Molecular and Empirical Formula Calculations, Lecture notes of Stoichiometry

Download Molecular and Empirical Formula Calculations and more Lecture notes Stoichiometry in PDF only on Docsity! 1 Homework #2 Chapter 3 Stoichiometry 23. Need to find average molar mass of X (MX) 𝑀𝑋 = (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋46 ) (𝑀 𝑋46 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋47 ) (𝑀 𝑋47 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋48 ) (𝑀 𝑋48 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋49 ) (𝑀 𝑋49 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋50 ) (𝑀 𝑋50 ) 𝑀𝑋 = (0.0800)(45.95269 𝑎𝑚𝑢) + (0.0730)(46.951764 𝑎𝑚𝑢) + (0.7380)(47.947947 𝑎𝑚𝑢) + (0.0550)(48.94784 𝑎𝑚𝑢) + (0.0540)(49.944792 𝑎𝑚𝑢) = 47.88 𝑎𝑚𝑢 Find the atomic weight on the periodic table that matches 47.88 amu. Ti titanium 26. Need to find the mass present of 151Eu and 153Eu Know MEu = 151.96 amu fraction of 151Eu = x 𝑚 𝐸𝑢151 = 150.9196 𝑎𝑚𝑢 fraction of 153Eu = 1-x 𝑚 𝐸𝑢153 = 152.9209 𝑎𝑚𝑢 𝑀𝐸𝑢 = (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝐸𝑢151 ) (𝑀 𝐸𝑢151 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝐸𝑢153 ) (𝑀 𝐸𝑢153 ) 151.96 𝑎𝑚𝑢 = 𝑥(150.9196 𝑎𝑚𝑢) + (1 − 𝑥)(152.9209 𝑎𝑚𝑢) −0.96 𝑎𝑚𝑢 = (−2.001 𝑎𝑚𝑢)𝑥 𝑥 = 0.48 Therefore 48% 151Eu and 52% 153Eu 27. Need to find the molar mass of 185Ru(𝑀 𝑅𝑢185 ) 𝑀𝑅𝑢 = (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑅𝑢187 ) (𝑀 𝑅𝑢187 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑅𝑢185 ) (𝑀 𝑅𝑢185 ) 186.207 𝑎𝑚𝑢 = (0.6260)(186.956 𝑎𝑚𝑢) + (1 − 0.6260)𝑀 𝑅𝑢185 𝑀 𝑅𝑢185 = 185 𝑎𝑚𝑢 28. Need to find the average molar mass of X (MX) 𝑀𝑋 = (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋𝑎 ) (𝑀 𝑋𝑎 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋𝑏 ) (𝑀 𝑋𝑏 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋𝑐 ) (𝑀 𝑋𝑐 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋𝑑 ) (𝑀 