נושא 5. 1 Prof. Zvi C. Koren

Size: px

Start display at page:

Download "נושא 5. 1 Prof. Zvi C. Koren"

Annabel Clark
5 years ago
Views:

1 נושא 5 סטויכיאומטריה: כימות כימי 1 Prof. Zvi C. Koren

2 Stoichiometry Stoicheion + (element) metron (measure) Weight relations in chemical rxns. based on conservation of matter For any rxn., The absolute value of each coefficient is meaningless by itself! BUT, the RATIOS are HOLY!!! Examples: אם-אז IFThen: relationship כימות כימי 2Mg(s) + O 2 (g) 2MgO(s) 2 atoms 2x 6.02x10 23 atoms 2 moles of atoms 48.6 g 1 molecule 6.02x10 23 molecules 1 mole of molecules 32.0 g 2 molecules 2x 6.02x10 23 molec. 2 moles of molecules 80.6 g 2 Prof. Zvi C. Koren

3 Stoichiometric Calculations: The Approach Stoichiometry (rxn) mole B gram mole MW (or AW) mole A gram A Helpful Tips for Solving Problems: 1. Write the Balanced rxn!! 2. Think in Moles!!! Simple formula: n m MW 3. Setup a Flow-Chart whereby g mol mol 4. Always include Units and Substance Name 3 Prof. Zvi C. Koren

4 Stoichiometric Calculations: Examples Calculate the number of grams of MgO produced from g Mg. 2Mg(s) + O 2 (g) 2MgO(s) grams grams? AW, MW moles Stoichiometric Ratio: Rxn moles MW 1mol Mg g Mg g Mg MW factor 2 mol MgO g MgO 2 mol Mg 1mol MgO Rxn factor or Stoichiometric factor MW factor = g MgO So, Remember, All Roads Go Through Moles!!! 4 Prof. Zvi C. Koren

5 The Limiting Reagent (or Limiting Reactant) Limits quantities reacting and produced Completely reacts (if rxn goes to completion) 3CCl 4 (l) + 2 SbF 3 (s) 3 CCl 2 F 2 (l) + 2 SbCl 3 (s) 150. g CCl 4 and 100. g SbF 3 are in a flask. How many grams of CCl 2 F 2 are produced? 1. Calculate the number of moles of each reactant: 1mol CCl 150. g CCl 4 4 1mol SbF = mol CCl g CCl g SbF 3 3 = mol SbF g SbF Assume all of one reactant completely reacts and determine how much of the other is needed. If all of CCl 4 reacts, how many moles of SbF 3 are needed? 2 mol SbF IF mol CCl 4 3 = mol SbF 3 mol CCl 3 needed. BUT, we DON T have it!!! 4 If all of SbF 3 reacts, how many moles of CCl 4 are needed? 3 mol CCl IF mol SbF 3 4 = mol CCl 4 needed. AND, we DO have it!!! 2 mol SbF3 3. Identify the Limiting Reagent: SbF 3 is the limiting reagent! (CCl 4 is the reagent in excess ) 4. Calculate all quantities based on the limiting reagent: 3 mol CCl mol SbF 3 2F g CCl 2F2 2 mol SbF = 101 g CCl 1mol CCl F 2 F Prof. Zvi C. Koren

6 Analyzing Mixtures of Similar Compounds For example, consider NaHCO 3 (s) and Na 2 CO 3 (s). Both react with HCl(aq): NaHCO 3 (s) + HCl(aq) NaCl(aq) + H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) Na 2 CO 3 (s) + 2HCl(aq) 2NaCl(aq) + H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) Do NOT combine the two rxns.!!! Problem: Calculate the percent of Na 2 CO 3 in the mixture when a 10.0-g mixture of sodium bicarbonate and sodium carbonate react with an excess of hydrochloric acid to produce g of NaCl, collected after evaporation. Solution: Let the wonders of algebra solve all of your problems: Let x = grams of Na 2 CO 3 in mixture 10.0 x = grams of NaHCO 3 in mixture n NaCl,total = (n NaClbicarbonate ) + (n NaClcarbonate ) MW of NaCl = (n bicarbonate ) + 2 (n carbonate ) 10.0 x MW of NaHCO 3 = + Also simplify by using n = m/mw x 2 MW of Na CO 2 3 x = 3.50 g % Na 2 CO 3 in mixture = (3.50 g/ 10.0 g) 100 = 35.0 % 6 Prof. Zvi C. Koren

