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

Here is mine 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 the 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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physical-starrkingschool.net stoichiometry elemental-analysis
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edited Feb 26 "18 at 2:50

pentavalentcarbon
request Sep 18 "14 in ~ 21:11

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## 1 prize 1

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\$egingroup\$
\$\$ceC_\$a\$H_\$b\$ + \$left(a+frac b4 ight)\$O2 -> \$a\$CO2 + \$fracb2\$ H2O\$\$

Suppose you had actually \$n\$ moles of hydrocarbon, climate we have \$acdot n\$ moles of \$ceCO2\$ and also \$fracb2cdot n\$ moles of \$ceH2O\$ splitting their moles we"ll acquire \$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 in ~ 4:39

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answered Sep 19 "14 in ~ 3:41

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