Question #1827a

1 Answer
Feb 21, 2017

Answer:

Empirical formula #C_5##H_10##O_3#

Explanation:

Step 1 calculate the mass of Carbon in #CO_2#
#CO_2# = #0.01962 # moles
There is #1# mole of #C# in #CO_2#
And if hydrocarbon completely burns and all the #C# from the compound becomes #CO_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_2O#
Given #H_2O# = #0.01961 #moles
There are #2# moles of #H# in #H_2O#
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#