7 Percent Yield Many rxns do NOT go to completion. There is a chemical energy barrier involved. For example: actual yield % Yield = x 100 theoretical yield 2 C 7 H 6 O 3 (s) + C 4 H 6 O 3 (l) 2 C 9 H 8 O 4 (s) + H 2 O(l) salicylic acetic acetyl acid anhydride salicylic acid Aspirin If from g of the acid, 6.26 g of aspirin is produced, what is the % yield for the rxn? Theoretical yield or maximum yield (assuming rxn goes to completion): 1mol acid 2 mol aspirin g aspirin g acid g acid 2 mol acid 1mol aspirin = g aspirin 6.26 g % yield = x g = 33.3 % 7 Prof. Zvi C. Koren

8 Molarity = M = n / V 3.0 M KMnO 4 Solution Concentrations Molarity, Formality, molality, Normality, % w/w, % w/v 3.0 molar potassium permanganate solution # of moles of solute per liter of solution # of mmoles of solute per ml of solution 3.0 moles of KMnO 4 per L of solution 3.0 mmoles of KMnO 4 per ml of solution Dilutions Use n = M V Prepare ml of a M KMnO 4 solution from a 3.0-M stock solution? moles/l mmoles/ml n solute in new solution = M new V new = (0.100 mol KMnO 4 /L)( L) = = n solute from stock solution = M stock V stock = (3.0 mol KMnO 4 /L) V old = n = M stock V stock M new V new (3.0 M) V stock = (0.100 M) ( L) V stock = L = 16.7 ml Preparation of a Diluted Solution: Remove 16.7 ml of the 3.0-M stock solution Add enough water ( ml ) to produce 500. ml of solution 8 Prof. Zvi C. Koren

Solution Stoichiometry Titrations: Acid-Base & Redox n = M V End-Point or Equivalence Point of an Acid-Base or Redox rxn: Point at which all of of the Acid reacts with all of the Base or all of the

Problem 1: Reaching an end-point How many ml of 0.300 M sodium hydroxide are needed to titrate 25.0 ml of 0.400 M oxalic acid? Buret Solution: Write the Rxn.

9 Solution Stoichiometry Titrations: Acid-Base & Redox n = M V End-Point or Equivalence Point of an Acid-Base or Redox rxn: Point at which all of of the Acid reacts with all of the Base or all of the oxidant reacts with all of the reductant. Acid-Base Titrations Add indicator (or use ph meter) to indicate when the end point of the rxn has been reached: color changes (base acid). Problem 1: Reaching an end-point How many ml of M sodium hydroxide are needed to titrate 25.0 ml of M oxalic acid? Buret Solution: Write the Rxn.: 2 NaOH(aq) + H 2 C 2 O 4 (aq) Na 2 C 2 O 4 (aq) + 2 H 2 O(l) Stoichiometric Flow-Chart: n=m V rxn V=n/M Erlenmeyer M A,V A moles acid moles base V base flask n ACID = M A V A = (0.400 M)( L) = mol = 2 n ACID = 2 ( mol) = mol n BASE V BASE = n BASE / M BASE = mol / M = L = 66.7 ml 9 Prof. Zvi C. Koren

