Chapter 3A-Stoichiometry TAVSS.doc

CHEM 161-2007 CHAPTER 3A ? STOICHIOMETRY: CHEMICAL CALCULATIONS PRACTICE PROBLEMS DR. ED TAVSS Atoms, ions, moles and molecular wts. Empir. & molec. formulas ATOMS, IONS, MOLES AND MOLECULAR WTS. 5 Chem 161-2006 Exam I Hill, Petrucci et al., 4th edition Chapter 3 ? Stoichiometry: Chemical Calculations Atoms, ions, moles and molecular weights How many oxygen atoms are there in 2.25 moles of BaSO4? A. 9.00 B. 2.41 × 1024 C. 5.42 × 1024 D. 1.35 × 1024 E. 3.39 × 1023 2.25 mol x (6.022 x 1023 molecules/mol) x (4 oxygen atoms/molecule) = 5.42 x 1024 atoms O. C 7 Chem 161-2005 Exam I Zumdahl 6th edition Chapter 3A Atoms, ions, moles and molecular weights What is the % N (by mass) in ammonium nitrate? A. 17.5% B. 21.2% C. 53.8% D. 22.2% E. 35.0% NH4NO3 = 14.01 + (4 x 1.008) + 14.01 + (3 x 16.00) = 80.05 ((2 x 14.01)/80.05) x 100 = 35.00% 3 Chem 161-2005 Final exam Chapter 3 - Stoichiometry Moles and molecular weights A 0.0567 g sample of a compound contains 5.865 x 1019 molecules. What is the molar mass of this compound? A. 128 EMBED Equation.3 B. 256 EMBED Equation.3 C. 384 EMBED Equation.3 D. 452 EMBED Equation.3 E. 582 EMBED Equation.3 Compound 0.0567 g 5.865 x 1019 molecules ?MW Plan: molecules ? moles ? MW 5.865 x 1019 molecules x (1 mol/(6.022 x 1023 molecules)) = 9.739 x 10-5 moles 0.0567 g/(9.739 x 10-5 moles) = 582 g/mol 27 Chem 161-2005 Final exam Chapter 3 - Stoichiometry Moles and molecular weights Which one of the following is true regarding the mass of one molecule of H2O? Mass(amu) Mass(g) A. 18 18 B. 18 1.0 C. 1.1 x 1025 1.1 x 10-23 D. 18 3.0 x 10-23 E. 3.0 x 10-23 18 MW H2O = 18.02 g/mol (18.02 g/mol)/(6.022 x 1023 molecules/mol) = 2.99 x 10-23 g/molecule The mass in amu?s is the same as the mass per mole, except that since it is the mass per atom, it is described as amu?s rather than grams. Hence the mass of 1 atom of H2O is 18.02 amu?s (whereas the mass of 1 mol of H2O is 18.02 g). 46 Chem 161-2005 Final exam Chapter 3 - Stoichiometry Moles and molecular weights How many molecules of H2 are required to make 0.44 g of propane, C3H8(g)? A. 6.02 x 10-23 molecules B. 2.4 x 1022 molecules C. 6.0 x 1021 molecules D. 2.4 x 1025 molecules E. 3.0 x 1022 molecules H2 + C ? C3H8 4H2 + 3C ? C3H8 0.44g Plan: gC3H8 ? molC3H8 ? molH2 ? moleculesH2 0.44g x (1 mol/44.11g) x (4 mol H2/mol C3H8) x (6.022 x 1023 molecules H2/mol H2) = 2.40 x 1022 molecules of H2 50 Chem 161-2005 Final exam Chapter 3 - Stoichiometry Moles and molecular weights What mass of sodium sulfate will contain 5.00 x 1022 oxygen atoms? A. 12.5 g B. 16.2 g C. 8.39 g D. 3.26 g E. 2.95 g Na2SO4 1 mol Na2SO4 = 4 mol O Plan: atomsO ? molesO ? molesNa2SO4 ? gNa2SO4 5.00 x 1022 O atoms x (1 mol O/6.022 x 1023 O atoms) x (1 mol Na2SO4/4 mol O atom) x 142.05 g/mol = 2.95g Na2SO4 12 Chem 161-2005 Exam I Zumdahl 6th edition Chapter 3A Atoms, ions, moles and molecular weights How many sulfur atoms are there in 0.566g of Na2S2O3? A. 1.20 ( 1024 atoms B. 2.16 ( 1021 atoms C. 4.31 ( 1021 atoms D. 0.00716 atoms E. 2 atoms MW Na2S2O3 = (2 x 22.99) + (2 x 32.06) + (3 x 16.00) = 158.1 g/mol 0.566 g x (1 mol/158.1 g) x (6.022 x 1023 molecules Na2S2O3/mol Na2S2O3) x (2 atoms S/molecule Na2S2O3) = 4.31 x 1021 atoms of sulfur 21 Chem 161-2005 Exam I Zumdahl 6th edition Chapter 3 Atoms, ions, moles and molecular weights The estimated population of the planet is currently about 6.5 ( 109 people. How many moles of people is that? A. 3.9 ( 1033 mol B. 6.5 mol C. 1.1 ( 10-14 mol D. 1.1 mol E. 3.9 mol 6.5 x 109 people x (1 mole/6.022 x 1023 people) = 1.08 x 10-14 mol 22 Chem 161-2005 Exam I Zumdahl 6th edition Chapter 3 Atoms, ions, moles and molecular weights A 0.0567g sample of a compound is found to contain 9.74 ( 10-5 mols. What is the molar mass of the compound? A. 1.72 ( 10-3 g/mol B. 5.52 g/mol C. 552 g/mol D. 582 g/mol E. 172 g/mol mol = g/MW MW = g/mol 0.0567g/(9.74 x 10-5 mol) = 5.82 x 102 g/mol = MW 12. Chem 161-2004 Exam I Zumdahl 6th edition Chapter 3 Moles and Molar Masses For which one of the compounds below does 0.256 mol have a mass of 12.9 g? A. C2H4O B. CO2 C. C2H6 D. CH2Cl2 E. CH3Cl Plan: mol ? g MW A. C2H4O 44.05 B. CO2 44.01 C. C2H6 30.07 D. CH2Cl2 84.93 E. CH3Cl 50.48 Plan: g ? mol C2H4O: 12.9 g x (1 mol/44.05 g) = 0.293 mol CO2: 12.9 g x (1 mol/44.01 g) = 0.293 mol C2H6: 12.9 g x (1 mol/30.07 g) = 0.429 mol CH2Cl2: 12.9 g x (1 mol/84.93 g) = 0.152 mol CH3Cl: 12.9 g x (1 mol/50.48 g) = 0.256 mol 19. Chem 161-2004 Exam I Zumdahl 6th edition Chapter 3 Moles and Molar Masses A given sample of a compound XeFn (where n is a whole number) has a mass of 0.311 g and contains 9.03 x 1020 molecules. The value of n is A. 2 B. 3 C. 4 D. 5 E. 1 XeFn 0.311 g 9.03 x 1020 molecules We know the atomic weight of Xe and F from the periodic table. If we found the MW of XeFn, then we could determine the value of n. Plan: molecules ? moles ? MW 9.03 x 1020 molecules x (1 mol/6.022 x 1023 molecules) = 0.0014995 mol 0.311 g/0.0014995 mol = 207.40 g/mol = MW Xe = 131.30 g/mol 