Question #1827a

Feb 21, 2017

Empirical formula ${C}_{5}$${H}_{10}$${O}_{3}$

Explanation:

Step 1 calculate the mass of Carbon in $C {O}_{2}$
$C {O}_{2}$ = $0.01962$ moles
There is $1$ mole of $C$ in $C {O}_{2}$
And if hydrocarbon completely burns and all the $C$ from the compound becomes $C {O}_{2}$
Hence moles of $C$ in the hydrocarbon compound = $0.01962$ moles
mass of $C$ = $0.01962$ mol x $12$ g/ mol = $0.23577$ g

Step $2$ calculate the mass of Hydrogen in ${H}_{2} O$
Given ${H}_{2} O$ = $0.01961$moles
There are $2$ moles of $H$ in ${H}_{2} O$
so moles of $H$ in the compound = $0.03922$ moles
mass of $H$ = $0.03922$moles x $1.008$g/mol = $0.039530$ g

Step $3$ calculate the mass of $O$
mass $O$ = Total mass of simple compound – sum of masses of $C$ and $H$
Total mass of $H$ + $C$ = $0.27530$ g
Given mass of sample = $0.4647$ g
mass of $O$ = ($0.4647$ g - $0.27530$ g) = $0.18940$ g
moles of $O$ = $0.18940$ g /$16$ g/mol = $0.01183$ moles

Step $4$ calculate molar ratio
Molar ratio of all the elements ($H$ , $C$,and $O$) in the compound
molar ratio of $C$ : $H$ : $O$ = $0.01962$ : $0.03922$ : $0.01183$
smallest number $0.01183$
divide the ratio by the smallest number
molar ratio of $C$ : $H$ : $O$ = $1.657$ : $3.313$ : $1.000$
Moltiply with $3$ to get simplest whole number ratio
Final ratio = $C$ : $H$ : $O$ = $5$ : $10$ : $3$

Hence Empirical formula ${C}_{5}$${H}_{10}$${O}_{3}$