1 / 52

Stoichiometry

Stoichiometry. Chapter 3 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/. Stoichiometry - Ch. 3.

xarles
Download Presentation

Stoichiometry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Stoichiometry Chapter 3 E-mail: benzene4president@gmail.com Web-site: http://clas.sa.ucsb.edu/staff/terri/

  2. Stoichiometry - Ch. 3 1. Chlorine has two naturally occurring isotopes 35Cl (34.9688 amu) and 37Cl (36.9659 amu). The average atomic mass of chlorine is 35.453 amu. What are the relative abundances of each isotope?

  3. Stoichiometry - Ch. 3 Average Atomic Mass = (fraction of isotope A)(mass of isotope A) + (fraction of isotope B)(mass of isotope B) + etc.

  4. Stoichiometry - Ch. 3 Mole Model 1 mole of atoms => 6.022x1023 atoms => atomic mass in grams 1 mole of molecules => 6.022x1023 molecules => sum of the atomic masses in grams For example: Al => 6.022x1023 Al atoms/mol => 26.98 g Al/mol Al PCl5 => 6.022x1023 PCl5 molecules/mol => 30.97 g P/mol PCl5 and 5(35.45) g Cl/mol PCl5 => 208.22 g PCl5/mol PCl5

  5. Stoichiometry - Ch. 3 2. Mole Model 1 mole NBr3 has _____moles of N and ____moles of Br 1 mole NBr3 has ____ g of N and ____ g of Br and ____ g total 1 mole NBr3 has ____ molecules 1 mole of NBr3 has ____ N atoms and ____ Br atoms and ____ total atoms

  6. Stoichiometry - Ch. 3 3. How many mg are in 0.63 moles of H2SO4?

  7. Stoichiometry - Ch. 3 4. For which of the following compounds does 1.0 g represent 3.32 × 10-2 mol? a. NO2 b. H2O c. C2H6 d. NH3 e. CO

  8. Stoichiometry - Ch. 3 5. The mass of 0.82 mol of a diatomic element is 131.3 g. Identify the element.

  9. Stoichiometry - Ch. 3 6. How many grams of calcium nitrate will contain the same number of nitrogen atoms that are in 32.6 g of iron(III) cyanide?

  10. Stoichiometry - Ch. 3 7. What is the percent by mass of sulfur in aluminum sulfate?

  11. Stoichiometry - Ch. 3 Percent by mass = (mass of 1 part)x100 (Total mass)

  12. Stoichiometry - Ch. 3 8. An alkali metal oxide contains 83.01% metal by mass. Determine the identity of the metal.

  13. Stoichiometry - Ch. 3 9. Tryptophan is an amino acid well known for its sleep inducing properties. Tryptophan is 64.7% carbon, 5.9% hydrogen, 13.7% nitrogen and 15.7% oxygen. What is the empirical formula for tryptophan?

  14. Stoichiometry - Ch. 3 Empirical Formula => The lowest whole number molar ratio of the elements in a compound 1. Convert given values into moles 2. Divide all moles by the smallest mole value 3. If you have all whole numbers you have the EF – if not try multiplying them all by 2 or 3 etc.

  15. Stoichiometry - Ch. 3 10. The empirical formula for xylene is C4H5 and xylene has a molar mass of 106.16 g/mol. Determine the molecular formula for xylene.

  16. Stoichiometry - Ch. 3 Molecular Formula => The actual molar ratio of the elements in a compound – it is some multiple of the empirical formula (x1, x2 etc) 1. Derive empirical formula 2. Determine the empirical mass 3. (Molar mass)/(empirical mass) = multiple 4. Multiply the empirical formula by the multiple

  17. Stoichiometry - Ch. 3 11. A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.8635 g of CO2 and 0.1767 g of H2O. What is the empirical formula of the compound?

  18. Stoichiometry - Ch. 3 12. A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced 0.213 g of CO2 and 0.0310 g of H2O. In another experiment, 0.103 g of the compound produced 0.0230 g of NH3 (assume all of the N ends up in the ammonia). What is the empirical formula for the compound?

  19. Stoichiometry - Ch. 3 13. Write out a balanced chemical reaction for the following: a. Lithium metal reacts with water to produce lithium hydroxide and hydrogen b. Ammonia reacts with oxygen to yield nitrogen monoxide and water

  20. Stoichiometry - Ch. 3 14. Consider the following unbalanced reaction: H2(g) + Fe3O4(s) Fe (s) + H2O (l) a. How many moles of hydrogen gas are required to make 12.8 moles of iron metal? b. How many grams of water can be produced from 9.64 g of iron oxide?

