Portage Chemistry 103 Final Exam Study Guide., Exams of Chemistry

Download Portage Chemistry 103 Final Exam Study Guide. and more Exams Chemistry in PDF only on Docsity! PORTAGE LEARNING CHEMISTRY 103 FINAL EXAM STUDY GUIDE MODULES 1-6 MODULE 1 – EXAM Question 1 – Section 1.1 10 / 10 pts Complete the two problems below: 1. Convert 1005.3 to exponential form and explain your answer. 2. Convert 4.87 x 10-6 to ordinary form and explain your answer. Your Answer: 1.) Convert 1005.3 to exponential form and explain your answer: All number values must be between 1-10. Because this value is greater than 1, the exponent remains positive. In order for this to happen, we have to move the decimal three places to the LEFT thus, the exponential form of 1005.3 is: 1.0053 X 10 3 2.) Convert 4.87 x 10-6 to ordinary form and explain your answer. All number values must be between 1-10. Because this value has a negative exponent, the final answer will be less than 1. In order for this to happen, we have to move the decimal six places to the LEFT thus: 0.00000487 Question 2 10 / 10 pts Complete the two problems below: Using the following information, do the conversions shown below, showing all work: 1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts 1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints kilo (= 1000) milli (= 1/1000) centi (= 1/100) deci (= 1/10) 1. 3.6 pounds = ? oz 2. 2680 ml = ? liters Your Answer: 1.) 3.6 pounds = 57.6 oz 1lb = 16oz 3.6(16) / 1 =57.6 oz 2.) 2680 ml = 2.68 liters New/Old Liters/Milliliters 1L = 1,000mL 2,680/1,000 =2.68 Question 3 10 / 10 pts Do the conversions shown below, showing all work: 1. 78oC = ? oK 2. 248oF = ? oC 3. 427oK = ? oF Your Answer: Kelvin is the larger Fahrenheit is the middle Celcius is the smaller 1.) 78oC = 351 oK Add: 273+78 = 351K 2.) 248oF = 120 oC Subtract 32 then divide by 1.8 248-32= 216 Now divide 216/1.8 =120 3.) 427oK = 309.2 oF Kelvin to Fahrenheit needs to convert K to C then continue equation. 427-273 = 154oC Celsius to Fahrenheit needs to multiply first then add 32 154 x 1.8 = 277.2 add 32 = 309.2 Question 4 10 / 10 pts Non-Binary Ionic Compound because there is a metal and poly-atomic group. The name of this compound is: aluminum carbonate 3.) H2CrO4 Hydrogen and poly-atomic group = non-binary acid meaning the polyatomic suffix ending in "ATE" turns to "IC". The name of this compound is: chromic acid 1. PF5 - binary molecular = phosphorus pentafluoride 2. Al2(CO3)3 - nonbinary ionic = aluminum carbonate 3. H2CrO4 - nonbinary acid = chromic acid Question 10 9 / 10 pts Write the formula for each of the following chemical compounds explaining the answer with appropriate charges and/or prefixes and/or suffixes. 1. Chromium (III) sulfide 2. Hydroselenic acid 3. Barium iodide Your Answer: 1.) Chromium (III) sulfide Chromium - metal has +3 charge: Cr+3 Sulfur - nonmetal has a -2 charge: S-2 Binary Ionic Compound Cr+3 S-2 Have to balance out charges: CrS3 2.) Hydroselenic acid Hydrogen turns this into an acid Hydrogen has +1 charge: H1 Selenium has a -2 charge: Se-2 Binary Acid H1Se-2 Have to balance out charges: H2Se 3.) Barium iodide Barium has +2 charge Iodine has a -1 charge Binary ionic compound