207.40 - 131.30 = 76.1 g/mol = mass of F 76.1/19.00 = 4.01 mol of F Therefore, XeFn = XeF4 24. Chem 161-2004 Exam I Zumdahl 6th edition Chapter 3 Moles and Molar Masses Fluorine has only one stable isotope F-19 of relative atomic mass 19.0. The mass of a fluorine atom is A. 3.16 x 10-23 g B. 19.0 g C. 6.02 x 10-23 g D. 38.0 g E. 1.14 x 10-21 g Plan: AW ? mass of one atom (19.0 g/mol) x (1 mol/6.022 x 1023 atoms) = 3.16 x 10-23 g 7 Chem 161-2004 Exam I Zumdahl 6th edition Chapter 3A Atoms, ions, moles and molecular weights What is the % N (by mass) in ammonium nitrate? A. 17.5% B. 21.2% C. 53.8% D. 22.2% E. 35.0% NH4NO3 = 14.01 + (4 x 1.008) + 14.01 + (3 x 16.00) = 80.05 ((2 x 14.01)/80.05) x 100 = 35.00% 12 Chem 161-2004 Exam I Zumdahl 6th edition Chapter 3A Atoms, ions, moles and molecular weights How many sulfur atoms are there in 0.566g of Na2S2O3? A. 1.20 ( 1024 atoms B. 2.16 ( 1021 atoms C. 4.31 ( 1021 atoms D. 0.00716 atoms E. 2 atoms MW Na2S2O3 = (2 x 22.99) + (2 x 32.06) + (3 x 16.00) = 158.1 g/mol 0.566 g x (1 mol/158.1 g) x (6.022 x 1023 molecules Na2S2O3/mol Na2S2O3) x (2 atoms S/molecule Na2S2O3) = 4.31 x 1021 atoms of sulfur 12 Chem 161-2004 Exam I Zumdahl 6th edition Chapter 3A Atoms, ions, moles and molecular weights How many sulfur atoms are there in 0.566g of Na2S2O3? A. 1.20 ( 1024 atoms B. 2.16 ( 1021 atoms C. 4.31 ( 1021 atoms D. 0.00716 atoms E. 2 atoms MW Na2S2O3 = (2 x 22.99) + (2 x 32.06) + (3 x 16.00) = 158.1 g/mol 0.566 g x (1 mol/158.1 g) x (6.022 x 1023 molecules Na2S2O3/mol Na2S2O3) x (2 atoms S/molecule Na2S2O3) = 4.31 x 1021 atoms of sulfur 21 Chem 161-2004 Exam I Zumdahl 6th edition Chapter 3A Atoms, ions, moles and molecular weights The estimated population of the planet is currently about 6.5 ( 109 people. How many moles of people is that? A. 3.9 ( 1033 mol B. 6.5 mol C. 1.1 ( 10-14 mol D. 1.1 mol E. 3.9 mol 6.5 x 109 people x (1 mole/6.022 x 1023 people) = 1.08 x 10-14 mol 22 Chem 161-2004 Exam I Zumdahl 6th edition Chapter 3A Atoms, ions, moles and molecular weights A 0.0567g sample of a compound is found to contain 9.74 ( 10-5 mols. What is the molar mass of the compound? A. 1.72 ( 10-3 g/mol B. 5.52 g/mol C. 552 g/mol D. 582 g/mol E. 172 g/mol mol = g/MW MW = g/mol 0.0567g/(9.74 x 10-5 mol) = 5.82 x 102 g/mol = MW 18. CHEM 161- 2004 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY MOLES AND MOLECULAR WEIGHTS Calculate the mass of a sample of butane, C4H10, that contains 7.50 x 1022 carbon atoms. A. 0.452 g B. 3.62 g C. 14.5 g D. 29.0 g E. 1.81 g MW C4H10 = (4 x 12.01) + (10 x 1.01) = 58.14 g Since there are 4 carbon atoms per molecule, then 7.50 x 1022 carbon atoms would be in (7.50 x 1022)/4 = 1.875 x 1022 C4H10 molecules (1.875 x 1022molecules)/(6.022 x 1023 molecules/mol) = 0.03114 mol C4H10 0.03114 mol C4H10 x 58.14 g/mol = 1.81 g C4H10 23. CHEM 161- 2004 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY MOLES AND MOLECULAR WEIGHTS Which statements are true about the mole? X. A mole of any element contains exactly 6.022 x 1023 atoms. Y. A mole of Carbon-12 has a mass of exactly 12 grams. Z. A mole of water has a mass of approximately 18.0 amu. A. X only B. Y and Z only C. Y only D. X, Y and Z E. X and Y only X. Whereas a mole of Na contains exactly 6.022 x 1023 atoms, a mole of O2 contains 12.044 x 1023 atoms. Therefore, ?A? is not true. Y. A mole of Carbon-12 having a mass of exactly 12 grams is the SI definition of a mole. Therefore, this is true. Z. A mole of water has a mass of approximately 18.0 grams. A molecule of water has a mass of approximately 18.0 amu?s. Therefore, ?C? is not true. ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 3RD WEEK CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 3-29 Calculate the mass of 500. atoms of iron (Fe). Fe = 55.85 g/mol 6.022 x 10^23 atoms/mol 500 atom x (1 mol/(6.022 x (10^23) atom)) x 55.85 g/mol = 4.64 x 10-20 g Fe ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 3RD WEEK CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 3-35 Calculate the molar mass of the following substances. a. NH3 N = 14.01 H = 1.01 3 H = 3 x 1.01 (14.01 + (3 x 1.01)) = 17.04 ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 3RD WEEK CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 3-37 mod. How many moles, molecules and atoms are present in 1.00 g of NH3? 