worksheet_06_103 (1)

1 CHEMISTRY 103 – WORKSHEET #6 – Module 4 Stoichiometry (Part III) Do the topics appropriate for your course Prepared by Dr. Tony Jacob https://clc.chem.wisc.edu (Resource page) MOLARITY: concentration of solutions; abbreviated as M ( “molar”) Solute : chemical that dissolves; Solvent : liquid solute dissolves in; Solution : Solvent + Solute 1. Molarity calculations (ID this type of problem: 1 chemical; 1 concentration) : use: Example 1: How many grams Na 2 CO 3 are needed to prepare 650ml of a 2.25M Na 2 CO 3 solution? Answer 1: 155g Na 2 CO 3 {1 chemical and 1 concentration ® use ; ; x = 1.4625mol Na 2 CO 3 ; } 2. Dilution Problems (ID this type of problem: 1 chemical; 2 concentrations) : use: M 1 V 1 = M 2 V 2 equation used only with dilutions (1 = initial, 2 = final; V 2 = total volume = V 1 + water) Example 2: 2.00ml of 2.5 x 10 -4 M Zn(NO 3 ) 2 (aq) was diluted in a volumetric flask to 25.00ml. 5.00ml of this new solution was removed with an Eppendorf pipette and diluted again in a new volumetric flask to 25.00ml. What is the final concentration of the Zn(NO 3 ) 2 ? Answer 2: 4.0 x 10 -6 M (1 chemical and 2 concentrations ® a dilution; use M 1 V 1 = M 2 V 2 two times); M 1 = 2.5 x 10 -4 M; V 1 = 2.00ml; M 2 = x; V 2 = 25.00ml; (2.5 x 10 -4 M)(2.00) = M 2 (25.00); M 2 = 2.0 x 10 -5 M; use M 1 V 1 = M 2 V 2 again: M 1 = 2.0 x 10 -5 M; V 1 = 5.00ml; M 2 = x; V 2 = 25.00ml; (2.0 x 10 -5 M)(5.00) = M 2 (25.00); M 2 = 4.0 x 10 -6 M; Note: you can use ml in the dilution equation as long as both volumes are in ml 3. Stoichiometry (ID this type of problem: 2 chemicals reacting) : can be referred to as neutralization, titration, completely reacts – both reactants run out; use flowchart Example 3: If it required 56.0ml of a 0.250M H 2 SO 4 solution to neutralize 26.5ml NaOH, what was the original concentration of the NaOH? Answer 3: 1.06M NaOH (2 chemicals ® stoichiometric problem; M A ® M B question; 3 steps from the above flow chart) Step 1: find mol H 2 SO 4 : M = mol/L; Step 2: convert mol H 2 SO 4 to mol NaOH using a balanced reaction; [hint: the stoichiometric ratio between an acid and base can be determined by inspection without writing the reaction; the ratio must be: 1 H + to 1 OH - or in other words, the same number of H + and OH - ; since H 2 SO 4 has 2 H + and NaOH has 1 OH - , there needs to be 1 H 2 SO 4 (2 H + ) and 2 NaOH (2 OH - ): H 2 SO 4 (aq) + 2NaOH(aq) ® 2H 2 O(l) + Na 2 SO 4 (aq) Step 3: calculate the concentration using M NaOH = mol NaOH /L NaOH ; M = mol solute L solution M = mol L 2.25M = x mol (650ml)(1L /1000ml) 1 . 4625 mol Na 2 CO 3 106 . 0 g Na 2 CO 3 1 mol Na 2 CO 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 155 . 03 g Na 2 CO 3 atoms or molecules B atoms or molecules A grams A grams B moles A moles B 1 mol B = 6.022 x 10 B 23 1 mol A = 6.022 x 10 A 23 molar mass A molar mass B molarity A molarity B M = mol /L A A A M = mol /L B B B chemical formula or chemical reaction mol H 2 SO 4 = M x L = 0.250M (56.0ml)(1L) 1000ml ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.0140mol H 2 SO 4 0 . 0140 mol H 2 SO 4 2 mol NaOH 1 mol H 2 SO 4 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0 . 0280 mol NaOH M NaOH = 0.0280mol NaOH (26.5ml)(1L /1000ml) = 1.057M

