Question

Cobalt (II) chloride exits in hydrated forms, the dehydrate and hexahydrate forms. A chemist grabs cobalt (II) chloride hexahydrate from the shelf and wants to dehydrate the salt. If 5.00g of hexahydrate were heated to drive off waters of hydration, how many grams of the anhydrous salt would remain?

Answer 1

2.48 grams.

Explanation

Start by finding the formula mass for cobalt (II) chloride and cobalt (II) chloride hexahydrate.

As a transition element in the middle d-block of the periodic table, cobalt can form ions with a plurality of charges. It is thus necessary to indicate its charge in systematic names of cobalt-containing formula.  

The cation "cobalt" in the systematic name of the salt has the Roman numeral "(II)" attached to it in brackets. As a metal, cobalt forms positively-charged ion. The one here thu has charge of +2.

Chloride ions have charges -1. Charges cancel out to produce neutral compounds. Each cobalt cation in this salt would thus pair with two chloride anions. Hence the empirical formula: $\text{Co}\text{Cl}_2$.

The prefix "hexa-" in the name cobalt (II) chloride hexahydrate indicates that every formula unit of this salt contains six units of water. The hydrated salt thus has an empirical formula of $\text{Co}\text{Cl}_2\cdot (\text{H}_2\text{O})_6$.

Given the relative atomic mass for each of the elements, as seen on a modern periodic table of the elements:

• Cobalt- 58.93 $\text{g}\cdot \text{mol}^{-1}$
• Chloride- 35.45 $\text{g}\cdot \text{mol}^{-1}$
• Hydrogen- 1.008 $\text{g}\cdot \text{mol}^{-1}$
• Oxygen- 16.00 $\text{g}\cdot \text{mol}^{-1}$

Thus the formula mass of each compound

• Cobalt (II) chloride $\text{Co}\text{Cl}_2$- 129.83 $\text{g}\cdot \text{mol}^{-1}$
• Cobalt (II) chloride hexahydrate $\text{Co}\text{Cl}_2\cdot (\text{H}_2\text{O})_6$- 262.12 $\text{g}\cdot \text{mol}^{-1}$

Cobalt (II) chloride hexahydrate $\text{Co}\text{Cl}_2\cdot (\text{H}_2\text{O})_6$ decomposes under heat to produce cobalt (II) hexahydrate and water. Hence the equation:

$\text{Co}\text{Cl}_2\cdot (\text{H}_2\text{O})_6 \stackrel{\Delta}{\to}\text{Co}\text{Cl}_2 +6 \; \text{H}_2\text{O}$

Therefore

• Molar ratio: $n(\text{Co}\text{Cl}_2) : n(\text{Co}\text{Cl}_2 \cdot (\text{H}_2\text{O})_6 = 1 : 1$
• Mass ratio: $m(\text{Co}\text{Cl}_2) : m(\text{Co}\text{Cl}_2 \cdot (\text{H}_2\text{O})_6) = 262.12 : 129.83$

The mass ratio $m(\text{Co}\text{Cl}_2) : m(\text{Co}\text{Cl}_2 \cdot (\text{H}_2\text{O})_6)$ indicates that 262.12 grams of cobalt (II) chloride hexahydrate decomposes to produce 129.83 grams of its corresponding anhydrous salt. Accordingly, heating 5.00 grams of the hexahydrate would  produce 2.48 grams of its anhydrate.