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Step-by-Step Solution
Step 1: Understand the Given Information
We have a complex A with the composition H12O6Cl3Cr (in elementary form,
it shows Cr, Cl, and H2O). When treated with concentrated H2SO4,
the complex loses 13.5% of its original mass. We aim to find the actual formula of this chromium complex.
Step 2: Let the Number of Water Molecules Lost Be x
Suppose during the reaction with conc. H2SO4, x moles of water
($H_{2}O$) are lost from each mole of the complex. Our task is to determine x.
Step 3: Express the 13.5% Mass Loss in Terms of x
Let the initial molar mass of the complex (before losing water) be M.
Then the mass of water lost (x molecules) is $x \times 18$ (since the molar mass of
each water molecule is 18 g/mol).
The percentage loss is 13.5%, so we can set up the equation:
$$
13.5 = \left[\frac{x \times 18}{M}\right] \times 100
$$
Hence,
$$
\frac{x \times 18}{M} = 0.135
$$
Step 4: Calculate M (the Molar Mass of the Complex Before Losing Water)
The complex initially has 6 molecules of water (as indicated by H12O6),
3 chlorine atoms, and 1 chromium atom within its coordination sphere and as potential water
of crystallization. To make a good estimate, let us consider the formula “$[Cr(H_2O)_6]Cl_3$”
as a candidate and see how it modifies. The total mass of $[Cr(H_2O)_6]Cl_3$ would be:
$$
1 \, (\text{Cr}) \times 52 \;+\; 3 \, (\text{Cl}) \times 35 \;+\; 6 \, (\text{H}_2\text{O}) \times 18 = 52 + 105 + 108 = 265
$$
However, from the simpler representation in the problem, the sum of
atomic components (6 waters + 3 chlorines + 1 chromium) is:
$$
M = (6 \times 18) + (3 \times 35) + 52 = 108 + 105 + 52 = 265
$$
So indeed before losing water, the formula’s total mass is 265 g/mol
(consistent with something akin to having 6 water molecules, 3 chlorines,
and 1 chromium in some arrangement).
Step 5: Solve for x
Substitute $M = 265$ into the percentage equation:
$$
13.5 = \left[\frac{x \times 18}{265}\right] \times 100
$$
$$
\frac{x \times 18}{265} = 0.135
$$
Solving for x:
$$
x \times 18 = 0.135 \times 265
$$
$$
x \times 18 = 35.775
$$
$$
x = \frac{35.775}{18} \approx 1.99 \approx 2
$$
Thus, approximately two molecules of water are lost.
Step 6: Deduce the Formula of the Complex
Since two water molecules are lost upon treatment with conc. H2SO4,
the complex must have two water molecules as water of crystallization. The consistent structure
that fits these data is:
$$
[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O
$$
This shows:
• 4 water molecules directly bound to Cr in the coordination sphere,
• 2 chlorine atoms also bound to Cr,
• 1 chloride ion outside the coordination sphere, and
• 2 molecules of water of crystallization.
