Question

Estrone, which contains only carbon, hydrogen, and oxygen, is a female sexual hormone that occurs in the urine of pregnant women. Combustion analysis of a 1.893-g sample of estrone produced 5.543g of CO2 and 1.388g H2O. The molar mass of estrone is 270.36g/mol .Find the molecular formula for estrone.Express your answer as a chemical formula.

Answer 1

The formula of Estrone =
`C_(18)H_(22)O_2`

Step-by-step explanation:

Mass of water obtained = 1.388 g

Molar mass of water = 18 g/mol

Moles of
`H_2O` = 1.388 g /18 g/mol = 0.07711 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.07711 = 0.1542 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.1542 x 1.008 = 0.1555 g

Mass of carbon dioxide obtained = 5.543 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of
`CO_2` = 5.543 g /44.01 g/mol = 0.126 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.126 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.126 x 12.0107 = 1.5133 g

Given that the Estrone only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C - Mass of H

Mass of the sample = 1.893 g

Mass of O in sample = 1.893 - 1.5133 - 0.1555 = 0.2242 g

Molar mass of O = 15.999 g/mol

Moles of O = 0.2242 / 15.999 = 0.01401 moles

Taking the simplest ratio for H, O and C as:

0.1542 : 0.01401 : 0.126

= 11 :1 : 9

The empirical formula is =
`C_9H_(11)O`

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 9×12 + 11×1 + 16= 135 g/mol

Given, Molar mass = 270.36 g/mol

So,

Molecular mass = n × Empirical mass

270.36 = n × 135

⇒ n = 2

The formula of Estrone =
`C_(18)H_(22)O_2`