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Stoichiometry Molarity and more

Stoichiometry Molarity and more. Some questions require scrap paper/whiteboards.

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Stoichiometry Molarity and more

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  1. StoichiometryMolarityand more Some questions require scrap paper/whiteboards.

  2. The following diagram represents the collection of elements formed by the decomposition of a compound. The blue spheres represent nitrogen atoms and the red spheres represent oxygen atoms. What was the empirical formula of the original compound? • N6O12 • N3O6 • NO2 • NO • NO3

  3. What was the empirical formula of the original compound? • N6O12 • N3O6 • NO2 • NO • NO3 • We cannot know the molecular formula from the information given (eliminates 1 & 2). • We only know the ratio of N’s to O’s in the original compound.

  4. The following diagram represents the collection of carbon dioxide and water formed by the decomposition of a hydrocarbon. What was the empirical formula of the original hydrocarbon? • C4H16 • C2H4 • C2H8 • CH4 • CH2

  5. What was the empirical formula of the original hydrocarbon? C4H16 C2H4 C2H8 CH4 CH2 • While the diagram indicates 4 carbons, and you might think there could have been 1 C4H16, 2 C2H8, or 4 CH4. • However, the maximum number of H’s that can attach to C’s is CnH2n+2. Thus to achieve the 1:4 C:H ratio, both the empirical and molecular formula must have been CH4.

  6. The following diagram represents a high-temperature reaction between CH4 and H2O. Based on this diagram, write a balanced chemical equation to represent this reaction (Use an = for the arrow).

  7. The following diagram represents a high-temperature reaction between CH4 and H2O. Based on this diagram, write a balanced chemical equation to represent this reaction. • 2CH4 + 2H2O → 2CO+ 6H2 • It is more appropriate to write chemical equations in the lowest whole number ratio. • CH4 + H2O → CO+ 3H2

  8. Nitrogen and hydrogen react to form ammonia (NH3). Consider the model of the mixture shown below. Draw a representation of the product mixture, assuming the reaction goes to completion. Which color sphere best represents N and which color for H? Blue = N, White = H White = N, Blue = H Break out the scrap paper/white board to sketch a response.

  9. Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3). Consider the model of the mixture shown below. Blue spheres = N and white spheres = H. Draw a representation of the product mixture, assuming the reaction goes to completion. • N2 + 3H2 → 2NH3 • 8N’s, 4N2 require 24 H’s, 12H2 for a complete reaction. • Only 9H2 are present, thus H2 limits. • 9H2 require 3N2, one N2 in excess, and 6NH3 are produced.

  10. Nitrogen monoxide and oxygen gas react to form nitrogen dioxide. Consider the model of the mixture shown below. Blue spheres = N and red spheres = O. Draw a representation of the product mixture, assuming the reaction goes to completion. Break out the whiteboards to sketch a response.

  11. Nitrogen monoxide and oxygen react to form nitrogen dioxide. Consider the model of the mixture shown below. Blue spheres = N and white spheres = O. Draw a representation of the product mixture, assuming the reaction goes to completion. • 2NO + O2 → 2NO2 • 8NO require 4O2 for a complete reaction. • 5O2 are present, thus O2 is in excess and NO limits. • 8NO require 4O2, one O2 in excess, and 8NO2 are produced.

  12. Molarity Review • Making Solutions • Calculating Molarity

  13. The molarity of an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500 mL of solution would be No calculator • 2.1 M • 0.20 M • 0.05 M • 50 M • 0.0084 M • 8.4 M

  14. moles Molarity, M = V (L) The molarity an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500. mL of solution would be • 2.1 M • 0.20 M • 0.05 M • 50 M • 0.0084 M • 8.4 M

  15. The molarity of the sodium ions in an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500 mL of solution. No calculator • 2.1 M • 4.2 M • 0.20 M • 0.10 M • 0.05 M

  16. The molarity of the sodium ions an aqueous solution made from 4.2 g of sodium fluoride dissolved to make 500 ml of solution. • 2.1 M • 4.2 M • 0.20 M of the solution • the molarity of each ion is present in a 1:1 ratio with the “molecule” and thus has the same molarity, 0.20 M for each ion, Na+ and F− • NaF → Na+ + F− • 0.10 M • 0.05 M

  17. In a 0.010 M solution of K2SO4 the total concentration of all the ions is No calculator • 0.010 M • 0.020 M • 0.030 M • 0.070 M • impossible to determine.