𝑋𝑑 ) 𝑀𝑋 = (0.0140)(203.973 𝑎𝑚𝑢) + (0.2410)(205.9745 𝑎𝑚𝑢) + (0.2210)(206.9759 𝑎𝑚𝑢) + (0.5240)(207.9766 𝑎𝑚𝑢) = 207.22 𝑎𝑚𝑢 Find the atomic weight on the periodic table that matches 207.22 amu. Lead (Pb) 2 34. Note: Grey information is given a) Moles of Sample 4.24 𝑔 𝐶6𝐻6 ( 1 𝑚𝑜𝑙 𝐶6𝐻6 78.12 𝑔 𝐶6𝐻6 ) = 0.0543 𝑚𝑜𝑙 𝐶6𝐻6 Molecules in Sample 0.0543 𝑚𝑜𝑙 𝐶6𝐻6 ( 6.02214×1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶6𝐻6 1 𝑚𝑜𝑙 𝐶6𝐻6 ) = 3.27 × 1022 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶6𝐻6 Atoms in Sample 3.27 × 1022 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶6𝐻6 ( 12 𝑎𝑡𝑜𝑚𝑠 1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶6𝐻6 ) = 3.92 × 1023 𝑎𝑡𝑜𝑚𝑠 b) Mass of Sample 0.224 𝑚𝑜𝑙 𝐻2𝑂 ( 18.02 𝑔 𝐻2𝑂 1 𝑚𝑜𝑙 𝐻2𝑂 ) = 4.04 𝑔 𝐻2𝑂 Molecules in Sample 0.224 𝑚𝑜𝑙 𝐻2𝑂 ( 6.02214×1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐻2𝑂 1 𝑚𝑜𝑙 𝐻2𝑂 ) = 1.35 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐻2𝑂 Total Atoms in Sample 1.35 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐻2𝑂 ( 3 𝑎𝑡𝑜𝑚𝑠 1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐻2𝑂 ) = 4.04 × 1023 𝑎𝑡𝑜𝑚𝑠 c) Moles of Sample 2.71 × 1022 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶𝑂2 ( 1 𝑚𝑜𝑙 𝐶𝑂2 6.02214×1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶𝑂2 ) = 0.0450 𝑚𝑜𝑙 𝐶𝑂2 Mass of Sample 0.0450 𝑚𝑜𝑙 𝐶𝑂2 ( 44.01 𝑔 𝐶𝑂2 1 𝑚𝑜𝑙 𝐶𝑂2 ) = 1.98 𝑔 𝐶𝑂2 Total Atoms in Sample 2.71 × 1022 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶𝑂2 ( 3 𝑎𝑡𝑜𝑚𝑠 1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶𝑂2 ) = 8.13 × 1022 𝑎𝑡𝑜𝑚𝑠 d) Molecules in Sample 3.35 × 1022 𝑎𝑡𝑜𝑚𝑠 𝐶𝐻3𝑂𝐻 ( 1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶𝐻3𝑂𝐻 6 𝑎𝑡𝑜𝑚𝑠 ) = 5.58 × 1021 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶𝐻3𝑂𝐻 Moles of Sample 5.58 × 1021 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶𝐻3𝑂𝐻 ( 1 𝑚𝑜𝑙 𝐶𝐻3𝑂𝐻 