Steps to success

Dimensional Analysis Steps to success: Chemistry calculations using dimensional analysis Introduction School and college students tend to find stoichiometric (quantitative) calculations in chemistry difficult. This is sometimes attributed to weaknesses in handling ratio and proportion but the difficulty may also lie in the number of different factors that have to be correctly assembled in order to arrive at a correct answer. When presented with the following type of problem "One apple costs 40 pence. How much would three apples cost?" little difficulty is normally encountered by students, and chemistry teachers are often at a loss to explain why chemistry calculations give difficulty since they too are based on ratio. However, the typical stoichiometric calculation is more complex than the simple ratio calculation above. A chemistry problem might consist of the student being supplied with a balanced equation along with the requirement to calculate a mass of product from a given mass of reactant. Solving this problem will involve converting masses into moles, then using a mole ratio and finally converting moles back into masses, something a little more complicated than pricing apples. But the way chemical calculations are taught in UK schools differs little from the apple approach and relies heavily on students having the ability to 'see' the proportional relationship between the numbers. To take a simple example, after doing some work on mole ratios, the student might finally arrive at: "24 Grams of A gives 95 grams of B so 10 grams of A will give how many grams of B?" The answer being 95 x 10/24 grams of B is obvious to some but gives difficulty to others. Errors often occur at setting up the ratio stage since the numbers lose their units in the calculation. There is also no way of checking whether the ratio has been set up properly other than by inspection of the magnitude of the answer obtained, a step often omitted by poor problem solvers. What we need is an algorithm that would allow chemical calculations to be laid out correctly, that would avoid the ratio difficulty, and would allow error checking. A method does exist: it is known as Dimensional Analysis or the Factor Label Method and its use in a range of stoichiometric calculations is explained in what follows here. The fact that numbers in chemical calculations represent quantities is often overlooked. Dimensional Analysis (or DA) emphasises the quantities and units involved in chemical calculations not just the numbers. The quantities in question are defined by units or dimensions (g, kg, cm3 etc) linked to chemical entities so one Wilson Flood Page 1 9/24/2009 Dimensional Analysis gram of sodium is not equivalent to one gram of chlorine, for example. Understanding this simple idea drives the DA method. It would be advisable to introduce the DA method gradually in the course of a chemistry programme, starting with the relatively straightforward mole calculations in introductory courses and then moving on to more complicated examples in advanced courses. Teaching the method would follow the usual procedure of worked example, followed by guided practice on the part of students. It might also be advisable to introduce the method in a non-chemistry context in the first instance as shown in the explanatory example. As with calculations generally, at first acquaintance it is good practice to lay out the calculation stepwise in some detail, but, as experience is gained, calculation steps may be combined to the extent to which students feel comfortable. In some of the examples that show the method in use I have explained each step in some detail as a teacher might do but in some of the later examples I have combined steps to avoid tedium. However, for students, each new type of calculation would need to be explained in detail so no steps should be omitted on first encounter. Wilson Flood Page 2 9/24/2009 Dimensional Analysis The method explained Dimensional analysis (DA) can be used to carry out a wide range of different calculations. The method is useful in setting out calculations and in checking for setup errors because if the unit of the final quantity is incorrect, then the relationship of the numbers in the calculation and hence the answer must also be incorrect. Using dimensional analysis usually involves four steps: 1. In the calculation, you first identify what you are given (called here the 'given quantity') and what you want to find out (called the 'desired quantity'). 