2 1. What volume in liters of a 0.20M Na 2 CO 3 solution contains 53g of Na 2 CO 3 ? (Helpful information: molar mass Na 2 CO 3 = 106g/mol) a. 2.5L b. 0.40L c. 0.50L d. 1.6L e. 2.0L 2. In lab, you obtain a 5.00 x 10 -3 M solution of ferroin. You take 2.00ml using an Eppendorf pipet, add it to a 10.00ml volumetric flask, and dilute to the mark with deionized water. You then do one more dilution using the same volumes listed above. What is the final concentration of the ferroin solution? a. 5.00 x 10 -3 M b. 1.00 x 10 -3 M c. 2.00 x 10 -4 M d. 0.125M e. 0.025M 3. If 0.50mol Li 3 N is dissolved into 2.0L water, what is the concentration of all ions in solution? a. 0.25M b. 0.50M c. 1.0M c. 2.0M e. 4.0M 4. Four solutions are shown below. Each circle represents an equal quantity of solute. i ii iii iv I. Which solution has the greatest concentration ? a. i b. ii c. iii d. iv II. Which solution has the smallest concentration ? a. i b. ii c. iii d. iv III. Which solutions have the same concentration ? a. i, ii b. ii, iii c. i, iii d. ii, iv e. iii, iv 1.0L 1.0L 0.50L 0.25L

4 CHEMISTRY 103 – WORKSHEET #6 - ANSWERS Stoichiometry (Part III) Do the topics appropriate for your lecture Prepared by Dr. Tony Jacob https://clc.chem.wisc.edu (Resource page) ANSWERS 1. a { ; } 2. c {use M 1 V 1 = M 2 V 2 2 times; (5.00 x 10 -3 M)(2.0ml) = M 2 (10.00ml); M 2 = 1.00 x 10 -3 M; use this value as M 1 for the next dilution; (1.00 x 10 -3 M)(2.0ml) = M 2 (10.00ml); M 2 = 2.00 x 10 -4 M} 3. c { ; Li 3 N ® 3Li + (aq) + N 3- (aq); 4 mol ions per 1 mol of Li 3 N; ; } 4. I. d {calculate the concentrations: i. M = mol/L = 8mol/1.0L = 8M; ii. M = 4mol/1.0L = 4M; iii. M = 4mol/0.50L = 8M; iv. M = 4mol/0.25L = 16M} II. b III. c 5. a {mol A ® molarity B from the flowchart; ; M = mol/L ® solve for L; } 6. c {Molarity A ® Molarity B from the flowchart; mol Ba(OH) 2 = M x L = 0.15M x 0.045L = 0.00675mol Ba(OH) 2 ; ; } 7. d {M A ® mol B from the flowchart; need to determine molar mass of H 2 A; ; find mol from titration data: mol = M x L = 0.1550M x 0.04525L = 7.0138 x 10 -3 mol NaOH; convert mol NaOH to mol H 2 A; ; calculate molar masses of each option: “a”: molar mass H 2 SO 4 = 98g/mol; “b”: molar mass H 2 S = 34g/mol; “c”: molar mass H 2 CO 3 = 62g/mol; “d”: molar mass H 2 SO 3 = 82g/mol; “e”: molar mass H 2 C 2 O 4 = 90g/mol; result matches “d”, H 2 SO 3 } 53g Na 2 CO 3 1mol Na 2 CO 3 106g Na 2 CO 3 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.50mol Na 2 CO 3 L = mol M = 0.50mol Na 2 CO 3 0.20M Na 2 CO 3 = 2.5L [ions] = mol ions L 0.50mol Li 3 N 4mol ions 1mol Li 3 N ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 2.0mol ions [ions] = 2.0mol ions 2.0L = 1.0M ions 2.50mol Pb(NO 3 ) 2 2mol NaCl 1mol Pb(NO 3 ) 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 5.00mol NaCl L = mol M = 5.00mol NaCl 0.500M NaCl = 10.0L 0.00675mol Ba(OH) 2 2mol H 3 PO 4 3mol Ba(OH) 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.0045mol H 3 PO 4 [H 3 PO 4 ] = mol H 3 PO 4 L = 0.0045mol H 3 PO 4 0.085L = 0.05294M molar mass = grams H 2 A mol H 2 A 7.0138 x10 − 3 mol NaOH 1mol H 2 A 2mol NaOH ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 3.5069 x10 − 3 mol H 2 A molar mass = grams H 2 A mol H 2 A = 0.2879g 3.5069 x 10 − 3 mol = 82.10g / mol