Ba+2 I-1 Have to balance out the charges: BaI2 1. Chromium (III) sulfide - ide - binary of Cr+3 and S-2 = Cr2S3 2. Hydroselenic acid - hydro = binary acid of H + Se-2 = H2Se 3. Barium iodide - ide = binary of Ba+2 and I-1 = BaI2 MODLUE 2 – EXAM Question 1 10 / 10 pts Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal. 1. (NH4)2CO3 2. C8H6NO4Br Your Answer: (NH4)2CO3 N 14.01(2) =28.02 H 1.008(8) =8.064 C 12.01 O 16.00(3) = 48.00 MW=96.09 C8H6NO4Br C 12.01(8) = 96.08 H 1.008(6) = 6.048 N 14.01 O 16.00 (4) = 64.00 Br 79.90 MW=260.04 Question 2 8 / 10 pts Show the calculation of the number of moles in the given amount of the following substances. Report your answer to 3 significant figures. 1. 12.0 grams of (NH4)2CrO4 2. 15.0 grams of C8H8NOI 1. Moles = grams / molecular weight = 12.0 / 152.08 = 0.0789 mole 2. Moles = grams / molecular weight = 15.0 / 261.05 = 0.0575 mole Error in the calculation of molecular weight of compound #1. There are only 2 nitrogen atoms in this compound. Question 3 10 / 10 pts Show the calculation of the number of grams in the given amount of the following substances. Report your answer to 1 place after the decimal. 1. 1.20 moles of Al2(SO4)3 2. 1.04 moles of C7H5NOBr Your Answer: 1) 1.20 moles of Al2(SO4)3 Al 26.98(2) = 53.96 S 32.07 (3) = 96.21 O 16.00 (12) = 192.00 = 342.17 x 1.20 = 342.2 x 1.20 = 410.6 GRAMS 2) 1.04 moles of C7H5NOBr C 12.01(7) = 84.70 H 1.008(5) = 5.04 N 14.01 O 16.00 Br 79.90 =199.65 = 199.02 x 1.04 = 207.0 GRAMS Question 4 8 / 10 pts Show the calculation of the percent of each element present in the following compounds. Report your answer to 2 places after the decimal. 1. (NH4)2CrO4 2. C8H8NOI 1. %N = 2 x 14.01/152.08 x 100 = 18.42% %H = 8 x 1.008/152.08 x 100 = 5.30% %Cr = 1 x 52.00/152.08 x 100 = 34.19% %O = 4 x 16.00/152.08 x 100 = 42.08% 2. %C = 8 x 12.01/261.05 x 100 = 36.81% Redox Table KMnO4 K +1 Mn +7 O -2, O4=-8 Sum=0 MnO2 Mn +4 O -2, O2=-4 Sum=0 H2S H +1, H2=+2 S -2 Sum=0 H2O H +1, H2=+2 O -2 Sum=0 S Sum=0 KOH K +1 O -2 H +1 Sum=0 +1 +7 -2. +1 -2 0 +4 -2. +1 -2. +1-2+1 KMnO4 + H2S → S + MnO2 + H2O + KOH +1. +7 -8. +2 -2. 0. +4. -4. +2 -2. +1-2+1 Mn has change of 3 (oxidation) ; coefficient is 2 3x2=6 S has change of 2 (reduction) ; coefficient is 3 2x3=6 LCM is 6 2KMnO4 + 3H2S → 3S + 2MnO2 + H2O + KOH L: K2, Mn2, O8, H6, S3 R: K1, Mn2, O6, H3, S3 Equation is unbalnaced! _______________________________________________ 2KMnO4 + 3H2S → 3S + 2MnO2 + 2H2O + 2KOH L: K2, Mn2, O8, H6, S3 R: K2, Mn2, O8, H6, S3 Equation is balanced! KMnO4 + H2S → S + MnO2 + H2O + KOH KMnO4: K is metal in group I = +1, each O is -2 (total is -8), so Mn is +7 MnO2: Each O is -2 (total is -4), so Mn is +4 H2S: H is +1 (total is +2), so S is -2 S: is uncombined so S = 0 Question 9 10 / 10 pts Show the balancing of the following redox equation, including the determination of the oxidation number (charge) of ONLY the atoms which are changing. Ca(OH)2 + NaOH + ClO2 + C → NaClO2 + CaCO3 + H2O Your Answer: Ca(OH)2 + NaOH + ClO2 + C → NaClO2 + CaCO3 + H2O Redox Table Ca(OH)2 Ca +2 O -2, O(2) = -4 H +1, H2= +2 Sum=0 NaClO2 Na +1 Cl +3 O -2, O2=-4 Sum=0 NaOH Na +1 O -2 H +1 Sum=0 CaCO3 Ca +2 C +4 O -2, O3= -6 Sum=0 ClO2 Cl +4 O -2, O2= -4 Sum=0 H2O H +1, H2=+2 O -2 Sum=0 C Sum = 0 +2 -2 +1 +1 -2 +1 +4 -2 0 +1 +3 -2 +2 +4 -2 +1 -2 Ca(OH)2 + NaOH + ClO2 + C → NaClO2 + CaCO3 + H2O +2. -4 +2. +1. -2 +1. +4 -4. 