  21. Stoichiometry - Ch. 3 Methodology for Reaction Stoichiometry Problems 1. Write a balanced chemical reaction 2. Convert given value(s) into moles (you may have to ID the limiting reagent) 3. Use reaction coefficients as a molar ratio 4. Convert moles of your unknown into the desired units

  22. Stoichiometry - Ch. 3 15. Identify the limiting reagent: 4 NH3 + 7 O2 4 NO2 + 6 H2O a. 3 moles of ammonia with 5 moles of oxygen b. 10 g of ammonia with 30 g of oxygen

  23. Stoichiometry - Ch. 3 Identifying Limiting Reagents: 1. Convert all given values into moles 2. Divide each mole value by the coefficient 3. The smallest number identifies the LR Limiting Reagent ⇒ Limits the amount of product that is produced due to running out 1st The limiting reagent is used to determine the maximum yield of product/s aka the theoretical yield and the maximum consumption of reactants

  24. Stoichiometry - Ch. 3 16. Phosphorus can be prepared from calcium phosphate by the following reaction: Ca3(PO4)2 + SiO2 + C  CaSiO3 + P4 + CO (unbalanced) a. How many grams of carbon monoxide can be produced from a mixture of 10g of each reactant? b. How many grams of excess reactant remains?

  25. Stoichiometry - Ch. 3 17. Ammonia is produced from nitrogen and hydrogen. When 23.1 g of nitrogen and 7.40 g of hydrogen are allowed to react, 21.8 g of ammonia is produced. Calculate the percent yield?

  26. Stoichiometry - Ch. 3 Percent yield = (Actual yield)x(100) Theoretical yield

  27. Stoichiometry - Ch. 3 18. How many grams of fluorine are needed to produce 83 g of phosphorus trifluoride, if the reaction has 63.2% yield? P4 + F2 PF3 (unbalanced)

  28. Answer Key – Ch. 3 1. Chlorine has two naturally occurring isotopes 35Cl (34.9688 amu) and 37Cl (36.9659 amu). The average atomic mass of chlorine is 35.453 amu. What are the relative abundances of each isotope? Average Atomic Mass = (fraction of isotope A)(mass of isotope A) + (fraction of isotope B)(mass of isotope B) + etc. Percents add up to 100 or fractions add up to 1 => if the fraction of 35Cl is x then the fraction of 37Cl is 1-x 35.453 amu = (34.9688 amu)(x) + (36.9659 amu)(1-x) => x = 0.758 => 75.8 % 35Cl and 24.2 % 37Cl

  29. Answer Key – Ch. 3 2. Mole Model 1 mole NBr3 has 1 mole of N 3 moles of Br 1 mole NBr3 has 14.01 g of N and3(79.9) g of Br and253.71 g total 1 mole NBr3 has 6.022x1023 molecules 1 mole of NBr3 has 6.022x1023 N atoms and3(6.022x1023) Br atoms and4(6.022x1023) total atoms

  30. Answer Key – Ch. 3 3. How much does 0.63 moles of H2SO4 weigh in grams? Given ⇒ 0.63 moles of H2SO4 Unknown ⇒ grams of H2SO4 Molar mass ofH2SO4 ⇒ 98.07 g/mol 0.63 mol H2SO4 x 98.07 g H2SO4 = 62 g H2SO4 1 mol H2SO4

  31. Answer Key – Ch. 3 4. For which of the following compounds does 1.0 g represent 5.55 × 10-2 mol? a. NO2 b. H2O c. C2H6 d. NH3 e. CO Molar mass is useful in identifying a compound Molar mass = 1.0 g/ 5.55 × 10-2 mol or 18.02 g/mol which is the molar mass of H2O

  32. Answer Key – Ch. 3 5. The mass of 0.82 mol of a diatomic molecule is 131.3 g. Identify the molecule. a. F2 b. Cl2 c. Br2 d. I2 e. Xe Xe is not an option since the substance is diatomic ⇒ the substance has a molar mass = 131.3 g/0.82 mol or 160. g/mol which is the molar mass of Br2

  33. Answer Key – Ch. 3 6. How many grams of calcium nitrate will contain the same number of nitrogen atoms that are in 32.6 g of iron(III) cyanide? 32.6 g Fe(CN)3 x 3 mol N/mol Fe(CN)3 6.022x1023 N atoms x = 133.91 g Fe(CN)3 /mol Fe(CN)3 1 mol of N 4.40 x1023 N atoms 1 mol of N 164.1 g Ca(NO3)2/mol Ca(NO3)2 4.40 x1023 N atoms x x = 6.022x1023 N atoms 2 mol N/mol Ca(NO3)2 60.0 g Ca(NO3)2