17.04 g/mol 1 g x (1mol/17.04 g) = 0.0587 mol 6.022 x1023 molecules/mol 0.0587 mol x (6.022 x1023 molecules/mol) = 3.53 x 1022 molecules 4 atoms/molecule 3.53 x 1022 molecules x 4 atoms/molecule = 14.14 x 1022 atoms CHEM 161-2001 SUMMER-EXAM I + ANSWERS ZUMDAHL CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 13. How many oxygen atoms are there in 3.00 g of sodium dichromate, Na2Cr2O7? A. 0.0801 atoms B. 9.85 x 1020 atoms C. 6.90 x 1021 atoms D. 4.83 x 1022 atoms E. 4.21 x 1024 atoms Strategy: grams ? moles ? molecules ? atoms MW Na2Cr2O7 = (2 x 22.99) + (2 x 52.00) + (7 x 16.00) = 261.98 g 3.00 g Na2Cr2O7 x (1 mol Na2Cr2O7/261.98 g Na2Cr2O7) x j(6.022 x 1023 molecules Na2Cr2O7/mol Na2Cr2O7) x (7 oxygen atoms/1 molecule Na2Cr2O7) = 4.83 x 1022 oxygen atoms 12. CHEM 161- 2002 EXAM I + ANSWERS ZUMDAHL CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS If you bought 396 g of gold (atomic weight = 197) for $601, each atom of gold would cost in dollars A. $4.96 x 10-22 B. $1.01 x 10-21 C. $299 D. Choose this choice if none of the others is correct E. $1.50 396 g x (1 mol Au/197 g Au) x (6.022 x 1023 atoms Au/mol Au) = 1.21 x 1024 atoms Au $601/1.21 x 1024 atoms Au = $4.97 x 10-22/atom Au CHEM 161-1999 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WTS. 9. What is the mass of a sample of sodium dichromate, Na2Cr2O7, that contains 2.6 x l020 atoms of oxygen? A. 0.024 g. B. 0.036 g. C. 0.0 l0g. D. 0.016 g. E. 0.048 g. 14 CHEM 161-2000 EXAM I ZUMDAHL CHAPTER 3 - STOICHIOMETRY ATOMS, MOLES AND MOLAR MASSES (MOLECULAR WEIGHTS) How many oxygen atoms are contained in 2.74 g of Al2(SO4)3? A. 12 B. 6.02 x 1023 C. 7.22 x 1024 D. 5.79 x 1022 E. 8.01 x 10-3 15 CHEM 161-2000 EXAM I ZUMDAHL CHAPTER 3 - STOICHIOMETRY ATOMS, MOLES AND MOLAR MASSES (MOLECULAR WEIGHTS) The percent oxygen by mass in borax, Na2B4O7(10 H2O, is A. 42% B. 71% C. 17% D. 40% E. 29% CHEM 161-2000 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 26. The mass percent of H2O in the hydrate Na2SO4?XH2O is 47.0 %. What is X? (X is the number of moles of water for each mole of Na2S04) A. 3 B. 4 C. 5 D. 6 E. 7 Na2SO4 · XH2O 53g 47g 53/141.98 = 0.3733 moles Na2SO4 47/18.02 = 2.6082 moles H2O 0.3733 : 2.6082 = 1 : 7 24. Chem 161-2003 Exam I Chapter 3 - Stoichiometry Atoms, ions, moles and molecular weights The total number of atoms in 0.639 mol of C15H21NO2 is A. 3.85 x 1023 B. 6.02 x 1023 C. 1.50 x l025 D. 1.46 x l025 E. 9.42 x 1023 mol ? molecules ? atoms 0.639 mol C15H21NO2 x 6.022 x 1023 molecules/mol x 39 atoms/molecule = 1.50 x 1025 atoms CHEM 161-2001-HOURLY EXAM I + ANSWERS ZUMDAHL CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 11. Consider two elements X and Y. The compound XY2 has equal masses of X and Y. What is the formula of a compound in which the mass of X is four times the mass of Y? A. XY4 B. X2Y C. X2Y8 D. XY E. X4Y Grams = moles x AW Therefore, since AW is a constant for a give element, if the mass is increased by 4 then the moles are increased by 4. Hence, increasing the mass of X four times results in X4Y2. Other compounds in which the mass of X is four times the mass of Y are X2Y1 and X8Y4. CHEM 161-2001-HOURLY EXAM I + ANSWERS ZUMDAHL CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 12. The mass of 0.00227 mol of the compound XOF3 is 0.236 g. The atomic weight of element X is: A. 31.0 B. 69.0 C. 177 D. 46.9 E. 59.0 Strategy: Find MW XOF3. Subtract AW of O and F from the MW to get AW of X 0.236 g of XOF3/0.00227 mol XOF3 = 103.96 g XOF3/mol XOF3 = MWXOF3 103.96 - (16 + (3 x 19)) = 30.96 CHEM 161-2001-HOURLY EXAM I + ANSWERS ZUMDAHL CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 15. Chromium (element 24) and oxygen form the polyatomic ion, Cr2O72-. The total number of electrons in this polyatomic ion is: A. 16 B. 2 C. 58 D. 106 E. 162 The atomic number = the number of protons The number of protons = the number of electrons, adjusted to charge, if any. Cr atomic number = 24 O atomic number = 8 (2 x 24) + (7 x 8) +2 = 106 electrons. CHEM 161-2001-HOURLY EXAM I + ANSWERS ZUMDAHL CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 19. The number of sulfate ions in 43.1 g of aluminum sulfate is: A. 5.54 x l023 B. 2.09 x l023 C. 7.57 x l022 D. 1.51 x l023 E. 2.28 x 1023 Al2(SO4)3 ? 2 Al3+ + 3(SO4)2- Strategy: gAS ? molesAS ? moleculesAS ? ionsS MW Al2(SO4)3 = (26.98 x 2) + 3(32.07 + [4 x 16.00]) = 342.17 g 43.1 g AS x (1 mol AS/342.17 g AS) x (6.022 x 1023 molecules AS/mol AS) x (3 SO4 ions/molecule AS) = 2.28 x 1028 S ions 3. CHEM 161- 2002 EXAM I + ANSWERS CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS A given sample of xenon fluoride (Atomic Weights: xenon = 131.3, fluorine = 19.00) contains molecules of the formula XeFn, where n is a whole number. Given that 9.03 x 1020 molecules of this compound have a mass of 0.311 g, the value of n is A. 4 B. 1 C. 2 D. 5 E. 3 Strategy: Determine the moles of XeFn present, and then the MW of XeFn. Since we know the atomic weight of Xe, then we can determine the value of ?n?, so that X + Fn = the molecular weight. 