Question

A compound contains only Fe and O. A 0.2729 g sample of the compound was dissolved in 50 mL of concentrated acid solution, reducing all the iron to Fe2+ ions. The resulting solution was diluted to 100 mL and then titrated with a 0.01621 M KMnO4 solution. The unbalanced chemical equation for reaction between Fe2+ and MnO−4 is given below.

MnO−4(aq)+Fe2+(aq)→ Mn2+(aq)+Fe3+(aq) (not balanced)
The titration required 42.17 mL of the KMnO4 solution to reach the pink endpoint.What is the empirical fomula of the compound?

Answer 1

The titration of a compound consisting of iron and oxygen with KMnO4 revealed that the moles of iron to oxygen were in a ratio of approximately 2:3. This empirical formula indicates the compound is iron (III) oxide, with the formula Fe2O3.

Step-by-step explanation:

To find the empirical formula of a compound composed of iron (Fe) and oxygen (O), we need to determine the molar amounts of these elements. Given that during the titration 42.17 mL of 0.01621 M KMnO4 solution was used, we can calculate the moles of KMnO4:

Moles of KMnO4 = 0.01621 mol/L × 0.04217 L = 6.832 × 10⁻´ moles KMnO4

The balanced equation for the redox reaction between KMnO4 and Fe2+ under acidic conditions is:

MnO4⁻ (aq) + 5Fe²⁺ (aq) + 8H⁺ (aq) → Mn²⁺ (aq) + 5Fe³⁺ (aq) + 4H2O(l)

According to the stoichiometry of the balanced equation, 1 mole of KMnO4 reacts with 5 moles of Fe2+. Thus, the moles of Fe in the original sample is 5 times the moles of KMnO4:

Moles of Fe = 5 × 6.832 × 10⁻´ moles = 3.416 × 10⁻³ moles Fe

To find the mass of Fe, we multiply moles of Fe by its molar mass (55.845 g/mol):

Mass of Fe = 3.416 × 10⁻³ moles × 55.845 g/mol = 0.1906 g

To determine the moles of oxygen, we subtract the mass of Fe from the total mass of the compound and then divide by the molar mass of oxygen (15.999 g/mol):

Moles of O = (0.2729 g - 0.1906 g) / 15.999 g/mol = 5.144 × 10⁻³ moles O

Next, we find the mole ratio of Fe to O by dividing the moles of each element by the smallest number of moles:

Fe : O ratio = 3.416 × 10⁻³ moles Fe / 3.416 × 10⁻³ moles

= 1 : (5.144 × 10⁻³ moles O / 3.416 × 10⁻³ moles Fe)

1 : 1.505

The subscripts for O x 2 and Fe x 2:

Fe2O3

Thus, the empirical formula for the iron oxide compound is Fe2O3, which is known as iron (III) oxide.

Answer 2

FeO

Step-by-step explanation:

The balanced reaction equation is

MnO−4(aq)+5Fe2+(aq)→ Mn2+(aq)+5Fe3+(aq)

Number of moles of KMnO4 reacted = 0.01621 M × 42.17/1000 = 6.84 × 10^-4 moles

Since

1 mole of MnO4^- reacted with 5 moles of Fe^2+

6.84 × 10^-4 moles of MnO4 will react with 6.84 × 10^-4 moles × 5 = 34.2 × 10^-4 moles of Fe^2+

Mass of Fe^2+ reacted = 34.2 × 10^-4 moles of Fe^2+ × 56gmol-1 = 0.192 g of Fe

Hence % of iron in the sample = 0.192/0.2729 × 100 = 70.4%

% of oxygen = (0.2729 -0.192)/0.2729 × 100 = 29.6%

Empirical formula of the compound is obtained by

70.4/56, 29.6/16

1.257, 1.85

Divide through by the lowest ratio

1.257/1.257, 1.85/1.257

1, 1

Hence the empirical formula is FeO