0 +1 +3 -4 +2 +4 -6 +2 -2 C has change of 4 (reduction) ; coefficient is 1 4x1=4 Cl has change of 1 (oxidation) ; coefficient is 4 1x4=4 LCM is 4 Ca(OH)2 + NaOH + 4ClO2 + 1C → 4NaClO2 + 1CaCO3 + H2O L: Ca1, O11, H3, Na1, Cl4, C1 R: Ca1, O12, H2, Na4, Cl4, C1 Equation is unbalnaced! _______________________________________________ Ca(OH)2 + 4NaOH + 4ClO2 + 1C → 4NaClO2 + 1CaCO3 + 3H2O L: Ca1, O14, H6, Na4, Cl4, C1 R: Ca1, O14, H6, Na4, Cl4, C1 Equation is balanced! Ca(OH)2 + NaOH + ClO2 + C → NaClO2 + CaCO3 + H2O ClO2: Each O is -2 (total is -4), so Cl is +4 NaClO2: Na is metal in group I = +1, each O is -2 (total is -4), so Cl is +3 C: is uncombined so C = 0 CaCO3: Ca is metal in group II = +2, each O is -2 (total is -6), so C is +4 Since Cl (on left side) is +4 and Cl (on right side) is +3: Cl changes by 1 Since C (on left side) is 0 and C (on right side) is +4: C changes by 4 Multiply Cl compounds by 4 and C compounds by 1 and after balancing other atoms = 1 Ca(OH)2 + 4 NaOH + 4 ClO2 + 1 C → 4 NaClO2 + 1 CaCO3 + 3 H2O Question 10 7 / 10 pts Show the balanced equation and the calculation of the number of moles and grams of CO2 formed from 11.9 grams of O2. Show your answers to 3 significant figures. C6H14 + O2 → CO2 + H2O Your Answer: C6H14 + O2 → CO2 + H2O L: C6, H14, O2 R: C1, H2, O2 Equation is unbalanced! 2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l) by using the following thermochemical data: ΔHf0 C2H6 (g) = -84.0 kJ/mole, ΔHf0 CO (g) = -110.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/mole Your Answer: ΔHrxn = Σ n ΔHf0 (products) - Σ m ΔHf0 (reactants) -110.5 + (-285.8) - (-84.0) ΔHrxn= -312.3kJ 2 C2H6 (g) + 5 O2 (g) → 4 CO (g) + 6 H2O (l) ΔHf0 C2H6 (g) = -84.0 kJ/mole, ΔHf0 CO (g) = -110.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/mole ΔHrxn = 2(+84.0) + 5(0) + 4(-110.5) + 6(-285.8) = - 1988.8 kJ/mole Question 6 0 / 10 pts Show the calculation of the new temperature of a gas sample has an original volume of 740 ml when collected at 710 mm and 35oC when the volume becomes 460 ml at 1.20 atm. (Pi x Vi ) / Ti = (Pf x Vf ) / Tf 740 ml/1000 = 0.740 liters = Vi 710 mm/760 = 0.934 atm = Pi 460 ml/1000 = 0.460 liters = Vf 1.20 atm = Pf 35oC + 273 = 308oK = Ti (0.934) x (0.740) / 308 = (1.20) x (0.460) / Tf Tf = 246 oK Question 7 0 / 10 pts Show the calculation of the molecular weight of a gas sample with a mass of 0.456 grams which has a volume of 230 ml when collected at 29oC and 740 mm. P x V = (g/MW) x R x T 740 mm/760 = 0.974 atm = P R = 0.0821 V = 230 ml/1000 = 0.230 liters 29oC + 273 = 302oK = T 0.456 = g (0.974) x (0.230) = (0.456/MW) x (0.0821) x (302) MW = 50.5 Question 8 7 / 10 pts Show the calculation of the volume of H2O gas formed by the combustion of 18.6 grams of C6H6 at 30oC and 1.10 atm? The combustion of benzene (C6H6) takes place by the following reaction equation. 