A compound containing carbon, hydrogen, and oxygen. Combustion analysis of a 4.30- g sample of butyric acid produced 8.59g CO2 and 3.52 H2O. Find the empirical formula.

The empirical formula represents the simplest whole-number ratio of various atoms in a compound hence here n=2 and the empirical formula is C2H4O.

Steps wise:

 

Mass of CO2  given in the problem is 8.5900g , convert the mass to moles by dividing mass in grams with the molar mass of CO2 

8.59 / 44 = 0.193 moles of CO2

In CO2 there are 1 mole of C and 2 moles of O .So, mass of C in CO2 is

.193 moles CO2 X 1 moles C / 1 mole CO2 = .193 moles of C

now convert moles of C to mass ( multiple the moles with MM of  C)

.193 moles of C x 12 .01 = 2.34 g of C

Similarly find the mass of H in given mass of H2O .

3.52 g of H2O/ 18 of H2O x 2 moles of H / 1 mole H2O x 1g H =  0.394 g of H

The total mass of compound:4.3000g

Mass of O = 1.5630g: a mass of the compound – mass of (C+H)
Moles of O = 0.0977mmoles: / atomic weight of O to get the number of moles
check the mass of C+H+O:4.3000g

moles of C = 0.1952
moles of H = 0.3911
moles of O = 0.0977

divide by smallest to get the ratio
C :2.00
H :4.00
O :1.00

Empirical formula is C2H4O

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