Question

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?

Answer 1

Answer: The empirical formula for the given compound is
`FeC_(47)H_(66)O_(26)`

Step-by-step explanation:

• The chemical equation for the combustion of compound having carbon, hydrogen, iron and oxygen follows:

`Fe_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O`

where, 'w', 'x', 'y' and 'z' are the subscripts of Iron, carbon, hydrogen and oxygen respectively.

We are given:

Mass of
`CO_2=3.1737g`

Mass of
`H_2O=0.90829g`

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

So, in 3.1737 g of carbon dioxide,
`(12)/(44)* 3.1737=0.865g` of carbon will be contained.

• For calculating the mass of hydrogen:

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

So, in 0.90829 g of water,
`(2)/(18)* 0.90829=0.101g` of hydrogen will be contained.

• For calculating the mass of iron:

Percent of Fe in
`Fe_2O_3` =
`\frac{(2* \text{molar mass of Fe}}{\text{molar mass of }Fe_2O_3}* 100`

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:

`\%\text{ mass of iron in }Fe_2O_3=(2* 55.85)/(159.69)* 100=69.94\%`

So, the amount of iron present in 0.1230 g of
`Fe_2O_3=(69.94)/(100)* 0.1230=0.0860g` of iron.

• Mass of oxygen in the compound = (1.6968) - (0.865 + 0.101 + 0.0860) = 0.6448 g

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

• Step 1: Converting the given masses into moles.

Moles of Carbon =
`\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=(0.865g)/(12g/mole)=0.072moles`

Moles of Hydrogen =
`\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=(0.101g)/(1g/mole)=0.101moles`

Moles of Oxygen =
`\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=(0.6448g)/(16g/mole)=0.0403moles`

Moles of Iron =
`\frac{\text{Given mass of iron}}{\text{Molar mass of iron}}=(0.0860g)/(55.85g/mole)=0.00153moles`

• Step 2: Calculating the mole ratio of the given elements.

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 =
`(0.072)/(0.00153)=47.05\approx 47`

For Hydrogen =
`(0.101)/(0.00153)=66.01\approx 66`

For Oxygen =
`(0.0403)/(0.00153)=26.33\approx 26`

For Iron =
`(0.00153)/(0.00153)=1`

• Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is
`Fe_1C_(47)H_(66)O_(26)=FeC_(47)H_(66)O_(26)`