9.03 x 1020 x (1 mole/6.022 x 1023 molecules) = 1.4995 x 10-3 moles of XeFn 0.311 g XeFn/1.4995 x 10-3 moles of XeFn = 207.40 = MW XeFn AWXe = 131.3 131.3 + Fn = 207.40 Fn = 76.13 76.13/19 = 4.01 Therefore, n = 4, and the molecule is XeF4 CHEM 161-2001-HOURLY EXAM I + ANSWERS ZUMDAHL CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 24.The mass of an average grape is 4.1 g. The mass of 1.0 mol grapes: A. is 4.1 x l023 g B. cannot be calculated from the data C. is 2.5 x l024 g D. is 2.5 Gg E. is 6.02 x l023 g (6.022 x 1022 grapes/mol of grapes) x (4.1 g/grape) = 2.47 x 1024 g/mol of grapes CHEM 161- 2002 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 18. If you bought 396 g of gold (atomic weight = 197) for $601, each atom of gold would cost in dollars A. $4.96 x 10-22 B. $299 C. $1.01 x 10-21 D. $1.50 E. Choose this choice if none of the others is correct Strategy: Convert g of gold to moles of gold to atoms of gold; then calculate cost per atom. (396 g Au/197 gmol-1) x (6.022 x 1023 atommol-1) = 12.107 x 1023 atoms = 1.21 x 1024 atoms $601/1.21 x 1024 atom = $497 x 10-24/atom = $4.97 x 10-22/atom 19. CHEM 161- 2002 EXAM I + ANSWERS CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS Which one of the following statements is true A. If one ping pong ball has a mass of 5.0 g, one mole of ping pong balls has a mass of 3.0 x 1022 g 5.0 g/ball x (6.023 x 1023 balls/mole) = 3.01 x 1024 balls/mole B. The mass of one mole of 14C is approximately 14 amu The mass of one atom of 14C is approximately14 amu C. A one mole sample of 14C does not have a mass of exactly 14 g. This is true. One mole of 12C was arbitrarily assigned to have a mass of exactly 12 g. The difference between 12C and 14C is two neutrons. One mole of neutrons weighs a little more than 1 g. Hence, the mass of 14C is a little over 14 g. D. If 6.0 x 1023 ping pong balls have a mass of 3.0 x 1024 g, then one ping pong ball has a mass of 5 amu If 6.0 x 1023 ping pong balls have a mass of 3.0 x 1024 g, then one ping pong ball has a mass of 3.0 x 1024 amu E. One mole of any substance contains 6.02 x 1023 atoms One mole of any substance contains 6.02 x 1023 things. If the thing is a molecule than one mole contains 6.02 x 1023 molecules. If the thing is atoms than one mole contains 6.02 x 1023 atoms. CHEM 161- 2002 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY ATOMS, IONS, MOLES AND MOLECULAR WEIGHTS 21. The number of atoms in 1.000 mol of aluminum sulfate is A. 6.022 x 1023 B. 1.700xl024 C. 3.011xl024 D. 1.024xl025 E. Choose this choice if none of the others is correct Al2(SO4)3 = 17 atoms/molecule Strategy: Convert moles to molecules; then molecules to atoms. 1.000 mol x (6.022 x 1023 molecules/mol) x (17 atoms/molecule) = 1.02 x 1025 atoms 10. Chem 161-2003 Exam I Chapter 3 - Stoichiometry Moles and molecular weights Which of the following represents the greatest number of molecules? A. 1.5 mol N2 + 1.5 mol Ne B. 66 g O2 C. 3.5 mol of S2 D. 1 mol of S8 E. 1.5 x 1024 molecules of CO2 mol ? molecules Ne is an atom, not a molecule. 1.5 mol N2 x (6.022 x 1023 molecules/mol) = 9.03 x 1023 molecu g ? mol ? molecules 66 g O2/32 gmol-1 x (6.022 x 1023 molecules/mol) = 1.24 x 1024 mol ? molecules 3.5 mol S2 x (6.022 x 1023 molecules/mol) = 2.10 x 1024 mol ? molecules 1 mol S8 x (6.022 x 1023 molecules/mol) = 6.022 x 1023 = 0.60 x 1024 1.5 x 1024 molecules of CO2 = 1.5 x 1024 molecules of CO2 12. Chem 161-2003 Exam I Chapter 3 - Stoichiometry Moles and molecular weights The mass of one molecule of ClF3 is A. 92.4 g B. 1.54x10-22 g C. 9.24x10-23 g D. 5.57 x 1025 g E. 5.44x10-22 g molecules ? mol ? mass 1 molecule x (1 mol/6.023 x 1023 molecules) x (35.453 + (18.998 x 3)gmol-1) = 1.53 x 10-22 g 161EXAMIV1 FALL 2003 4 18. Chem 161-2003 Final exam Chapter 3 - Stoichiometry Moles and molecular weights It is recommended that we drink 8 glasses of water a day. Each glass contains 240.0 g of water. If the human body contains 1.00 x 1012 cells and each cell is to get an equal amount of water, how many molecules of water does each cell get each day? A. 1.44 x 1024 B. 6.40 x 1023 C. 6.42 x 1013 D. 9.10 x 1016 E. 3.00 x 1014 Plan: glasses of water ? mass of water ? mass of water per cell ? moles of water per cell ? molecules of water per cell 8 glasses x 240.0 g water glass-1 = 1920 g water 1920 g water/(1.00 x 1012 cell) = 1.92 x 10-9 g water/cell 1.92 x 10-9 g watercell-1/18.02 g watermol-1 = 1.0655 x 10-10 mol water cell-1 1.0655 x 10-10 mol water cell-1 x (6.022 x 1023 molecules mol-1) = 6.416 x 1013 molecules of water per cell EMPIR. & MOLEC. FORMULAS 14 Chem 161-2006 Final Exam Chapter 3 ? Stoichiometry: Chemical Calculations Empirical and molecular formulas A compound of element X and chlorine has the formula XCl3. This compound is 67.2% Cl by mass. Identify X. A. Al B. Cr C. N D. Fe E. Tb XCl3 0.672 g Cl 0.328 g X mol Cl = g/AW = 0.672g/35.45gmol-1 = 0.01896 mol Cl 0.01896 mol Cl x (1mol X/3 mol Cl) = 0.006319 mol X mol = g/AW AW = g/mol = 0.328g/0.006319 mol = 51.91 g/mol = Cr B 2 Chem 161-2006 Final Exam Chapter 3 ? Stoichiometry: Chemical Calculations Empirical and molecular formulas The combustion of a 5.000g sample of a hydrocarbon is burned in excess oxygen, yielding 15.0g of carbon dioxide. What is the empirical formula of the hydrocarbon? A. C2H5 B. C3H8 C. CH3 D. C3H9 E. C4H9 HC + O2 ? CO2 5.000g XS 15.0 g gCO2 ? molCO2 ? molCinCO2 ? gCinCO2 ? gCinHC ? gHinHC ? ? molCinHC molHinHC Empirical formula The key concept is that the g of C in CO2 is the same as the grams of C in HC, by the law of conservation of mass. 15.0 g CO2/44.01gmol-1 = 0.3408 mol CO2 0.3408 mol CO2 x (1 mol C/1mol CO2) = 0.3408 mol C in CO2 0.3408 mol C x 12.01g C/mol = 4.093 g C in CO2 By the law of conservation of mass, there must be 4.093 g C in the unknown HC. Therefore, the g of H in HC = 5.000 ? 