1 REACTION STOICHIOMETRY 1792 JEREMIAS RICHTER The amount of substances produced or consumed in chemical reactions can be quantified 4F-1 (of 14)

2 INFORMATION FROM CHEMICAL EQUATIONS 2H 2 + O 2 → 2H 2 O 2 molecules 2 moles 0.84 moles 0.028 moles 1 molecule 1 moles 0.42 moles 0.014 moles 2 molecules 2 moles 0.84 moles 0.028 moles 4F-2 (of 14) The moles that react and form do so in the ratio of the balanced equation

3 INFORMATION FROM CHEMICAL EQUATIONS 2H 2 + O 2 → 2H 2 O 0.60 moles -0.60 moles 0.00 moles 0.40 moles -0.30 moles 0.10 moles 0.00 moles +0.60 moles 0.60 moles 4F-3 (of 14) starting reacting ending 0.50 moles -0.40 moles 0.10 moles 0.20 moles -0.20 moles 0.00 moles +0.40 moles 0.40 moles starting reacting ending 2H 2 + O 2 → 2H 2 O The moles that react and form do so in the ratio of the balanced equation

4 MASS CALCULATIONS Calculate the mass of oxygen needed to burn 5.00 g of propanol, C 3 H 8 O. C 3 H 8 O + O 2 →CO 2 + H 2 O 34 4½ 4F-4 (of 14)

5 MASS CALCULATIONS Calculate the mass of oxygen needed to burn 5.00 g of propanol, C 3 H 8 O. C 3 H 8 O + O 2 →CO 2 + H 2 O 68 5.00 g x g 2 mol9 mol 9 2 2 mol C 3 H 8 O = 9 mol O 2 5.00 g C 3 H 8 O x 9 mol O 2 _________________ 2 mol C 3 H 8 O = 12.0 g O 2 x 1 mol C 3 H 8 O ____________________ 60.094 g C 3 H 8 O x 32.00 g O 2 ______________ 1 mol O 2 4F-5 (of 14)

6 Calculate the mass of carbon dioxide produced from the 5.00 g of propanol. C 3 H 8 O + O 2 →CO 2 + H 2 O 68 5.00 gx g 2 mol6 mol 9 2 2 mol C 3 H 8 O = 6 mol CO 2 5.00 g C 3 H 8 O x 6 mol CO 2 _________________ 2 mol C 3 H 8 O = 11.0 g CO 2 x 1 mol C 3 H 8 O ____________________ 60.094 g C 3 H 8 O x 44.01 g CO 2 ________________ 1 mol CO 2 4F-6 (of 14)

7 C 3 H 8 O + O 2 →CO 2 + H 2 O68 5.0 g6.0 g 9 2 12.0 g11.0 g 4F-7 (of 14)

8 LIMITING REACTANT CALCULATIONS LIMITING REACTANT – The reactant that is completely used up in a reaction 4F-8 (of 14)

9 x 12 piecesx 1 sandwich _______________ 1 piece = 12 sandwiches 22 slicesx 1 sandwich _______________ 2 slices = 11 sandwichesactual amount produced limiting reactant 12 pieces22 slices 4F-9 (of 14)

10 Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride. Ba + WF 3 →BaF 2 + W 25.0 g 26.0 g 3 mol 2 mol 323 x g 2 mol 2 4F-10 (of 14)

11 25.0 g Ba x 2 mol W ____________ 3 mol Ba = 22.3 g W x 1 mol Ba _______________ 137.3 g Ba x 183.8 g W ______________ 1 mol W Ba + WF 3 → 25.0 g 26.0 g 3 mol 2 mol 23BaF 2 + W3 x g 2 mol 2 Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride. 4F-11 (of 14)

12 25.0 g Ba x 2 mol W ____________ 3 mol Ba = 22.3 g W x 1 mol Ba _______________ 137.3 g Ba x 183.8 g W _______________ 1 mol W Ba + WF 3 → 25.0 g 26.0 g 3 mol 2 mol 23BaF 2 + W3 x g 2 mol 2 Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride. 26.0 g WF 3 x 2 mol W _____________ 2 mol WF 3 = 19.8 g W x 1 mol WF 3 _________________ 240.8 g WF 3 x 183.8 g W _______________ 1 mol W WF 3 is the limiting reactant 19.8 g W are produced 4F-12 (of 14)

