Chapter 6 Chemistry Notes

By the time you finish studying ch 6 and working the

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Unformatted text preview: Composition Calculating Elemental Composition & Empirical Formula Calculating Molecular Formula Mass Percent Composition When 4.01 g of mercury is heated strongly in air, the resulting oxide weighs 4.33 g. Calculate the empirical formula of the compound? A. B. C. D. E. 80 8 Hg O 200.59 16.00 Hg2O 1mol Hg 4.01g Hg = 0.0200 mol Hg HgO 200.59 g Hg Hg2O3 1 mol O 0.32g O = 0.0200 mol O HgO2 16.00 g O The problem cannot be solved with the info. provided 1:1 ratio Example: Mass % Composition Galena is PbS. Think Mighty Mole to the rescue! Pb mining, just like Galena, IL; Galena, AK; Galena, IN; Galena, KS; Galena, MD; Galena, MO; Galena, OH; Galena, OR and Galena Hill, CA In 5.00g PbS, how many g Pb? Formula Mass (PbS) = 1(207.2)+1(32.07)=239.27 g/mol PbS 82 16 5.00 g PbS x 1 mol = 0.0209 mol PbS 239.27 g Pb S 0.0209 mol PbS x 1 mol Pb = 0.0209 Mol Pb 207.2 32.07 1 mol PbS Chemical 0.0209 mol Pb x 207.2 g = 4.33 g Pb Formula 1 mol Moles What is mass % composition of Pb in PbS? 4.33 g Pb x 100 = 86.6% Pb by mass 5.00 g PbS Grams 6 9/23/2013 Empirical Formulas from Experimental Data Hydrocarbons contain only C and H and can be combusted (burned) in O2 to yield CO2 and H2O. CxHy + excess O2 X CO2 + Y/2 H2O A sample of a hydrocarbon is fully oxidized to yield 61.6 g CO2 and 28.4 g H2O. What is the empirical formula of the hydrocarbon? Molar mass (CO2) = 1(12.01)+2(16.00)=44.01 g/mol Molar mass (H2O) = 2(1.01) + 1(16.00)=18.02 g/mol Example: Species ID by Composition Common Iron oxides include Hematite (Fe2O3) and Magnetite (Fe3O4). An oxide sample with a mass 2.64 g contains 1.91 g Fe. Is the sample hematite (Fe2O3) or magnetite (Fe3O4)? FM (Fe2O3) = 2(55.85)+3(16.00) = 159.70 g/mol Fe2O3 FM (Fe3O4) = 3(55.85)+4(16.00) = 231.55 g/mol Fe3O4 2 mol Fe 55 g Fe 1mol Fe2O3 1 mol Fe2O3 mol Fe 159.70 g Moles CO2 = 61.6 g = 1.40 moles CO2 Moles H2O = 28.4 g = 1.58 moles H2O 44.01 g/mol 18.02 g/mol 1.40 mole CO2 x 1 mol C = 1.40 mole C 1 mol CO2 1.58 mole H2O x 2 mol H = 3.15 mole H 1 mol H2O Empirical Formula: C1.4H3.15 C1H2.25 C1H9/4 C4H9 3 mol Fe 55 g Fe 1mol Fe3O 4 x100 = 72.36% 1 mol Fe3O 4 mol Fe 231.55 g Chemical Formula Sample = (1.91g/2.64 g)*100 = 72.36% Fe Moles Grams x100 = 69.94% ∴Sample is __________ based on mass fraction. ∴ http://www.youtube.com/watch?v=me5Zzm2TXh4 26 Fe 55.85 8 O 16.00 Chemical Formula Moles Grams "Chemical Skills" that are summarized at the end of the chapter pp. 88-? 7...
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This note was uploaded on 02/04/2014 for the course CHEM 004:007 taught by Professor Russellarsen during the Fall '13 term at University of Iowa.

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