Empirical and Molecular Formulas - Chemistry I | CHEM 1100, Study notes of Chemistry

Download Empirical and Molecular Formulas - Chemistry I | CHEM 1100 and more Study notes Chemistry in PDF only on Docsity! Exam Room Assignments (These are also posted on our LMS site) 1110-01 DCC 318 1110-04 DCC 330 1100-09 Sage 3303 1100-11 DCC 308 Extra Time Students Walker 3221 Reminder: “High School Knowledge Exam” is on Wednesday (September 9th) at 8am. Make sure that you bring a #2 pencil and a scientific calculator (NO Graphing Calculators Allowed) New SI (Supplemental Instruction) Session Information: Tuesday September 8th: Review for HSK Exam, 8-10pm in DCC 337 Starting on September 15th: SI sessions will be Tuesdays and Thursdays, 6:30 – 8:30pm in DCC 239 (Note that the information I posted in my first lecture has changed) Two main ways to determine empirical formula: 1. Elemental Analysis – gives us information about the mass percent of elements in a compound. Example: Analysis of Zircon, a diamond-like mineral, shows that it contains 34.91% O, 15.32% Si, 49.76% Zr. The molar mass of zircon is known to be less than 200 g/mol. Determine the molecular formula of Zircon. (Example of Elemental Analysis) Gwdir.demon.co.uk/jo/minerals/zircon.htm Yss 0. C466- 2.18 0 os. = [mk er - Emi QO o.syse Hal O.SYSF Ors4s ~ OY multy! th whol, avSi Oy moa maco © 183.3 hal Conprwert) EQ 0 fret Ard Om oo Sornphe bg /, 0004 samphe ~ (o.w2lgc 4 011045 H) = im =O OlFm O ‘ 4. S4ep a O.os!lt mk C fo0lr OC :l0O3mel HZ O.0!h = boom H O.0I1Ft mk Ofo.o1% =: [mt O CHO . a enpurwcad feermalts cB. Fa gmk tor moss 4, ompond vo bie [00 - [20 glow 0) Grmutan 0.27540 Hydroxylamine nitrate contains 29.17 mass% N, 4.20 mass % H, and 66.63 mass % O. Determine its empirical formula. (A) HNO (B) H2NO2 (C) HN6O16 (D) HN16O7 (E) H2NO3 Terephthalic acid, used in the production of polyester fibers and films, is composed of carbon, hydrogen, and oxygen. When 0.6943 g of terephthalic acid was subjected to combustion analysis it produced 1.471 g CO2 and 0.226 g H2O. What is its empirical formula?  A. C2H3O4 B. C3H4O2 C. C5H12O4 D. C4H3O2 E. C2H2O Example: When 24.3 g of magnesium is burned in 32.0 g of oxygen, which reactant, if any, will be in excess? Aluminum reacts with oxygen to produce aluminum oxide which can be used as an adsorbent, desiccant or catalyst for organic reactions. 4Al(s) + 3O2 (g)  2Al2O3 (s) A mixture of 82.49 g of aluminum (Molar Mass: 26.98 g/mol) and 117.65 g of oxygen (Molar Mass: 32.00 g/mol) is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete. A. Oxygen is the limiting reactant; 19.81 g of aluminum remain. B. Oxygen is the limiting reactant; 35.16g of aluminum remain. C. Aluminum is the limiting reactant; 16.70 g of oxygen remain. D. Aluminum is the limiting reactant; 35.16 g of oxygen remain. E. Aluminum is the limiting reactant; 44.24 g of oxygen remain. Check out this online interactive “movie” about limiting reagents … www.mhhe.com/physsci/chemistry/essentialchemistry/flash/limitr15.swf Example: Barium sulfate, BaSO4, is made by the following reaction: Ba(NO3)2 (aq) + Na2SO4 (aq)  BaSO4 (s) + 2NaNO3 (aq) An experiment was begun with 75.00 g of Ba(NO3)2 and an excess of Na2SO4. After collecting and drying the product, 64.45 g of BaSO4 was obtained. Calculate the theoretical yield and percent yield of BaSO4. Methanol (CH4O) is converted to bromomethane (CH3Br) as follows: CH4O + HBr → CH3Br + H2O If 12.23 g of bromomethane are produced when 5.00 g of methanol is reacted with excess HBr, what is the percentage yield? A. 82.6% B. 100% C. 40.9% D. 245% E. 121% iw we Solutions (se ld) out ° lohat in a free | ars hemogentano wor . mayerurge composed ye Or MOE Sub stances (Herts) ‘So LAC » Substance preant 1 Lote hase wed um. amon & oo he cubsteunce) Solvent: vent: Sho stance Seen an greater amnt. © Gaguams solu , solvent is A 2) www.fotosearch.com, www. michigan.gov Parts per Million (ppm) Equivalent to: 1 mg/L or 1 μg/mL Parts per billion (ppb) Equivalent to: 1 μg/L or 1 ng/mL golvent + volun fysolate Example: How many grams of NaCl are in 74.0 mL of 0.520 M NaCl?