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Stoichiometric Calculations

Stoichiometric Calculations. Review of Fundamental Concepts. Formula Weight

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Stoichiometric Calculations

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  1. Stoichiometric Calculations

  2. Review of Fundamental Concepts Formula Weight It is assumed that you can calculate the formula or molecular weights of compounds from respective atomic weights of the elements forming these compounds. The formula weight (FW) of a substance is the sum of the atomic weights of the elements from which this substance is formed from.

  3. Find the formula weight of CaSO4.7H2O

  4. The Mole Concept

  5. Assuming approximate atomic weights of 1.00, 16.00, 23.00, 35.5, and 12.00 atomic mass units for hydrogen, oxygen, sodium, chlorine atom, and carbon, respectively. The number of moles contained in a specific mass of a substance can be calculated as: mol = g substance/FW substance The unit for the formula weight is g/mol

  6. In the same manner, the number of mmol of a substance contained in a specific weight of the substance can be calculated as mmol = mol/1000 Or, mmol = mg substance/FW substance

  7. The number of mmolof Na2WO4 (FW = 293.8 mg/mmol) present in 500 mg of Na2WO4 can be calculated as ? mmol of Na2WO4 = 500 mg/293.8 (mg/mmol) = 1.70 mmol

  8. The number of mgcontained in 0.25 mmol of Fe2O3 (FW = 159.7 mg/mmol) can be calculated as ? mg Fe2O3 = 0.25 mmol Fe2O3 x 159.7 (mg/mmol) = 39.9 mg Therefore, either the number of mg of a substance can be obtained from its mmols or vice versa.

  9. Calculations Involving Solutions Molarity Molarity of a solution can be defined as the number of moles of solute dissolved in 1 L of solution. This means that 1 mol of solute will be dissolved in some amount of water and the volume will be adjusted to 1 L. The amount of water may be less than 1 L as the final volume of solute and water is exactly 1 L.

  10. Calculations Involving Molarity Molarity = mol/L = mmol/mL This can be further formulated as Number of moles = Molarity X volume in Liters, or mol = M (mol/L) x VL Number of mmol = Molarity x volume in mL , or mmol = M (mmol/mL) x VmL

  11. Find the molarity of a solution resulting from dissolving 1.26 g of AgNO3 (FW = 169.9 g/mol) in a total volume of 250 mL solution. First find mmol AgNO3 = 1.26*103 mg AgNO3 / 169.9 mg/mmol = 7.42 mmol Molarity = mmol/mL M = 7.42 mmol/250 mL = 0.0297 mmol/mL

  12. We can find the molarity directly in one step using dimensional analysis ? mol AgNO3 / L = (1.26 g AgNO3 / 250 mL) x ( mol AgNO3/169.9 g AgNO3) x (1000 mL/1L) =0.0297 M

  13. Let us find the number of mg of NaCl per mL of a 0.25 M NaCl solution First we should be able to recognize the molarity as 0.25 mol/L or 0.25 mmol/mL. Of course, the second term offers what we need directly ? mg NaCl in 1 mL = (0.25 mmol NaCl/mL) x (58.5 mg NaCl/mmol NaCl) = 14.6 mg NaCl/mL

  14. Find the number of grams of Na2SO4 required to prepare 500 mL of 0.1 M solution. First, we find mmoles needed from the relation mmol = M (mmol/mL) x VmL mmol Na2SO4 = 0.1 mmol/mL x 500 mL = 50 mmol mmol = mg substance/FW substance ?mg Na2SO4 = 50 mmol x 142 mg/mmol = 7100 mg or 7.1 g

  15. One can use dimensional analysis to find the answer in one step as ? g Na2SO4 = (0.1 mol Na2SO4/1000 mL) x 500 mL x (142 g Na2SO4/mol) = 7.1 g

  16. When two or more solutions are mixed, one can find the final concentration of each ion. However, you should always remember that the number of moles ( or mmoles ) is additive. For example: Find the molarity of K+ after mixing 100 mL of 0.25 M KCl with 200 mL of 0.1 M K2SO4.

  17. The idea here is to calculate the total mmol of K+ and divide it by volume in mL. mmol K+ = mmol K+ from KCl + mmol K+ from K2SO4 = 0.25 mmol/ml x 100 mL + 2x0.1 mmol/mL x 200 mL = 65 mmol Molarity = 65 mmol/(200 + 100) mL = 0.22 M Note that the concentration of K+ in 0.1M K2SO4 is 2x0.1M (i.e. 0.2 M)

  18. Lecture 9 Stoichiometric Calculations Normality Density Calculations

  19. Normality We have previously talked about molarity as a method for expressing concentration. The second expression used to describe concentration of a solution is the normality. Normality can be defined as the number of equivalents of solute dissolved in 1 L of solution. Therefore, it is important for us to define what we mean by the number of equivalents, as well as the equivalent weight of a substance as a parallel term to formula weight.

  20. An equivalent is defined as the weight of substance giving an Avogadro’s number of reacting units. Reacting units are either protons or hydroxides (in acid base reactions) or electrons (in oxidation reduction reactions). For example, HCl has one reacting unit (H+) when reacting with a base like NaOH but sulfuric acid has two reacting units (two protons) when reacting completely with a base.