10 Acid-Base Problems (continued) Problem 2: Standardization of a solution If g of solid sodium carbonate requires ml of hydrochloric acid for titration to the equivalence point, what is the molar concentration of HCl? Solution: Write the Rxn.: basic salt acid Na 2 CO HCl H 2 CO NaCl Stoichiometric Flow-Chart: MW g salt/base rxn M=n/V moles salt/base moles acid M acid g Na n Na2CO3 = m / MW = 2CO3 = mol base g / mol Na2CO3 2 mol acid n HCl = mol base = mol acid 1mol base M HCl = n / V = ( mol) / ( L) = M HCl 10 Prof. Zvi C. Koren

11 Redox Titrations Redox Problems (same stoichiometric principles as acid-base problems) Problem: In an acidic solution, g of an iron(ii)-containing ore requires ml of M KMnO 4 to reach the equivalence point, what is the weight percentage of iron in the ore? [The two redox products are Fe 3+ and Mn 2+.] Solution: Write the Rxn.: MnO 4 + Fe > Mn 2+ + Fe 3+ MnO 4 purple + 8H + + 5Fe 2+ colorless Mn Fe 3+ colorless colorless + 4H 2 O Stoichiometric Flow-Chart: n=mv rxn MW M KMnO4,V KMnO4 moles MnO 4 moles Fe 2+ grams Fe 2+ n KMnO4 = M V = ( M)( L) = x 10 4 mole MnO mol Fe g Fe m x 10-4 Fe2+ = mole MnO mol MnO4 1mol Fe g Fe % Fe in ore = x ore = 13.3 % Fe in ore g 11 Prof. Zvi C. Koren 2 = g Fe 2+

12 Solution Concentrations (continued) Molarity, Formality, molality, Normality, % w/w, % w/v Normality = N = eq / V # of equivalents of solute per liter of solution # of meqs of solute per ml of solution eqs/l meqs/ml 1 equivalent of a substance is that quantity that reacts with (or produces) 1 mole of H + (in acid-base rxns.) or 1 mole of e (in redox rxns.), etc. In general: )צ ור ון( 1 eq 1 mol transferred species In addition: Equivalent Weight (EW) = Weight (Mass) of 1 eq: g/eq Recall: Molecular Weight (MW) = Weight (Mass) of 1 mol: g/mol EW vs. MW # g/eq = (# g/mol)(1 mol/# eqs) EW MW Koren s N vs. M Alphabet # eq/l = (# mol/l)(# eqs/mol) N M Rule In general, MW & M are absolute; EW & N are relative (depend on rxn.) 12 Prof. Zvi C. Koren

13 Examples: Normality (continued) H 2 SO 4 as an acid (full rxn): 1 mol H 2 SO 4 2 mol H + 1 mol H 2 SO 4 2 eqs H 2 SO 4 MW (H 2 SO 4 ) = 98 g/mol. EW (H 2 SO 4 ) = (98 g/mol)(1 mol/2 eqs) = 49 g/eq For a 3.0 M H 2 SO 4 solution Normality = (3.0 mol/l)(2 eqs/mol) = 6.0 N Na 2 Cr 2 O 7 reduced to CrCl 3 : Na 2 Cr 2 O 7 + 6e ---> 2CrCl 3 : 1 mol Na 2 Cr 2 O 7 6 mol e 1 mol Na 2 Cr 2 O 7 6 eqs Na 2 Cr 2 O 7 MW (Na 2 Cr 2 O 7 ) = 262 g/mol. EW (Na 2 Cr 2 O 7 ) = (262 g/mol)(1 mol/6 eqs) = 43.7 g/eq For example, for a 3.0 M Na 2 Cr 2 O 7 solution Normality = N Ca(OH) 2 : Ca(OH) 2 (aq) Ca OH 1 mol Ca(OH) 2 2 eq Ca(OH) 2 H 2 O oxidized to H 2 O 2 : 2H 2 O H 2 O 2 + 2e 1 mol H 2 O 1 eq H 2 O 13 Prof. Zvi C. Koren