13 Determine the percentage of magnesium and silver in an alloy of the two metals. A 6.50 gram sample of the alloy reacts with 14.5 grams of hydrogen chloride. Mg + HCl →MgCl 2 + H 2 x g14.5 g 1 mol2 mol 2 1 mol Mg = 2 mol HCl 14.5 g HCl x 1 mol Mg ______________ 2 mol HCl = 4.834 g Mg x 1 mol HCl ________________ 36.458 g HCl x 24.31 g Mg ______________ 1 mol Mg 4F-13 (of 14) 4.834 g Mg x 100 _______________ 6.50 g alloy = 74.4% Mg100% - 74.4% = 25.6% Ag

14 A 1.17 gram sample of the unknown is dissolved in water, treated with lead (II) ions, and 1.42 grams of precipitate are collected. Pb 2+ + I - →PbI 2 x g 2 mol 2 1.42 g 1 mol 1.42 g PbI 2 x 2 mol I - _____________ 1 mol PbI 2 = 0.7818 g I - x 1 mol PbI 2 _________________ 461.0 g PbI 2 x 126.9 g I - _____________ 1 mol I - 0.7818 g I - ___________________ 1.17 g sample = 66.8% I - in the sample x 100 4F-14 (of 14) Determine the percentage by mass of iodide in a solid unknown.

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16 MOLES FROM SOLUTION DATA Find the moles of potassium carbonate contained in 275 mL of a 0.300 M potassium carbonate solution. M = n ___ V MV= n x 0.275 L solution = 0.0825 mol K 2 CO 3 0.300 mol K 2 CO 3 ______________________ L solution 4G-1 (of 12)

17 Find the moles of each ion in the 0.300 M potassium carbonate solution. 0.0825 mol K 2 CO 3 = 0.165 mol K + x 2 0.0825 mol K 2 CO 3 = 0.0825 mol CO 3 2- x 1 4G-2 (of 12)

18 10.0 mL of 0.450 M BaCl 2 are mixed with 20.0 mL of 0.300 M K 2 SO 4. (a)Find the moles of each ion in the solution. x 0.0100 L solution = 0.004500 mol BaCl 2 0.450 mol BaCl 2 ______________________ L solution  0.00450 mol Ba 2+ and 0.00900 mol Cl - x 0.0200 L solution = 0.006000 mol K 2 SO 4 0.300 mol K 2 SO 4 ______________________ L solution  0.0120 mol K + and 0.00600 mol SO 4 2- 4G-3 (of 12)

19 10.0 mL of 0.450 M BaCl 2 are mixed with 20.0 mL of 0.300 M K 2 SO 4. (b)Find the moles of each ion after any reaction. Ba 2+ + SO 4 2- →BaSO 4 Initial moles Reacting moles Final moles 0.004500.00600 0 -0.00450-0.00450+0.00450 00.001500.00450  0 mol Ba 2+ 0.00150 mol SO 4 2- 0.00900 mol Cl - 0.0120 mol K + 4G-4 (of 12)

20 10.0 mL of 0.450 M BaCl 2 are mixed with 20.0 mL of 0.300 M K 2 SO 4. (c)Find the final molarities of each ion in the solution. = 0 M Ba 2+ 0 mol Ba 2+ ______________________ 0.0300 L solution = 0.0500 M SO 4 2- 0.00150 mol SO 4 2- _______________________ 0.0300 L solution = 0.300 M Cl - 0.00900 mol Cl - _______________________ 0.0300 L solution = 0.400 M K + 0.0120 mol K + _______________________ 0.0300 L solution 4G-5 (of 12)

