Question

Problem Page Cobalt(II) chloride forms several hydrates with the general formula , where is an integer. If the hydrate is heated, the water can be driven off, leaving pure behind. Suppose a sample of a certain hydrate is heated until all the water is removed, and it's found that the mass of the sample decreases by . Which hydrate is it? That is, what is ?

Answer 1

The question has missing information, the complete question is:

Cobalt(II) chloride forms several hydrates with the general formula CoCl₂.xH₂O, where x is an integer. If the hydrate is heated, the water can be driven off, leaving pure CoCl₂ behind. Suppose a sample of a certain hydrate is heated until all the water is removed, and it's found that the mass of the sample decreases by 22.0%. Which hydrate is it? That is, what is x?

Answer:

CoCl₂.26H₂O

Explanation:

The molar masses of the compounds that forms the hydrate are:

Co = 59 g/mol

Cl = 35.5 g/mol

H = 1 g/mol

O = 16 g/mol

The molar mass of CoCl₂ is 130 g/mol and of H₂O is 18 g/mol, thus for the hydrate, it will be 130 + 18x g/mol.

Let's suppose 1 mol of the compound. Thus, the mass of the hydrate is: 130 + 18x, and the mass of CoCl₂ will be 130 g. Because the mass decreassed by 22.0% :

0.22*(130 + 18x) = 130

130 + 18x = 590.91

18x = 460.91

x ≅ 26

Thus, the hydrate is CoCl₂.26H₂O

Answer 2

Answer:

It is cobalt (II) chloride dihydrate, CoCl2.2H2O.

X = 2

Explanation:

We can calculate the value of x using the formula:

RMM of nH2O/RMM of anhydrous salt = mass of water expelled or absorbed/mass of anhydrous salt

Where RMM of nH2O= relative molecular mass of nH2O =n[ (1 X 2) + 16 ]= 18n where n = number of moles of water of crystallization = x

RMM of anhydrous salt= RMM of CoCl2= 58.9+(35.5 X 2) = 129.9

Since the mass of the hydrated salt decreased by 22%, and assuming we have 100g of original sample,

Mass of water expelled = 22g

And (100-22)g of anhydrous salt will remain after heating off all the water, so,

Mass of anhydrous salt = 78g

Substituting the values into the formula,

18n/129.9 = 22/78

18n X 78 = 129.9 X 22

1404n = 2857.8

n = 2857.8/1404

n = 2.035470085

n = 2 ( to the nearest whole number )

You can check if your answer is correct by using this formula to calculate the percentage of water of crystallization in CoCl2.2H2O:

% of water of crystallization = (mass of water expelled/mass of hydrated salt) X 100

Where mass of water expelled = mass of 2H2O = 2.035470085 X 18 = 36.63846154

Mass of hydrated salt = mass of CoCl2.2H2O = 129.9 + 36.63846154 = 166.5384615.

Substituting into the formula we have

% of water of crystallization = ( 36.63846154/166.5384615) X 100.

% of water of crystallization = 0.22 X 100 = 22%.

Notice that we used the actual value of the number of moles of water of crystallization in calculating the percentage to minimize errors and avoid approximate percentage composition values. But we used 2 for the formula of the hydrated salt because we had to round off the value to the nearest whole number