ing Grade 3/10/2018 1 1 :59 PM O 84.2/100 () 3/10/2018 09:13 PM Print CaculatrPe
ID: 1025944 • Letter: I
Question
ing Grade 3/10/2018 1 1 :59 PM O 84.2/100 () 3/10/2018 09:13 PM Print CaculatrPeriocic Tuble Question 39 of 40 Sapling Learning Complete combustion of 4.40 g of a hydrocarbon produced 14.3 g of CO2 and 4.40 g of HO. What is the empirical formula for the hydrocarbon? Insert subscripts as necessary There is a hint available Viow the hint bottom divider on the dvider bar again to hide the hint Close O Previous Give Up & View Solution 9 Check Answer 0 Next Exit about us careers prvacy policy terms of use contact usExplanation / Answer
Molar mass ( g/mole): CO2= 44 and H2O=18
moles= mass/molar mass
moles : CO2= 14.3/44=0.325 and H2O= 4.4/18=0.244
the reactions involving formation of CO2 and H2O are C+ O2------->CO2 and H2+0.5O2----->H2O
1 mole of CO2 requires 1 mole of C and 1 mole of H2O requires 1 mole of H2 as per the reactions given
hence moles of C required =0.325 and that of H2= 0.244, moles of H= 0.244*2=0.488
mass of Carbon = 0.325*12= 3.9 and mass of H= 0.488*1= 0.488
total mass of hydrocarbon = 3.9+0.488=4.4 same as that is given
molar rati of C : H= 0.325:0.488 dividing by 0.325, the smallest number
the ratio becomes 0.325/0.325:0.488/0.325= 1:1.5= 1:3/2= 2:3
so the molecular formula is C2H3
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.