LABORATORY CALCULATIONS. You are expected to be competent at the following and should tick off in the table when you feel competent:

Transcription

1 LABORATORY CALCULATIONS These exercises are designed to give you practice in the basic mathematical concepts required for successful completion of the course. The maths is generally not difficult but the application of the techniques to the chemical world may challenge even the best mathematician. Lots of practice will ensure you are competent and therefore able to successfully complete the assessments. You are expected to be competent at the following and should tick off in the table when you feel competent: Concept Week Week Week Yes No Yes No Yes No 1. Use of your calculator 2. Estimation of result 3. Rounding 4. Order of operations 5. Scientific notation 6. Directed numbers 7. Significant numbers 8. Substitution into formula 9. Rearranging formula 10. Unit conversions 11. Graphing 12. Formula weights 13. Mole calculations 14. Concentration calculations 15. Stoichiometry 16. Putting it all together The first assessment covers concepts The second assessment covers concepts

2 Use of the calculator All calculators operate slightly differently but all should give the same answer for a particular calculation. If the answer is not correct then it is most likely that the operator, you, has incorrectly entered data or not followed appropriate mathematical logic. You should know how the following functions operate on your calculator square root log 10 (normally called log or log10) 10 x (often on the same button as log10) memory (store, add, recall) EXP for scientific notation The best way to work our how your calculator operates is to read the manual but they are very thick and this will most likely not occur. The next best way is to try some calculations that you know the expected answer. Exercise 6 +2 x 3 = (12) 5-4 x 3 = (-7) 16 x x 3 = (70) 2

3 Estimation of result The calculator result is only as good as the operator ie if you enter an incorrect number or do not use correct mathematical logic the answer will be wrong. You should have an idea of the expected answer, not just trust the calculator. Basic process for estimation 1. Round all numbers to one non-zero digit 2. Calculate the values inside any brackets and round the results. 3. Cancel out any zeroes where possible. 4. Calculate estimate. Example Estimate the answer to the calculation from Example 1.2. ( ) x x10 1. Round all the numbers ( 60 2)x x10 2. Calculate inside the brackets and round result. 60 x x10 3. Remove spare zeroes. 6x0. 5 6x1 4. Calculate estimate. 0.5 (the exact answer was 0.541). Now try these (i) x = (ii) 24,300 x x 298 = ( ) x iii) =

4 Rounding To round off to a stated accuracy the last figure to be retained is corrected depending on the digit which follows. A typical rounding system states that if the following digit is 5, 6, 7, 8 or 9 then the last figure is taken up one. Example Round off correct to the nearest whole number Round off correct to two decimals places Round off correct to the nearest thousand Exercise 1. Round off to the nearest whole number Round off correct to the nearest figure as shown in brackets) (2 d.p) (3 d.p) (4 d.p) (2 d.p) (tens) (hundreds) (hundreds) (ten thousands) (millions)

5 Order of Operations The rules for order of operations are: 1. Brackets first 2. and from left to right second 3. + and from left to right third Exercise 1. 5 x 4-1 = = x 2 = 4. 2 x (3 + 6) = 5. 6 x 4 2 = = 7. (2 + 3 x 6) + 1 = 8. 4 x 2 + (3 + 1) = = 10. [(3 + 2) + 4 x 2] + 2 x 2 = x x (4x2) 56 7 = = = = = = x8 = = (20 3 x 6) 2 = x3 2 = 5

6 Scientific Notation Scientific notation is a method for writing very large and very small numbers in the form of a number between 1 and 10 and a power of 10. Example Avagadro s Number, the number of entities in a mole is a number that the mind has difficulty understanding Written in Scientifc notation the number is 6.02 x If the index number is positive it indicates that the number is large and the decimal point has been moved right to left If the index number is negative it indicates the original number is small (less than 1) and the decimal point has been moved left to right Exercises Write in scientific notation Write the original number for x x x x x

7 Directed numbers To multiply or divide Two positive numbers give a positive number Two negative humbers give a positive number A positive and a negarive number gives a negative number (-2) = 2. (-3) + 3 = 3. (-5) x 6 = = x -3 = 6. Given a = -1, b= 2, c = -4 d = -3 Evaluate a b c = d 2 = a 3 = a + b + c + d = 2b c = ab cd = -(-d) = 7

8 Significant Figures Significant figures are defined as follows: 1. All non-zero digits are significant 2. Zeros at the end of a whole number or at the beginning of a decimal are not significant 3. Zeroes between non-zero digits are significant 4. Zeroes at the end of a decimal are significant Examples has 3 significant figures has 3 significant figures has 2 significant figures has 3 significant figures Exercise 1. Give the number of significant figures in each of these numbers Rewrite each of these numbers i) correct to 2 sig fig ii) correct to 1 sig fig