2. You then identify what are called conversion factors (CF). Conversion factors must show units. They are ratios that show what is equivalent to what; for example, one mole of sodium is equivalent to 23 grams of sodium. In most stoichiometric calculations an additional CF is the mole ratio (MR). 3. Now you arrange the given quantity and the conversion factors in the calculation so that identical units cancel leaving you with the desired quantity. The given quantity is always on the top line (the numerator) at the left of the calculation and the desired quantity will remain on the top line at the end of the calculation. 4. Finally, perform the calculation. Almost all stoichiometric calculations can be performed using this algorithm. As will be seen, titration calculations require an additional conversion factor. Basically all DA calculations are laid out like this: given quantity x conversion factors = desired quantity The number of conversion factors needed depends on the calculation to be performed. As a simple memory aid this can be simplified to : given X CFs = desired Units In DA it is units and the chemical entities associated with them which cancel in setting up the final calculation. This emphasises an important point in chemistry which is worth repeating. Stoichiometric calculations are not about numbers, they are not even about quantities, they are about quantities of substances. Wilson Flood Page 3 9/24/2009 Dimensional Analysis An explanatory example To illustrate the use of the method here is a simple example based on the following problem: A piece of metal is 11.7 inches in length. Convert this length into centimetres. (Use 1 in = 2.54 cm) Using the four steps 1. Identify the given quantity and the desired quantity given quantity desired quantity 2. The equivalence is 1 in : 2.54 cm and in the calculation the CF (conversion factor) could be written as one of two possible ratios ie 1 in 2.54 cm or 2.54 cm 1 in 11.7 in ? cm Both are correct but which one should we use in the calculation? The answer to the question is determined by the fact that the units of the desired quantity must be on the top line of the fraction if the calculation is to make any sense and also by the fact that all other units (dimensions) must cancel. Therefore the correct CF in this case is 2.54 cm 1 in 3. We now have our given quantity from step 1 and we have the conversion factor from step 2 so the calculation is set up as follows Length in centimetres = 11.7 in x 2.54 cm 1 in Cancelling identical units leaves us with the basic calculation to perform. 4. Length in cm = 11.7 x 2.54 cm = 29.7 cm 1 Wilson Flood Page 4 9/24/2009 Dimensional Analysis Note that the units must be included when laying out the calculation since cancelling identical units leads us to the desired answer. This also gives a method of error checking since if we are not left with the unit of the desired quantity on the top line then the conversion factors must have been set up wrongly. Wilson Flood Page 5 9/24/2009 Dimensional Analysis Stoichiometric calculations Simple mass-mole conversions Example 1 What is the mass (in grams) of 0.40 moles of sodium ? given quantity desired quantity 0.40 mol Na - NB: not just mol ? g Na - NB: not just g The equivalence is 1 mol Na : 23 g Na so the conversion factor could be 1 mol Na 23 g Na or 23 g Na 1 mol Na The right hand conversion factor is chosen in order to cancel identical units and leave the desired units of grams of sodium on the top line. Setting up the calculation: given quantity x conversion factor = desired quantity Mass = 0.40 mol Na x 23 g Na 1 mol Na = 0.40 x 23 g Na 1 = 9.2 g Na Example 2 How many moles of water are in 0.600 kg of water? given quantity desired quantity Conversion Factors conversion factor (1), CF(1) 1 mol H2O : 18g H2O 0.600 kg H2O ? mol H2O Since molar mass is usually measured in grams we will need a conversion factor to convert kg H2O to g H2O ie conversion factor (2), CF(2) 1 kg H2O : 1000 g H2O = 600 g H2O So, grams of water = 0.600 kg H2O x 