Approximately 0.330g of an unknown hydrocarbon molecule (M.W.128.2) reacts with

Question

Approximately 0.330g of an unknown hydrocarbon molecule (M.W.128.2) reacts with oxygen to form 1.133g of CO2 and 0.185g of water. a) what is the empirical formula for this molecule? b) what is the molecular formula for this molecule? c) what is the percent by mass composition of carbon in this molecule? Approximately 0.330g of an unknown hydrocarbon molecule (M.W.128.2) reacts with oxygen to form 1.133g of CO2 and 0.185g of water. a) what is the empirical formula for this molecule? b) what is the molecular formula for this molecule? c) what is the percent by mass composition of carbon in this molecule? a) what is the empirical formula for this molecule? b) what is the molecular formula for this molecule? c) what is the percent by mass composition of carbon in this molecule?

Explanation / Answer

a)
let in compound number of moles of C and H be x and y respectively

Number of moles of CO2 = mass of CO2 / molar mass CO2
    = 1.133 / 44
    = 0.02575 mol
Since 1 mol of CO2 has 1 mol of C
Number of moles of C in CO2= 0.02575 mol
so, x = 0.02575 mol

Number of moles of H2O = mass of H2O / molar mass H2O
    = 0.185 / 18
    = 0.0103 mol
Since 1 mol of H2O has 2 mol of H
Number of moles of H =0 .0103 *2 = 0.0206 mol
so, y = 0.0206 mol
so:
C = 0.02575 mol
H = 0.0206 mol

divide by smallest number:
C = 0.02575 /0.0206 =1.25
H = 0.0206 /0.0206 =1

Get as whole number:
C = 1.25*4 = 5
H=1*4 =4
So, compound is C5H4

b)
Emipirical formula mass = 5*12 + 4*1 = 64 g
Molar mass = 128.2 g
Multiplying factor = 128.2/64 = 2
Molcular formula is C10H8

c)
Total mass = 128.2 g
Mass of C =12*10 =120 g
Mass % of C = 120*100 /128.2 = 93.6 %

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