6.02214×1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶𝐻3𝑂𝐻 ) = 0.00927 𝑚𝑜𝑙 𝐶𝐻3𝑂𝐻 Mass of Sample 0.00927 𝑚𝑜𝑙 𝐶𝐻3𝑂𝐻 ( 32.05 𝑔 𝐶𝐻3𝑂𝐻 1 𝑚𝑜𝑙 𝐶𝐻3𝑂𝐻 ) = 0.297 𝑔 𝐶𝐻3𝑂𝐻 35. a) 20.0 𝑚𝑔 𝐶8𝐻10𝑁4𝑂2 ( 1 𝑔 1000 𝑚𝑔 ) ( 1 𝑚𝑜𝑙 𝐶8𝐻10𝑁4𝑂2 194.22 𝑔 𝐶8𝐻10𝑁4𝑂2 ) = 1.03 × 10−4 𝑚𝑜𝑙 𝐶8𝐻10𝑁4𝑂2 b) 2.72 × 1021 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶2𝐻5𝑂𝐻 ( 1 𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻 6.02214×1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶𝐻3𝑂𝐻 ) = 0.00452 𝑚𝑜𝑙 𝐶2𝐻5𝑂𝐻 c) 1.50 𝑔 𝐶𝑂2 ( 1 𝑚𝑜𝑙 𝐶𝑂2 44.01 𝑔 𝐶𝑂2 ) = 0.0341 𝑚𝑜𝑙 𝐶𝑂2 Mass of Sample Moles of Sample Molecules in Sample Total Atoms in Sample a) 4.24 g C6H6 0.0543 mol 3.27×1022 molec 3.92×1023 atoms b) 4.04 g 0.224 mol H2O 1.35×1023 molec 4.04×1023 atoms c) 1.98 g 0.0450 mol 2.71×1022 molec CO2 8.13×1022 atoms d) 0.297 g 0.00927 mol 5.58×1021 molec 3.35×1022 atoms in CH3OH 5 Determine moles of O 0.0480 𝑔 𝑂 ( 1 𝑚𝑜𝑙 𝑂 16.00 𝑔 𝑂 ) = 0.00300 𝑚𝑜𝑙 𝑂 Divide through by smallest number of moles (0.00300 mol) Mercury Oxygen 0.003000 𝑚𝑜𝑙 0.00300 𝑚𝑜𝑙 = 1 0.00300 𝑚𝑜𝑙 0.00300 𝑚𝑜𝑙 = 1 Empirical Formula HgO Compound 2 Heating 0.4172 g of compound 2 results in 0.016 g O being produced (mass of O is the mass lost) 𝑚𝐻𝑔 = 𝑚𝑡𝑜𝑡𝑎𝑙 − 𝑚𝑂 = 0.4172 𝑔 − 0.016 𝑔 = 0.401 𝑔 Determine moles of Hg 0.401 𝑔 𝐻𝑔 ( 1 𝑚𝑜𝑙 𝐻𝑔 200.59 𝑔 𝐻𝑔 ) = 0.00200 𝑚𝑜𝑙 𝐻𝑔 Determine mol of O 0.016 𝑔 𝑂 ( 1 𝑚𝑜𝑙 𝑂 16.00 𝑔 𝑂 ) = 0.00100 𝑚𝑜𝑙 𝑂 Divide through by smallest number of moles (0.00100 mol) Mercury Oxygen 0.002000 𝑚𝑜𝑙 0.00100 𝑚𝑜𝑙 = 2 0.00100 𝑚𝑜𝑙 0.00100 𝑚𝑜𝑙 = 1 Empirical Formula Hg2O 54. Assume 100 g sample therefore 56.79 g C, 6.56 g H, 28.37 g O, and 8.28 g N Calculate the number of moles of C, H, O, and N Carbon 56.79 𝑔 𝐶 ( 1 𝑚𝑜𝑙 𝐶 12.01 𝑔 𝐶 ) = 4.729 𝑚𝑜𝑙 𝐶 Hydrogen 6.56 𝑔 𝐻 ( 1 𝑚𝑜𝑙 𝐻 1.01 𝑔 𝐻 ) = 6.50 𝑚𝑜𝑙 𝐻 Oxygen 28.37 𝑔 𝑂 ( 1 𝑚𝑜𝑙 𝑂 16.00 𝑔 𝑂 ) = 1.773 𝑚𝑜𝑙 𝑂 Nitrogen 8.28 𝑔 𝑁 ( 1 𝑚𝑜𝑙 𝑁 14.01 𝑔 𝑁 ) = 0.591 𝑚𝑜𝑙 𝐻 Divide through by