4.093 = 0.907 g H 4.093g C/12.01gmol-1 = 0.3408 mol C 0.907g H/1.01 = 0.898 mol H 0.898 mol H/0.3408 mol C = 2.635 mol H:1.000 mol C = 7.90 mol H:3 mol C = C3H7.90 = C3H8 Another way of looking at this is by examining the stability of the options. Alkanes are CHn+2; alkenes are CHn; alkynes are CHn-2. Hence, all stable hydrocarbons have an even number of hydrogens. Neither A, C, D or E fit into this category. Hence, they are all unstable products. The only one that is stable is C3H8. B 23 Chem 161-2006 Exam I Hill, Petrucci et al., 4th edition Chapter 3 ? Stoichiometry: Chemical Calculations Empirical and molecular formulas Combustion analysis of 2.456g of some compound produced 3.043g of CO2, 0.3115g of H2O, and 0.4844g of N2. What is the empirical formula of this compound? A. C4HNO2 B. CH2N2O C. C2HN D. C2HN2O E. C2HNO2 CkHmNnOp + O2 ? CO2 + H2O + N2 2.456g 3.043g 0.3115g 0.4844g Plan: grams1 ? mol1 ? mol2 ? grams2 3.043gCO2/(44.01gCO2/molCO2) x (1molC/1molCO2) x 12.01gC/molC = 0.8304g C 0.3115gH2O/(18.02gH2O/molH2O) x (2molH/1molH2O) x 1.01gH/molH = 0.0349gH 0.4844gN2/(28.02gN2/moN2) x (2molN/1molN2) x 14.01gN/molN = 0.4844gN Total analyzed mass of product = 0.8304 + 0.0349 + 0.4844 = 1.3537g But there are 2.456g of sample. Balance must be oxygen. 2.456g ? 1.3537g = 1.1023g O Calculate moles: moles = g/MW 0.8304gC/(12.01gC/molC) = 0.0691 mol C 0.0349gH/(1.01gH/molH) = 0.0346 mol H 0.4844gN/(14.01gN/molN) = 0.0346 mol N 1.1023gO/(16.00gO/molO) = 0.0689 mol O C0.0691H0.0346N0.0346O0.0689 = C2H1N1O2 = C2HNO2 = ?E? E 14 Chem 161-2006 Exam I Hill, Petrucci et al., 4th edition Chapter 3 ? Stoichiometry: Chemical Calculations Empirical and molecular formulas, and percent composition What is the mass of sodium in 30.0 g of Na2HPO4? A. 4.86 g B. 0.423 g C. 0.211 g D. 9.72 g E. 1.30 g MWNa2HPO4 = (2 x 22.99) + (1 x 1.01) + (1 x 30.97) + (4 x 16.00) = 141.96g Plan: gNa2HPO4 ? molNa2HPO4 ? mol Na ? g Na 30.0gNa2HPO4/(141.96gNa2HPO4/molNa2HPO4) x (2molNa/1mol Na2HPO4) x (22.99gNa/molNa) = 9.72g Na D 11 Chem 161-2006 Exam I Hill, Petrucci et al., 4th edition Chapter 3 ? Stoichiometry: Chemical Calculations Empirical and molecular formulas Which of the following statements best describes an empirical formula? A. It is the experimentally measured number of each atom in a molecule. B. The ratio of charges of ions in ionic compounds. C. The ratio of compounds in a mixture. D. The simplest, whole number ratio of elements in a compound. E. The smallest number needed to balance a chemical equation. A. No. Although an empirical formula comes from an experimental measurement, the mass, not the empirical formula numbers, is what is obtained from this experimental measurement. The empirical formula numbers is then calculated from the mass. B. No. For example, in the formula Na2SO3 the ratio of charges of ions is 2:1, which says nothing about the ratio of the atoms, which is 3:1:2. C. No. We are interested in the ratio of atoms in a compound, not the ratio of compounds in a mixture of compounds. D. Yes, e.g., C2H3 is an empirical formula. E. No. An empirical formula has nothing to do with balancing an equation. D 6 Chem 161-2005 Hourly Exam II Chapter 3 - Stoichiometry Empirical and molecular formulas Combustion of a hydrocarbon produces 3.000 g of CO2 and 1.637 g of H2O. What is the empirical formula of the hydrocarbon? A. CH2 B. CH3 C. C2H3 D. C3H4 E. C3H8 CxHy + O2 ? CO2 + H2O 3.00g 1.637g Strategy: gCO2 ? molCO2 ? molC 3.00 g CO2 x (1 mol CO2/44.01gCO2) x (1 mol C/1 mol CO2) = 0.0682 mol C Strategy: gH2O ? molH2O ? molH 1.637 g H2O x (1 mol H2O/18.02 g H2O) x (2 molH/1molH2O) = 0.1817 mol H Mol ratio of H:C = 2.66. Mol ratio of H:C in C3H8 = 2.66. 17 Chem 161-2005 Hourly Exam II Chapter 3 - Stoichiometry Empirical and molecular formulas Which one of the following could not be an empirical formula? A. C3H4 B. C12H17NO C. C4H8O D. C55H72MgN4O5 E. C12H15N3 An empirical formula is a formula containing the lowest common denominator of atoms, expressed in integers. A, B, C and D are possible empirical formulas, because the relative number of atoms cannot be reduced any further and still be integers. However, E, C12H15N3, can be simplified further to C4H5N, so E is not an empirical formula. 