9 Substitution into a formula Often the calculations you will have to do use provided formulae which are supposed to make the process of getting the right answer easier. A formula is simply a calculation where the numbers are replaced by symbols, such as the formula for moisture content in Example 1.1. EXAMPLE You are analysing soil for its moisture (water) content. You weigh out soil into a basin weighing g, until the balance records a weight of g. After 2 hours of drying, the basin and dried soil weighs g. Moisture content(%) m m w w m m d b x100 where m b is the mass of the empty basin = m d the mass of the basin with dried soil = m w the mass of the basin with wet soil = Moisture content(%) x % Other formulae often used in the laboratory include: C 1 V 1 = C 2 V 2 the dilution equation C 1 is the concentration of the first (more concentrated) solution V 1 is the volume of the first (more concentrated) solution C 2 is the concentration of the second (diluted) solution V 2 is the volume of the second (diluted) solution C is the concentration is g/l, m the mass in grams V the volume in L. m C a concentration equation V Exercise Substitute the data into the concentration equation to determine the value for C 1. m = 15g, V = 2L 2. m = 0.5g, V = 0.05L 3. m = 525g, V = 200L 9

10 Rearranging formula to solve equations If the unknown in an equation is not the subject of the formula than the formula must be rearranged in order to obtain an answer. The process is simple as long as the rules are followed: Equations may be simplified and solved by: 1. adding the same number to both sides 2. subtracting the same number from both sides 3. multiplying both sides by the same number 4. dividing both sides by the same number Example x + 8 = 4 x = 8 8 = 4 8 x = -4 (to remove the 8 we need to subtract 8, but from both sides) 6 x d = 42 (to remove the 6 we need to divide both sides by 6) 6 x d = d = 7 Exercise Solve the following equations 1. c + 18 = t 8 = a = t = 4 5. y 2.8 =

11 The same process can be used when substituting into formula Example 1 m C rearrange to make m the subject of the formula V In order to remove the V we need to multiply each side by V C x V = m x V V C x V = m Example 2 C 1 V 1 = C 2 V 2 rearrange to make V 1 the subject of the formula In order to remove C 1 from the left hand side we need to divide both sides by C 1 C 1 V 1 = C 2 V 2 C 1 C 1 then V 1 = C 2 V 2 C 1 Exercise Rearranging the following formula to make the symbol in brackets the subject. (a) m C (V) V (b) C 1 V 1 = C 2 V 2 (C 2 ) (c) T c = 0.556T f 17.8 (T f ) (d) m C (m) V 11

12 General Revision of Concepts Given the following data and formulae, calculate the values of the unknown, reporting the result to an appropriate number of significant figures Formula Data (a) C 1 V 1 = C 2 V 2 C 1 = 1.2 g/l, V 1 = 5 ml, V 2 = 250 ml (b) PV = nrt P = x 10 5, V = 0.234, R = 8.314, T = (c) T c = (T f 32) T f = (d) C 8000 (b 2a) V b = 65.2, a = 21.3, V = 10.0 (e) C 1 V 1 = C 2 V 2 C 1 = 1000 mg/l, C 2 = 50 mg/l, V 2 = 250 ml 12

13 Unit conversions There are many different nits that could be used to state a measurement. For example, if your height is 1.8 m, you could also say you are 180 cm, or 1800 mm, or 1.8 x 10 6 um or km. And then there are all the non-metric measures for length: 5.91 feet, 70.9 inches, 1.12 x 10-3 miles etc. Which is correct? They all are! It is important to be able to convert units to provide numbers that are reasonable. The process for linear, area and volume conversions is similar, You should practice the method regularly to ensure that for even the most difficult conversions the answer is correct. The metric prefix system (not complete) Prefix Abbreviation How Many Base Units How Many in One Base Unit nano n micro milli m centi c deci d kilo k mega M The process will work for all conversions..you just need to practice Example 1. How many metres are in 24.7 km 24.7 km = 24.7 km x 1000 m (km will cancel out) km thus 24.7 km = 24.7 x 1000 m 2. How many grams in 356 mg 356 mg = 356 mg x 1 g. (mg will cancel out) 1000 mg 356 mg = g The same system will work for imperial conversions as well 13

