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Solve the following problem: 1.00 g of a hydrated salt contains 0.2014 g of iron, 0.1153 g of sulfur, 0.2301 g of oxygen, and 0.4532 g of water of crystallisation. Find the empirical formula. - Chemistry

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Question

Solve the following problem:

1.00 g of a hydrated salt contains 0.2014 g of iron, 0.1153 g of sulfur, 0.2301 g of oxygen, and 0.4532 g of water of crystallisation. Find the empirical formula. (At. wt.: Fe = 56; S = 32; O = 16)

Sum

Solution

Given: Atomic mass of Fe = 56, S = 32, and O = 16
Mass of iron, sulfur, oxygen, and water = 0.2014 g, 0.1153 g, 0.2301 g, and 0.4532 respectively.

To find: The empirical formula of the compound

Calculation: Since the mass of crystal is 1 g, the % iron, sulfur, oxygen and water = 20.14%, 11.53%, 23.01% and 45.32% respectively.

Moles of Fe = `"% of Fe"/"Atomic mass of Fe"=20.14/ 56` = 0.360 mol

Moles of S = `"% of S"/"Atomic mass of S"=11.53/32` = 0.360 mol

Moles of O = `"% of O"/"Atomic mass of O"=23.01/16` = 1.438 mol

Moles of water = `"% of water"/"Atomic mass of water"=45.32/18` = 2.518 mol

Hence, the ratio of number of moles of Fe : S : O : water is

`0.360/0.360` = 1, `0.360/0.360` = 1, `1.438/0.360` = 4 and `2.518/0.360` = 7

Hence, empirical formula is FeSO4.7H2O.

Empirical formula of the compound = FeSO4.7H2O

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Determination of Molecular Formula
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Chapter 2: Introduction to Analytical Chemistry - Exercises [Page 25]

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Balbharati Chemistry [English] 11 Standard
Chapter 2 Introduction to Analytical Chemistry
Exercises | Q 4. (J) | Page 25

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