Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 36.86g CO2 and 10.06g H2O. The molar mass of estriol is 288.38g/mol .

Find the molecular formula of the unknown compound

molecular formula = C12H16O8

Explanation:

NB Mm CO2= 44g/mol

Mm H2O= 18g/mol

Moles of CO2 = 36.86/44=0.84mol

0.84mole of CO2 has 0.84 mol of C

Moles of H2O = 10.06/18= 0.56mol

1mol of H20 contains 1mol of O and 2 mol H,

Hence there are 0.56mol O and (0.56×2)mol H

Hence the compound contains

C= 0.84 mol H= 1.12mol O=0.56mol

Divide through by smallest number

C= 0.83/0.56= 1.5mol

H= 1.12/0.55= 2mol

O= 0.56/0.56= 1mol

Multiply all by 2 to have whole number of moles = 3:4:2

Hence empirical formula= C3H4O2

(C3H4O2)n = 288.38

[(12×3) + 4+(16×2)]n= 288.38

72n=288.38

n= 4

:. Molecular formula=(C3H4O2)4= C12H16O8

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o