2 C6H6 (g) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) Your Answer: MW : 78 32 44 18 2 C6H6 (g) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) 18.6g/78(C6H6 (g) = 0.2384 moles of C6H6 0.2384 (2 moles of C6H6/6 moles of H2O) =0.07867 PV=nRT V=nRT/P 0.07867(0.0821)(30+273) / 1.10atm 1.957/1.10 V= 1.78 Liters H2O (MW = 78) (MW = 32) (MW = 44) (MW = 18) 2 C6H6 (g) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) 18.6 grams 16.18 liters ↓ by V = nRT/P = (0.7155)(0.0821)(303)/1.10 ↑ 0.2385 mol → 6/2 x 0.2385 mol The ratio of these compounds is 6/2. Question 9 10 / 10 pts Show the calculation of the mole fraction and partial pressure of each gas in a mixture which consists of 3.00 moles of He, 1.50 moles of H2 and 2.50 moles of CO2 if the total pressure of the mixture is 740 mm. Express your answer in mm. XHe = 3.00 / (3.00 + 1.50 + 2.50) = 0.4286 XH2 = 1.50 / (3.00 + 1.50 + 2.50) = 0.2143 XCO2 = 2.50 / (3.00 + 1.50 + 2.50) = 0.3571 PHe = XHe (740 mm) = 0.4286 (740 mm) = 317 mm PH2 = XH2 (740 mm) = 0.2143 (740 mm) = 159 mm PCO2 = XCO2 (740 mm) = 0.3571 (740 mm) = 264 mm Question 10 10 / 10 pts Show the calculation of the molecular weight of an unknown gas if the rate of effusion of Neon gas (Ne) is 1.86 times faster than that of an unknown gas. Your Answer: (1.86)2 x 20.18 3.4596 x 20.18 = 69.81 MWunknown= 69.81 MODULE 4 – EXAM Question 1 10 / 10 pts Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the Fe26 atom. Your Answer: Fe26= 1s2 2s2 2p6 3s2 3p6 4s2 3d6 Question 2 10 / 10 pts Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the Si14 atom. Your Answer: Si14= 1s2 2s2 2p6 3s2 3p2 Question 3 10 / 10 pts Write the subshell electron configuration (i.e.1s2 2s2, etc.) for the Cl17 atom and identify which are valence (outer shell) electrons and determine how many valence electrons there are. Your Answer: Cl17= 1s2 2s2 2p6 3s2 3p5 Valence electrons are 3s2 3p5 and there are seven (7) total valence electrons Question 4 5 / 10 pts * For the following question, use the "Insert Math Equation" tool (indicated by the icon on the toolbar and then choose arrows from the window which opens). Using up and down arrows, write the orbital diagram for the Co27 atom. Co27 = 1s2 2s2 2p6 3s2 3p6 4s2 3d7 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ ↑ The orbital diagram should have spaces that clearly illustrate the levels and the orbitals. Question 5 6 / 10 pts * For the following question, use the "Insert Math Equation" tool (indicated by the icon on the toolbar and then choose arrows from the window which opens). Using up and down arrows, write the orbital diagram for the Co27 atom and identify which are unpaired electrons and determine how many unpaired electrons there are. Co27 = 1s2 2s2 2p6 3s2 3p6 4s2 3d7 ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ ↑ = 3 unpaired electrons The orbital diagram should have spaces so that levels and orbitals are clearly illustrated. Question 6 10 / 10 pts Your Answer: H-O 3.5-2.1 = 1.4 (0.5-1.6) = polar H-O bond is POLAR O-P 3.5-2.1 = 1.4 (0.5-1.6) = polar O-P bond is POLAR H-O bond electronegativity difference = 3.5 - 2.1 = 1.4 1.6 - 0.5 bond is Polar P-O bond electronegativity difference = 3.5 - 2.1 = 1.4 1.6 - 0.5 bond is Polar Question 3 10 / 10 pts On a piece of scratch paper, draw the Lewis structure for the ClO4-1 ion. Then choose the correct