21 20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH. (a)Find the moles of each ion in the solution. x 0.0200 L solution = 0.007000 mol HCl 0.350 mol HCl ___________________ L solution  0.00700 mol H + and 0.00700 mol Cl - x 0.0300 L solution = 0.007500 mol NaOH 0.250 mol NaOH ______________________ L solution  0.00750 mol Na + and 0.00750 mol OH - 4G-6 (of 12)

22 20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH. (b)Find the moles of each ion after any reaction. H + + OH - → H 2 O Initial moles Reacting moles Final moles 0.007000.007500 -0.00700-0.00700+0.00700 00.000500.00700  0 mol H + 0.00050 mol OH - 0.00700 mol Cl - 0.00750 mol Na + 4G-7 (of 12)

23 20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH. (c)Find the final molarities of each ion in the solution. = 0 M H + 0 mol H + ______________________ 0.0500 L solution = 0.010 M OH - 0.00050 mol OH - _______________________ 0.0500 L solution = 0.140 M Cl - 0.00700 mol Cl - _______________________ 0.0500 L solution = 0.150 M Na + 0.00750 mol Na + _______________________ 0.0500 L solution 4G-8 (of 12)

24 20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH. (d)Find the mass of water produced by the reaction. = 0.126 g H 2 O 0.007000 mol H 2 O x 18.016 g H 2 O __________________ 1 mol H 2 O 4G-9 (of 12)

25 DILUTION CALCULATIONS When a solution is diluted, only solvent is added, so the moles of solute are unchanged mol solute (concentrated) = mol solute (diluted) M C V C = M D V D 4G-10 (of 12)

26 Calculate the volume of 6.00 M ammonia needed to prepare 250. mL of a 0.100 M ammonia solution. M C V C = M D V D V C = M D V D _______ M C = (0.100 M)(250. mL) ________________________ (6.00 M) = 4.17 mL M C = V C = 6.00 M ? M D = V D = 0.100 M 250. mL 4G-11 (of 12)

27 Calculate the volume of water that must be added to 5.00 mL of concentrated hydrochloric acid (12.1 M) to make the acid 3.00 M. M C V C = M D V D M C V C = V D _______ M D = (12.1 M)(5.00 mL) _______________________ (3.00 M) = 20.2 mL M C = V C = 12.1 M 5.00 mL M D = V D = 3.00 M ? 20.2 mL - 5.00 mL ____________ Volume of dilute solution Volume of concentrated solution 15.2 mLWater that must be added 4G-12 (of 12)

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29 TITRATION – A technique in which one solution is used to analysis another Buret: a solution of 1 reactant of known concentration REACTIONS IN SOLUTION Flask: another reactant of unknown concentration, mass, etc. STANDARD SOLUTION – A solution of known concentration 4H-1 (of 13)

30 The mass of sodium bicarbonate in an antacid tablet is to be determined. ACID-BASE INDICATOR – A weak organic acid or base that changes color in acidic or basic solutions The tablet is dissolved in water, an acid-base indicator added, and 21.5 mL of a 0.300 M hydrochloric acid solution produces a color change. NaHCO 3 + HCl →NaCl + H 2 O + CO 2 x g 21.5 mL 0.300 M 1 mol x 1 mol NaHCO 3 _________________ 1 mol HCl 0.300 mol HCl x 0.0215 L solution __________________ L solution = 0.542 g NaHCO 3 x 84.008 g NaHCO 3 _______________________ mol NaHCO 3 4H-2 (of 13)

31 x 1 mol O.A.D. ____________________ 126.08 g O.A.D. A sodium hydroxide solution is to be standardized. 34.2 mL of the sodium hydroxide solution are required to neutralize a solution made with 0.619 grams of solid H 2 C 2 O 4. 2H 2 O (m = 126.08 g/mol). NaOH + H 2 C 2 O 4 →. 2H 2 ONa 2 C 2 O 4 + H 2 O 34.2 mL x M 0.619 g 2 mol 1 mol x 2 mol NaOH _________________ 1 mol O.A.D. 0.619 g O.A.D. = 0.287 M NaOH x 1 _______________________ 0.0342 L solution 2 24 _ 4H-3 (of 13)

