When 1.6968 g of an organic iron compound containing fe, c, h, and o was burned in o2, 3.1737 g of co2 and 0.90829 g of h2o were produced. in a separate experiment to determine the mass percent of iron, 0.5446 g of the compound yielded 0.1230 g of fe2o3. what is the empirical formula of the compound?

The empirical formula for the given compound is

Explanation:

where, 'w', 'x', 'y' and 'z' are the subscripts of Iron, 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

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

So, in 3.1737 g of carbon dioxide, of carbon will be contained.

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

So, in 0.90829 g of water, of hydrogen will be contained.

Percent of Fe in =

Molar mass of iron = 55.85 g/mol

Molar mass of iron (III) oxide = 159.69 g/mol

Putting values in above equation, we get:

So, the amount of iron present in 0.1230 g of of iron.

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

Moles of Carbon =

Moles of Hydrogen =

Moles of Oxygen =

Moles of Iron =

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00153 moles.

For Carbon =

For Hydrogen  =

For Oxygen  =

For Iron  =

The ratio of Fe : C : H : O = 1 : 47 : 66  : 26

Hence, the empirical formula for the given compound is

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