combustion analysis of hydrocarbons?

Question

a hydrocarbon is subjected to combustion analysis. a sample of the compound gave 0.314g CO2 and 0.129g H2O.
a)calculate the mass of carbon and hydrogen in the sample

b) find the no of moles of carbon and hydrogen in the sample of the compound

c)what is the empirical formulae

Answer 1

0.314gCO2 x 1molCO2/44gCO2 x 1molC/1molCO2 = 0.00714 moles C

0.129gH2O x 1molH2O18gH2O x 2molH/1molH2O = 0.0143 mole H

0.00714/0.00714 = 1.0

0.0143/0.00714 = 2.0

The empirical formula is CH2

Answer 2

that's been 2 many years when you consider that I honestly have carried out the style of equation. yet once I keep in mind, indexed right here are the stairs: you know you have some style of CH hydrocarbon burning in O2. H2O and CO2 shows perfect combustion when you consider which you haven't any longer have been given any nitrogen compounds being considered. this implies which you do no longer lose any H, C, or O to different compounds being fashioned different than those in the equation. you should make certain the proportions of each element . You try this via figuring the atomic weights. you have X C atoms and Y O atoms and their entire mass provides as much as 33.01 g. you know which you have two times as many O's in the CO2. Your equation would be something like (X)CH+(Y)O2 = (A)H2O + (B)CO2 the place the letters in parenthesis are the proportions of each molecule. i know that's somewhat imprecise, even nonetheless that's been 2 many years.