1000 g H2O 1 kg H2O We can now convert g H2O to mol H2O using CF(1) Moles of H2O = 600 g H2O x 1 mol H2O = 18 g H2O Wilson Flood Page 6 600 mol H2O = 33.3 mol H2O 18 9/24/2009 Dimensional Analysis The two separate steps can be combined into a single calculation. The given quantity (kg H2O) and the desired quantity (mol H2O) must be on the top line so the conversion factors must be arranged as follows: CF(1) - 1000 g H2O 1 kg H2O kg H2O x and CF(2) 1 mol H2O 18 g H2O giving CF(1) x CF(2) (kg g) (g mol) mol H2O as shown below 0.600 kg H2O x 1000 g H2O x 1 mol H2O 1 kg H2O 18 g H2O = 33.3 mol H2O = 0.600 x 1000 x 1 mol H2O 1 18 A common type of calculation asks students to calculate the number of chemical entities in a mass of substance. Example 3 How many gold atoms are present in a ring of pure gold which has a mass of 5 g? (Take: Avogadro's Number = 6 x 1023 ; relative atomic mass of Au = 197) given quantity desired quantity 5 g Au ? Au atoms We need two conversion factors: the first will convert 'g Au' to 'mol Au' and the second will convert 'mol Au' to 'atoms Au'. Conversion factor (1), CF(1) Conversion factor (2), CF(2) Arranging these we get Number of atoms of Au = 5 g Au x 1 mol Au x 6 x 1023 atoms Au 197 g Au 1 mol Au = 5 x 1 x 6 x 1023 atoms Au 197 1 = 1.52 x 1022 atoms Au 197 g Au : 1 mol Au 1 mol Au : 6 x 1023 atoms Au Wilson Flood Page 7 9/24/2009 Dimensional Analysis Calculations involving reacting masses and/or moles only In any chemical calculation involving balanced equations, moles and masses the arithmetic employs the same steps which are: given mass given moles desired moles desired mass The calculation could be one of four types     given and desired are moles given is mass and desired is moles given is moles and desired is mass given and desired are masses. As with any calculation a correctly balanced equation is required. Taking mass information:     mass calculations as our example we need the following a correctly balanced equation a given quantity mole-mass conversion factors (ie molar masses) mole-mole conversion factor (ie mole ratio) in order to find a desired quantity. Mass mole, mole mass, and mole mole calculations use exactly the same procedure but have fewer conversion factors. It would be normal practice to have masses expressed in grams but other units can be used so long as the appropriate conversion factors are employed (see Example 2 above). Wilson Flood Page 8 9/24/2009 Dimensional Analysis Calculations involving masses The next example shows the method applied to a typical calculation. The method is identical for all calculations of this type. Example 4 How many grams of hydrogen would be produced by completely reacting 1.7 grams of sodium with water? 2Na + 2H2O 2 moles given quantity desired quantity 2NaOH + H2 1 mole 1.7 g Na ? g H2 2 moles of H2 3 grams of H2 The calculation routine is 1 grams of Na moles of Na The calculation can be done as separate steps or as a composite calculation. To illustrate the method we will look at the stepwise procedure first. Step 1 – Calculate the number of moles of Na that is equivalent to 1.7 grams of Na given quantity desired quantity 1.7 g Na ? mol Na conversion factor(1), CF(1) - 23 g Na (molar mass) : 1 mol Na This must have the form 1 mol Na 23 g Na to cancel with the given quantity. so moles of Na = 1.7 g Na x 1 mol Na 23 g Na = 0.074 mol Na Step 2 – Calculate the number of moles of H2 produced from 0.074 moles of Na given quantity desired quantity 0.074 mol Na ? mol H2 2 mol Na : 1 mol H2 mole ratio conversion factor (2), MR(2) (from balanced equation) This must have the form 1 mol H2 to cancel with the unit of the given quantity. 