Here is my work:
$$eginarraycccccc & ceC_aH_b & ce-> & ceCO2 & + & ceH2O \ extmasses (g) & 1.05 & & 3.30 & & 1.35endarray$$
eginalign*ceCO2 & ightarrow ceC \44~mathrmg & ightarrow 12~mathrmg \3.30~mathrmg & ightarrow xendalign*
$$x = 0.9~mathrmg,~ extmoles that C = frac0.912 = 0.075$$
eginalign*ceH2O & ightarrow ce2H \18~mathrmg & ightarrow 2~mathrmg \1.35~mathrmg & ightarrow yendalign*
$$y = 0.15~mathrmg,~ extmoles that H = frac0.151 = 0.15$$
$$ extempirical formula~ceC_0.075/0.075H_0.15/0.075 -> CH2$$
$$frac7014 = 5$$
$$ extmolecular formula is~ceC5H10$$
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edited Feb 26 "18 in ~ 2:50
pentavalentcarbon
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asked Sep 18 "14 in ~ 21:11
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$egingroup$
$$ceC_$a$H_$b$ + $left(a+frac b4 ight)$O2 -> $a$CO2 + $fracb2$ H2O$$
Suppose you had $n$ mole of hydrocarbon, then we have $acdot n$ mole of $ceCO2$ and $fracb2cdot n$ mole of $ceH2O$ splitting their mole we"ll gain $2fracab$:$$2frac ab=frac330/44135/18=frac7.57.5=1implies frac ab=frac12$$So the empirical formula is $ceCH2$Now because that actual formula $frac7014=5$Yes the formula is $ceC5H10$.
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edited Sep 19 "14 at 4:39
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reply Sep 19 "14 in ~ 3:41
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