worksheet_04_103

1 CHEMISTRY 103 – WORKSHEET #4 – Module 4 Stoichiometry (Part I) Do the topics appropriate for your course Prepared by Dr. Tony Jacob https://clc.chem.wisc.edu (Resource page) MOLE = 6.022 x 10 23 units MOLAR MASS: mass of 1mol of substance MASS PERCENT of elements within a compound Example 1 : What is the mass% of N in (NH 4 ) 3 PO 4 ? Answer 1 : 28.19% 1. Assume 1 mol ® total mass = molar mass = 149.12g/mol 2. Find mass of N in 1mol of compound: 3mol N x (14.01g N/1mol N) = 42.03g N 3. ; EMPIRICAL FORMULA: simplest formula/ratio between atoms. Determining EF from mass%: 1. Assume 100g 2. % ® g 3. g ® mol (if grams given instead of mass% ® start at step 3) 4. Write chemical formula; divide by the smallest number of moles 5. Fractions: 1 / 2 (0.5) ® x 2; 1 / 3 or 2 / 3 (0.33, 0.66) ® x 3; 1 / 4 or 3 / 4 (0.25, 0.75) ® x 4 Example 2 : What is the empirical formula for a compound that contains 43.64% by mass phosphorous and 56.36% oxygen? Answer 2 : P 2 O 5 1. assume 100g 2. % ® grams: P: 43.64% x 100g = 43.64g P; O: 56.36% x 100g = 56.36g O 3. g ® mol: ; 4. Write empirical formula: P 1.409 O 3.523 and divide by smallest number of mol: 1.490: 5. Fraction (i.e., 0.500 = 1 / 2 ): multiply by 2: P 1 x 2 O 2.500 x 2 ® P 2 O 5 = empirical formula MOLECULAR FORMULA: the exact formula of a compound (e.g., N 2 O 4 : molecular formula (MF) = N 2 O 4 and empirical formula (EF) = NO 2 ) Can only determine empirical formula from mass%; need molar mass to determine molecular formula Determine Molecular Formula from Empirical Formula and Molar Mass 1. Determine ratio : 2. Take empirical formula and multiply each subscript by the number from the above ratio Example 3 : From Example 2 above: What is the molecular formula if the molar mass is 283.88g/mol? Answer 3 : P 4 O 10 1. Determine ratio: molar mass molecular formula (given) = 283.88g/mol; molar mass empirical formula = 2(30.97) + 5(16.00) = 141.94g/mol; (ratio should always be a whole number) 2. Multiply empirical formula by 2: P (2 x 2) O (5 x 2) = P 4 O 10 = molecular formula mass%A = massA total mass x 100% mass%N = mass N total mass x100% mass%N = 42.03 149.12 x 100% = 28.19% 43.64g P 1mol P 30.97g P ! " # $ % & = 1.409mol P 56.36g O 1mol O 16.00g O ! " # $ % & = 3.523mol O P 1.409 1.409 O 3.523 1.409 = P 1 O 2.500 molar mass molecular formula molar mass empirical formula molarmass molecular formula molarmass empiricalformula = 283.88 141.94 = 2