  34. Answer Key – Ch. 3 7. What is the percent composition for aluminum sulfate? Molar mass for Al2(SO4)3 ⇒342.17 g/mol % Al = (2(26.98)/342.17) x 100 = 15.77 % Al % S = (3(32.07)/342.17) x 100 = 28.12 % S % O = (12(16)/342.17) x 100 = 56.11 % O

  35. Answer Key – Ch. 3 8. An alkali metal oxide contains 83.01% metal by mass. Determine the identity of the metal. Since an alkali metal has a charge of 1+ the chemical formula for the metal oxide is M2O If the molar mass of the metal is denoted by x ⇒ 83.01 = ((2x)/(2x + 16))(100) x = 39.1 Potassium is the unknown metal

  36. Answer Key – Ch. 3 9. Tryptophan is an amino acid well known for its sleep inducing properties. Tryptophan is 64.7% carbon, 5.9% hydrogen, 13.7% nitrogen and 15.7% oxygen. What is the empirical formula for tryptophan? First we need to get a moles for the molar ratio If we have 100 g sample of typtophan ⇒ 64.7 g C/12.01 g/mol = 5.347 mol C 5.9 g H/1.008 g/mol = 5.853 mol H 13.7 g N/14.01 g/mol = 0.978 mol N 15.7 g O/16 g/mol = 0.981 mol O continue to next slide...

  37. Answer Key – Ch. 3 9. …continued Divide each by the smallest mole value to simplify the ratio ⇒ 5.347 mol C/0.978 = 5.5 mol C 5.853 mol H/0.978 = 6 mol H 0.978 mol N/0.978 = 1 mol N 0.981 mol O/0.978 = 1 mol O Double each to get whole numbers ⇒ 11 mol C:12 mol H :2 mol N:2 mol O Empirical Formula ⇒ C11H12N2O2

  38. Answer Key – Ch. 3 10. The empirical formula for xylene is C4H5 and xylene has a molar mass of 106.16 g/mol. Determine the molecular formula for xylene. First determine the molar mass of the empirical formula ⇒ 4(12.01)g/mol + 5(1.008)g/mol = 53.08 g/mol Divide molar mass of empirical formula into molar mass of the compound ⇒ (106.16g/mol)/(53.08g/mol) = 2 Multiply the empirical formula by 2⇒ Molecular formula ⇒ C8H10

  39. Answer Key – Ch. 3 11. A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.8635 g of CO2 and 0.1767 g of H2O. What is the empirical formula of the compound? CxHyOz + O2 CO2 + H2O All of the carbon in the compound will end up in the CO2 and all of the hydrogen will end up in the water the oxygen is unpredictable so we need determine how much C and H there is in our compound ? …continue to next slide

  40. Answer Key – Ch. 3 11. …continued C ⇒ 0.8635 g CO2 x H ⇒ 0.1767 g H2O x O ⇒ 0.4647 g cmpd – 0.2356 g C – 0.01977 g H = 0.2093 g O …continue to next slide 12.01 g C = 0.2356 g C 44.01 g CO2 2.016 g H = 0.01977 g H 18.016 g H2O

  41. Answer Key – Ch. 3 11. …continued Covert each to moles and divide by smallest value ⇒ (0.2356 g C)/(12.01 g/mol) = 0.01962 mol C/0.01308 = 1.5 mol C (0.01977 g H)/(1.008 g/mol) = 0.01961 mol H/0.01308 = 1.5 mol H (0.2093 g O)/(16 g/mol) = 0.01308 mol O/0.01308 = 1 mol O Double each value 3 mol C:3 mol H:2 mol O Empirical formula ⇒ C3H3O2

  42. Answer Key – Ch. 3 12. A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced 0.213 g of CO2 and 0.0310 g of H2O. In another experiment, 0.103 g of the compound produced 0.0230 g of NH3. What is the empirical formula for the compound? CaHbNcOd + O2 CO2 + H2O + ? ? CaHbNcOd + ?  NH3 + ? …continue to next slide

  43. Answer Key – Ch. 3 12. …continued C ⇒ 0.213 g of CO2 x H ⇒ 0.0310 g of H2O x N ⇒ 0.0230 g of NH3 x 12.01 g C = 0.05813 g C (1st experiment) 44.01 g CO2 2.016 g H = 0.003469 g H (1st experiment) 18.016 g H2O 14.01 g N = 0.01892 g N (2nd experiment) 17.034 g NH3 Since the nitrogen was determined from a different experiment we can use % by mass to figure out the mass of nitrogen in the 1st experiment …continue to next slide