  18. In a 0.010 M solution of K2SO4 the total concentration of the ions is • 0.010 M • 0.020 M • 0.030 M • When dissolves, 3 ions are produced. • K2SO4 → 2K+ + SO42− • resulting in 3 ions per particle dissolved • 0.020 M solution of K+ ions and 0.010 M solution of SO42− ions • 0.070 M • impossible to determine.

  19. When making a 1.0 M aqueous solution of NaCl. Select all that apply. • It is best to put 58.44 g of NaCl in an empty 2000 mL beaker and add 1000 mL of water with a graduated cylinder to dissolve the NaCl. • It is best to put 29.22 g of NaCl in an empty 2000 mL beaker and use a graduated cylinder to add 1000 mL of water to dissolve the NaCl. • It is best to put 58.44 g of NaCl in an empty volumetric flask and then add 1000 mL of water. • It is best to put 58.44 g of NaCl in an empty volumetric flask, add some water and dissolve, then fill the volumetric flask up to the 1 L mark.

  20. moles Molarity, M = V (L) When making a 1.0 M aqueous solution of NaCl. Select all that apply. • It is best to dissolve 58.44 g of NaCl in a 2000 mL beaker and add 1000 mL of water with a graduated cylinder. • It is best to dissolve 58.44 g of NaCl in a 2000 mL beaker and use a graduated cylinder to add 1000 mL of water. • It is best to put 58.44 g of NaCl in a volumetric flask and then add 1000 mL of water. • It is best to dissolve 58.44 g of NaCl in some water in a volumetric flask and then fill the volumetric flask up to the 1 L mark. • This of course will allow for the room that the dissolved salt will take up as part of the total volume of the solution. This is volume of the solution, not volume of water added.

  21. Mass Percent Problems

  22. What is the mass of copper(II) sulfate (molecular weight = 159.6 g/mol) in 40. mL of 2.0 M copper(II) sulfate? No Calculator • 3.2 g • 5.5 g • 8.8 g • 13 g • 16 g • 32 g

  23. What is the mass of copper(II) sulfate (molecular weight = 159.6 g/mol) in 40. mL of 2.0 M copper(II) sulfate? No Calculator Learn to estimate using easy math • 3.2 g • 5.5 g • 8.8 g • 13 g 0.1 mol would mean 16g, half that or 0.05 mol would be ~8g, thus the 0.08 mol must be in between≃13g • 16 g • 32 g 0.04 L x 2M = 0.08 mol CuSO4

  24. How many grams of zinc nitrate(189 g/mol) contain 48 grams of oxygen atoms? No Calculator • 95 g • 125 g • 145 g • 165 g • 189 g • none of the above

  25. How many grams of zinc nitrate(189 g/mol) contain 48 grams of oxygen atoms? No Calculator • 95 g Look for easy math. Zn(NO3)2 3 O’s = 48g, since 2 NO3‘s per zinc nitrate, you only need 0.5 mol of zinc nitrate = half of molar mass • 125 g • 145 g • 165 g • 189 g • None of the above

  26. A 2.00 gram mixture of calcium carbonate and calcium chloride are treated with an excess of hydrochloric acid and 0.66 grams of carbon dioxide (44 g/mol) are produced. What is the percent of CaCO3 (100 g/mol) by mass in the original mixture? No Calculator • 25% • 30% • 50% • 75% • 90% Write a net ionic equation to represent the hydrochloric acid + calcium carbonate reaction