smallest mol amount (0.591 mol) Carbon Hydrogen 4.729 𝑚𝑜𝑙 0.591 𝑚𝑜𝑙 = 8.00 6.50 𝑚𝑜𝑙 0.591 𝑚𝑜𝑙 = 11.0 Oxygen Nitrogen 1.773 𝑚𝑜𝑙 0.591 𝑚𝑜𝑙 = 3.00 0.591 𝑚𝑜𝑙 0.591 𝑚𝑜𝑙 = 1.00 Empirical Formula C8H11O3N 6 59. Assume 100 g sample therefore 26.7 g P, 12.1 g N, and 61.2 g Cl Calculate the number of moles of P, N, and Cl Phosphorus 26.7 𝑔 𝑃 ( 1 𝑚𝑜𝑙 𝑃 30.97 𝑔 𝑃 ) = 0.862 𝑚𝑜𝑙 𝑃 Nitrogen 12.1 𝑔 𝑁 ( 1 𝑚𝑜𝑙 𝑁 14.01 𝑔 𝑁 ) = 0.864 𝑚𝑜𝑙 𝑁 Chlorine 62.2 𝑔 𝐶𝑙 ( 1 𝑚𝑜𝑙 𝐶𝑙 35.46 𝑔 𝐶𝑙 ) = 1.75 𝑚𝑜𝑙 𝐶𝑙 Divide through by smallest mol amount (0.862mol) Phosphorus Nitrogen Hydrogen 0.862 𝑚𝑜𝑙 0.862 𝑚𝑜𝑙 = 1.00 0.864 𝑚𝑜𝑙 0.862 𝑚𝑜𝑙 = 1.00 1.75 𝑚𝑜𝑙 0.862 𝑚𝑜𝑙 = 2.03 Empirical Formula PNCl2 𝑀𝑃𝑁𝐶𝑙2 = 114.06 𝑔 𝑚𝑜𝑙 Find molecular formula 580 𝑔 𝑚𝑜𝑙 114.06 𝑔 𝑚𝑜𝑙 = 5.09 Multiply empirical formula by 5 P5N5Cl10 60. CxHyOz + O2 CO2 + H2O (Combustion Reaction) All of the carbon in the compound goes into forming CO2. Calculate 𝑛𝐶 41.98 𝑚𝑔 𝐶𝑂2 ( 1 𝑔 1000 𝑚𝑔 ) ( 1 𝑚𝑜𝑙 𝐶𝑂2 44.01 𝑔 𝐶𝑂2 ) ( 1 𝑚𝑜𝑙 𝐶 1 𝑚𝑜𝑙 𝐶𝑂2 ) = 9.539 × 10−4 𝑚𝑜𝑙 𝐶 All of the hydrogen in the compound goes into forming H2O. Calculate 𝑛𝐻 6.45 𝑚𝑔 𝐻2𝑂 ( 1 𝑔 1000 𝑚𝑔 ) ( 1 𝑚𝑜𝑙 𝐻2𝑂 18.02 𝑔 𝐻2𝑂 ) ( 2 𝑚𝑜𝑙 𝐻 1 𝑚𝑜𝑙 𝐻2𝑂 ) = 7.16 × 10−4 𝑚𝑜𝑙 𝐻 To find the mass of O, find the mass of C and H and subtract them from the overall mass of the compound. Calculate 𝑚𝐶 𝑚𝐶 = 9.539 × 10−4 𝑚𝑜𝑙 𝐶 ( 12.01 𝑔 𝐶 1 𝑚𝑜𝑙 𝐶 ) = 0.01146 𝑔 𝐶 Calculate 𝑚𝐻 𝑚𝐻 = 7.16 × 10−4 𝑚𝑜𝑙 𝐻 ( 1.01 𝑔 𝐻 1 𝑚𝑜𝑙 𝐻 ) = 7.23 × 10−4 𝑔 𝐻 Calculate 𝑚𝑂 𝑚𝑂 = 𝑚𝐶𝑥𝐻𝑦𝑂𝑧 − 𝑚𝐶 − 𝑚𝐻 𝑚𝐶𝑥𝐻𝑦𝑂𝑧 = 19.81 𝑚𝑔 ( 1 𝑔 1000 𝑚𝑔 ) = 0.01981 𝑔 𝑚𝑂 = 0.01981 𝑔 − 0.01146 𝑔 − 7.23 × 10−4 𝑔 = 0.00763 𝑔 Calculate 𝑛𝑂 0.00763 𝑔 ( 1 𝑚𝑜𝑙 𝑂 16.00 𝑔 𝑂 ) = 4.77 × 10−4 𝑚𝑜𝑙 𝑂 7 Divide through by smallest mole amount (3.64×10-4mol) Carbon Hydrogen Oxygen 9.539×10−4𝑚𝑜𝑙 4.77×10−4 𝑚𝑜𝑙 = 2.00 7.16×10−4𝑚𝑜𝑙 4.77×10−4 𝑚𝑜𝑙 = 1.5 4.77×10−4𝑚𝑜𝑙 4.77×10−4 𝑚𝑜𝑙 = 1.00 Multiple through by 2 to get whole numbers Empirical Formula C4H3O2 