2 mol Na so moles of H2 produced = 0.074 mol Na x 1 mol H2 2 mol Na = 0.037 mol H2 Wilson Flood Page 9 9/24/2009 Dimensional Analysis Step 3 – Calculate the number of grams of H2 equivalent to 0.037 moles of H2 given quantity desired quantity 0.037 mol H2 ? g H2 1 mol H2 : 2 g H2 (molar mass) conversion factor(3), CF(3) This must have the form 2 g H2 to give the desired unit on the top line. 1 mol H2 = 0.037 mol H2 x 2 g H2 1 mol H2 = 0.074 g H2 So, grams of H2 produced Once some practice has been gained with the method students should be able to combine the steps in a single calculation as shown below. In setting out the conversion factors for a single calculation it is good practice to start with the "given" and work towards the "desired". conversion factor (1), CF(1) mole ratio conversion factor (2), MR(2) (from balanced equation) conversion factor (3), CF(3) 23 g Na : 1 mol Na 2 mol Na : 1 mol H2 1 mol H2 : 2g H2 Arranging so that identical units cancel, grams of H2 produced is given by 1.7 g Na x 1 mol Na 23 g Na CF(1) 1.7 x x 1 mol H2 2 mol Na MR(2) 1 2 x 2 g H2 1 mol H2 CF(3) 1 x 23 x 2 g H2 1 = 0.074 g H2 It is identical units of chemical entities that cancel not the numbers and only if the conversion factors are set up correctly can you end up with grams of H 2 remaining and on the top line. With practice the method becomes routine since every calculation is essentially a variation on the same theme. Identifying what goes on the top and bottom lines in the CFs may cause difficulty at first but this can be made easier if a systematic approach is taken. In the example above, CF(1) must have 'g Na' on the bottom line to cancel with the given quantity; CF(3) must have 'g H2' on the top line to give the desired quantity; finally, the mole ratio conversion factor is determined from CF(1) and CF(3). Although doing it this way the mole ratio CF is last to be determined it is probably least confusing if you place the "desired quantity" CF furthest to the right in the calculation layout whenever possible. Wilson Flood Page 10 9/24/2009 Dimensional Analysis Using the grid method – an alternative way of laying out the calculation Taking the example of the combined calculation above it is easy to see that some students could get lost in cancelling the units. A grid method has been developed to reduce this problem. With the grid method, instead of the multiplication signs and traditional numerators and denominators (top line and bottom line terms) the various quantities are placed in a grid that looks like this: The number of cells in the grid will depend on the calculation to be performed. An advantage is that the grid can be built up as needed. First, the given quantity is always placed at the top left hand corner of the grid: 1.7 g Na Then each of the conversion factors (and mole ratio) is added until you arrive at: 1.7 g Na 1 mol Na 1 mol H2 2 g H2 23 g Na 2 mol Na 1 mol H2 Like units of like entities are cancelled and the calculation completed. No multiplication signs are shown so it needs to be explained that top line numbers are multiplied together then bottom line numbers are multiplied together. The bottom line figure is then divided into the top line figure, as is the usual practice. Many of the examples that follow will show the grid method in use. Wilson Flood Page 11 9/24/2009 Dimensional Analysis Calculations involving mass and volume of gas Example 5 – similar to the previous example but different types of CF are needed Some zinc was reacted completely with excess hydrochloric acid. What mass of zinc (in grams) would be required to produce 500 cm3 of hydrogen (under standard conditions ie 298 K and 100 kPa)? (Take 1 mole of hydrogen = 24.0 dm3 ; relative atomic mass of Zn = 65.4 ). Zn + 1 mole 2HCl ZnCl2 500 cm3 H2 ? g Zn + H2 1 mole given quantity desired quantity Start converting from the "given" and work towards the "desired". The calculation routine is 1 vol H2 (cm3) vol H2 (dm3) conversion factor (1) conversion factor (2) mole ratio (3) conversion factor (4) Step 1 vol H2 (dm3) Step 2 mol H2 = 0.500 dm3 H2 x 1mol H2 24 dm3 H2 = 0.0208 mol H2 = 500 cm3 H2 x 1 dm3 H2 1000 cm3 H2 = 0.500 dm3 H2 - 2 mol H2 3 : : : : mol Zn 4 mass Zn 1000 cm3 H2 24 dm3 H2 1 mol H2 1 mol Zn 1 dm3 H2 1 mol H2 1 mol Zn 65.4 g Zn Step 3 mol Zn = 0.0208 mol H2 x 1mol Zn 1 mol H2 = 0.0208 mol Zn Step 4 mass Zn(g) = 0.0208 mol Zn x 65.4 g Zn 1 mol Zn = 1.36 g Zn Wilson Flood Page 12 9/24/2009 Dimensional Analysis In a composite calculation, using the grid method and arranging so that identical units cancel, grams of Zn required is given by 500 cm3 H2 1 dm3 H2 1000 cm3 H2 CF(1) 500 1 1000 1 24 1 mol H2 24 dm3 H2 CF(2) 1 1 1 mol Zn 1 mol H2 MR(3) 65.4 g Zn 1 65.4 g Zn 1 mol Zn CF(4) = 1.36 g Zn Wilson Flood Page 13 9/24/2009 Dimensional Analysis Calculations involving empirical formulae Example 6 A compound containing only