Use the Refereaces to access important values if needed for this questien Furnac
ID: 1028940 • Letter: U
Question
Use the Refereaces to access important values if needed for this questien Furnace 0: 11,0 alnorber (02 absorber 13,64 grams of CO, and 5.584 grams of H,O were When 4346 grams of a hydrocarbon, C,Hy were burned in a combustion analysis apparatus, 13,64 grams of CO in a separnatc experiment, the molar mass of the compound was found to be 28.05 g/ the hrydrocarbon. Determine the empirical formala and the molocular Enter the elemests in the arder presented in the question empirical formala molocular formala Retry Entire Group 9 more group attempts remaining Cengage LearningCengage Technical Support 7 0Explanation / Answer
mass of hydrocarbon= 4.346 grams
mass of CO2= 13.64 grams
mass of H2O= 5.584 grams
molar mass of CO2=44 grams
molar mass of H2O= 18.0gram
% by mass of Carbon = 12/44 xmass of CO2/mass of htdrocarbon x100
%by mass of Carbon = 12/44 x13.64/4.346 x100
= 85.59%
% by mass of Carbon= 85.59%
% by mass of Hydrogen = 2/18 xmass of H2O/mass of hydrocarbon x100
= 2/18 x 5.584/4.346 x100
= 14.28%
% by mass of Hydrogen = 14.28%
Element % by mass atomic mass relative number simple ratio
C 85.59 12.0 85.59/12=7.1325 7.1325/7.1325 = 1.0
H 14.28 1.0 14.28/1.0=14.28 14.28/7.1325= 2.00
emperical formula of a compound = C1H2
emperical formula mass = 14.0
molar mass = 28.05 gram/mole
n= molar mass/emoerical formula mass
n= 28.05/14.0 = 2.00
Molecular formula = n xemperical formula
Molecular formula = 2 xCH2 = C2H4
Molecular formula of a hydrocarbon = C2H4.
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