  21. Therefore, we say that the equivalent weight of HCl is equal to its formula weight and the equivalent weight of sulfuric acid is one half its formula weight. In the reaction where Mn(VII), in KMnO4, is reduced to Mn(II) five electrons are involved and the equivalent weight of KMnO4 is equal to its formula weight divided by 5.

  22. N = eq/L or N = meq/mL Number of eq = Normality x VL = (eq/L) x L Number of meq = Normality x VmL = (meq/mL) x mL Also, number of equivalents = wt(g)/equivalent weight (g/eq)  meq = mg/eqw

  23. Equivalent weight = FW/n meq = mg/eqw substitute for eqw = FW/n gives: meq = mg/(FW/n), but mmol = mg/FW, therefore: meq = n * mmol  dividing both sides by volume in mL, we get: N = n M Where n is the number of reacting units ( protons, hydroxides, or electrons ) and if you are forming factors always remember that a mole contains n equivalents. The factor becomes (1 mol/n eq) or (n eq/1 mol).

  24. Find the equivalent weights of NH3 (FW = 17.03), H2C2O4 (FW = 90.04) in the reaction with excess NaOH, and KMnO4 (FW = 158.04) when Mn(VII) is reduced to Mn(II). Solution Ammonia reacts with one proton only Equivalent weights of NH3 = FW/1 = 17.03 g/eq

  25. Two protons of oxalic acid react with the base Equivalent weights of H2C2O4 = FW/2 = 90.04/2 = 45.02 g/eq Five electrons are involved in the reduction of Mn(VII) to Mn(II) Equivalent weights of KMnO4 = FW/5 = 158.04/5 = 31.608 g/eq

  26. Find the normality of the solution containing 5.300 g/L of Na2CO3 (FW = 105.99), carbonate reacts with two protons. Normality is the number of equivalents per liter, therefore we first find the number of equivalents eq wt = FW/2 = 105.99/2 = 53.00 eq = Wt/eq wt = 5.300/53.00 = 0.1000 N = eq/L = 0.1000 eq/1L = 0.1000 N

  27. The problem can be worked out simply as below ? eq Na2CO3 /L = (5.300 g Na2CO3 /L ) x (1 mol Na2CO3 /105.99 g Na2CO3 ) x (2 eq Na2CO3 /1 mol Na2CO3) = 0.1 N A further option is to find the number of moles first followed by multiplying the result by 2 to obtain the number of equivalents.

  28. The other choice is to find the molarity first and the convert it to normality using the relation: N = n M No of mol = 5.300 g/(105.99 g/mol) M = mol/L = [5.300 g/(105.99 g/mol)]/ 1L N = n M = 2 x [5.300 g/(105.99 g/mol)]/ 1L = 0.1000

  29. Find the normality of the solution containing 5.267 g/L K2Cr2O7 (FW = 294.19) if Cr6+ is reduced to Cr3+. The same as the previous example N = eq/L, therefore we should find the number of eq where eq = wt/eq wt, therefore we should find the equivalent weight; where eq wt = FW/n. Here; each Contributes three electrons and since the dichromate contains two Cr atoms we have 6 reacting units

  30. Eq wt = (294.19 g/mol)/(6 eq/mol) Eq = 5.267 g/ (294.19 g/mol)/(6 eq/mol) N = eq/L = (294.19 g/mol)/(6 eq/mol)/1L = 0.1074 eq/L Using the dimensional analysis we may write ? eq K2Cr2O7 /L = (5.267 g K2Cr2O7 /L) x (mol K2Cr2O7 /294.19 g K2Cr2O7 ) x (6 eq K2Cr2O7 /mol K2Cr2O7 ) = 0.1074 eq/L

  31. Again one can choose to calculate the molarity then convert it to normality mol = 5.267 g/(294.19 g/mol) M = mol/L = [5.267 g/(294.19 g/mol)]/L N = n M N = (6 eq/mol)x [5.267 g/(294.19 g/mol)]/L = 0.1074 eq/L

  32. Density Calculations In this section, you will learn how to find the molarity of solution from two pieces of information (density and percentage). Usually the calculation is simple and can be done using several procedures. Look at the examples below:

  33. Example What volume of concentrated HCl (FW = 36.5g/mol, 32%, density = 1.1g/mL) are required to prepare 500 mL of 2.0 M solution. Always start with the density and find how many grams of solute in each mL of solution. Density = g solution/mL

  34. Remember that only a percentage of the solution is solute . mg HCl/ml = 1.1 x 0.32 x103 mg HCl / mL The problem is now simple as it requires conversion of mg HCl to mmol since the molarity is mmol per mL M = mmol HCl/mL = 1.1x0.32 x103 mg HCl/(36.5 mg/mmol) = 9.64 M

  35. Now, we can calculate the volume required from the relation MiVi (before dilution) = MfVf (after dilution) 9.64 x VmL= 2.0 x 500mL  VmL = 10.4 mL This means that 10.4 mL of the concentrated HCl should be added to distilled water and the volume should then be adjusted to 500 mL