Lewis structure for ClO4-1 from the options listed below. Your Answer: B Question 4 10 / 10 pts On a piece of scratch paper, draw the Lewis structure for H2SO3. Then choose the correct Lewis structure for H2SO3 from the options listed below. Your Answer: D Question 5 0 / 10 pts Determine the electron geometry and explain your answer for the S atom in H2S. The S atom in H2S has 4 groups of electrons around it in its Lewis structure, therefore, its electron geometry would be tetrahedral. Question 6 0 / 10 pts Determine the hybridization and explain your answer for one of the C atoms in H2CCH2. Your Answer: The C atoms in H2CCH2 has 3 groups of electrons around it in its Lewis structure, therefore, its hybridization would be sp2. Question 7 5 / 10 pts Determine the shape and explain your answer for HCN. Your Answer: The C atom in HCN has 2 groups of electrons around it in its Lewis structure, therefore, its electron geometry would be linear and since there are 2 atoms around the central C atom, the shape would be linear. The shape of a molecule is determined by the number of atoms around the central atom, and by the number of electron groups around the central atom. Both need to be considered when determining the shape of the molecule, and both should be noted in your answer. Question 8 5 / 10 pts H = 2.1 Li = 1.0 Be = 1.5 B = 2.0 C = 2.5 N = 3.0 O= 3.5 F = 4.0 Na = 1.0 Mg = 1.2 Al = 1.5 Si = 1.8 P = 2.1 S = 2.5 Cl = 3.0 K = 0.8 Ca = 1.0 Ga = 1.6 Ge = 1.8 As = 2.0 Se = 2.4 Br = 2.8 Use the electronegativities above and your knowledge of the shape of HCCH to determine the molecular polarity of HCCH explaining your answer in detail. Your Answer: C = 2.5 H = 2.1 (between 1.6-0.5 = polar) Thus, H-C 2.5-2.1 = 0.4 NONPOLAR C-H 2.5-2.1 = 0.4 NONPOLAR The shape of HCCH is linear and since all of its C-H and C-C bonds are nonpolar, it would be nonpolar since it has all nonpolar bonds. Electronegativity differences determine the bond polarity, not the molecularity polarity. Question 9 10 / 10 pts Is O2 Polar, Ionic or Nonpolar and List and Explain whether it is Soluble or Insoluble in Water? Your Answer: O2 is nonpolar and is INSOLUBLE O=3.5 and there are two oxygen atoms so 3.5-3.5 = 0 (nonpolar) and Oxygen is insoluble because it is a gas. O2 has one nonpolar bond which makes it Nonpolar and since it is Nonpolar it is Insoluble in water. Question 10 7 / 10 pts Arrange the following compounds in a vertical list from highest boiling point (top) to lowest boiling point (bottom) and explain your answer on the basis of whether the substance is Polar, Nonpolar, Ionic, Metallic or Hydrogen bonding: LiF, Co, Cl2, HF, HI Please note in this question you are not being asked to list BPs but the compounds in a list from highest to lowest BP on the basis of the type of compound. LiF (ionic) = Co (metallic) HF (Hydrogen Bonding) HI (Polar) Cl2 (Nonpolar) Metallic and ionic are equally high. Non-polar has the lowest boiling point. MODULE 6 – EXAM Question 1 10 / 10 pts Air is a homogeneous mixture solution made up of 20% oxygen gas and 80% nitrogen gas. Which gas is the solvent and which is the solute. Explain your answer. Your Answer: The lower quantity dissolves into the higher quantity. 