13 Chem 161-2005 Final exam Chapter 3 - Stoichiometry Empirical and molecular formulas An oxide of iron weighing 6.752 g reacts with carbon monoxide to yield carbon dioxide and 4.72 g of elemental iron. What is the empirical formula of the iron oxide? A. FeO B. FeO2 C. Fe2O D. Fe3O4 E. Fe2O3 FexOy + CO ? CO2 + Fe 6.752g 4.72g Due to the law of conservation of mass, the FexOy must contain 4.72 g of Fe. Therefore, it also contains 2.032 oxygen Fe O 4.72g 2.032g 4.72 g Fe x (1 mol/55.85 g) = 0.0845 mol Fe 2.032 g O x (1 mol/16.00 g) = 0.127 mol O 0.127 mol O/0.0845 mol Fe = 1.50 O: 1.00 Fe = 3.00 O: 2.00 Fe = Fe2O3 30 Chem 161-2005 Final exam Chapter 3 - Stoichiometry Empirical and molecular formulas A compound contains only C, H and O. Combustion of 10.0 g of the compound produces 14.7 g CO2 and 6.02 g H2O. What is the empirical formula of the compound? A. C2H3O B. CH2O2 C. C3H4O D. C3H5O E. CH2O CxHyOz + O2 ? CO2 + H2O 10.0 g 14.7g 6.02g It is assumed that combustion is done with an excess of oxygen. Hence, determine quantity of carbon and quantity of hydrogen, and then determine quantity of oxygen by difference. Plan: gCO2 ? molCO2 ? molC ? gC 14.7 g CO2 x (1 molCO2/44.01 gCO2) x (1 mol C/1 mol CO2) x 12.01 gC/molC = 4.01 g C Plan: gH2O ? molH2O ? molH ? gH 6.02 g H2O x (1 molH2O /18.02 gH2O) x (2 molH/1 mol H2O) x 1.01 gH/molH = 0.6748 g H Plan: Determine g of oxygen by subtraction of grams of C and H from total. 10.0 g CxHyOz ? (4.01 + 0.6748) = 5.32 g O C H O 4.01g 0.6748g 5.32g C: 4.01g x (1mol/12.01g) = 0.334 mol C H: 0.6748g x (1 mol/1.01g) = 0.668 mol H O: 5.32g x (1 mol/16.00g) = 0.333 mol O Therefore, C0.334H0.668O0.333 = C1H2O1 = CH2O 17. Chem 161-2004 Exam I Zumdahl 6th edition Chapter 3 Empirical and Molecular Formulas Vitamin C contains the elements C, H, and O. It is 40.9% C and 4.58% H by mass. Its molar mass is about 180. The molecular formula for Vitamin C is A. C2H3O2 B. C5H8O7 C. C7H4O6 D. C4H6O4 E. C6H8O6 C = 40.9% H = 4.58% O = 100.00 - (40.9 + 4.58) = 54.52% Plan: g ? mol C: 40.9 g x (1 mol/12.01 g) = 3.41 mol H: 4.58 g x (1 mol/1.01 g) = 4.53 mol O: 54.52 g x (1 mol/16.00 g) = 3.408 mol Empirical formula: C3.41H4.53O3.408 Empirical formula in integers: C1.00H1.329O1.000 C3.00H3.99O3.00 C3H4O3 Empirical weight: 88.1 Molecular formula: 180/88.1 = 2.04 Therefore: C6H8O6 22. Chem 161-2004 Exam I Zumdahl 6th edition Chapter 3 Empirical and Molecular Formulas A chloride of rhenium contains 63.6% rhenium by mass. The formula of this compound is A. Re2Cl3 B. ReCl7 C. ReCl2 D. ReCl3 E. ReCl RexCly 0.636 g Re 1.000 - 0.636 = 0.364 g Cl Plan: g ? mol Re: 0.636 g x (1 mol/186.21 g) = 0.00342 mol Cl: 0.364 g x (1 mol/35.45 g) = 0.01027 mol Empirical ratio of moles: 0.01027:0.00342 = 3.00 Cl:1.00 Re ReCl3 13. CHEM 161- 2004 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS A compound containing only the elements carbon and hydrogen is burned in air, forming 0.660 grams of CO2 and 0.270 grams of H2O. The empirical formula of the hydrocarbon is: A. CH4 B. CH3 C. C2H3 D. CH2 E. CH Plan: massCO2 ? molesCO2 ? molesC massH2O ? molesH2O ? molesH 0.660 gCO2 x (1 molCO2/44.01 g) x (1 molC/1 molCO2) = 0.015 mol C 0.270 gH2O x (1 molH2O/18.02 g) x (2 molH/1 mol H2O) = 0.030 mol H Empirical formula: C0.015H0.030 = CH2 ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 3RD WEEK CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS 3-71 A compound that contains only nitrogen and oxygen is 30.4% N by mass; the molar mass of the compound is 92 g/mol. What is the empirical formula of the compound? What is the molecular formula of the compound? 30.4% nitrogen 100.0% - 30.4% = 69.6% oxygen N0.304 gO0.696 g Molecular formulas are based on mole relationships, not mass relationships. So mass must be converted into moles. 0.304 g x (1 mol/14.01 g) = 0.0217 mol N 0.696 g x (1 mol/16.00 g) = 0.0435 mol O N0.0217molO0.0435mol 0.0435/0.0217 = 2.0046/1.0000 = N1O2 = NO2 = empirical formula The molecular weight of N1O2 = 46.01. But the molar mass of the stated compound is 92. Therefore, it is (92/46.01 =) 2.000 times the mass of N1O2, or N2O4. Molecular formula = N2O4 CHEM 161- 2002 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS 12. A compound consisting of only Mg, P, and O is found to contain 21.9% by mass Mg, and 27.8% by mass P. The empirical formula of the compound is A. Mg3(PO4)2 B. Mg2P2O7 C. MgP2O4 D. Mg2P3O8 E. MgPO4 Find the % O by subtracting the % Mg and % P from 100%. Then convert all percentages to moles. Then convert moles to integers. 100 - (21.9 + 27.8) = 50.3% O Mg21.9P27.8O50.3 21.9 g Mg/24.305 g mol-1 = 0.90 mol Mg 27.8 g P/30.974 g mol-1 = 0.898 mol P 50.3 g O/16.00 gmol-1 = 3.14 mol O Mg0.90P0.898O3.14 Divide all by the smallest amount, which is 0.898. Mg1.002P1.000O3.50 We need integers. Mg2P2O7 CHEM 161-1998 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS 6. The bitter-tasting compound quinine is a component of tonic water and is used as a protection against malaria. When a sample of mass 0.487 g was burned, 1.321 g of CO2, 0.325 g of H2O, and 0.0421 g of N2 were produced. What is the empirical formula of quinine? A. C10H20N2O B. C10H32NO2 C. C9H23N2O2 D. C10H12NO E. C9H14N2O Quinine ? CO2 + H2O + N2 0.487 g 1.321 g 0.325 g 0.0421 g Plan: g C in CO2: gCO2 ? molCO2 ? mol C ? gC MWCO2 = 12.01 + (16.00 x 2) = 44.01 1.321 g CO2 x (1molCO2/44.01 g CO2) x (1 mol C/1mol CO2) x (12.01 g C/mol C) = 0.360 g C g H in H2O: gH2O ? molH2O ? mol H ? gH 0.325 g H2O x (1 mol H2O/18.02 g H2O) x (2 mol H/1mol H2O) x (1.008 g H/1 mol H) = 0.0364 g H g N in N2: gN2 ? mol N2 ? mol N ? g N It?s not necessary to calculate the mass of N because 0.0421 g N2 = 0.0421 g N, i.e., it?s not relevant what the form of N is. However, if we want to follow the standard calculation procedure: 0.0421 g N2 x (1 mol N2/28.02 g N2) x (2 mol N/1 mol N2) x (14.01 g N/mol N) = 0.0421 g N Oxygen not analyzed for. Calculate ?O? by difference. 