  36. How many mL of concentrated H2SO4 (FW = 98.1 g/mol, 94%, d = 1.831 g/mL) are required to prepare 1 L of 0.100 M solution? mg H2SO4 / mL = 1.831*0.94*103 mg /mL Now we only need to convert mg to mmol M = mmol/mL = [(1.831 x 0.94 x 103 mg) / (98.1 mg/mmol)] / mL = 17.5 M

  37. To find the volume required to prepare the solution MiVi (before dilution) = MfVf (after dilution) 17.5 x VmL = 0.100 x 1000 mL VmL = 5.71 mL which should be added to distilled water and then adjusted to 1 L.

  38. Density * percentage * 103 M = Formula Weight An Easy Short-Cut The percentage is a fraction: (i.e. a 35% is written as 0.35)

  39. Analytical Versus Equilibrium Concentration When we prepare a solution by weighing a specific amount of solute and dissolve it in a specific volume of solution, we get a solution with specific concentration. This concentration is referred to as analytical concentration. However, the concentration in solution may be different from the analytical concentration, especially when partially dissociating substances are used.

  40. An example would be clear if we consider preparing 0.1 M acetic acid (weak acid) by dissolving 0.1 mol of the acid in 1 L solution. Now, we have an analytical concentration of acetic acid (HOAc) equals 0.1 M. But what is the actual equilibrium concentration of HOAc? We have HOAc = H+ + OAc- The analytical concentration ( CHOAc ) = 0.1 M CHOAc = [HOAc]undissociated + [OAc-] The equilibrium concentration = [HOAc]undissociated.

  41. For good electrolytes which are 100% dissociated in water the analytical and equilibrium concentrations can be calculated for the ions, rather than the whole species. For example, a 1.0 M CaCl2 in waterresults in 0 M CaCl2, 1.0 M Ca2+, and 2.0 M Cl- since all calcium chloride dissociates in solution. For species x we express the analytical concentration as Cx and the equilibrium concentration as [x].

  42. Lecture 10 Stoichiometric Calculations, cont….. Expressing Concentrations

  43. Dilution Problems In many cases, a dilution step or steps are involved in analytical procedures. One should always remember that in any dilution the number of mmoles of the initial (concentrated) solution is equal to the number of mmoles of the diluted solution. This means: MiVi (concentrated) = MfVf (dilute)

  44. Prepare 200 mL of 0.12 M KNO3 solution from 0.48 M solution. MiVi (concentrated) = MfVf (dilute) 0.48 x VmL = 0.12 x 200 VmL = (0.12 x 200)/0.48 = 50 mL Therefore, 50 mL of 0.48 M KNO3 should be diluted to 200 mL to obtain 0.12 M solution

  45. A 5.0 g Mn sample was dissolved in 100 mL water. If the percentage of Mn (At wt = 55 g/mol) in the sample is about 5%. What volume is needed to prepare 100 mL of approximately 3.0x10-3 M solution. First we find approximate mol Mn in the sample = 5.0 x (5/100) g Mn/(55 g/mol) = 4.5x10-3 mol Molarity of Mn solution = (4.5x10-3 x 103 mmol)/100 mL = 4.5x10-2 M

  46. The problem can now be solved easily using the dilution relation MiVi (concentrated) = MfVf (dilute) 4.5x10-2 x VmL = 3.0x10-3 x 100 VmL = (3.0x10-3 x 100)/4.5x10-2 = 6.7 mL Therefore, about 6.7 mL of the Mn sample should be diluted to obtain an approximate concentration of 3.0x10-3 M solution.

  47.  What volume of 0.4 M Ba(OH)2 should be added to 50 mL of 0.30 M NaOH in order to obtain a solution that is 0.5 M in OH-. We have to be able to see that the mmol OH- coming from Ba(OH)2 and NaOH will equal the number of mmol of OH- in the final solution, which is mmol OH- from Ba(OH)2 + mmol OH- from NaOH = mmol OH- in final solution

  48. The mmol OH- from Ba(OH)2 is molarity of OH- times volume and so are other terms. Molarity of OH- from Ba(OH)2 is 0.8 M (twice the concentration of Ba(OH)2, and its volume is x mL. Now performing the substitution we get 0.8 * x+ 0.30 * 50 = 0.5 * (x+ 50) x = 33 mL

  49. Part per hundred, % Part per thousand, ppt Part per million, ppm Part per billion, ppb Solid solutes in solid samples Solid solutes in solutions Liquid solutes in solutions Expressing Concentrations

  50. For Solid Solutes in solid samples % (w/w) = [weight solute (g)/weight sample (g)] x 100 ppt (w/w) = [weight solute (g)/weight sample (g)] x 1000 ppm (w/w) = [weight solute (g)/weight sample (g)] x 106 ppb (w/w) = [weight solute (g)/weight sample (g)] x 109 A ppm can be represented by several terms like the one above, (mg solute/kg sample), ( g solute/106g sample), etc..

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