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A sample of what was thought to be gold was removed from a mine in San Francisco during the gold rush. After analysis, it was determined that it was actually fool's gold, or pyrite. The sample contained 17.6 grams of iron and 103 grams of sulfur. What is the percentage composition of each element in pyrite?

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Answer:

14.594% Iron and 85.406% Sulfur

Step-by-step explanation:

103g (S) + 17.6g (Fe) = 120.6g total mass

103/120.6 × 100 = 85.406% Sulfur

17.6/120.6 × 100 = 14.594% Iron

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