00:01
In this question we will be solving for the percent composition.
00:05
So here we are given chromite and we are to solve for the molar mass of the chromite first to solve for the percent composition.
00:17
So molar mass of f .e.
00:20
Cr 204 is equal to f .e, which is 55 .85 grams per mole plus 2 .4 plus 2 .40 is equal to f .e, which is 55 .85 grams per mole, plus 2 .2.
00:30
Times 52 grams per mole plus 4 times 16 grams per mole this is equal to 223 .85 grams per mole.
00:44
Since molar mass is the sum of the molar masses of the elements present in the compound.
00:53
Therefore, our present fe here is equal to molar mass of f3, divided by fe, cr 204 times 100%.
01:08
Substituting we have 55 .85 divided by 223 .85 times 100%.
01:18
This is equal to 24 .95 % fe.
01:25
Next, for percent cr we have in the compound we have 2cr divided by fecr 204 times 100%...