Question

0.658 g of a compound containing only carbon, hydrogen, and oxygen is burned in excess O2. CO2(1.285 g) and H20 (0.658g) are produced. The mol mass of the compound is determined by mass spectrometry to be 90 g/mole Determine the empirical and molecular formulas. 20:0-8453x(12.01s /uu.o 032 029/1B.A mal 1 2

Answer 1

The empirical formula is =
`C_2H_5O`

The formula of the compound =
`C_4H_(10)O_2`

Step-by-step explanation:

Explanation:

Mass of water obtained = 0.658 g

Molar mass of water = 18 g/mol

Moles of
`H_2O` = 0.658 g /18 g/mol = 0.03656 moles

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

Moles of H = 2 x 0.03656 = 0.07311 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.07311 x 1.008 = 0.07369 g

Mass of carbon dioxide obtained = 1.285 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of
`CO_2` = 1.285 g /44.01 g/mol = 0.029197 moles

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

Moles of C = 0.029197 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.029197 x 12.0107 = 0.350676 g

Given that the compound 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 = 0.658 g

Mass of O in sample = 0.658 - 0.350676 - 0.07369 = 0.233634 g

Molar mass of O = 15.999 g/mol

Moles of O = 0.233634 / 15.999 = 0.014603 moles

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

0.07311 : 0.014603 : 0.029197

= 5 : 1 : 2

The empirical formula is =
`C_2H_5O`

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 = 2×12 + 5×1 + 1×16= 45 g/mol

Molar mass = 90 g/mol

So,

Molecular mass = n × Empirical mass

90 = n × 45

⇒ n = 2

The formula of the compound =
`C_4H_(10)O_2`