14 Perform the following conversions (a) 57 metres to centimetres (b) 20 metres to millimetres (c) ug to g (d) 1.24 hectares to m 2 (e) 72 cm 3 to m 3 (f) 23 m to feet, given 1 m = 3.28 ft. (g) 2682 acres to hectares, given 1 acre = ha (h) 5983 Pa to psi, given 1 psi = 6891 Pa (i) 1.24 pounds to grams, given that 1 pound = 454 g Some more to try (a) 457 nm to cm. (b) khz to MHz. (c) 25 mg to kg (d) 2.54 x 10 4 inches to miles, given 36 inches = 1 yard and 1760 yards = 1 mile. And some more difficult ones to try Convert: (a) 6932 ug/l to g/ml. (b) mg/g to mg/kg. (c) 181 ug/ml to g/l. 14

15 And these also Questions 1 foot (ft) = metre (m) 1 pound (1b) = kilogram (kg) 1 gallon (gal) = 4.55 litres (L) 1 acre (ac) = 4047 square metres (m 2 ) 1 foot (ft) = 12 inches (in) 1 inch = 2.54 cm nanometres to metres ,500 grams to kilograms MJ to kj mm 2 to m dm 3 to cm m 3 to ML x 10 3 L to kl mm 3 to cm g/l to mg/ml g/ml to kg/m ,740 kj/kg to J/g feet to metres lb/ft 3 to kg/m m 2 to acres millilitres per minute cubic centimetres per hour square inches to square centimetres litres per second to gallons per minute. 15

16 Graphing Graphs are pictorial versions of numerical data. They are intended to tell the story in the data in a visual way. The can also allow certain types of calculations to be done more easily. Types of graphs There are many different types of graphs, each with a particular purpose. Some can be used for all types of data, others will only suit particular types of data sets. When you study spreadsheets in computing, you will learn how to construct graphs using a computer program. The main graph types are: column bar pie line X-Y scatter Examples of each are shown in Figure 4.1. In this chapter, we will concentrate on X- Y scatter graphs. Column Bar Line Pie X-Y scatter Graph types You may think that line and X-Y scatter graphs are the same - they look similar, but there is one important difference. In column, bar and line graphs, each data point is equally spread from the next one, whereas in X-Y scatter, they are separated by differences in the X-value. Figure 4.2 shows this more clearly. 16

17 Reading Reading Data X Y points spread across horizontal axis according to X-values points evenly spread across horizontal axis X-Y Scatter Line FIGURE 4.2 The difference between line and X-Y scatter graphs X-Y scatter graphs are used where the X-value is as important as the Y-value. This is particular the case in so-called calibration graphs, where an instrument response is checked against solutions of known concentrations. In line graphs, the X-value is only important as an identifier, eg months of the year. 4.2 How to draw graphs The are a number of points to consider when drawing graphs, so that they are correct, informative and not misleading: which axis is which - the y- (usually vertical) axis plots the measurement while the x- (usually horizontal) axis plots the set value (the solution concentration) - this does not apply to pie charts; for bar charts, the y-axis is the horizontal axis labelling - axes should be labelled the name and units of the data variable labelled and sensible axis scale - where an axis scale is important (the y-axis in all cases and the ax-axis in X-Y graphs), the scale should be clearly labelled and sensible - make sure you fill up the sheet of graph paper as bets as possible, but don t choose a scale where 1 cm = 3.45 units just to fill the page! consistent scale - the scale (what each cm, for example, is worth) should be the same across the whole in most circumstances (see Figure 4.3) (a) Calibration Graph (b) Calibration Graph Concentration Concentration FIGURE 4.3 (a) Correct scale (b) incorrect scale plain paper vs graph paper if you have a computer, then you can produce professional-looking graphs; however, the grid on graph paper is useful where you need to make a measurement from the graph computer vs hand-drawn draw the graph yourself on graph paper if you have to make a measurement from the graph, eg a calibration graph 17

18 Response 4.3 Calibration graphs As mentioned above, these are X-Y scatter graphs, used to calibrate an instrument measuring its response to standards (solutions of known concentration) for the purpose of using the graph to determine the concentration of samples. Normally, a calibration graph will be a straight line, but the data you have from the standards will probably not be perfect. A line of best fit is a straight line drawn through the data points being as close to each as possible. A computer program, such as Excel, will not only graph your data, but also work out the correct line of best fit. A calibration graph does not: join the dots, plot samples they have an unknown concentration EXAMPLE 4.1 Plot a calibration graph from the data below. Conc. Respons e Graph on next page. CLASS EXERCISE 4.2 Draw a calibration graph from the data below on the paper provided. Concentration Response (mg/l) Graph for Example Conc. 18

19 Response How to find the concentration of a sample from the calibration graph If you are using graph paper, then you draw a horizontal line across from the vertical axis at the response value for the sample. When this hits the calibration graph, drop a vertical line to the concentration axis. Where the line hits the axis is the sample concentration. EXAMPLE 4.2 Using the calibration graph from Example 4.1, determine the concentration of a sample with a response of conc. = 2.5 Conc. CLASS EXERCISE 4.3 Determine the concentration of the sample with an instrument response of 0.44, using your graph from Exercise

