You are given a sample that contains a mixture of the following two arsenic-containing minerals: duranosite (A?
You are given a sample that contains a mixture of the following two arsenic-containing minerals: duranosite (As4S) and getchellite (AsSbS3). You start with a sample of this mixture that has a mass of 61.23 grams. You perform a reaction that converts the arsenic (As) in these minerals into the form iron(III) arsenate (FeAsO4), and 79.18 grams of this product are obtained.
What is the percent by mass of duranosite in the original sample?
Answers & Comments
Verified answer
First of all, there is no such mineral as duranosite. There is a mineral realgar, whose formula is As4S4. However I will use the hypothetical mineral.
Secondly the correct formula for Iron (III) arsenate is Fe3(AsO4)2
I will use the following atomic weights
O 16, S 32.1, Fe 55.8, As 74.9,Sb 121.7
The molecular weight of the iron arsenate is 3 x55.8 + 2 x 74.9 + 8 x 16 = 445.2
The proportion of arsenic in it = (2 x 74.9)/445.2 = 33.65 %
The amount of arsenic in the final product is .3365 x 79.18 = 26.64 grams.
The molecular weight of the mythical As4S is 4 x 74.9 + 32.1 = 331.7
The proportion of arsenic is (4x74.9)/331.7 = 90.3%
The molecular weight of the getchellite is 74.9 + 121.7 + 3x 32.1 = 292.9
The proportion of arsenic is 74.9/292.9 = 25.6%
Now we can make two equations
Let the amount of the mythical duranosite be d grams
and the amount of the rare but real getchelite be g grams
We know that d+g = 61.23
Also we know from the arsenic calculation that
0.903d + .256 g = 26.64
Therefore 0.903d + .256(61.23-d) = 26.64
0.903d + 15.67 - .256d= 26.64
.647d= 10.97
d= 16.96 grams
Therefore Duranosite is 16.96/61.23 = 27.7% of the original mixture