32 A 10.0 mL aliquot of a solution containing V 2+ ions is acidified, and 32.7 mL of a 0.115 M MnO 4 - solution produces a light purple color. If the V 2+ was oxidized to V 5+, determine the molarity of the V 2+ in the original solution. 4H-4 (of 13) ( ) x 3 ( ) x 5 MnO 4 - (aq) + V 2+ (aq) → MnO 4 - → Mn 2+ +7 -2 +2 V 2+ → V 5+ + 4H 2 O 8H + +5e - + + 3e - Mn 2+ + V 5+ +2+5

33 A 10.0 mL aliquot of a solution containing V 2+ ions is acidified, and 32.7 mL of a 0.115 M MnO 4 - solution produces a light purple color. If the V 2+ was oxidized to V 5+, determine the molarity of the V 2+ in the original solution. 4H-5 (of 13) 3MnO 4 - → 3Mn 2+ 5V 2+ → 5V 5+ + 12H 2 O 24H + +15e - + + 15e - 15e - + 24H + + 3MnO 4 - + 5V 2+ →3Mn 2+ + 12H 2 O + 5V 5+ + 15e - MnO 4 - (aq) + V 2+ (aq) → +7 -2 +2+2+5 Mn 2+ + V 5+

34 A 10.0 mL aliquot of a solution containing V 2+ ions is acidified, and 32.7 mL of a 0.115 M MnO 4 - solution produces a light purple color. If the V 2+ was oxidized to V 5+, determine the molarity of the V 2+ in the original solution. 10.0 mL x M 32.7 mL 0.115 M 5 mol3 mol x 5 mol V 2+ ________________ 3 mol MnO 4 - 0.115 mol MnO 4 - x 0.0327 L solution _____________________ L solution = 0.627 M V 2+ x 1 _______________________ 0.0100 L solution 4H-6 (of 13) 3MnO 4 - (aq) + 5V 2+ (aq) → 3Mn 2+ (aq) + 12H 2 O (l) + 5V 5+ (aq)24H + (aq) +

35 molar mass H 2 X = grams H 2 X _______________ moles H 2 X molar mass H 2 X = 0.109 grams H 2 X ______________________ ? moles H 2 X Calculate the molar mass of a diprotic acid if 0.409 grams of it are neutralized by 19.50 mL of a 0.287 M sodium hydroxide solution. 4H-7 (of 13)

36 H 2 X + NaOH →H(OH) + Na 2 X 0.409 g x mol 19.50 mL 0.287 M 1 mol 2 mol 2 x 1 mol H 2 X _______________ 2 mol NaOH 0.287 mol NaOH x 0.01950 L sol’n _____________________ L sol’n = 0.002798 mol H 2 X 2 Calculate the molar mass of a diprotic acid if 0.409 grams of it are neutralized by 19.50 mL of a 0.287 M sodium hydroxide solution. 0.409 g H 2 X ________________________ 0.002798 mol H 2 X = 146 g/mol 4H-8 (of 13)

37 The molarity of an aluminum hydroxide solution is to be determined. An acid-base indicator is added to 10.0 mL of the aluminum hydroxide solution, and 12.5 mL of 0.300 M hydrochloric acid produces a color change. Al(OH) 3 + HCl →AlCl 3 + H(OH) 10.0 mL x M 12.5 mL 0.300 M 1 mol3 mol 3 x 1 mol Al(OH) 3 _________________ 3 mol HCl 0.300 mol HCl x 0.0125 L solution __________________ L solution = 0.125 M Al(OH) 3 x 1 ____________ 0.0100 L 3 4H-9 (of 13)

38 ( ) x 8 Cinnabar ore contains S 2- ions. A 1.534 g sample of the ore is dissolved in acid, then all the S 2- is oxidized by 20.4 mL of a 0.110 M Cr 2 O 7 2- solution. Determine the percentage of S 2- in cinnabar ore. 4H-10 (of 13) ( ) x 3 K 2 Cr 2 O 7 (aq) + S 2- (aq) → K + (aq) + Cr 2 O 7 2- (aq) + S 2- (aq) → Cr 2 O 7 2- → Cr 3+ +1 +6 -2 -2 S 2- → S 8 2 8 + 7H 2 O14H + +6e - + + 16e - Cr 3+ + S 8 +30