  27. A 2.00 gram mixture of calcium carbonate and calcium chloride are treated with an excess of hydrochloric acid and 0.66 grams of carbon dioxide (44 g/mol) are produced. What is the percent of CaCO3 (100 g/mol) by mass in the original mixture? No Calculator CaCO3 + H+ ➙ H2O + CO2 + Ca2+ • 25% • 30% • 50% • 75% Thus 0.015 mol of CaCO3 to start = 1.5 g and 1.5 is 75% of 2 g total • 90%

  28. In which compound below is the mass ratio of copper to sulfur closest to 2:1? No Calculators • CuS2 • CuS • Cu2S • Cu2S3 • Cu3S2

  29. In which compound below is the mass ratio of copper to sulfur closest to 2:1? Expect easy math. MM Cu=63.6 and S=32. For calculation purposes assume 60 and 30 • CuS2 ~1:1 • CuS ~2:1 • Cu2S ~4:1 • Cu2S3 ~4:3 • Cu3S2 ~6:4 No Calculators

  30. A certain compound contains only one sodium atom and is 5% sodium by mass. What is the molar mass of the compound? No Calculators • 460 g/mol • 230 g/mol • 110 g/mol • 55 g/mol • none of the above

  31. A certain compound contains only one sodium atom and is 5% sodium by mass. What is the molar mass of the compound? No Calculators • 460 g/mol • make the math easy. molar mass of Na is 23, so 23 is 10% of 230, and thus 5% of 460. • 230 g/mol • 110 g/mol • 55 g/mol • none of the above

  32. Analysis of a tellurium oxide compound indicated 84.22 % tellurium. The molar mass is between 580 and 610 g/mole.Determine the molecular formula. Yes, Calculators • TeO • Te2O3 • Te2O • TeO2 • TeO3 • Te3O • Te4O • TeO4 • Te4O6

  33. Analysis of a tellurium oxide compound indicated 84.22 % tellurium. The molar mass is between 580 and 610 g/mole.Determine the molecular formula. The Empirical Formula is Te2O3. Te2O3 MM = 303.2 g/mol thus must have to x2, so the molecular formula is Te4O6.

  34. Determine the empirical formula of a nerve gas that gave the following analysis: 39.10% C, 7.67% H, 26.11% O, 16.82% P, 10.32 % F. Yes, Calculators • The Formula is CvHwOxPyFz. Write the formula into the short answer (don’t worry about the numbers being subscripts)

  35. Determine the empirical formula of a nerve gas that gave the following analysis: 39.10% C, 7.67% H, 26.11% O, 16.82% P, 10.32 % F. The Empirical Formula is C6H14O3PF

  36. Combustion Analysis • Determining Empirical Formula • by Combustion

  37. Combustion analysis is used to determine the amount of carbon, hydrogen, and oxygen in a combustible compound. • Measure the mass of compound to be combusted. • Measure the mass of water produced. • Measure the mass of carbon dioxide produced • Oxygen will make up any remaining mass in original compound. • CxHyOz + O2 ➙ H2O + CO2

  38. A combustion device was used to determine the empirical formula of an organic compound. A 0.6349 g sample was burned and produced 1.603 g of carbon dioxide and 0.2810 g of water and no other oxides. Determine the empirical formula for the compound. • watch out for compounds that may contain oxygen. • change to moles of each element • determine mass of C, mass of H • check to see if there is any “missing mass” that would be oxygen Yes, Calculators

  39. A combustion device was used to determine the empirical formula of an organic compound. A 0.6349 g sample was burned and produced 1.603 g of carbon dioxide and 0.2810 g of water and no other oxides. Determine the empirical formula for the compound. C7H6O2