20 Response You can see in Example 4.2 that the plain paper and computer combination is not ideal for measuring the sample concentration. If you are using a computer then it is better to use an equation, which describes the straight line (Equation 4.1). response y int ercept concentration Eqn 4.1 slope where the slope is the rise run of the straight line, and the y-intercept is where the line hits the vertical axis (when the concentration is zero). The computer program will calculate the slope and y-intercept for you, and you will b shown how to do this in the computing subject where you cover spreadsheets. However, we will use the equation here, given the necessary data. EXAMPLE 4.3 For the graph in Example 4.1, the slope is 38.6 and the y-intercept is 1.0. What is the concentration for the sample with a response of 100. conc. 100 ( 1) CLASS EXERCISE 4.4 Calculate the sample concentration for the following: Slope y-intercept Concentration Exercise 4 page Exercise 5 page Calculating the slope of a line from graph paper The slope is equal to the vertical rise divided by the horizontal run. This is not the rise and run as measured by your ruler, but from the measured values from the scale. EXAMPLE 4.4 Calculate the slope of the graph in Example 4.1 using rise/run rise = = run = 4-2 = Conc. Slope rise run CLASS EXERCISE

21 Calculate the slope of your graph in Exercise 4.2. Questions 1. Determine the sample concentrations for the following. (a) Standard Conc. Response (mg/l) Sample 0.36 (b) Standard Conc. % Response Sample (c) Std Conc. (mole/l) Response x x x x Sample 0.40 (d) Standard Conc. Response mg/l) Sample

22 1. (e) Standard Conc. (%) Response Sample Calculate the sample concentration, given the following line-of-best fit data. Slope y-intercept Sample Reading (a) (b) (c) Calculate the slope of the your graphs in Q1. 22

23 Formula weights The formula weight of a pure chemical substance is the sum of the atomic weights of the elements that make up the substance. The Table gives you the atomic weights of common elements: if you need the weight of a less common element, use a periodic table. Atomic weights of selected elements Element At. Weight Element At. Weight Element At. Weight Hydrogen Aluminium Iron Carbon Phosphorous Copper Nitrogen Sulfur Bromine Oxygen Chlorine Iodine Sodium Potassium Barium Magnesium Calcium Lead What is the formula weight of copper nitrate Cu(NO 3 ) 2? You have to include all atoms, not just one of each element, so in this substance there are: 1 copper, 2 nitrogen, and 6 oxygen atoms FW = (1 x 63.54) + (2 x 14.01) + (6 x 16.00) = au Exercise Calculate the formula weights of the following: (a) sodium phosphate, Na 3 PO 4 (b) glucose, C 6 H 12 O 6 (c) magnesium nitrate, Mg(NO 3 ) 2 (d) ammonium iron (II) sulfate, (NH 4 ) 2 Fe(SO 4 ) 2 23

24 The Mole The mole is an important concept in Chemistry. It is the Unit used by chemists and engineers to determine the quantities involved in reactions. There are a number of relationships that would have been explained in your Chemistry subject. These include: Avagadro s Number Mass Formula weight (or formula mass or molar mass or molecular mass) Concentration Volume of gas Complete the mole map below : No of Entities Mass (g) Moles Vol of Gas (L) Concentration (mol/l) 24

25 THE MASS OF ONE MOLE OF A SUBSTANCE IS EQUAL TO ITS FORMULA WEIGHT. moles mass (g) FW Exercise Calculate the number of moles in the following (the FWs were calculated previously (a) g of sodium phosphate (b) g of glucose (c) g of magnesium nitrate (d) 1.81 g of ammonium iron (II) sulfate Calculate the mass of the following: (e) moles of sodium phosphate (f) 5.09 x 10-4 moles of glucose (g) 2.5 moles of magnesium nitrate (h) 7.84 x 10-5 moles of ammonium iron (II) sulfate 25

26 Some more to try 1. What is the formula weight of: (a) NaNO 3 (b) Al 2 (SO 4 ) 3 (c) CuSO 4. 5H 2 O 2. How many moles are in the following? (a) 0.25 g of NaNO 3 (b) 34.2 mg of CaCO 3 (c) 21.2 ml of M HCl (d) 250 ml of 2.5 M H 2 SO 4 26