carbon, hydrogen and oxygen was analysed and was found to have the following percentage composition by mass: carbon hydrogen oxygen 39.9 % 6.67% 53.4 % Determine the empirical formula of the compound. (Relative atomic masses: C = 12.0; H = 1.00; O = 16.0) The number of moles of each element in 100 g of compound would be moles C = 39.9 g C x 1 mol C = 3.33 mol C 12.0 g C moles H = 6.67 g H x 1 mol H = 6.67 mol H 1.00 g H moles O = 53.4 g O x 1 mol O = 3.34 mol O 16.0 g O The ratio of C:H:O is 3.33:6.67:3.34 which simplifies to 1:2:1. Hence the empirical formula is CH2O The method can be easily adapted to calculate empirical formulae from combustion data or to calculate percentage composition from formulae. Example 7 0.255 g of a compound containing only C, H, and O gave on complete combustion 0.561 g of CO2 and 0.306 g of H2O. Calculate the empirical formula of the compound. Selecting appropriate CFs we first calculate the moles of C and H in the sample. mol C = 0.561 g CO2 x 1 mol CO2 x 1 mol C = 0.0128 mol C 44.0 g CO2 1 mol CO2 mol H = 0.306 g H2O x 1 mol H2O x 2 mol H 18 g H2O 1 mol H2O = 0.0340 mol H Wilson Flood Page 14 9/24/2009 Dimensional Analysis To calculate the moles of O in the sample we must first calculate the mass of C and the mass of H in the sample. The remainder will be the mass of O, From this we can calculate the number of moles of O. mass of C = 0.0128 mol C x 12.0 g C 1 mol C mass of H = 0.0340 mol H x = 0.154 g C 1.01 g H = 0.0343 g H 1 mol H (We could also have carried out the calculation by calculating the masses of C and H in the CO2 and H2O respectively and then converted these into moles.) The mass of O in the sample is therefore mass of sample - (mass of C + mass of H) = 0.255 g - (0.154 g + 0.0343 g) = 0.067 g O mol O = 0.067 g O x 1 mol O 16.0 g O = 0.0042 mol O The ratio of C:H:O is 0.0128 : 0.0340 : 0.0042 ie 3 : 8 : 1 approx Hence the empirical formula is C3H8O Example 8 Calculate the percentage composition by weight of N in NH4NO3. In effect we must calculate the mass of N in 100 g NH4NO3. We must also take note that there are two N atoms in the formula. Selecting appropriate CFs the calculation looks like this: 100 g NH4NO3 x 1 mol NH4NO3 x 2 mol N x 14 g N 60 g NH4NO3 1 mol NH4NO3 1 mol N Therefore the percentage composition by weight of N is 35%. = 35 g N (This calculation could have been carried out stepwise. First, we could have calculated the mass of nitrogen in one mole (60 g) of ammonium nitrate and then converted this mass into a percentage.) Wilson Flood Page 15 9/24/2009 Dimensional Analysis Calculations involving limiting and excess reactants Calculations of this type have two "given" quantities. Example 9 A strip of zinc metal weighing 2.50 g is placed in an aqueous solution containing 2.00 g of silver nitrate. The reaction that occurs is Zn(s) + 2AgNO3(aq) 2Ag(s) + Zn(NO3)2(aq) (a) Determine which reactant is in excess. (b) Calculate how many grams of silver will be formed. (a) Determine which reactant is in excess. given quantity desired quantity CF and given quantity desired quantity CF mol Zn = 2.00 g AgNO3 ? mol AgNO3 1 mol AgNO3 : 170 g AgNO3 2.50 g Zn 1 mol Zn 65.4 g Zn mol Ag NO3 = 2.00 g AgNO3 1 mol AgNO3 170 g AgNO3 The mole ratio CF for complete reaction is 1 mol Zn : 2 mol AgNO3. Moles of AgNO3 needed to react with 3.82 x 10-2 mol Zn would be 3.82 x 10-2 mol Zn 2 mol AgNO3 1 mol Zn Since there is only 1.18 x 10-2 mol of AgNO3 available it is the case that AgNO3 is the limiting reactant and so the zinc metal is in excess. = 7.64 x 10-2 mol AgNO3 = 1.18 x 10-2 mol AgNO3 = 3.82 x 10-2 mol Zn 2.50 g Zn ? mol Zn 1 mol Zn : 65.4 g Zn Wilson Flood Page 16 9/24/2009 Dimensional Analysis (b) Calculate how many grams of silver will be formed. The mass of silver formed will be determined by the limiting reactant which was found to be AgNO3. given quantity desired quantity mole ratio CF CF g Ag formed = 1.18 x 10-2 mol AgNO3 ? g Ag 1 mol AgNO3 : 1 mol Ag 1 mol Ag : 108 g Ag 1.18 x 10-2 mol AgNO3 1 mol Ag 1 mol AgNO3 = 1.27 g Ag 108 g Ag 1 mol Ag Wilson Flood Page 17 9/24/2009 Dimensional Analysis Calculations involving solutions Calculations involving solutions demand a slightly different approach from mass – mole conversions since solutions must be described in terms of