  40. A combustion device was used to determine the empirical formula of an organic compound. A 0.6349 g sample was burned and produced 1.603 g of carbon dioxide and 0.2810 g of water and no other oxides. Determine the empirical formula for the compound. Alternatively you could solve for the missing mass first C7H6O2

  41. A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis.Combustion of a 32.00 mg sample yielded 78.99 mg of carbon dioxide and 19.96 mg of water.Analysis for nitrogen showed that the compound contained 4.62 % N by mass.Calculate the empirical formula. • The Formula is CwHxOyNz. • Type the formula into the short answer prompt – don’t worry about subscripts. Yes, Calculators

  42. A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis.Combustion of a 32.00 mg sample yielded 78.99 mg of carbon dioxide and 19.96 mg of water.Analysis for nitrogen showed that the compound contained 4.62 % N by mass.Calculate the empirical formula. • Calculate millimoles of C from CO2 • and millimoles of H from H2O • and millimoles of N from % • from mass of the above 3, determine mass of remaining element; must be oxygen, then convert to millimoles and calculate the empirical fomula Yes, Calculators

  43. A confiscated white substance, suspected of being cocaine, was purified by a forensic chemist and subjected to elemental analysis.Combustion of a 32.00 mg sample yielded 78.99 mg of carbon dioxide and 19.96 mg of water.Analysis for nitrogen showed that the compound contained 4.62 % N by mass. Calculate the empirical formula. • C17H21O4N

  44. The combustion analysis of 19.8mg of an organic acid produced 39.6 mg of carbon dioxide and 16.2 mg of water.The molar mass is ~88 g/mole.Determine the molecular formula and type it in. Yes, Calculators

  45. The combustion analysis of 19.8 mg of an organic acid produced 39.6 mg of carbon dioxide and 16.2 mg of water.The molar mass is ~88 g/mole. Determine the molecular formula. • thus C2H4O1 MM=44 thus C4H8O2 • Now let’s draw a possible structural formula of this organic acid. Hint, put all the C’s in a row, and put oxygens on the same C (do on whiteboard or scrap paper).

  46. O = -C-O-H CH3CH2CH2 A possible structural formula of an organic acid with the formula C4H8O2 • All organic acids have a -COOH group. • Thus the formula below would be an option. • Name?

  47. O = -C-O-H CH3CH2CH2 A possible structural formula of an organic acid with the formula C4H8O2 • All organic acids are • carbonroot-oic acid • This 4-carbon acid would be butanoic acid.

  48. Dianabol is one of the anabolic steroids that has been used by some athletes to increase the size and strength of their muscles. It is similar to the male hormone testosterone. Some studies indicate that the desired effects of the drug are minimal, and the side effects, which include sterility, behavior changes, increased risk of liver cancer and heart disease, keep most people from using it. The molecular formula of Dianabol, which consists of carbon, hydrogen, and oxygen, can be determined using the data from two different experiments. In the first experiment, 14.765 g of Dianabol is burned, and 43.257 g CO2 and 12.395 g H2O are formed. In the second experiment, the molecular mass of Dianabol is found to be 300.44 g/mole.  What is the molecular formula for Dianabol • Again, type in a numerical answer in order CwHx(and)Oy.

  49. The molecular formula of Dianabol, which consists of carbon, hydrogen, and oxygen, can be determined using the data from two different experiments. In the first experiment, 14.765 g of Dianabol is burned, and 43.257 g CO2 and 12.395 g H2O are formed. In the second experiment, the molecular mass of Dianabol is found to be 300.44 g/mole.  What is the molecular formula for Dianabol • C20H28O2

  50. One of the additives in unleaded gasoline that replaced tetraethyl lead in leaded gasoline is called MTBE. When 15.078 g MTBE is burned completely, 37.640 g CO2 and 18.489 g H2O form.In a separate experiment the molecular mass of MTBE is found to be 88.150 g/mole. What is the molecular formula for MTBE? • Again, type in a numerical answer in order CwHx(and)Oy(if necessary).

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