27 Concentration calculations The concentration of something is how much of that something there is in a given amount of material. To be slightly more technical, we call the something we are interested in the analyte, and the material containing it the sample. Concentration is always a ratio of the analyte to the sample, as in Equation 3.2: amount of analyte concentrat ion Eqn 3.2 amount of sample The units for the amounts will vary depending on the type of sample (solid, liquid or gas) and the situation of the analysis. In general, mass of analyte conc. in solid sample Eqn 3.3 mass of sample moles OR mass of analyte conc. in liquid sample (solution) Eqn 3.4 volume of sample Solutions A solution is a mixture of a solute (the dissolved substance) in a liquid (normally, but not always), known as the solvent. The concentration is the amount of the solute per unit of volume of the solution, or the ratio of amount to volume. Common solution concentration units Concentration unit Amount of solute Volume of solution molarity (mole/l) moles litres g/l mass grams litres g/100 ml (%w/v) mass grams 100 ml mg/l (ppm) mass milligrams litres ug/l (ppb) mass micrograms litres ml/100 ml (%v/v) volume - ml 100 ml EXAMPLE 1. Calculate the concentration of a solution (in g/100 ml) if 1.45 g is dissolved in 250 ml of solution. Amount unit is g: 1.45 g Volume unit is 100 ml: = 2.5 (100 ml) Concentration = amount volume = = 0.58 g/100 ml 2. What mass of solute is required to prepare 300 ml of 250 mg/l? Volume unit is L: 300 ml = 0.3 L Amount unit will be mg. Amount = concentration x volume = 250 x 0.3 = 75 mg 27

28 3. What volume of solution is required to make a 15 ml/100 ml (%v/v) solution from 25 ml of solute? Amount unit is ml: 25 Volume unit will be 100 ml lots. Volume = amount concentration = = 1.67 x 100 ml = 167 ml. CLASS EXERCISE Calculate the solute concentration in the following solutions in the specified unit: Amount Volume Unit Concentration (a) 3.25 g 75 ml g/l (b) 27.3 mg 500 ml mg/l (c) g 21.4 ml g/100 ml (d) 6.9 ml 250 ml ml/100 ml 2. What amount of solute is required to prepare the following solutions: (a) 1.25 L of 0.46 g/l (b) 50 ml of 4.56 g/100 ml (c) 250 ml of 0.1 mole/l (d) 1500 ml of 1 ml/100 ml 3. What volume of solution (to the nearest multiple of 10) is required to make the following solutions from the given solute amount? Amount Concentration Volume (a) 350 mg 200 mg/l (b) 45 g 20 g/100 ml (c) 0.55 g 1.0 g/l (d) 25 ml 2 ml/100 ml 28

29 Solids The concentration of a chemical in a solid is just as important as that in a liquid. However, you won t normally be making up a solid mixture of a known concentration, but rather trying to work out the concentration in a solid sample from an analysis, eg the moisture content in soil, the lead content in fallout. The basic format for concentration in a solid is the same, but this time the amount of sample is a mass, not a volume. Concentrat ion mass of analyte mass of sample The two common concentration units for solids are: g/100 g (also known as %w/w) mg/kg (also known as ppm, used for low concentrations) CLASS EXERCISE 3.6 Complete the following table. Concentration unit Mass unit of analyte Mass unit of sample g/100 g mg/kg EXAMPLE Calculate the concentration if there is g of analyte in g of solid sample in g/100 g. Mass of analyte should be in g: Mass of sample should be in 100 g lots: = Concentrat ion What is the concentration of lead in fallout if a g sample was found to have 1.45 ug of lead? Mass of analyte should be in mg: 1.45 ug = mg Mass of sample should be in kg lots: = Concentrat ion How much water is in a 50 g sample of soil which has a moisture concentration of 8.75 g/100 g? This time, the unknown is the mass of analyte, not the concentration. Same equation, but a bit re-arranging. Mass of sample should be in 100 g lots: = 0.5 mass of analyte Mass of analyte = 8.75 x 0.5 = g 29

30 4. How much soil should be weighed out if it has an approximate lead concentration of 1 mg/kg, and the analysis requires 0.1 mg lead? This time, the unknown is the mass of sample. Mass of analyte should be in mg: mass of sample Mass of sample = = 0.02 kg = 20 g. CLASS EXERCISE Calculate the analyte concentration in the following solid samples in the specified unit: Analyte mass Sample mass Unit Concentration (a) g g g/100 g (b) 14.6 mg g mg/kg (c) 456 mg g g/100 g (d) 181 ug g mg/kg 2. What mass of analyte is contained in the following: (a) g of 123 mg/kg (b) 500 g of 1.05 g/100 g (c) g of 222 mg/kg (d) 75 g of 1%w/w 3. What mass of sample is required to contain: (a) 10 mg from 50 mg/kg sample (b) 1 g from 12.5 g/100 g 3. What mass of sample is required to contain: (c) 500 ug from 200 mg/kg (d) 0.1 g from 5% w/w 30

