Question

# An organic compound containing only C, H, and possibly O was subjected to combustion analysis. A...

An organic compound containing only C, H, and possibly O was subjected to combustion analysis. A sample weighing 0.5857 g yielded 1.140 g CO2 and 0.233 g H2O. What is the empirical formula of the compound?

let in compound number of moles of C, H and O be x, y and z respectively

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 1.14/44

= 2.591*10^-2

Number of moles of H2O = mass of H2O / molar mass H2O

= 0.233/18

= 1.294*10^-2

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 2.591*10^-2

so, x = 2.591*10^-2

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*1.294*10^-2 = 2.589*10^-2

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 0.5857 - 2.591*10^-2*12 - 2.589*10^-2*1

= 0.2489

number of mol of O = mass of O / molar mass of O

= 0.2489/16.0

= 1.556*10^-2

so, z = 1.556*10^-2

Divide by smallest:

C: 2.591*10^-2/1.556*10^-2 = 5/3

H: 2.589*10^-2/1.556*10^-2 = 5/3

O: 1.556*10^-2/1.556*10^-2 = 1

multiply by 3 to get simplest whole number ratio:

C: 5/3 * 3 = 5

H: 5/3 * 3 = 5

O: 1 * 3 = 3

So empirical formula is:C5H5O3

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