Question

An unknown compound contains only C , H , and O . Combustion of 9.30 g of this compound produced 22.7 g CO 2 and 9.29 g H 2 O . What is the empirical formula of the unknown compound? Insert subscripts as needed. empirical formula:?

Answer 1

The empirical formula is =
`C_4H_8O`

Step-by-step explanation:

Mass of water obtained = 9.29 g

Molar mass of water = 18 g/mol

Moles of
`H_2O` = 9.29 g /18 g/mol = 0.51611 moles

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

Moles of H = 2 x 0.51611 = 1.03222 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 1.03222 x 1.008 = 1.0407 g

Mass of carbon dioxide obtained = 22.7 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of
`CO_2` = 22.7 g /44.01 g/mol = 0.5158 moles

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

Moles of C = 0.5158 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.5158 x 12.0107 = 6.1950 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 = 9.30 g

Mass of O in sample = 9.30 - 6.1950 - 1.0407 = 2.0643 g

Molar mass of O = 15.999 g/mol

Moles of O = 2.0643 / 15.999 = 0.12903 moles

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

1.03222 : 0.12903 : 0.5158

= 8 : 1 : 4

The empirical formula is =
`C_4H_8O`