14 Normality & Titrations In general, for chemical rxns between two reactants: n 1 n 2 M 1 V 1 M 2 V 2 For example: H 2 SO 4 (aq) + 2NaOH(aq) Na 2 SO 4 (aq) + 2H 2 O(l) But, the following is ALWAYS true (for acid-base or redox rxns, etc.): Koren s Jealousy Equation: # eqs 1 = # eqs 2 N 1 V 1 = N 2 V 2 Acid-Base Example: Question: In the above acid-base rxn, what is the normality and molarity of the acid if 20.0 ml of the acid are titrated with 40.0 ml of M base to reach the end-point? Answer: For NaOH: N = M. N A V A = N B V B N A (20.0 ml) = (0.200 N)(40.0 ml) N A = N M A = M Redox Example: For the rxn: MnO 4 + 8H + + 5Fe 2+ Mn Fe H 2 O N V N 2V 2 N vs. M, N 2 vs. M 2 MnO 4 MnO 4 Fe Fe MnO 4 MnO 4 Fe Fe 14 Prof. Zvi C. Koren

15 More Solution Concentrations Terminology for dissolution: a solute is dissolved by a solvent to form a solution. Name Definition Formula Units Molarity # of moles of solute per L of solution M = mols solute /L sol n M Normality # of equivalents of solute p. L of sol n N = eqs solute /L sol n N Density of Sol n (d or, rho) mass of the sol n per volume of sol n d sol n = m sol n /V sol n g/ml Molality # of moles of solute per kg of solvent m = mols solute /kg solvent In dilute aqueous solutions: m M m Mole fraction moles of a component p. total moles X i = n i /n sol n, ΣX i = 1 (none) Mole percent mole fraction of a comp. as a percent X i 100 % Weight percent weight of solute/weight of sol n, as % w/w % = m i /m sol n 100 weight of solute/volume of sol n, as % w/v % = g i /ml sol n 100 % # of solute parts p. million sol n parts ppm = g i /g sol n = mg/kg Parts per million In dilute aqueous solutions: ppm mg/l ppm # of solute parts p. billion sol n parts ppb = ng i /g sol n = g/kg Parts per billion In dilute aqueous solutions: ppb g/l ppb Volume percent vol. of pure solute/vol. of sol n, as % v/v %=V i,pure /V sol n 100 % Proof Double the Volume % (for whiskey) Proof = 2(v/v %) Proof Questions: What is ppt? pph? 15 Note: Must always write the units and the substance, e.g., 2.0 g solute. Prof. Zvi C. Koren

16 Selected Concentration Examples: (1) 1.2 kg ethylene glycol (HOCH 2 CH 2 OH), an antifreeze, is added to 4.0 kg water. Calculate (for ethylene glycol): mole fraction, molality, weight/weight%. [Answers: X = 0.080, m = 4.8 m, w/w % = 23 %] (2) 560 g NaHSO 4 are dissolved in 4.5x10 5 L water at 25 o C. Calculate the Na + concentration in parts per million. [Answer: 0.24 ppm] (3) 10.0 g of sucrose (C 12 H 22 O 11 ) are dissolved in 250. g of water. Calculate (for sugar): X, m, w/w %. [Answers: X = , m = m, w/w % = 3.85 %.] (4) Sea water has a sodium ion concentration of 1.08 x 10 4 ppm. If the Na is present in the form of dissolved sodium chloride, how many grams of NaCl are in each liter of sea water? (Density of sea water is 1.05 g/ml.) [Answer: 28.7 g NaCl/L] (5) A M aqueous solution of ethylene glycol has a density of 1.09 g/ml. What is the molality of the solution? [Answer: m] 16 Prof. Zvi C. Koren

Solution Concentration

Solution Concentration solution: homogeneous mixture of substances present as atoms, ions, and/or molecules solute: component present in smaller amount solvent: component present in greater amount Note: