case 1: Iron (3) sulphide contains 46.3% by mass of sulphur, and 53.7% by mass of Fe.
case 2: Also Iron (2) sulphide contain 36.50% by mass of S and 63.50% by mass of Fe.a) Calculate the mass of sulphur which combine with 1gm of Iron in case 1 and in case 2 above
b) what is the ratio of mass of sulphur which combined with 1 gm of Fe from Iron (3) sulphide (case 1) to the ratio of Sulphur which combined with 1gm of Fe from Iron (2) Sulphur, in (Case 2 ).
To calculate the mass of sulfur that combines with 1 gram of iron in both Case 1 and Case 2, we need to use the given percentages by mass of sulfur and iron.
a) Case 1:
In Iron (3) sulphide (Fe2S3), the mass percent of sulfur is 46.3%.
So, we can calculate the mass of sulfur as follows:
Mass of sulfur = 46.3% * 1 gram
= (46.3/100) * 1 gram
= 0.463 grams
b) Case 2:
In Iron (2) sulphide (FeS), the mass percent of sulfur is 36.50%.
Using the same calculation method:
Mass of sulfur = 36.50% * 1 gram
= (36.50/100) * 1 gram
= 0.365 grams
To calculate the ratio of the mass of sulfur that combines with 1 gram of iron in Case 1 (Iron (3) sulphide) to the ratio in Case 2 (Iron (2) sulphide):
Ratio = (Mass of sulfur in Case 1) / (Mass of sulfur in Case 2)
= 0.463 grams / 0.365 grams
≈ 1.27
Therefore, the ratio of the mass of sulfur that combines with 1 gram of iron in Case 1 to the ratio in Case 2 is approximately 1.27.