80% Nitrogen gas has more abundance compared to the 20% Oxygen gas therefore, the solvent is nitrogen while the solute is oxygen. The (solute) oxygen is dissolving into the (solvent) nitrogen. Nitrogen is the solvent since it is the major component (80%) and oxygen is the solute since it is the minor component (20%). Question 2 0 / 10 pts Water (H2O) with a molecular weight of 18 has a boiling point of 100oC and methane (CH4) with a somewhat similar molecular weight of 16 has a boiling point of -161oC. This difference is much higher than can be explained by the polarity of water. Explain what causes this difference. Hydrogen bonds which are strong intermolecular attractive forces form between water molecules requiring much more energy to separate the molecules than for other molecules. Question 3 10 / 10 pts Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare. List from lowest freezing point to highest freezing point. BeBr2 ¸AlBr3, Mg3(PO4)2, KBr Your Answer: LOWEST FREEZING POINT Mg3(PO4)2 = Mg3 + (PO4)2 = 5 0.100 x 1.86 x 5 = 0.93 0oC - 0.93 = -0.93oC AlBr3 = Al + Br3 = 4 0.100 x 1.86 x 4 = 0.74 Moles = 0.250 x 0.200 = 0.0500 Moles = (gsolute / MW) 0.0500 = (gsolute / 148.325) gsolute = 0.0500 x 148.325 = 7.42 g Question 8 10 / 10 pts Show the calculation of the volume of 0.667 M solution which can be prepared using 37.5 grams of Ba(NO3)2. Your Answer: ml solution = gsolute / mwsolute / molarity x 1000 MW = 261.55 Ba = 137.53 N x2 = 14.01 x 2 = 28.02 O x6 = 16.00 x 6 = 96.00 37.5/261.55/0.667 = 0.2149565 x 1000 = 214.96mL or 215mL molessolute = gsolute / MW molessolute = 37.5 g / 261.55 = 0.1434 mol Molarity = moles / (mL /1000) 0.667 = 0.1434 / (mL / 1000) mL / 1000 = 0.1434 / 0.667 = 0.2150 mL = 0.2150 x 1000 = 215 mL Question 9 10 / 10 pts Show the calculation of the boiling point of a solution made by dissolving 28.9 grams of the nonelectrolyte C8H16O8 in 250 grams of water. Kb for water is 0.51, BP of pure water is 100oC. Calculate your answer to 0.01oC. ∆tb = Kb x m Your Answer: Molality (m)= (grams of solute ÷ molecular weight of solute) / (grams solvent ÷1000) ∆tb = Kb x m BPsolution = BPsolvent + ∆tb MW C8H16O8 = 240.208 C8 = 96.08 H16 = 16.128 O8 = 128.00 (28.9 / 240.208) / (250 /1000) 0.1203123 / 0.250 = 0.4812492m ∆tb = Kb x m 0.51 x 0.4812492 = 0.245437 or 0.25∆tb 100oC + 0.25oC = 100.25oC molality = (gsolute / MW) / (gsolvent / 1000) molality = (28.9 / 240.208) / (250 / 1000) = 0.481 m ∆tb = Kb x m = 0.51 x 0.481 = 0.245oC BPsolution = BPsolvent - ∆tb = 100oC + 0.245 = 100.25oC Question 10 10 / 10 pts Show the calculation of the molar mass (molecular weight) of a solute if a solution of 13.9 grams of the solute in 200 grams of water has a freezing point of -1.40oC. Kffor water is 1.86 and the freezing point of pure water is 0oC. Calculate your answer to 0.1 g/mole. tf = Kf m Your Answer: MWsolute = (Kf × gsolute ×1000) / (gsolvent × ∆tf) MWsolute = (1.86 x 13.9 x 1000) / (200g x 1.40) 25,854 / 280 = 92.3g/mole ∆tf = Kf x m molality = ∆tf / Kf = 1.40 / 1.86 = 0.753 m molality = (gsolute / MW) / (gsolvent / 1000) 0.753 = (moles) / (200 / 1000) Moles = 0.753 x 0.200 = 0.1506 0.1506 = (13.9 / MW) MW = 13.9 / 0.1506 = 92.3