both their volume and their concentration. Here are some examples. Most calculations of this type also need to use the 1 dm3 : 1000 cm3 equivalence. Example 10 How many moles of sodium hydroxide (NaOH) would be present in 150 cm 3 of a NaOH solution of concentration 0.600 mol dm-3 ? given quantity 150 cm3 soln of 0.600 mol NaOH 1 dm3 soln Using the grid convention, this is written as follows: 150 cm3 soln 0.600 mol NaOH 1 dm3 soln desired quantity conversion factor - ? mol NaOH 1 dm3 soln : 1000 cm3 soln Therefore the calculation becomes mol NaOH = 150 cm3 soln x 0.600 mol NaOH x 1 dm3 soln 1 dm3 soln 1000 cm3 soln 150 x 0.600 1 x 1 mol NaOH 1000 = = 0.090 mol NaOH Using the grid method it looks like this 150 cm3 soln 1 dm3 soln 1000 cm3 soln mol NaOH = 0.600 mol NaOH 1 dm3 soln Wilson Flood Page 18 9/24/2009 Dimensional Analysis Example 11 How many grams of sodium hydroxide (NaOH) would be needed to make 180 cm3 of a NaOH solution of concentration 0.600 mol dm-3 ? given quantity desired quantity conversion factor (1) conversion factor (2) mol NaOH needed = 180 cm3 soln of 0.600 mol NaOH 1 dm3 soln ? g Na 1 dm3 soln : 1000 cm3 soln 1 mol NaOH : 40.0 g NaOH 0.600 mol NaOH 1 dm3 soln = 180 0.600 1 = 0.108 mol NaOH 1 1000 1 dm3 soln 1000 cm3 soln 180 cm3 soln mol NaOH g NaOH needed = 0.108 mol NaOH 40 g NaOH 1 mol NaOH = 4.32 g NaOH Up until Examples 10 and 11 it had been possible to place cancelling units side by side in the calculation but this is not possible with calculations involving solutions. Note also that the presence of a solution is indicated by "soln" in the unit. Since the chemical entity (NaOH) is included with the "mol" unit it is not necessary to include it in the "soln" unit also. While it might seem pedantic to write "soln" each time it helps to avoid confusion when dealing with calculations involving both volumes of solutions and volumes of gases (see Example 14). In Examples 12 and 14 which follow the calculations have been performed both in the stepwise and in the composite manner. Wilson Flood Page 19 9/24/2009 Dimensional Analysis Calculations involving volumetric analysis Any volumetric calculation, acid-base or redox, may be carried out. Example 12 20.0 cm3 of a solution of NaOH is exactly neutralised by 34.7 cm3 of a solution of H2SO4 of concentration 0.500 mol dm-3. Calculate the concentration of the NaOH solution in mol dm -3 . Balanced equation : H2SO4 + 2NaOH 1 mole 2 moles mole ratio CF given quantity - Na2SO4 + 2H2O 1 mol H2SO4 : 2mol NaOH 34.7 cm3 soln of 0.500 mol H2SO4 1 dm3 soln ? concentration of NaOH ie mol NaOH 1dm3 soln desired quantity - Step 1 - calculate moles of H2SO4 using CF - 1 dm3 soln : 1000 cm3 soln mol H2SO4 = 34.7 cm3 soln 0.500 mol H2SO4 1 dm3 soln 1 dm3 soln 1000 cm3 soln = 0.0174 mol H2SO4 Step 2 - calculate moles of NaOH reacting using the mole ratio mol NaOH = 0.0174 mol H2SO4 2 mol NaOH 1 mol H2SO4 = 0.0348 mol NaOH Wilson Flood Page 20 9/24/2009 Dimensional Analysis Step 3 - calculate the concentration CF 1 dm3 soln :1000 cm3 soln of the NaOH solution using the There were 0.0348 mol of NaOH in 20 cm3 of solution Therefore the number of moles of NaOH in 1 dm3 of solution (ie the concentration) = 0.0348 mol NaOH 20 cm3 soln 1000 cm3 soln 1 dm3 soln = 0.0348 20 1000 1 mol NaOH dm3 soln ie the concentration is 1.74 mol dm-3 = 1.74 mol NaOH 1 dm3 soln With practice, the three steps can be combined into a single calculation. When this is done, the two dm3/cm3 CFs cancel out and can be omitted, giving mol NaOH 1 dm3 soln = 34.7 cm3 soln 0.500 mol H2SO4 1 dm3 soln = 1.74 mol NaOH ie 1.74 mol dm-3 1 dm3 soln 2 mol NaOH 1mol H2SO4 20.0 cm3 soln This is certainly a quick way of doing a volumetric calculation but is only for the highly practised. The volume quantities relate to the inverse of the mole ratio CF, in other words "mol NaOH" appears on the top line and the volume of NaOH solution appears on the bottom line. Once again note that the volume unit uses the term "soln" but, for the method to work, does not distinguish between different solutions. This is because the chemical entity is already identified in the mole unit. However, it is necessary to use "soln" in order to distinguish between volumes of gas and volumes of solution as Example 14 shows. In Example 13 