Combustion analysis of 0.900 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.564 g of CO2 and 0.851 g of H2O. What is the empirical formula of the compound?

The empirical formula for the given compound is

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of

Mass of

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 1.564 g of carbon dioxide, of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.851 g of water, of hydrogen will be contained.

Mass of oxygen in the compound = (0.900) - (0.426 + 0.0945) = 0.3795 g

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon =

Moles of Hydrogen =

Moles of Oxygen =

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0237 moles.

For Carbon =

For Hydrogen =

For Oxygen =

Multiplying the mole fraction by '2' to make it a whole number

Mole fraction of carbon = (1.5 × 2) = 3

Mole fraction of hydrogen = (4 × 2) = 8

Mole fraction of oxygen = (1 × 2) = 2  

The ratio of C : H : O = 3 : 8 : 2

Hence, the empirical formula for the given compound is

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