  44. Answer Key – Ch. 3 12. …continued % N = ((0.01892 g N)/(0.103 g cmpd)) x 100= 18.37 % N So in the 1st sample ⇒ (0.1837)x(0.157g cmpd) = 0.02883 g N Oxygen is the remainder ⇒ 0.157 g cmpd – 0.01583 g C – 0.003469 g H – 0.02884 g N =0.06656g O Covert each to moles and divide by the smallest value (0.05813 g C)/(12.01 g/mol) = 0.00484 mol C/0.002058 = 2.35 mol C (0.003469 g H)/(1.008 g/mol) = 0.003441 mol H/0.002058 = 1.67mol H (0.02883 g N)/(14.01 g/mol) = 0.002058 mol N/0.002058 = 1mol N (0.06656 g O)/(16 g/mol) = 0.00416 mol O/0.002058 = 2 mol O multiply by 3to get the empirical formula⇒ C7H5N3O6

  45. Answer Key – Ch. 3 13. Write out a balanced chemical reaction for the following: a. Lithium metal reacts with water to produce lithium hydroxide and hydrogen 2Li + 2 H2O  2 LiOH + H2 b. Ammonia reacts with oxygen to yield nitrogen monoxide and water 4 NH3 + 5 O2 4 NO + 6 H2O

  46. Answer Key – Ch. 3 14. Consider the following unbalanced reaction: H2(g) + Fe3O4(s) Fe (s) + H2O (l) a. How many moles of hydrogen gas are required to make 12.8 moles of iron metal? First ⇒ balance the reaction 4 H2 (g) + Fe3O4 (s) 3Fe (s) + 4H2O (l) 12.8 moles Fe x 4 mol H2 = 17.1 mol H2 3 mol Fe

  47. Answer Key – Ch. 3 b. How many grams of water can be produced from 9.64 g of iron oxide? 9.64 g Fe3O4 x 1 mol Fe3O4 4 mol H2O 18.016 g H2O x x 235.55 g Fe3O4 1 mol Fe3O4 1 mol H2O = 2.95 g H2O

  48. Answer Key – Ch. 3 15. Identify the limiting reagent: 4 NH3 + 7 O2 4 NO2 + 6 H2O a. 3 moles of ammonia with 5 moles of oxygen divide each mole by the molar coefficient and look for smaller value (3 mol NH3)/4 = 0.75 vs. (5 mol O2)/7 = 0.71 ⇒ O2 is the LR b. 10 g of ammonia with 30 g of oxygen convert to moles then divide by molar coefficient (10 g NH3)/(17.034 g/mol) = 0.587 mol NH3/4 = 0.147 vs. (30 g O2)/(32 g/mol) = 0.938 mol O2/7 = 0.143 ⇒ O2 is the LR

  49. Answer Key – Ch. 3 16. Phosphorus can be prepared form calcium phosphate by the following reaction: Ca3(PO4)2 + SiO2 + C  CaSiO3 + P4 + CO (unbalanced) a. How many grams of carbon monoxide can be produced from a mixture of 10 g of each reactant? First ⇒ balance the reaction 2 Ca3(PO4)2 + 6 SiO2 + 10 C  6 CaSiO3 + P4 + 10 CO Next ⇒ determine the limiting reagent (10 g Ca3(PO4)2)/(310.18 g/mol) = 0.0322 mol Ca3(PO4)2/2 = 0.0161 (10 g SiO2)/(60.09 g/mol) = 0.166 mol SiO2/6 = 0.0277 (10 g C)/(12.01g/mol) = 0.833 mol C/10 = 0.0833 Ca3(PO4)2 is the LR … continue to next slide

  50. Answer Key – Ch. 3 Use LR to determine yield of CO ⇒ 0.0322 mol Ca3(PO4)2 x b. How many grams of excess reactant remains? The LR can be used to determine the consumption of the other reactants 0.0322 mol Ca3(PO4)2 x 28.01 g CO 10 mol CO x = 0.451 g CO 2 mol Ca3(PO4)2 10 mol CO 6 mol SiO2 60.09 g SiO2 x = 5.8 g SiO2 1 mol SiO2 2 mol Ca3(PO4)2 10 g – 5.8 g = 4.2 g SiO2 remain 10 mol C 12.01 g C 0.0322 mol Ca3(PO4)2 x x = 1.9 g C 1 mol C 2 mol Ca3(PO4)2 10 g – 1.9 g = 8.1 g C remain

More Related