which follows the method is used to calculate a volume of solution of known concentration. This example also illustrates how conversion factors work equally well if the entities involved are merely ions rather than full chemical formulae. Wilson Flood Page 21 9/24/2009 Dimensional Analysis Example 13 The redox reaction between manganate(VII) ions and iron(II) ions is MnO4+ 8H+ + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O What volume (in cm3) of manganate(VII) solution of concentration 0.020 mol dm -3 would react exactly with 25 cm3 of a solution of iron(II) which has a concentration of 0.40 mol dm-3? given quantity 25 cm3 soln of 0.4 mol Fe2+ 1 dm3 soln ? cm3 soln desired quantity - Moles of Fe2+ present = 25 cm3 soln 0.4 mol Fe2+ 1 dm3 soln 1 dm3 soln 1000 cm3 soln = 0.010 mol Fe2+ Moles of MnO4- reacting = 0.010 mol Fe2+ 1 mol MnO45 mol Fe2+ = 0.0020 mol MnO4The volume of solution (in cm3) of concentration 0.020 mol MnO4- containing 1 dm3 soln 0.0020 mol MnO4= 0.0020 mol MnO41 dm3 soln 0.02 mol MnO4= 100 cm3 soln 1000 cm3 soln 1 dm3 soln Wilson Flood Page 22 9/24/2009 Dimensional Analysis Calculations involving solutions and molar volume of gas Example 14 What volume (in dm3) of hydrogen (under standard conditions ie 298 K and 100 kPa) would be produced by completely reacting 75.0 cm 3 of hydrochloric acid of concentration 1.60 mol dm-3 with magnesium? (Take 1 mole of hydrogen = 24.0 dm3) Mg + 2HCl 2 moles mole ratio CF given quantity desired quantity H2 + 1 mole MgCl2 2 mol HCl : 1 mol H2 75.0 cm3 soln of 1.60 mol HCl 1 dm3 soln 3 ? dm H2 Step 1 - calculate the number of moles of HCl using CF 1 dm3 : 1000 cm3 Moles of HCl = 75.0 cm3 soln 1.60 mol HCl 1 dm3 soln 1 dm3 soln 1000 cm3 soln = 0.120 mol HCl Step 2 - calculate the equivalent number of moles of hydrogen using the mole ratio Moles of hydrogen = 0.120 mol HCl 1 mol H2 2 mol HCl = 0.0600 mol H2 Step 3 - calculate the volume of hydrogen produced using the CF 1 mol H2 : 24.0 dm3 H2 Vol of H2 = 0.0600 mol H2 24.0 dm3 H2 1 mol H2 = 1.44 dm3 H2 Wilson Flood Page 23 9/24/2009 Dimensional Analysis As in previous cases the separate steps can be combined into a single calculation. Vol H2 = 75 cm3 soln 1.6 mol HCl 1 dm3 soln 1 dm3 soln 1000 cm3 soln 1 mol H2 2 mol HCl 24 dm3 H2 1 mol H2 = 1.44 dm3 H2 Note that in this calculation the unit 'dm3 H2' is distinct from 'dm3 soln' and they do not cancel. Wilson Flood Page 24 9/24/2009 Dimensional Analysis Sample calculations Outline solutions to these calculations using the DA method are provided. The solutions have been determined using the following relative atomic masses. Ag Al C Cl Cr F 107.9 27.0 12.0 35.5 52.0 19.0 Fe H K Mg N Na 55.8 1.0 39.1 24.3 14.0 23.0 O P Pb S Ti 16.0 31.0 207.2 32.1 47.9 The Avogadro Constant was taken to be 6 x 1023. 1 2 What is the mass (in grams) of 0.42 mol of K2SO4? How many moles of aluminium carbonate, Al2(CO3)3 , are present in 5.72 g of the substance? What is the mass of 4 x 1022 molecules of H2O? Lead oxide is produced when lead sulphide is heated in oxygen. 2PbS + 3O2 2PbO + 2SO2 3 4 What mass of oxygen would react exactly with 27.1 g of PbS? 5 Titanium(IV) chloride can be converted to titanium by reacting it with an excess of magnesium. TiCl4 + 2Mg Ti + 2MgCl2 What mass of titanium could theoretically be obtained from 2.78 x 10 4 kg of titanium(IV) chloride? 6 Under conditions in which the molar volume of hydrogen is 24.0 dm 3, what volume of hydrogen would be required to produce 10 g of cyclohexane from cyclohexene? C6H10 7 + H2 C6H12 Under conditions in which the molar volume of ozone, O 3, is 22.4 dm3, what volume of ozone would be produced by reacting 17.4 g of fluorine with excess steam? 3F2 + 3H2O 6HF + O3 Wilson Flood Page 25 9/24/2009 Dimensional Analysis 8 What is the empirical formula of the organic compound, a sample of which was found to contain 0.451 g of carbon, 0.0376 g of hydrogen, and 0.202 g of oxygen only? How many moles of potassium hydroxide, KOH , would be present in 220 cm 3 of a solution of concentration 0.45 mol dm-3? How many grams of FeCl2 would be needed to make 500 cm 3 of a solution of concentration 0.250 mol dm-3? A sample of liquid containing phosphoric acid (and no other acids) was completely neutralised by 52.4 cm3 of 0.2 mol dm-3 sodium hydroxide solution. H3PO4 + 3NaOH Na3PO4 + 3H2O 9 10 11 What mass (in grams) of phosphoric acid was in the sample? 