2 CONVERTING BETWEEN MOL, GRAMS, AND ATOMS/MOLECULES Example 4 : How many grams of O are there in 71.0g P 2 O 5 ? Answer 4 : 40.0g O {gA ® gB : } FINDING “M” IN A FORMULA: A specific type of mole calculation; find M’s atomic mass. Example 5: 15.000g of M 2 O 3 is broken into its elements and produces 3.841g O 2 . What element is M? Answer 5: Ga Find the average atomic mass of M: and then look it up on the Periodic Table Step 1. Find grams M: g total = gM + gO ® gM = g total - gO = 15.000 – 3.841 = 11.159g M; Step 2. Find mol M from g O 2 : Step 3. ® Ga from the periodic table MASS SPECTROSCOPY: (skip if mass spectroscopy of compounds was not covered) Hard Mass Spectroscopy: Breaks the molecule into individual atoms; can determine empirical formula Soft Mass Spectroscopy: Ionizes the molecule but does not break the molecule apart; one peak is present and represents the molecule’s molar mass; used with Hard Mass Spectrum yields molecular formula Example 4: A chemical sample was analyzed by mass spectroscopy and both a “hard” and “soft” mass spectra were obtained. a. What is the empirical formula for the unknown compound? b. What is the molecular formula for the unknown compound? Answer 4: a. empirical formula = C 5 H 5 N {From the hard mass spectrum it can be determined that the compound contains H (mass = 1), C (mass = 12), and N (mass = 14). From the percent intensity the number of atoms of each element can be determined by dividing each element’s abundance by the smallest abundance: 9.1%. ; ;; this yields an empirical formula = C 5 H 5 N 1 } b. molecular formula = C 15 H 15 N 3 {From the soft mass spectrum the peak is the compound’s molar mass = 237g/mol; empirical formula’s molar mass = 5(C) + 5(H) + 1(N) = 5(12.01) + 5(1.008) + 1(14.02) = 79.11g/mol; determine ; ; multiply the empirical formula by 3: C (5 x 3) H (5 x 3) N (1 x 3) = C 15 H 15 N 3 } atoms or molecules B atoms or molecules A grams A grams B moles A moles B 1 mol B = 6.022 x 10 B 23 1 mol A = 6.022 x 10 A 23 molar mass A molar mass B chemical formula or chemical reaction 71.0g P 2 O 5 1mol P 2 O 5 142.0g P 2 O 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 5mol O 1mol P 2 O 5 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 16.0g O 1mol O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 40.0g O averageatomic mass M = grams M mol M 3.841g O 2 1mol O 2 32.00g O 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2mol O 1mol O 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2mol M 3mol O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.160mol M averageatomic mass M = grams M mol M = 11.159g M 0.160mol M = 69.744g / mol Intensity or Abundance m/z 1 12 14 45.5% 45.5% 9.1% Hard Spec Intensity or Abundance m/z 237 100% Soft Spec

4 5. What is the mass percent of sulfur in Al 2 (SO 4 ) 3 ? a. 9.40% b. 28.1% c. 56.1% d. 96.2% e. none of these 6. The reaction of 5.500g of Ni(s) metal with excess oxygen gas yields 7.000g of a nickel oxide compound. What is the empirical formula of the compound? a. NiO b. NiO 2 c. Ni 2 O 3 d. Ni 2 O 5 e. NiO 5 7. Which chemical has the smallest mass percent of O in it? (Hint: Can be done without calculations.) a. CO 3 2- b. Br 2 O c. CO 2 d. F 2 O e. PO 4 3- 8. 10.00g of MO is heated and converted into M and O 2 producing 1.044g of O 2 . What element is M? a. Ba b. S c. Fe d. As e. Na

5 9. 35.00g of an unknown chromium chloride compound, Cr x Cl y , is reacted with AgNO 3 and generates 95.03g AgCl(s). What is the formula of the chromium chloride compound? a. CrCl b. Cr 2 Cl c. CrCl 2 d. Cr 2 Cl 3 e. CrCl 3 10. From a mass spectrometer experiment, it was determined that the mass of one molecule of a pure isotopic version of oxygen gas was determined to be 5.98 x 10 -23 g. Oxygen exists as 3 stable isotopes: 16 O, 17 O, and 18 O; which isotopic version of oxygen gas is this molecule? a. 16 O 16 O b. 16 O 17 O c. 16 O 18 O d. 17 O 18 O e. 18 O 18 O Mass Spectroscopy-compounds (skip if mass spectroscopy of compounds was not covered) 11. A chemical sample was analyzed by mass spectroscopy and both “hard” and “soft” mass spectra were obtained. What is the molecular formula for the unknown compound? Hard Mass Spectrum Soft Mass Spectrum a. C 8 H 11 ON 2 b. C 16 H 18 O 4 N 2 c. C 8 H 9 O 2 d. C 8 H 9 O 2 N e. C 9 H 13 ON Intensity or Abundance m/z 1 12 14 45% 40% 5% 16 10%

