CxHy + (x+"{\\frac {y} {4}}")O2 = xCO2 + "{\\frac {y} {2}}"H2O
"{\\frac {n(CO_2)} {n(C_xH_y)}}=x={\\frac {1.20} {0.30}}=4"
x = 4
"{\\frac {n(H_2O)} {n(C_xH_y)}}={\\frac {y} {2}}={\\frac {1.35} {0.30}}=4.5"
y = 9
Then the empirical formula of CxHy = C4H9, answer a
(Calculations are right, but i can't claim this answer as a correct one because hydrocarbon C4H9 doesn't exist)
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