A sample of an iron sulfide contains $2.561 \cdot g$ $2.561g$ of metal, and $2.206 \cdot g$ $2.206g$ of sulfur. What is the empirical formula of iron sulfide?

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A sample of an iron sulfide contains $2.561 \cdot g$ $2.561g$ of metal, and $2.206 \cdot g$ $2.206g$ of sulfur. What is the empirical formula of iron sulfide?

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Answer 1 · 2016-06-10T18:12:42

$F e_{2} S_{3}$ $Fe_2S_3$

Explanation:

$Moles of iron$ $"Moles of iron"$ $=$ $=$ $\frac{2.561 \cdot g}{55.85 \cdot g \cdot m o l^{- 1}}$ $(2.561*g)/(55.85*g*mol^-1)$ $=$ $=$ $0.0459 \cdot m o l$ $0.0459*mol$

$Moles of sulfur$ $"Moles of sulfur"$ $=$ $=$ $\frac{2.206 \cdot g}{32.06 \cdot g \cdot m o l^{- 1}}$ $(2.206*g)/(32.06*g*mol^-1)$ $=$ $=$ $0.0688 \cdot m o l$ $0.0688*mol$

If we divide thru by the lower molar quantity, we get, $F e S_{1.5}$ $FeS_(1.5)$ . However, by definition, the empirical formula is the simplest WHOLE number that defines constituent elements in a species, so we double the formula to get WHOLE numbers:

$F e_{2} S_{3}$ $Fe_2S_3$ .

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A sample of an iron sulfide contains $2.561 \cdot g$ $2.561g$ of metal, and $2.206 \cdot g$ $2.206g$ of sulfur. What is the empirical formula of iron sulfide?

1 Answer

Explanation:

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A sample of an iron sulfide contains 2.561⋅g2.561*g of metal, and 2.206⋅g2.206*g of sulfur. What is the empirical formula of iron sulfide?

1 Answer

Explanation:

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A sample of an iron sulfide contains $2.561 \cdot g$ $2.561g$ of metal, and $2.206 \cdot g$ $2.206g$ of sulfur. What is the empirical formula of iron sulfide?