**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#