31 3.4 Dilution of solutions When a solution has more solvent added to it, it is said to be diluted. This means that the concentration has decreased, because the amount of solute the dissolved substance does not change, but the volume increases, as shown in Figures 3.1 and 3.2. add 100 ml extra solvent 1 g 1 g ml) 1 g/100 ml 1 g/200 ml (or 0.5 g/100 FIGURE 3.1 Dilution of a solution volume concentratio nn amount FIGURE 3.2 Summary of the effect of dilution on solution measures If the dilution is done accurately ie the volumes of the first (concentrated) solution and the volume of the second (diluted) solution are volumetric (pipette, burette, vol. flask) then the relationship between the concentrations of the two solutions is simple, as shown in Equation 3.1. C 1 V 1 = C 2 V 2 Eqn 3.5 where C 1 and C 2 are the concentrations of the two solutions, and V 1 and V 2 the volumes. The units for the concentrations must be the same, and likewise for the volumes. EXAMPLE 3.5 (a) What is the concentration of a solution made by diluting 5 ml of 0.1 mole/l to 250 ml? 0.1 x 5 = C2 x 250 C 1 = 0.1 M V 1 = 5 ml C 2 =? V 2 = 250 ml C 2 = mole/l (we know the unit for C 2 is mole/l because that is the unit for C 1 ). 31

32 (b) What volume of a 1000 mg/l solution is needed to make up 1 L of 50 mg/l. C 1 = 1000 mg/l V 1 =? C 2 = 50 mg/l V 2 = 1 L 1000 x V 1 = 50 x 1 V 1 = 0.05 L (or 50 ml). CLASS EXERCISE What is the concentration of the solution if: (a) 25 ml of 2.5 g/100 ml is diluted to 100 ml (b) 5 ml of 400 mg/l is diluted to 200 ml (c) 15 ml of 2.5 mole/l is diluted to 2.5 L (d) 50 ml of g/l is evaporated down to 10 ml 2 (a) How much 0.05 M solution is required to make 200 ml of 0.01 M? (b) How much 2000 mg/l solution is required to make 500 ml of 100 mg/l? (c) How much 25 %v/v solution is required to make 250 ml of 5 %v/v? (d) How much 50 g/l solution is required to make 20 L of 100 mg/l? 32

33 Dilution factors The dilution factor is a common term, which describes the ratio of the volumes of the diluted solution to the concentrated solution, as shown in Equation 3.6. It becomes an easy way of working the concentration of the concentrated solution if that of the diluted solution is analysed (Eqn 3.7). Volume of diluted solution Dilution factor Eqn 3.6 Volume of concentratedsolution Conc. of original solution = dilution factor x conc. of diluted solution Eqn 3.7 EXAMPLE 3.6 (a) What is the dilution factor is 25 ml of solution is diluted to 500 ml? Dilution factor = = 20. (b) What is the concentration of the original solution if it was diluted by a factor of 20 and the diluted solution had a concentration of 23.5 mg/l? Conc. of original solution = 20 x 23.5 = 470 mg/l CLASS EXERCISE What is the dilution factor in the following cases? (a) 10 ml is diluted to 250 ml (b) 25 ml is diluted to 1 L (c) 5 ml is diluted to 250 ml (d) 2.5 ml is diluted to 250 ml 2. Using the dilution factors in Q1, find the concentration of the original solutions, if the diluted solutions are analysed and found to have the following concentrations: (a) mole/l (b) 3.67 mg/l (c) 69 mg/l (d) %w/w Remember that these dilution calculations only work for concentrations, not masses! Aliquots of solutions If an accurate volume (aliquot) is taken from a known volume of solution (volumetric flask), and then analysed for the amount (mass or moles), we can work how much was in the total solution, by simple fractions. Remember that the concentration of the aliquot is the same as the total solution. Amount in total solution Amount in analysed solution x Volume of total solution Volume of analysed solution 33

34 EXAMPLE 3.7 What mass of lead is contained in 250 ml of solution, if a 25 ml aliquot is found to contain 15.3 mg? The 25 ml aliquot is one-tenth of the original solution, so the mass will be onetenth, ie 153 mg in the main solution. CLASS EXERCISE 3.10 What amount (moles or mass) is contained in the original solution if: (a) a 10 ml sample from a 500 ml solution contains 2.34 x10-3 moles (b) a 50 ml aliquot from a 200 ml solution contains g (c) a 25 ml aliquot from a 500 ml solution contains 99.3 mg (d) a 1 ml aliquot from a 20 ml solution contains 4.3 ug 3.6 Titration calculations When chemical substances react together, they do so by number of molecules, not mass of compounds. Therefore, you have to work in moles. Remind yourself of the two formulae for calculating the number of moles. = Number of moles = from mass from solution A titration is a method of analysis which you will use practically in your first laboratory subject. Titrations rely on knowing the number of moles of one of the reacting substances (the standard), and using that to work out the moles of the other substance (the analyte). If the standard is a pure solid, then you need to know the mass AND the formula weight. If the standard is a solution, then you need to know the molarity (mole/l) and the volume. 34

