Analytical Chem istry - DePauw University

540 Analytical Chemistry 2.0( 0. 0400 MKMnO ) × ( 0.03562 LKMnO )which means that the sample contains4 435 molNaCO142 . × 10− molKMnO ×42mol KMnO= 142 . × 10−3 molKMnO 42 2 4Thus, the %w/w Na 2 C 2 O 4 in the sample of ore is3134.00 gNaCO355 . × 10− molNaCO ×2 2 4molNaCO4= 355 . × 10 −32 2 42 2 4= 0.476molNaCO2 2 4gNaCO2 2 40.476 gNaCO2 2 4× 100 = 93.%0 w/ wNaCO0.5116 gsampleClick here to return to the chapter.Practice Exercise 9.212 2 4For a back titration we need to determine the stoichiometry betweenCr 2 O 72–and the analyte, C 2 H 6 O, and between Cr 2 O 7 2– and the titrant,Fe 2+ . In oxidizing ethanol to acetic acid, the oxidation state of carbonchanges from –2 in C 2 H 6 O to 0 in C 2 H 4 O 2 . Each carbon releases twoelectrons, or a total of four electrons per C 2 H 6 O. In reducing Cr 2 O 7 2– ,in which each chromium has an oxidation state of +6, to Cr 3+ , each chromiumloses three electrons, for a total of six electrons per Cr 2 O 7 2– . Oxidationof Fe 2+ to Fe 3+ requires one electron. A conservation of electronsrequires that each mole of K 2 Cr 2 O 7 (6 moles of e - ) reacts with six molesof Fe 2+ (6 moles of e - ), and that four moles of K 2 Cr 2 O 7 (24 moles of e - )react with six moles of C 2 H 6 O (24 moles of e - ).The total moles of K 2 Cr 2 O 7 reacting with C 2 H 6 O and with Fe 2+ is−( 0. 0200 MK Cr O ) × ( 0. 05000 LI ) = 100 . × 10−3 molK Cr O2 2 7 3The back titration with Fe 2+ consumes2 2 72+2+0.1014 molFe0.02148 LFe × ×2+LFe1 mol KCrO2 2 7−= 363 . × 10 4 molK Cr O2+2 2 76 molFeSubtracting the moles of K 2 Cr 2 O 7 reacting with Fe 2+ from the total molesof K 2 Cr 2 O 7 gives the moles reacting with the analyte.−100 . × 10 KCrO − 3.63×10molK Cr O3 −42 2 7 2 2 7= 63 . 7× 10 −4molK Cr O2 2 7The grams of ethanol in the 10.00-mL sample of diluted brandy is