AP Chem Test- Titration and Gravimetric Analysis p. 2 Name date 4. Empirical Formula A compound is analyzed and found to contain 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. The molecular weight of this compound is known to be approximately 140 g/mol. What is the empirical formula? What is the molecular formula? There are multiple steps. I expect to see clearly labeled work showing how you got your answers. 68.54 g C 1 mol C 5.706 mol C 12.011 g 8.63 g H 1 mol H 8.562 mol H 1.008 g 22.83 g O 1 mol O 1.427 mol O 15.999 g Mol C Mol H Mol O 5.706 mol C 8.562 mol H 1.427 mol O 1.427 1.427 1.427 3.999 6.000 1 Empirical formula C 4 H 6 O EF mass = 70.0478 molar mass = 140 g/mol = EF x 2 so molecular formula is C 8 H 12 O 2 5. Limiting Reactant. Limestone and marble both contains calcium carbonate. Calcium carbonate reacts with hydrochloric acid in a double replacement reaction that produces a gas. a) Write the balanced equation for this reaction. CaCO 3 (s) + 2 HCl (aq) CaCl 2 (aq) + CO 2 (g) + H 2 O (l) b) If 6.088 g CaCO 3 reacted with 15.0 ml of 0.40 M HCl. i) Determine the limiting reactant. Justify your answer with calculations. Pick one of the products and calculate the amount that could be produced if all of each reactant were used up 15 ml HCl 1 L 0.4 mol HCl 1 mol CO 2 0.0030mol CO 2 1000 ml 1 L 2 mol HCl 6.088 g CaCO 3 1 mol CaCO 3 1 mol CO 2 0.06083 mol CO 2 100.078 g 1 mol CaCO 3 Since the HCl would produce less mol CO 2, it is the limiting reactant it limits how much product can be produced. ii) Calculate the mass of the remaining reactant. Calculate the grams of CaCO 3 that reacted to form the 0.003 moles of CO 2 that can be produced when all the HCl reacts 0.0030mol CO 2 1 mol CaCO 3 100.078 g 0.30 g CaCO 3 1 mol CO 2 1 mol CaCO 3 Then subtract 6.088 g CaCO 3 originally present - 0.30 g CaCO 3 that reacted = 5.79 g CaCO 3 remain
6. Gravimetric Analysis. A salt contains only barium and one of the halide ions. A 0.1480 g sample of the salt was dissolved in water and an excess of sulfuric acid was added to form barium sulfate, which was filtered, dried and weighed. a) Explain what happens when the sulfuric acid is added to the solution with the barium halide. What will be observed? Write a balanced equation to describe what happens in the solution. Tell me about the ions, solubility and bonding. Sulfuric acid is a strong acid that completely dissociated into sulfate ions and hydrogen ions. The sulfate ions form a strong ionic bond with the barium ions forming the insoluble barium sulfate precipitate that settles to the bottom and can be filtered to isolate it from the solution. b) How does this reaction allow you to measure the moles of the barium ions that were in the barium halide initially? Since the barium ions end up in the barium sulfate precipitate that can be isolated, we can calculate the moles of barium using the mass of the barium sulfate precipitate. c) Why is the solution filtered? The solution is filtered to separate the solution from the solid. The solution flows through the pores in the filter and the solid precipitate remains behind in the filter. d) Why is the solid rinsed? The solid is rinsed to remove any of the solution that remains clinging to it. So that it will only be wet with water that can be removed by heating. e) Why is the filter dried in the oven? The filter is dried in the oven to remove the water that clings to the solid and the filter paper. The boiling point of the water is much lower than the boiling point of the ionic solid. f) What happened if you do not add enough of the sulfuric acid? Will this result in a higher or lower molar mass for the unknown halide? If you do not add enough sulfuric acid, not all of the barium ions will end up in the precipitate some will remain in solution. Thus the mass of your precipitate will be lower number and appear to indicate that there are fewer barium ions present than are actually there. This will lead to a lower calculated moles of halide. g) What happened if you mass your filter paper before you tear it instead of after? Will this result in a higher or lower molar mass for the unknown halide? If you mass your filter paper before you tear it instead of after, the mass of filter paper that you subtract in the end will be higher than the actual mass. Then you will think that there are less barium ions present.