35 EXAMPLE 3.8 Identify the analyte and standard in the following. (a) 25 ml of apple juice is analysed for acidity by titration with 26.7 ml of M NaOH. Both substances are in solution, so we need to know the molarity and volume for the standard: the NaOH is therefore the standard, and the apple juice acid is the analyte, because the molarity is not known. (b) 12.3 ml of AgNO 3 titrates g of pure NaCl. The molarity of the silver nitrate is not known, so it must be the analyte. The sodium chloride is pure, and its mass is known, so it is the standard. CLASS EXERCISE 3.11 Identify the standard and the analyte in the following. (a) (b) (c) (d) (e) (f) 25 ml of HCl solution is titrated with 18.2 ml of M NaOH. The calcium in 50 ml of tap water is titrated with 12.3 ml of EDTA g of pure sodium carbonate is titrated with 17.5 ml of HCl. The acidity in 25 ml of vinegar is analysed by titration with 25.3 ml of M NaOH. The KOH content in g of washing powder is titrated with 11.2 ml of M H 2 SO 4. The chloride content in 50 ml of blood serum is analysed by titration with 18.1 ml of M AgNO 3. Standard Analyte The titration calculation procedure The following method shows you how to calculate the number of moles of analyte in the titrated solution. What you do with it from there depends on what you are after: concentration, mass, dilutions etc. 1. Calculate moles of standard 2. Determine reaction ratio 3. Calculate moles of analyte from 1 & 2 If that seems easy, that s because it is! However, you need to be careful to make sure you are using the right bit of information in the right place. For example, if you were calculating the number of moles of NaOH standard delivered form the burette, then make sure it is the volume and molarity of the NaOH that you use. 35

36 Step 1: use the appropriate formula (mass FW for solids, CV for solutions) Step 2: this is critical; either write the balanced equation, or find out what the reaction ratio for analyte to standard is Step 3: look at the numbers in the reaction ratio: what do you have to multiply the standard coefficient by to get to the analyte number; use that multiplication for the actual moles of standard to get the moles of analyte EXAMPLE 3.9 Referring back to Example 3.8 (a) The reaction ratio is 1 acid: 2 NaOH (you weren t expected to know this). moles of NaOH (std) = x 26.7 x 10-3 (volume must be in litres) = x 10-3 The multiplication factor is 0.5 (2 1). moles acid (analyte) = 0.5 x x 10-3 = x 10-3 (b) The reaction ratio is 1:1. moles of NaCl (std) = = x 10-3 The multiplication factor is 1 (1 1). moles AgNO 3 (analyte) = 1 x x 10-3 = x 10-3 CLASS EXERCISE 3.12 Calculate the moles of analyte for the titrations in Exercise 3.11, given the following reaction ratios: (a) 1 HCl : 1 NaOH (b) 1 Ca : 1 EDTA (c) 2 HCl : 1 Na 2 CO 3 (d) 1 acid : 1 NaOH (e) 1 H 2 SO KOH (f) 1 AgNO 3 : 1 Cl 36