𝑀𝐶4𝐻3𝑂2 = 83.07 𝑔 𝑚𝑜𝑙 Find molecular formula 𝑀 = 𝑚 𝑛 = 41.5 𝑔 0.250 𝑚𝑜𝑙 = 166 𝑔 𝑚𝑜𝑙 166 𝑔 𝑚𝑜𝑙 83.07 𝑔 𝑚𝑜𝑙 = 2.00 Multiply empirical formula by 2: C8H6O4 61. CxHyOz + O2 CO2 + H2O (Combustion Reaction) All of the C in the compound goes into forming CO2. Therefore, 𝑛𝐶𝑂2 = 𝑛𝐶. 16.01 𝑚𝑔 𝐶𝑂2 ( 1 𝑔 1000 𝑚𝑔 ) ( 1 𝑚𝑜𝑙 𝐶𝑂2 44.01 𝑔 𝐶𝑂2 ) ( 1 𝑚𝑜𝑙 𝐶 1 𝑚𝑜𝑙 𝐶𝑂2 ) = 3.638 × 10−4 𝑚𝑜𝑙 𝐶 All of the H in the compound goes into forming H2O. Therefore, 𝑛𝐻2𝑂 = 2𝑛𝐻. 4.37 𝑚𝑔 𝐻2𝑂 ( 1 𝑔 1000 𝑚𝑔 ) ( 1 𝑚𝑜𝑙 𝐻2𝑂 18.02 𝑔 𝐻2𝑂 ) ( 2 𝑚𝑜𝑙 𝐻 1 𝑚𝑜𝑙 𝐻2𝑂 ) = 4.85 × 10−4 𝑚𝑜𝑙 𝐻 To find the mass of O, find the mass of C and H, and subtract them from the overall mass of the compound. Calculate 𝑚𝐶 𝑚𝐶 = 3.638 × 10−4 𝑚𝑜𝑙 𝐶 ( 12.01 𝑔 𝐶 1 𝑚𝑜𝑙 𝐶 ) = 0.004369 𝑔 𝐶 Calculate 𝑚𝐻 𝑚𝐻 = 4.85 × 10−4 𝑚𝑜𝑙 𝐻 ( 1.01 𝑔 𝐻 1 𝑚𝑜𝑙 𝐻 ) = 4.90 × 10−4 𝑔 𝐻 Calculate 𝑚𝑂 𝑚𝑂 = 𝑚𝐶𝑥𝐻𝑦𝑂𝑧 − 𝑚𝐶 − 𝑚𝐻 𝑚𝐶𝑥𝐻𝑦𝑂𝑧 = 10.68 𝑚𝑔 ( 1 𝑔 1000 𝑚𝑔 ) = 0.01068 𝑔 𝑚𝑂 = 0.01068 𝑔 − 0.004369 𝑔 − 4.90 × 10−4 𝑔 = 0.00582 𝑔 Calculate 𝑛𝑂 0.00582 𝑔 ( 1 𝑚𝑜𝑙 𝑂 16.00 𝑔 𝑂 ) = 3.64 × 10−4 𝑚𝑜𝑙 𝑂 Divide through by smallest mole amount (3.638×10-4 mol) Carbon Hydrogen Oxygen 3.638×10−4𝑚𝑜𝑙 3.638×10−4 𝑚𝑜𝑙 = 1.00 4.85×10−4𝑚𝑜𝑙 3.638×10−4 𝑚𝑜𝑙 = 1.33 3.64×10−4𝑚𝑜𝑙 3.638×10−4 𝑚𝑜𝑙 = 1.00 Multiple through by 3 to get whole numbers Empirical Formula C3H4O3 𝑀𝐶3𝐻4𝑂8 = 88.07 𝑔 𝑚𝑜𝑙 Find molecular formula 176.1 𝑔 𝑚𝑜𝑙 88.07 𝑔 𝑚𝑜𝑙 = 2.000 Multiply empirical formula by 2 C6H8O6 63. X+2Y  XY2 10 The problem tells you that the coefficients in front of each of the species are 1 Determine the molecular formula of aspirin CxHyOz + O2 CO2 + H2O (Combustion Reaction) All of the carbon in the compound goes into forming CO2. Therefore, 𝑛𝐶𝑂2 = 𝑛𝐶 . 