What is the molecular formula the a compound containing only carbon and also hydrogen if burning of \$1.05~mathrmg\$ that the link produces \$3.30~mathrmg~ceCO2\$ and \$1.35~mathrmg~ceH2O\$ and also its molar massive is about \$70~mathrmg\$?

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
asked Sep 18 "14 in ~ 21:11

shaimashaima
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\$\$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

RE60KRE60K
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