39 Cinnabar ore contains S 2- ions. A 1.534 g sample of the ore is dissolved in acid, then all the S 2- is oxidized by 20.4 mL of a 0.110 M Cr 2 O 7 2- solution. Determine the percentage of S 2- in cinnabar ore. 4H-11 (of 13) K 2 Cr 2 O 7 (aq) + S 2- (aq) → K + (aq) + Cr 2 O 7 2- (aq) + S 2- (aq) → 8Cr 2 O 7 2- → 16Cr 3+ +1 +6 -2 -2 24S 2- → 3S 8 + 56H 2 O112H + +48e - + + 48e - 48e - + 112H + + 8Cr 2 O 7 2- + 24S 2- →16Cr 3+ + 56H 2 O + 3S 8 + 48e - Cr 3+ + S 8 +30

40 Cinnabar ore contains S 2- ions. A 1.534 g sample of the ore is dissolved in acid, then all the S 2- is oxidized by 20.4 mL of a 0.110 M Cr 2 O 7 2- solution. Determine the percentage of S 2- in cinnabar ore. x g20.4 mL 0.110 M x 24 mol S 2- _________________ 8 mol Cr 2 O 7 2- 0.110 mol Cr 2 O 7 2- x 0.0204 L solution _______________________ L solution = 0.2159 g S 2- x 32.07 g S 2- _____________ mol S 2- 0.2159 g S 2- x 100 _______________ 1.534 g ore = 14.1% S 2- in the ore 4H-12 (of 13) 8Cr 2 O 7 2- (aq) + 24S 2- (aq) → 16Cr 3+ (aq) + 56H 2 O (l) + 3S 8 (s)112H + (aq) + 8 mol 24 mol

41 Calculate the molarity of a sulfuric acid solution if 25.0 mL of it are neutralized by 33.5 mL of a 0.240 M potassium hydroxide solution. H 2 SO 4 + KOH →H(OH) + K 2 SO 4 25.0 mL x M 33.5 mL 0.240 M 1 mol2 mol 2 x 1 mol H 2 SO 4 _________________ 2 mol KOH 0.240 mol KOH x 0.0335 L solution ___________________ L solution = 0.161 M H 2 SO 4 x 1 ____________ 0.0250 L 2 4H-13 (of 13)

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43 MOLAR MASSES AND STOICHIOMETRIC CONVERSIONS 2 mol C (12.01 g/mol)= 24.02 g 6 mol H (1.008 g/mol)= 6.048g 1 mol O (16.00 g/mol)=16.00 g 46.068 g Calculate the molar mass of ethanol, C 2 H 5 OH 4I-1 (of 8) 46.068 g C 2 H 5 OH = 1 mol C 2 H 5 OH

44 x 1 mol C 2 H 5 OH ______________________ 46.068 g C 2 H 5 OH Calculate the number of ethanol molecules in 25.0 mL of pure ethanol. The density of the ethanol is 0.789 g/mL. 25.0 mL C 2 H 5 OH x 6.022 x 10 23 molecules C 2 H 5 OH ________________________________________ 1 mol C 2 H 5 OH = 2.58 x 10 23 molecules C 2 H 5 OH x 0.789 g C 2 H 5 OH ____________________ 1 mL C 2 H 5 OH 0.789 g C 2 H 5 OH = 1 mL C 2 H 5 OH 4I-2 (of 8)

45 x 0.789 g C 2 H 5 OH ____________________ 1 mL C 2 H 5 OH x 1 mol C 2 H 5 OH ______________________ 46.068 g C 2 H 5 OH Calculate the number of carbon atoms in a 10.0 mL sample of pure ethanol. 10.0 mL C 2 H 5 OH x 2 mol C __________________ 1 mol C 2 H 5 OH = 2.06 x 10 23 atoms C x 6.022 x 10 23 atoms C ___________________________ 1 mol C 4I-3 (of 8)

