solution from a pure solid or liquid compound thus: (i) Calculate the mass of solute needed to make the required volume (eg 250 cm3) of solution. (ii) Weigh this amount accurately using an electronic balance. (iii) Dissolve this material in a limited amount of distilledwater, in a beaker. (iv) Transfer this solution to a volumetric flask. Wash any remaining solution from the beaker into the flask with distilled water. Then make the volume up to the mark (eg 250 cm3) by adding more distilled water. (v) Invert the flask several times to mix the solution thoroughly. In most (but not all) titrations, you will have one standard solution and one of unknown concentration. Either one may be the solution in the burette so read the question carefully. Titration method: An accurate volume of one solution is measured using a pipette and placed in a conical flask (with an indicator, if necessary). The other solution is added from a burette until the reaction is complete. The required volume delivered from the burette is called the titre. Volume measurements using this equipment involve measuring the level of a solution, which has a curved top surface (meniscus) because of attractive forces between the solution and the glass. Correct use of the equipment means the meniscus will be on top of the line: The burette is read to the nearest 0.05 cm3 before the titration and again after reaching the end point. Normally a trial titration is done first to get an estimated end-point. You should not include the volume reading from a trial when calculating a mean titre. Then several accurate titrations are carried out until they are in close agreement (ideally within 0.1 cm3). Calculating a mean titre: omit as “outliers” any readings that are more distantly separated from the others: e.g. 20.10 20.15 20.20 20.10: Discard the 20.15 and 20.20 readings and use the two identical 20.10 readings: mean titre = 20.10 cm3 20.20 20.10 20.40: Discard the 20.40 reading because it is further away (0.2) than the other two, which just differ by 0.1: mean = (20.20 + 20.10)/2 = 20.15 cm3. 23.30 23.20 23.40: Readings are equally spaced so keep them all: mean titre = (23.30 + 23.20 + 23.40) / 3 = 23.30 cm3. Remember that in most titrations, you make a big volume of solution of one of your reagents (eg 250 cm3) but you use just a fraction of this in the titration. So your calculating will usually involve scaling up from moles in titration to moles in total volume of solution. e.g. if you used 25.0 cm3 of your solution, from a total of 250 cm3, you can simply say the volume is 10x bigger , so multiply moles x10. However it’s not always so simple: eg if you used 21.7 cm3 of your solution from a total of 250 cm3, you will have to multiply moles x (250/21.7). Percentage Error Each piece of equipment has some limitation on how accurately it can measure a quantity, such as a mass or volume. If the limit of uncertainty is known it can be used to calculate percentage error on individual measurements, according to: % error = (maximum error / value measured) x 100 e.g.: A balance has an uncertainty (maximum error) on each mass reading of ±0.005g. If this balance is used to measure out 2.50g of a substance, the % error is: (0.005 / 2.50) x 100 = 0.2% Take care when a number you are using requires two measurements because then you have to add up the uncertainties from each measurement. For example, this happens: (i) In a titration, each burette reading may have an error of ±0.10 cm3. But each titre requires two burette readings (one at the beginning and one at the end). So the maximum error has to be doubled. Hence on a titre of 22.70 cm3, the % error is: ( (2 x 0.10) / 22.70 ) x 100 = 0.88% error (ii) When measuring a difference in mass (eg in an experiment to drive off water of crystallization) or a difference in temperature (in an experiment to measure ΔH), you have to make 2 readings (before and after), so you have to double the uncertainty on each individual reading when finding the % error on the difference. % errors are important because they tell you how confident you can be about conclusions from an experiment. Also, if a calculation requires several numbers, you should look at the % error on each. The one with the biggest error is the one for which you should try to find an improved (less uncertain) measurement method. Significant Figures Any calculation you carry out will give an answer with a degree of precision that depends on the precision of the numbers that were fed into the calculation. This should be reflected in the number of significant figures to which the answer is quoted. Thus, in a titration there are typically lots of numbers involved, e.g: Mass of solid used to make solution A = 12.45g 4 sig figs Volume of solution A made up in volumetric flask = 250 cm3 3 sig figs Volume of solution A used in titration (pipette) = 25.0 cm3 3 sig figs Volume of solution B used in titration (burette) = 27.85 cm3 4 sig figs Some are known to 3 sig figs, some to 4 sig figs. If all of these numbers are now used in the calculation, the answer cannot be more precise than any of the individual numbers used, therefore the answer should be given to 3 sig figs. In general: if you are asked to round an answer to “an appropriate number of significant of significant figures”, this should be the same number of sig figs as the least precisely known number that is used in the calculation. If a question does not have any instruction about rounding it is generally OK to round to 3 sig figs or more at the end, but do not round significantly at intermediate stages in your calculation: keep the numbers unrounded in your calculator if you can or else leave in lots of sig figs at intermediate stages. Thermal Decomposition (eg Water of Crystallisation experiments) If a hydrated compound is heated, it will lose its water of crystallization (or, occasionally, just some of it). From the change in mass when this happens, the formula of the hydrated compound can be deduced. The apparatus shown can be used for such an experiment In a thermal decomposition experiment like this, it is necessary to be sure when all of the decomposition has occurred (eg when all the water of crystallization has been driven off). The method used is called “heating to constant mass”: • The sample is heated, cooled and then weighed. Then this cycle is repeated until no further change in the mass occurs. This indicates that the reaction is complete. Practical Organic Chemistry Reflux is the method used to allow prolonged heating without losing material through evaporation: • Anything that vaporizes condensers in the vertical condenser and drips back into the round-bottomed flask. • The cold water running through the condenser goes in at the bottom and out at the top – otherwise it won’t fill the water jacket completely and the cooling will be less efficient. • Direct heating with a Bunsen may be used but the advantage of a water bath is better control: heating at a constant temperature can be achieved. Purification of Organic Liquids (i) Organic liquids which are insoluble in water can be separated from aqueous reagents by using a separating funnel: A separating can also be used to allow a water-insoluble organic liquid to be neutralized and/or washed with water: • aqueous alkali (eg sodium carbonate) will neutralize any remaining acid reagent (lots of organic reactions include acidic reagents or catalysts) • distilled water will remove other water-soluble contaminants. To do this: Add alkali - Stopper the funnel and shake it to mix up the layers - Invert the funnel and open the tap to release any gas formed - close the tap, turn funnel right way up again and allow layers to separate - remove stopper and run the aqueous layer away, to leave the neutralized organic layer.