Question

The gravimetric analysis of iron in an ore sample was performed by digesting the sample in strong acid, followed by precipitation of Fe(OH)3 using KOH, filtration of the crystalline iron hydroxide, and finally drying and ignition at high temperature to produce Fe2O3. After cooling in a dessicator, two different students got the following replicate results:

Student 1, Trial 1 – 0.2109 g of ore results in 0.04594 g iron (III) oxide

Student 1, Trial 2 – 0.2110 g of ore results in 0.04596 g iron (III) oxide

Student 1, Trial 3 – 0.2109 g of ore results in 0.04594 g iron (III) oxide

Student 1, Trial 4 – 0.2112 g of ore results in 0.04601 g iron (III) oxide

Student 2, Trial 1 – 0.2111 g of ore results in 0.04601 g iron (III) oxide

Student 2, Trial 2 – 0.2108 g of ore results in 0.04594 g iron (III) oxide

Student 2, Trial 3 – 0.2109 g of ore results in 0.04596 g iron (III) oxide

Student 2, Trial 4 – 0.2109 g of ore results in 0.04596 g iron (III) oxide

A.)Calculate the percent iron in the ore samples and report mean values for each Student

(Atomic Weights: Fe = 55.845, O = 15.9994)

Answer 1

The formula for iron oxide is Fe2O3

i e 1 mole Fe₂O₃ 2 mole Fe

Molar mass of Fe₂O₃ = ( 2x Molar mass of Fe ) + ( 3x Molar mass of O)

=( 2 x 55.845 ) + ( 3 x 15.9924)

= 159.667 gm / mol

i e   159.667 gm Fe₂O₃   111.69 gm Fe

0.04594 gm  Fe₂O₃ 111.69 x 0.04594 / 159.667 gm Fe

   0.03213 gm Fe

Percent Fe = ( 0.03213 / 0.2109 ) x 100 = 15.23 %

Trial 2

159.667 gm   Fe₂O₃ 111.69 gm Fe

0.04596 gm  Fe₂O₃ 111.69 x 0.04596/ 159.667 gm Fe

   0.03215 gm Fe

Percent Fe = ( 0.03215/ 0.2110) x 100 = 15.24 %

Trail 3 same result as trail 1

Trail 4

159.667 gm  Fe₂O₃   111.69 gm Fe

0.04601 gm  Fe₂O₃ 111.69 x 0.04601/ 159.667 gm Fe

   0.03218 gm Fe

Percent Fe = ( 0.03218/ 0.2112) x 100 = 15.24 %

Mean of percent Fe= 15.23+15.24+15.23+15.24/4

= 15.235 %

STUDENT 2

TRAIL 1

159.667 gm  Fe₂O₃   111.69 gm Fe

0.04601 gm  Fe₂O₃ 111.69 x 0.04601/ 159.667 gm Fe

   0.03218 gm Fe

Percent Fe = ( 0.03218/ 0.2111) x 100 = 15.24 %

TRAIL 2

159.667 gm  Fe₂O₃   111.69 gm Fe

0.04594 gm Fe₂O₃ 111.69 x 0.04594 / 159.667 gm Fe

   0.03213 gm Fe

Percent Fe = ( 0.03213 / 0.2108) x 100 = 15.24 %

TRAIL 3

159.667 gm  Fe₂O₃ 111.69 gm Fe

0.04596 gm Fe₂O₃ 111.69 x 0.04596/ 159.667 gm Fe

   0.03215 gm Fe

Percent Fe = ( 0.03215/ 0.2109) x 100 = 15.24 %

TRAIL 4 same result as TRAIL 3

Mean % Fe = 15.24+15.24+15.24+15.24/ 4

= 15.24 %