46 x 40. mL C 2 H 5 OH ____________________ 100 mL vodka x 1 mol C 2 H 5 OH ______________________ 46.068 g C 2 H 5 OH Calculate the number of ethanol molecules in 45.0 mL of 80. proof vodka. The density of the vodka is 0.92 g/mL. 45.0 mL vodka x 6.022 x 10 23 molecules C 2 H 5 OH ________________________________________ 1 mol C 2 H 5 OH = 1.9 x 10 23 molecules C 2 H 5 OH x 0.789 g C 2 H 5 OH ____________________ 1 mL C 2 H 5 OH 80. Proof vodka = 40.% C 2 H 5 OH by volume 100 mL vodka = 40. mL C 2 H 5 OH 4I-4 (of 8)

47 x 46.068 g C 2 H 5 OH _______________________ mol C 2 H 5 OH Calculate the mass of one ethanol molecule, in grams. = 7.650 x 10 -23 g x mol C 2 H 5 OH ________________________________________ 6.022 x 10 23 molecules C 2 H 5 OH 4I-5 (of 8) 1 molecule C 2 H 5 OH

48 x 1 mol O ____________ 16.00 g O A metal oxide with the formula M 2 O 3 is 29.0% oxygen by mass. Calculate the molar mass of metal M. 29.0 g O x 2 mol M __________ 3 mol O = 58.8 g/mol 71.0 g M _________________ 1.208 mol M 4I-6 (of 8) g M mol M Molar Mass of M = = 71.0 g M ? mol M = 1.208 mol M

49 1 mol Na (22.99 g/mol)= 22.99 g 1 mol N (14.01 g/mol)= 14.01 g 3 mol O (16.00 g/mol)=48.00 g 85.00 g THEORETICAL PERCENT COMPOSITION OF COMPOUNDS BY MASS NaNO 3 Calculate the percent composition by mass of sodium nitrate 22.99 g Na  100 ___________________ 85.00 g NaNO 3 % Na == 27.05 % Na 48.00 g O  100 ___________________ 85.00 g NaNO 3 % O == 56.47 % O 14.01 g N  100 ___________________ 85.00 g NaNO 3 % N == 16.48 % N 4I-7 (of 8)

50 BaCl 2. 2H 2 O 1 mol Ba (137.3 g/mol)= 137.3 g 2 mol Cl (35.45 g/mol)= 70.90 g 2 mol H 2 O (18.016 g/mol)=36.032g 244.232g Calculate the percentage by mass of water in barium chloride dihydrate 36.032 g H 2 O  100 ____________________________ 244.232 g BaCl 2. 2H 2 O % H 2 O == 14.75 % H 2 O 4I-8 (of 8)

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52 EMPIRICAL AND MOLECULAR FORMULA CALCULATIONS When a sample of a hydrocarbon is burned, 8.45 g CO 2 and 1.73 g H 2 O are produced. Calculate the empirical formula of the hydrocarbon. C x H y + O 2 →CO 2 + H 2 O 8.45 g CO 2 = 2.306 g C x 12.01 g C ________________ 44.01 g CO 2 1.73 g H 2 O= 0.1936 g Hx 2.016 g H ________________ 18.016 g H 2 O 4J-1 (of 9)

53 x 1 mol C _____________ 12.01 g C = 0.1920 mol C 2.306 g C x 1 mol H ____________ 1.008 g H = 0.1921 mol H 0.1936 g H 0.1920 mol C _________________ 0.1920 0.1921 mol H _________________ 0.1920 = 1.00 mol C= 1.00 mol H Empirical formula: CH EMPIRICAL AND MOLECULAR FORMULA CALCULATIONS When a sample of a hydrocarbon is burned, 8.45 g CO 2 and 1.73 g H 2 O are produced. Calculate the empirical formula of the hydrocarbon. 4J-2 (of 9)