2.20 𝑔 𝐶𝑂2 ( 1 𝑚𝑜𝑙 𝐶𝑂2 44.01 𝑔 𝐶𝑂2 ) ( 1 𝑚𝑜𝑙 𝐶 1 𝑚𝑜𝑙 𝐶𝑂2 ) = 0.0500 𝑚𝑜𝑙 𝐶 All of the hydrogen in the compound goes into forming H2O. Therefore, 𝑛𝐻2𝑂 = 2𝑛𝐻. 0.400 𝑔 𝐻2𝑂 ( 1 𝑚𝑜𝑙 𝐻2𝑂 18.02 𝑔 𝐻2𝑂 ) ( 2 𝑚𝑜𝑙 𝐻 1 𝑚𝑜𝑙 𝐻2𝑂 ) = 0.0444 𝑚𝑜𝑙 𝐻 To find the mass of O, find the mass of C and H, and subtract them from the overall weight of the compound. Calculate 𝑚𝐶 𝑚𝐶 = 0.0500 𝑚𝑜𝑙 𝐶 ( 12.01 𝑔 𝐶 1 𝑚𝑜𝑙 𝐶 ) = 0.5005 𝑔 𝐶 Calculate 𝑚𝐻 𝑚𝐻 = 0.0444 𝑚𝑜𝑙 𝐻 ( 1.01 𝑔 𝐻 1 𝑚𝑜𝑙 𝐻 ) = 0.0448 𝑔 𝐻 Calculate 𝑚𝑂 𝑚𝑂 = 𝑚𝐶𝑥𝐻𝑦𝑂𝑧 − 𝑚𝐶 − 𝑚𝐻 𝑚𝑂 = 1.00 𝑔 − 0.601 𝑔 − 0.0448 𝑔 = 0.35 𝑔 Calculate 𝑛𝑂 0.35 𝑔 ( 1 𝑚𝑜𝑙 𝑂 16.00 𝑔 𝑂 ) = 0.022 𝑚𝑜𝑙 𝑂 Divide through by smallest mol amount (0.022 mol) Carbon Hydrogen Oxygen 0.0500 𝑚𝑜𝑙 0.022 𝑚𝑜𝑙 = 2.23 0.0444 𝑚𝑜𝑙 0.022 𝑚𝑜𝑙 = 2.01 0.022 𝑚𝑜𝑙 0.022 𝑚𝑜𝑙 = 1.00 Multiple through by 4 to get whole numbers Empirical Formula C9H8O4 𝑀𝐶9𝐻8𝑂4 = 180.17 𝑔 𝑚𝑜𝑙 Since the molecular formula weight is between 170 𝑔 𝑚𝑜𝑙 and 190 𝑔 𝑚𝑜𝑙 the empirical and molecular formula are the same. Determine the formula or salicylic acid Salicylic acid + C4H6O3  aspirin (C9H8O4) + C2H4O2 Salicylic acid: C7H6O3 137. You need to determine the amount of NO(g) produced. You know that 2.00 mol of NH3 are reacted with 10.00 mol of O2 and 6.75 moles of O2 remains after the reaction goes to completion. This indicates that NH3 is the limiting reagent Determine the moles of O2 reacted 𝑛𝑂2(𝑟𝑒𝑎𝑐𝑡𝑒𝑑) = 𝑛𝑂2(𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙) − 𝑛𝑂2(𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑) = 10.0 𝑚𝑜𝑙 − 6.75 𝑚𝑜𝑙 = 3.25 𝑚𝑜𝑙 NH3 reacts by one of the following equations 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g) Therefore for equation 1 𝑛𝑂2 = 5 4 𝑛𝑁𝑂 and for equation 2 𝑛𝑂2 = 7 4 𝑛𝑁𝑂2 The total number of moles of O2 used was 3.25 giving 5 4 𝑛𝑁𝑂 + 7 4 𝑛𝑁𝑂2 = 3.25 𝑚𝑜𝑙 Similarly for equation 1 𝑛𝑁𝐻3 = 𝑛𝑁𝑂 and for equation 2 𝑛𝑁𝐻3 = 𝑛𝑁𝑂2 11 The total number of moles of NH3 used was 2.00 giving 𝑛𝑁𝑂 + 𝑛𝑁𝑂2 = 2.00 𝑚𝑜𝑙 Combine equations and solve for mole of NO 𝑛𝑁𝑂2 = 2.00 𝑚𝑜𝑙 − 𝑛𝑁𝑂 5 4 𝑛𝑁𝑂 + 7 4 (2.00 𝑚𝑜𝑙 − 𝑛𝑁𝑂) = 3.25 𝑚𝑜𝑙 5𝑛𝑁𝑂 + 14.0𝑚𝑜𝑙 − 7𝑛𝑁𝑂 = 13 𝑚𝑜𝑙 −2𝑛𝑁𝑂 = −1 𝑚𝑜𝑙 𝑛𝑁𝑂 = 0.50 𝑚𝑜𝑙