12 47.8 cm3 of a solution of potassium hydroxide of concentration 0.4 mol dm -3 exactly neutralised 25 cm3 of sulphuric acid solution. 2KOH + H2SO4 K2SO4 + 2H2O What was the concentration of the acid? 13 20 cm3 of a 0.04 mol dm-3 solution of silver(I) nitrate solution was added to 23 cm3 of a 0.15 mol dm-3 solution of magnesium chloride. 2AgNO3 + MgCl2 2AgCl + Mg(NO3)2 What mass (in g) of silver(I) chloride would be formed? 14 Cr2O72Fe2+ + 14H+ + 6ee2Cr3+ + 7H2O Fe3+ + 30 cm3 of an acidified dichromate(VI) solution with a concentration 0.024 mol dm-3 was titrated against a 0.3 mol dm-3 Fe2+ solution. What volume of Fe2+ solution would be required to reach the end point of this titration? 15 Under conditions in which the molar volume of ammonia is 22.4 dm 3, what volume of ammonia (in dm3) would be required to neutralise 60 cm3 of a 0.075 mol dm-3 solution of sulphuric acid? 2NH3 + H2SO4 (NH4)2SO4 Wilson Flood Page 26 9/24/2009 Dimensional Analysis Solutions to calculations Problem 1 Mass = 0.42 mol K2SO4 x 174.3 g K2SO4 1 mol K2SO4 Problem 2 Mol Al2(CO3)3 Problem 3 Step 1 Mol H2O = 4 x 1022 molecules H2O x Step 2 Mass H2O = 6.67 x 10-2 mol H2O x 18 g H2O 1 mol H2O 1 mol H2O 6 x 1023 molecules H2O = 5.72 g Al2(CO3)3 x 1 mol Al2(CO3)3 234 g Al2(CO3)3 (Note: "molecules" could be abbreviated to "molec" but not "mol" or "mols") Problem 4 Step 1 Mol PbS = 27.1 g PbS x 1 mol PbS 239.3 g PbS Step 2 Mol O2 = 0.113 mol PbS x Step 3 Mass O2 = 0.170 mol O2 x 32 g O2 1 mol O2 3 mol O2 2 mol PbS Wilson Flood Page 27 9/24/2009 Dimensional Analysis Problem 5 Step 1 Mass TiCl4 in g = 2.78 x 104 kg TiCl4 Step 2 Mol TiCl4 = 2.78 x 107 g TiCl4 x 1 mol TiCl4 189.9 g TiCl4 Step 3 Mol Ti = 1.47 x 105 mol TiCl4 x 1 mol Ti 1 mol TiCl4 x 1000 g TiCl4 1 kg TiCl4 Step 4 Mass Ti = 1.47 x 105 mol Ti x 47.9 g Ti 1 mol Ti (This answer could be converted to kg using the CF for g and kg.) Problem 6 Step 1 Mol C6H12 Step 2 Mol H2 Step 3 Vol H2 = 0.119 mol H2 x 24 dm3 H2 1 mol H2 = 0.119 mol C6H12 x 1 mol H2 1 mol C6H12 = 10 g C6H12 x 1 mol C6H12 84 g C6H12 Wilson Flood Page 28 9/24/2009 Dimensional Analysis Problem 7 Step 1 Mol F2 Step 2 Mol O3 Step 3 Vol O3 = 0.153 mol O3 x 22.4 dm3 O3 1 mol O3 = 0.458 mol F2 x 1 mol O3 3 mol F2 = 17.4 g F2 x 1 mol F2 38 g F2 Problem 8 Mol C = 0.451 g C x 1 mol C 12 g C 1 mol H 1gH 1 mol O 16 g O = 3:3:1 Mol H = 0.0376 g H x Mol O = 0.202 g O x C:H:O = 0.038 : 0.038 : 0.013 Problem 9 Mol KOH = 220 cm3 soln x 0.45 mol KOH x 1 dm3 soln 1 dm3 soln 1000 cm3 soln Problem 10 Step 1 Mol FeCl2 Step 2 Mass FeCl2 = 0.125 mol FeCl2 x 126.8 g FeCl2 1 mol FeCl2 = 500 cm3 soln x 0.250 mol FeCl2 x 1 dm3 soln 1 dm3 soln 1000 cm3 soln Wilson Flood Page 29 9/24/2009 Dimensional Analysis Problem 11 Step 1 Mol NaOH = 52.4 cm3 soln x 0.2 mol NaOH 1 dm3 soln x 1 dm3 soln 1000 cm3 soln Step 2 Mol H3PO4 Step 3 Mass H3PO4 Problem 12 Step 1 Mol KOH = 47.8 cm3 soln x 0.4 mol KOH 1 dm3 soln x 1 dm3 soln 1000 cm3 soln = 0.0035 mol H3PO4 x 98 g H3PO4 1 mol H3PO4 = 0.0105 mol NaOH x 1 mol H3PO4 3 mol NaOH Step 2 Mol H2SO4 Step 3 Concn H2SO4 = 0.0095 mol H2SO4 25 cm3 soln 0.038 mol dm-3 x 1000 cm3 soln 1 dm3 soln = 0.019 mol KOH x 1 mol H2SO4 2 mol KOH = Wilson Flood Page 30 9/24/2009 Dimensional Analysis Problem 13 Step 1 Mol AgNO3 = 20 cm3 soln x 0.04 mol AgNO3 1 dm3 soln 8 X 10-4 mol AgNO3 x 1 dm3 soln 1000 cm3 soln = Step 2 Mol MgCl2 = 23 cm3 soln x 0.15 mol MgCl2 1 dm3 soln x 1 dm3 soln 1000 cm3 soln = 3.45 x 10-3 mol MgCl2 AgNO3 is limiting reactant Step3 Mol AgCl = 8 x 10-4 mol AgNO3 x 2 mol AgCl 2 mol AgNO3 Step 4 Mass AgCl = 8 x 10-4 mol AgCl x 143.4 g AgCl 1 mol AgCl Problem 14 Step 1 Mol Cr2O72Step 2 Mol Fe2+ = 4.8 x 10-4 mol Cr2O72- x 6 mol Fe2+ 1 mol Cr2O72= 20 cm3 soln x 0.024 mol Cr2O721 dm3 soln x 1 dm3 soln 1000 cm3 soln Step 3 Vol soln = 2.88 x 10-3 mol Fe2+ x 1 dm3 soln x 0.3 mol Fe2+ 1000 cm3 soln 1 dm3 soln Wilson Flood Page 31 9/24/2009 Dimensional Analysis Problem 15 Step 1 Mol H2 SO4 = 60 cm3 soln x 0.075 mol H2SO4 1 dm3 soln x 1 dm3 soln 1000 cm3 soln Step 2 Mol NH3 Step 3 Vol NH3 = 9 x !0-3 mol NH3 x 22.4 dm3 NH3 1 mol NH3 = 4.5 x 10-3 mol H2SO4 x 2 mol NH3 1 mol H2SO4 Wilson Flood Page 32 9/24/2009