0.487 g - (0.360 + 0.0364 + 0.0421) = 0.0485 g O. Determine moles of C, H, N and O C: 0.360g x (1 mol/12.01 g) = 0.02998 mol H: 0.0364 g x (1 mol/1.008 g) = 0.03611 mol N: 0.0421 g x (1 mol/14.01 g) = 0.00301 mol O: 0.0485 g x (1 mol/16.00 g) = 0.00303 mol Empirical formula: Divide by the smallest moles, which is 0.00301 C9.96H12.00N1.00O1.01 = C10H12NO CHEM 161-1999 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS 23. Iron of mass 2.00 grams reacts with sulfur to form 4.296 grams of "fool's gold". What is the empirical formula of the "fool's gold"? A. Fe3S4 B. Fe2S3 C. Fe2S D. FeS E. FeS2 16. CHEM 161-2000 EXAM I ZUMDAHL CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS If 0.600 mol of element X is found to react completely with 4.80 g of O2 gas, what is the empirical formula of the resultant oxide? A. X2O B. XO2 C. X4O D. XO4 E. XO 17. CHEM 161-2000 EXAM I ZUMDAHL CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS The products of the combustion in excess O2 gas of 3.42 g of a compound containing only nitrogen and hydrogen are NO2 and H2O. If 9.82 g of NO2 are obtained, what is the empirical formula of the compound? A. NH B. NH2 C. N2H D. NH3 E. N2H6 CHEM 161-2001 FINAL EXAM WITH ANSWERS CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS 11. Phenol, a general disinfectant, has the mass percent composition: 76.57% C, 6.43% H and 17.00% 0. What is its empirical formula? A. C2H6O2 B. CH2O C. C3H3O D. C6H6O E. C6H6O3 C76.57H6.42O17.00 is grams. We need moles. 76.57/12 = 6.38; 6.43/1.01 = 6.37 17.00/16.00 = 1.06 This is equal to 6:6:1 CHEM 161-2001 SUMMER FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS 36. Combustion analysis of 1.244 g of an unknown compound produced 1.542g of CO2, 0.1578g ofH2O and 0.2452g of N2. What is the empirical formula of the compound? A. C4HO5N B. C4H2O4N C. C2HO2N D. C2HO5N E. C2HO2N2 CHEM 161-2001 SUMMER FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS 37. Which of the following could not be an empirical formula? A. Al2S3O9 B. C2H4 C. HBr D. CH E. Na2Cr2O7 CHEM 161-2001 SUMMER-EXAM I ZUMDAHL CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS Chem 161-Su01-Empirical and molecular formulas 6. What is the empirical formula of a compound that is 29.09% sodium, 40.56% sulfur and 30.35% oxygen by mass? A. Na2S2O3 B. Na3S4O3 C. Na3S3O2 D. Na2SO4 E. Na2SO3 Since only the ratio of the percentages is what is important, arbitrarily call the percentages grams. Hence, 29.09% sodium = 29.09 g sodium AWNa = 22.99 AWS = 32.066 AWO = 16.00 moles of Na = 29.09g Na x (1 mol Na/22.99 g Na) = 1.265 moles of S = 40.56 g S x (1 mol S/32.066 g S) = 1.265 moles of O = 30.35 g O x (1 mol O/16.00 g O) = 1.897 Divide by the smallest number: Na1.00S1.00O1.50 Multiply each set of numbers by 2 = Na2S2O3 = ?A? CHEM 161-2001 SUMMER-EXAM I ZUMDAHL CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS 20. A 1.500 g sample of compound containing only C, H, and O was burned completely. The only combustion products were 1.738 g CO2 and 0.711 g H2O. What is the empirical formula of the compound? A. CH2O2 B. C2H4O3 C. CH2O D. CHO E. CH4O Strategy: gCO2 ? moles CO2 ? moles C ? g C gH2O ? moles H2O ? moles H ? g H Total g - (gC + gH) = gO g C ? molesC g H ? moles H g O V moles O MWCO2 = 12.01 + (2 x 16.00) = 44.01 g MW H2O = (2 x 1.01) + 16.00 = 18.02 g 1.738 g CO2 x (1 mol CO2/44.01 g CO2) x (1 mol C/1 mol CO2) x (12.01 g C/mol C) = 0.4743 g C 0.711 g H2O x (1 mol H2O/18.02 g H2O) x (2 mol H/1 mol H2O) x (1.01 g H/mol H) = 0.0797 g H g O = 1.500 g total - (0.4743 g C + 0.0797 g H) = 0.946 g O moles C = 0.4743 g C x (1 mol C/12.01 g C) = 0.03949 moles C moles H = 0.0797 g H x (1 mol H/1.01 g C) = 0.07891 moles H moles O = 0.946 g O x (1 mol O/16.00 g O) = 0.0591 moles O Divide all by the lowest number: 0.03949/0.03949 = 1.00 mol C 0.07891/0.03949 = 1.998 mol H 0.0591/0.03949 = 1.50 mol O Multiply everything by 2 Empirical formula = C2H4O3 CHEM 161-2001-HOURLY EXAM I + ANSWERS ZUMDAHL CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS 17. A 1.17-g sample of a binary compound of nitrogen and hydrogen is burned completely in excess O2 to form NO2 and H2O. The mass of NO2 is found to be 3.17g. The empirical formula of the compound is: A. NH B. NH2 C. NH3 D. N2H5 E. N2H NH + O2 ? NO2 + H2O Strategy: massNO2 ? molNO2 ? molN ? massN ? massH ? molH molN & molH ? empirical formula MW NO2 = 46 AW N = 14 3.17 g NO2 x (1 mol NO2/46 g NO2) x (1 mol N/1 mol NO2) = 0.0689 mol N 0.0689 mol N x (14 g N/mol N) = 0.965 g N 1.17 g total - 0.965 g N = 0.205 g H 0.205 g H x (1 mol H/1.01 g H) = 0.203 mol H Empirical Formula: 0.203 mol H/0.0689 mol N = 2.95 H : 1 N = H3N or NH3 CHEM 161-2001-HOURLY EXAM I + ANSWERS ZUMDAHL CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS 22. A binary