Question

gravimetric determination of Iron as Fe2O3

A sample containg iron can be analyzed by precipitation of the hydrous oxide from basic solution, followed by ignition to Fe2O3:

Fe³⁺ + (2+x)H₂O ----> FeOOH* xH₂O(s) + 3H⁺

^{FeOOH * xH₂O ------> Fe₂O3} (s)

weigh three samples of unknown containg enough Fe to produce ~0.3g of Fe2O3

In my experiment,

unknow sample weight: 0.75g 0.80 0.70

Fe₂O₃ collected from unknow: 0.05g 0.04 g 0.03g

how do I calculate (1)% Fe in unknown sample

(2) average % Fe

(3)standard deviation

(4) relative standard deviation

determine (1) % Fe in unknown sample

(2)average % Fe

(3)standard deviation

(4)relative standard deviation

For gravimetric determination of Iron as Fe2O3, how do I calculate?

weighed sample of unknown containing Fe to produce ~0.3g of Fe2O3 (~0.6 - 1 g of unkonwn)

I got 0.75g from unknow sample.

I got 0.05g from Fe2O3 collected from unknow.

Answer 1

Molecular mass of Fe2O3= 159.69 gm/mol

Atomic mass of iron (Fe)=55.845.

159.69 gm has 2*55.845 gm of Fe

a.0.05gm has 2*55.845/159.69 * 0.05 =0.0349 gm. Fe ( The gravimetric factor for Fe2O3=2*55.845/159.69 = 0.6994.)

b.0.04 gm has 0.6994*0.04 = 0.0279 gm of Fe.

c.0.03 gm has 0.6994*0.03 = 0.0209 gm of Fe.

% composition of Fe in (a) = 0.0349/0.05 *100 = 68%

% composition of Fe in (b)= 69.75 %

% composition in (c)= 69.66%

Standard Deviation=0.98541023606076

Average yeild= 69.136666666667 %

Relative standard deviation = 1.43%

To calculate amount of sample for 0.3 gm of Fe2O3= 0.3gm Fe2O3*1 mole Fe2O3/159.688 gm Fe2O3 * 2mole FeOOH/ 1 mole Fe2O3 * 1 mole initial salt(NH4)2Fe(SO4)2.6H2O/1mole FeOOH * 1/0.75 = 1.75 gm of unknown.