h) The filter paper had a mass of 0.6006 g. The mass of the filter paper and solid precipitate was found to be 0.7666 g. What is the formula for the barium halide? There are multiple steps that I expect to see clearly shown and labeled. Mass of barium halide = mass solid and filter paper mass filter paper 0.7666 g 0.6006 g = 0.1660 g Calculate mass of barium in original barium halide 0.1660 g BaSO 4 1 mol BaSO 4 1 mol Ba 2+ ions 137.3 g Ba 2+ 0.09767 g Ba 2+ 233.36 g BaSO 4 1 mol BaSO 4 1 mol Ba 2+ ions Since the mass of barium halide is 0.1480 g and it contains 0.09767 g Ba 2+ there must be 0.05033 g of the halide. And 0.1660 g BaSO 4 1 mol BaSO 4 1 mol Ba 2+ ions 0.0007113 moles of Ba 2+ ions 2 mol X- ions 0.001423 mol X- ions 233.36 g BaSO 4 1 mol BaSO 4 1 mol Ba 2+ ions Therefore the molar mass of X- must be 0.05033 g / 0.001423 mol X- ions = 35.376 g/mol The halide must be chloride since this molar mass is closest to chlorine and the formula must be BaCl 2
AP Chem Test- Titration and Gravimetric Analysis p. 1 Name date 1. Bronsted-Lowry Model a) Label the Bronsted-Lowry acid-base pairs. Connect them. BASE ACID CONJUGATE ACID CONJUGATE BASE b) Justify your selection of the acid. This is the ACID, since it DONATED a proton (H+), you will notice in that in the product it has one LESS H than it had in the reactants. c) Justify your selection of the base. This is the BASE, since it ACCEPTED a proton (H+), you will notice in that in the product it has one MORE H than it had in the reactants. d) Explain how water can be both a Bronsted-Lowry acid and base. H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) The water molecule contains a H therefore it can be a proton donor (H+). The water also has a lone pair of electrons so it can accept an H+ by making a covalent bond with the proton. i. Write the equilibrium constant expression for this reaction and its value. Kw = [H+][OH-] = 1.0 x 10-14 ii. What is the ph of neutral water? ph= 7 iii. What is the hydrogen ion concentration? 1.0 x 10-7 iv. What is the hydroxide ion concentration? How do you get this number from the value of the Kw? 1.0 x 10-7 Since the water is neutral the []H+] = [OH-] and [H+][OH-] = 1.0 x 10-14
2. Titration and ph. Below is the titration curve for a weak acid with a strong base. a) Explain why the ph increases as the NaOH is added to acid. As OH- ions are added they react with H+ reducing the [H+] so that the ph goes up b) Calculate the [H+] after 20.0mLofNaOH has been added given a ph of 4.7. 10-4.7 = [H+] = 2.0 x 10-5 M H + c) At what ph does the [H+] =4.3 x 10-6 M? -log 4.3 x 10-6 M = ph = 5.4 d) The equivalence point is reached after 60.0mLof NaOH has been added. Describe what you know about the acid at the equivalence point. At the equivalence point, just enough OH- have been added to react with all the H+ ions. e) Write the balanced equation for the reaction of sulfurous acid and sodium hydroxide. Label the states. f) Write the total ionic equation. g) Write the net ionic equation. H 2 SO 3 (aq) + 2 NaOH (aq) Na 2 SO 3 (aq) + 2 H 2 O (l) H 2 SO 3 (aq) + Na + (aq) + OH - (aq) Na + (aq) + SO 3 2- (aq) + H 2 O (l) H 2 SO 3 (aq) + OH - (aq) SO 3 2- (aq) + H 2 O (l) h) The concentration of the NaOH was 0.300 M. If 15.0 ml of acid H 2 SO 3 was titrated, what was its original concentration? 60 ml NaOH 1 L 0.3 mol NaOH 1 mol H 2 SO 3 0.009 mol H 2 SO 3 1000mL 1 L 2mol NaOH 0.009 mol H 2 SO 3 1000mL 0.600 mol H 2 SO 3 15 ml 1 L 1 L sol n
3. Titration Calculation. The active ingredients of an antacid tablet with a mass of 2.81 grams contained only aluminum hydroxide. Complete neutralization of a sample of the tablet required 48.5 ml of 0.187 M hydrochloric acid. a) Write the balanced equation for the neutralization of the solid aluminum hydroxide by hydrochloric acid. b) Write the total ionic equation. 3 HCl (aq) + Al(OH) 3 (s) AlCl 3 (aq) + 3 H 2 O (l) H + (aq) + Cl - (aq) + Al(OH) 3 (s) Al 3* (aq) + Cl - (aq) + H 2 O (l) c) Write the net ionic equation. H + (aq) + Al(OH) 3 (s) Al 3* (aq) + H 2 O (l) d) Calculate the moles of hydrochloric acid required to neutralize the active ingredient: aluminum hydroxide. 48.5 ml HCl 1 L 0.187 mol HCl 1000mL 1 L 0.00907mol HCl e) Calculate the mass of aluminum hydroxide that must have been present in the antacid tablet. 0.0112 mol HCl 1 mol Al(OH) 3 78.001 g Al(OH) 3 3 mol HCl 1 mol Al(OH) 3.236 g Al(OH) 3 f) Calculate the percent by mass of active ingredient in the antacid tablet. 0.236 g Al(OH) 3 / 2.81 grams of tablet x 100 = 8.40 %