37 1. How much NaCl (FW 58.44) would be needed to make the following solutions? (a) 500 ml of 5 g/100 ml (b) 500 ml of 5 g/l (c) 50 ml of 200 mg/l (d) 250 ml of 0.1 mole/l 2. What is the concentration of 0.75 g of NaCl in 500 ml? (a) in g/100 ml (b) in g/l (c) in mole/l 3. What is the molarity (mole/l) of the following solutions? (a) 5 g of NaOH in 250 ml (b) 5 g of CuSO 4. 5H 2 O in 250 ml (c) x 10-3 moles of NaCl in 15 ml 4. What is the concentration of analyte in the following solid samples? (a) g of water in g of soil (as %w/w) (b) 6.78 mg of DDT in g of soil (as mg/kg) (c) x 10-3 moles of NaCl in g of soil (as %w/w) 5. (a) How much lead is in 250 kg of waste, which has a concentration of 23.4 mg/kg? (b) How much nitrate is in 2 g of soil which has a concentration of %w/w? 6. What is the concentration of the following diluted solutions: (a) 25 ml of 100 mg/l is diluted to 250 ml (b) 5 ml of M is diluted to 100 ml (c) 10 ml of 0.65 g/100 ml diluted to 250 ml 7. What volume of concentrated solution is required in the following: (a) 1 L of 10 mg/l from 1000 mg/l (b) 200 ml of 0.05 M from 0.1 M (c) 100 ml of 0.1 g/l from 20 g/l 8. (a) 20 ml of sample was diluted to 500 ml and analysed. The diluted solution was found to have a concentration of 2.45 mg/l. What was the concentration of the original solution? (b) A sample is diluted by a factor of 25. The diluted solution was found to have a concentration of g/100 ml. What was the concentration of the original solution? 9. (a) g of pure Na 2 CO 3 was used to standardise a solution of HCl ml of the HCl was required to reach endpoint. Calculate the molarity of the HCl (b) A sample of soil was titrated with 14.2 ml of M iron (II) solution to determine its chromium concentration. The reaction ratio is 3 moles of iron reacts with 1 mole of chromium. Calculate the mass of Cr in the soil. (c) The chloride content of IV saline solution was analysed by titration with silver nitrate solution. The reaction is 1 Cl:1 Ag. Calculate the concentration of NaCl in the sample as g/100 ml, if a 20 ml aliquot of sample was titrated by ml of M silver nitrate. (d) 10 ml of sulfuric acid is titrated with 15.8 ml of M NaOH. Given that 2 moles of NaOH react with 1 mole of sulfuric acid, determine the concentration of the acid as g/l. 37

38 Answers Page 3 Estimation (i) 0.17 ( S.F) (ii) 167 (165 3 S.F.) (iii) 80 ( S.F Page 4 Rounding Page 5 Order of Operations Page 6 Scientific Notation x x x x x x x x x Page 7 Directed Numbers ; 9; -1; -6; -16; -14; -3 Page 8 Significant Figures 1. 4; 4; 4; 1; (60000); 460 (500); (0.0004) (0.002); 3.0 (3); ( ) ( ); 2500 (2000); Page 9 Substitution into a formula g/l g/l g/l Page 10 Rearranging formula to solve equation Page 11 a) V = m/c b) C 2 = C 1 V 1 /V 2 c) T f = (T c ) /0.556 d) m = CV Page 12 General Revision of Concepts a) g/l b) 9.66 c) d) e) 12.5 ml Page 14 a) 5700 cm b) 20000mm c) g (45mg) d) m 2 e) 7.2 x 10-5 m 3 f) ft g) 1086 ha h) psi i) 563g Some more to try a) 4.57 x 10-5 cm b) 4.98s x 10-4 MHz c) 2.5 x 10-5 kg d) 0.40 mile Some more difficult ones a) x 10-6 g/ml b) 48.7 mg/kg c) g/l Page 15 1) 2.06 x 10-7 m 2) kg 3) 3450 kj 4) 8.56 x 10-4 m 2 5) cm 3 6) 500 ML 7) 4.56 kl 8) 2.35 cm 3 9) 50 mg/ ml 10) 1822 kgm 3 11) J/g 12) m 13) 1248kg/m 3 14)0.6177acres 15) 1200cm 3 /hr 16) cm 2 17) gal/min 38

39 Page 19 Page 23 a) au b) au c) au d) au Page 25 a) x 10-4 b) c) d) Page 26 1 a) 85 au b) 342.1au c) au 2 a) 2.94 x 10-3 b) 3.42 x 10-4 c) 2.18 x 10-3 mol d) mol Page 28 1 a) g/l b) 54.6 mg/l c) mg/100ml d) 2.76 ml/100 ml 2 a) 0.575g b) 2.28 g c) mol d) 15 ml 3 a) 1.75 L b) 225 ml c) 0.55 L d) 1250 ml Page 30 1 a) g/100g b) mg/kg c) g/100g d) mg/kg 2 a) mg b) 5.25 g c) mg d) 0.75 g 3 a) 200 g b) 8 g c) 2.5 g d) 2.0g Page 32 1 a) 0.625g/100mL b)10 mg/l c) mol/l d) g/l 2 a) 40 ml b) 25 ml c) 50 ml d) 40 ml Page 33 1 a) 25 b) 40 c) 50 d) a) 0.85 mol/l b) mg/l c) 3450 mg/l d) 10.3% w/w Page 34 a) mole b) 0.42 g c)1986 mg d)86 ug Page 35 a) Std NaOH Analyte HCl b) Std EDTA Analyte Ca c) Std Na 2 CO 3 Analyte HCl d) Std NaOH Analyte H + e) Std H 2 SO 4 Analyte KOH f) Std AgNO 3 Analyte Cl - Page 36 a) x 10-3 mol b) x 10-3 c) 2.96 x 10-3 d) 2.50 x 10-3 e) 1.20 x 10-3 f) 1.75x