7 7. b {fewer O atoms in the chemical formula will have less mass percent O; in addition, if the other non-oxygen atoms are larger atoms then this will also lower the mass percent of the O; by inspection, options “b” and “d” have fewer oxygen atoms (each have only one O atom) and therefore a lower mass percent O than options “a”, “c”, or “e”; since O is coupled with F in “d” and coupled with the much larger Br atom in option “b”, the mass percent of O in Br 2 O will be lower than the mass percent of O in F 2 O; the calculated mass percents ( not required ): ; assume 1 mol of each substance; “a”: ; “b”: ; “c”: ; “d”: ; “e”: ; } 8. a {determine molar mass of M and match to possible answers; ; g total = grams metal + grams oxygen; grams oxygen = 1.044g; grams metal = 10.00 – 1.044 = 8.956gM; find mol M from g O 2 : ; ® Ba; option “a”} 9. e {need moles of Cr and moles of Cl to answer this question; first find moles of Cl: ; find grams Cl: ; since there were 35.00g Cr x Cl y by subtracting off the Cl this will yield the grams of Cr present; grams Cr: 35.00g – 23.51g = 11.49g Cr; convert to moles Cr: ; write chemical formula: Cr 0.2210 Cl 0.6631 ; divide by the smallest number of moles: } 10. e { ; find mass of each isotope; “a”: 16 O 16 O = 16 + 16 = 32g/mol; “b”: 16 O 17 O = 16 + 17 = 33g/mol; “c”: 16 O 18 O = 16 + 18 = 34g/mol; “d”: 17 O 18 O = 17 + 18 = 35g/mol; “e”: 18 O 18 O = 18 + 18 = 36g/mol; therefore the molar mass calculated matches option “e” ® 18 O 18 O which has a mass number of 36} mass% O = mass O total mass ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x 100% 1mol CO 3 2 − 3mol O 1mol CO 3 2 − ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 16.0g O 1mol O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 48.0g O mass% O = 48.0g 60.01g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x 100% = 80.0% 1mol Br 2 O 1mol O 1mol Br 2 O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 16.0g O 1mol O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 16.0g O mass% O = 16.0g 175.8g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x 100% = 9.1% 1mol CO 2 2mol O 1mol CO 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 16.0g O 1mol O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 32.0g O mass% O = 32.0g 44.01g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x 100% = 72.7% 1mol F 2 O 1mol O 1mol F 2 O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 16.0g O 1mol O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 16.0g O mass% O = 16.0g 18.02g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x 100% = 29.6% 1mol PO 4 3 − 4mol O 1mol PO 4 3 − ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 16.0g O 1mol O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 64.0g O mass% O = 64.0g 94.97g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x 100% = 67.4% molarmass M = gM molM 1.044g O 2 1mol O 2 32.00g O 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2mol O 1mol O 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1mol M 1mol O ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.06525mol M molarmass M = 8.956g M 0.0652mol M = 137.26g / mol 95.03g AgCl 1mol AgCl 143.32g AgCl ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1mol Cl 1mol AgCl ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.6631mol Cl 0.6631mol Cl 35.45g Cl 1mol Cl ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 23.51g Cl 11.49g Cr 1mol Cr 52.00g Cr ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0.2210mol Cr Cr 0.2210 0.2210 Cl 0.6631 0.2210 = Cr 1 Cl 3.00 = CrCl 3 5.98 x 10 − 23 g molecule ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ 6.022 x 10 23 molecules 1mol ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = 36.01g / mol

8 11. d { Hard MS: It can be determined from the Hard MS that the compound contains H (mass = 1), C (mass = 12), N (mass = 14), and O (mass = 16). From the peak percent intensities the number of each element can be determined by dividing each element’s peak intensity by the smallest percent intensity; in this case: 5%. ; ; ; ; this yields an empirical formula = C 8 H 9 O 2 N 1 . Soft MS : The Soft MS’s peak describes the overall molar mass of the compound; the molar mass = 151g/mol; the empirical formula molar mass = 151g/mol; determine ; ; multiply the empirical formula by 1 to get the molecular formula: C (8 x 1) H (9 x 1) O (2 x 1) N (1 x 1) = C 8 H 9 O 2 N 1 } H = 45 % 5 % = 9 C = 40 % 5 % = 8 O = 10 % 5 % = 2 N = 5 % 5 % = 1 ratio = molarmass molecularformula molarmass empiricalformula ratio = 151g / mol 151g / mol = 1