compound of bromine and oxygen has a molecular weight of 128. Which of the following compounds has the same empirical formula as this compound? A. Br2O8 MW = 144 B. Br2O6 MW = 128 C. Br2O5 MW = 240 D. Br2O2 MW = 192 E. Br2O4 MW = 112 The empirical formula must be one -- which when multiplied by an integer -- provides the MW. Therefore, the empirical formula can only be 128, 64, 32, 16, etc. Answer: B 14. CHEM 161- 2002 EXAM I + ANSWERS CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS The empirical formula of the compound putrescine is C2H6N and its molecular weight is 88.2. Its molecular formula is A. Choose this choice if none of the others is correct B. C2H6N2 C. C4H8N2 D. C4H12N E. C4H10N2 C2H6N = empirical weight of 44 C4H12N2 = molecular weight of 88 23. CHEM 161- 2002 EXAM I + ANSWERS CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS A sample of a compound containing only C and H is burned completely to form 9.94 g CO2 and 3.05 g H2O. (Atomic Weights: H = 1.008, C = 12.01, O = 16.00) The empirical formula of the compound is A. CH B. C2H3 C. C2H2 D. C2H E. C3H4 9.94 g CO2 x (1 mol CO2/44 g CO2) x (1 mol C/1 mol CO2) = 0.226 mol C 3.05 g H2O x (1 mol H2O/18 g H2O) x (2 mol H2 x 1 mol H2O) = 0.339 mol H 0.339 mol H : 0.226 mol C = 1.5 H : 1 C Empirical formulas must contain integers, with carbon before hydrogen. Therefore, C2H3. ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 3RD WEEK CHAPTER 3 - STOICHIOMETRY EMPIRICAL AND MOLECULAR FORMULAS 3-121 mod. A compound contains only carbon, hydrogen, nitrogen and oxygen. Combustion of 0.157 g of the compound produced 0.213 g CO2 and 0.0310 g H2O. In another experiment, it is found that 0.157 g of the compound produces 0.0351 g NH3. What is the empirical formula of the compound? CHNO + XS O2 ? CO2 + H2O + NH3 Plan: (1) Find grams of C, H and N. Subtract this from total grams to get grams of O. (2) Find moles of C, H, N and O. (3) Find empirical formula. Carbon: 0.213 g CO2 x (12.01 g C/44.01 g CO2) = 0.0581 g C Hydrogen: 0.0310 g H2O x (2.02 g H/18.02 g H2O) = 0.003475 g H Nitrogen: 0.03506 g NH3 x (14.01 g N/17.03 g NH3) = 0.02884 g N Find oxygen by difference: 0.157 g CHNO - (0.0581 g C + 0.003475 g H + 0.02884 g N) = 0.06659 g O Now convert each of these quantities into moles: 0.0581 g C x (1 mol/12.01 g) = 0.004838 mol C 0.003475 g H x (1 mol/1.01 g) = 0.003441 mol H 0.02884 g N x (1 mol/14.01 g) = 0.0020585 mol N 0.06659 g O x (1 mol/16.00 g) = mol 0.0041619 mol O C0.004838H0.003441N0.002059O0.004162 Calculate ratio to the lowest number: 0.004838/0.0020585 = 2.3503 moles C 0.003441/0.0020585 = 1.6716 moles H 0.0020585 /0.0020585 = 1.0000 moles N 0.0041619 /0.0020585 = 2.0218 moles O Find whole integers: Try multiplying each by 2: 2.3503 C x 2 = 4.7006 1.6716 H x 2 = 3.3432 1.0000 N x 2 = 2.0000 2.0218 O x 2 = 4.0436 Still not integers. Try multiplying each by 3: 2.3503 C x 3 = 7.0509 1.6716 H x 3 = 5.0148 1.0000 N x 3 = 3.0000 2.0218 O x 3 = 6.0654 These numbers are within the approximately 1% of whole integers, and are therefore acceptable. Empirical formula: C7H5 22. Chem 161-2003 Exam I Chapter 3 - Stoichiometry Empirical and molecular formulas The molecular weight of a compound whose empirical formula is B2CH3 is measured to be 184. The molecular formula of this compound is A. choose this choice if none of the others is correct B. B10C5H15 C. B8C5H15 D. B8C4H12 E. B10C5H17 B2CH3 Empirical weight = (10.81 x 2)+(12.01 x 1)+(1.01 x 3) = 36.66 g 184/36.66 = 5.02 B2CH3 x 5 = B10C5H15 3. Chem 161-2003 Exam I Chapter 3 - Stoichiometry Empirical and molecular formulas A compound contains 70.9 % Xe, 20.5 % F, and 8.6 % O, by mass. The empirical formula of the compound is A. XeOF2 B. Xe2O2F2 C. XeO2F D. Choose this choice if none of the others is correct E. Xe2OF Plan: g ? moles Xe F O 70.9 20.5 8.6 mol = g/AW 70.9/131.30=0.5400 20.5/19.00 = 1.08 8.6/16.00=0.54 Xe1F2O1 8. Chem 161-2003 Final exam Chapter 3 - Stoichiometry Empir & molec. formulas When 5.00 g of a compound containing only C and H was burned completely in excess oxygen, 15.0 g of carbon dioxide was isolated. What is the empirical formula of the compound? A. C3H6 B. C4H7 C. C3H8 D. CH4 E. C2H5 CxHy + O2 ? CO2 + H2O 5.00 g XS 15.0 g Plan: gCO2 ? molCO2 ? mol C ? g C ? g H ? mol H 15.0 g/(44.01g/mol) = 0.34083 mol CO2 = 0.34083 mol C 0.34083 mol C x (12.01g/mol) = 4.093 g C 5.00 g total - 4.093 g C = 0.907g H 0.907 g H/(1.01 g/mol) = 0.898 mol H 0.34083 mol C: 0.898 mol H = C1H2.635 Multiply by 2 = C2H5.269 Multiply by 3 = C3H7.905 Multiply by 4 = C4H10.54 C3H7.905 is the closest to the options (i.e., C3H8) 24. Chem 161-2003 Final exam Chapter 3 - Stoichiometry Empir & molec. formulas What is the empirical formula of a compound that contains 15.8% Al, 28.1% S, and 56.1% O by mass? A. Al2(SO4) 3 B. AlSO2 C. Al2S9O12 D. Al2SO3 E. AlSO3 Al S O 0.158g 0.281g 0.561g 0.158g/26.982gmol-1 = 0.005856 mol Al 0.281g/32.066gmol-1 = 0.008763 mol S 0.561g/16.000gmol-1 = 0.035060 mol O Al1S1.496O5.987 = Al2S3O12 = Al2(SO4)3 PAGE PAGE 1