54 A borane is any compound composed of boron and hydrogen. When a sample of a borane is burned, 12.89 g B 2 O 3 and 12.27 g H 2 O are produced. Calculate the empirical formula of the borane. B x H y + O 2 →B 2 O 3 + H 2 O 12.89 g B 2 O 3 = 4.0029 g B x 21.62 g B ________________ 69.62 g B 2 O 3 12.27 g H 2 O = 1.3730 g H x 2.016 g H ________________ 18.016 g H 2 O 4J-3 (of 9)

55 x 1 mol B _____________ 10.81 g B = 0.37030 mol B 4.0029 g B x 1 mol H _____________ 1.008 g H = 1.3621 mol H 1.3730 g H 0.37030 mol B ___________________ 0.37030 1.3621 mol H _________________ 0.37030 = 1.000 mol B= 3.678 mol H Empirical formula: B 3 H 11 A borane is any compound composed of boron and hydrogen. When a sample of a borane is burned, 12.89 g B 2 O 3 and 12.27 g H 2 O are produced. Calculate the empirical formula of the borane. 4J-4 (of 9) x 3

56 When a 3.84 g sample of a compound containing C, H, and N is combusted, 7.34 g CO 2 and 2.51 g H 2 O are collected. Calculate the empirical formula of the compound. C x H y N z + O 2 →CO 2 + H 2 O + N a O b 7.34 g CO 2 = 2.003 g C x 12.01 g C ________________ 44.11 g CO 2 2.51 g H 2 O = 0.2809 g H x 2.016 g H _____+__________ 18.016 g H 2 O 3.84 g C x H y N z - 2.003 g C - 0.2809 g H ________________________ 1.5561 g N 4J-5 (of 9)

57 When a 3.84 g sample of a compound containing C, H, and N is combusted, 7.34 g CO 2 and 2.51 g H 2 O are collected. Calculate the empirical formula of the compound. x 1 mol C _____________ 12.01 g C = 0.1668 mol C 2.003 g C x 1 mol H ____________ 1.008 g H = 0.2787 mol H 0.2809 g H 0.1668 mol C _________________ 0.1111 0.2787 mol H _________________ 0.1111 = 1.50 mol C= 2.51 mol H Empirical formula: C 3 H 5 N 2 x 1 mol N _____________ 14.01 g N = 0.1111 mol N 1.5561 g N 0.1111 mol N ________________ 0.1111 = 1.00 mol N 4J-6 (of 9)

58 When a 2.75 g sample of a compound containing C, H, and O is combusted, 5.49 g CO 2 and 2.25 g H 2 O are collected. Calculate the empirical formula of the compound. C x H y O z + O 2 →CO 2 + H 2 O 5.49 g CO 2 = 1.498 g C x 12.01 g C ________________ 44.11 g CO 2 2.25 g H 2 O = 0.2518 g H x 2.016 g H _________________ 18.016 g H 2 O 2.75 g C x H y O z - 1.498 g C - 0.2518 g H ________________________ 1.0002 g O 4J-7 (of 9)

59 When a 2.75 g sample of a compound containing C, H, and O is combusted, 5.49 g CO 2 and 2.25 g H 2 O are collected. Calculate the empirical formula of the compound. x 1 mol C ____________ 12.01 g C = 0.1247 mol C 1.498 g C x 1 mol H ____________ 1.008 g H = 0.2498 mol H 0.2518 g H 0.1247 mol C _________________ 0.06251 0.2498 mol H _________________ 0.06251 = 1.99 mol C= 4.00 mol H Empirical formula: C 2 H 4 O x 1 mol O _____________ 16.00 g O = 0.06251 mol O 1.0002 g O 0.06251 mol O __________________ 0.06251 = 1.00 mol O 4J-8 (of 9)

60 C2H4OC2H4O 2 mol C (12.01 g/mol)= 24.02 g 4 mol H (1.008 g/mol)=4.032g 1 mol O (16.00 g/mol)= 16.00 g 44.052 g 90 g/mol ________________ 44.052 g/mol ≈ 2 Molecular formula: C 4 H 8 O 2 Calculate the molecular formula of the previous compound if it has a molar mass of about 90 g/mol. 4J-9 (of 9)