Question

A \(2.634-\mathrm{g}\) sample containing impure \(\mathrm{CuCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) was heated. The sample mass after heating to drive off the water was \(2.125 \mathrm{g} .\) What was the mass percent of \(\mathrm{CuCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) in the original sample?

Short answer

The mass percent of \(\mathrm{CuCl}_2 \cdot 2\mathrm{H}_2\mathrm{O}\) in the sample is 100\%.

Step-by-step solution

Calculate Mass of Water Lost

First, find the mass of water that was driven off by subtracting the mass of the sample after heating from the original mass. \[\text{Mass of water} = 2.634 \, \text{g} - 2.125 \, \text{g} = 0.509 \, \text{g}\]

Calculate Mass of Pure \(\mathrm{CuCl}_2 \cdot 2\mathrm{H}_2\mathrm{O}\)

The mass loss is due to the loss of water only, so the compound left is pure \(\mathrm{CuCl}_2\). The original compound mass lost was water, so the mass of \(\mathrm{CuCl}_2 \cdot 2\mathrm{H}_2\mathrm{O}\) is still in the heated sample. Therefore, the mass of the entire \(\mathrm{CuCl}_2 \cdot 2\mathrm{H}_2\mathrm{O}\) before the sample was heated is equal to the initial mass.\[\text{Mass of \(\mathrm{CuCl}_2 \cdot 2\mathrm{H}_2\mathrm{O}\)} = 2.125 \, \text{g (after heating)} + 0.509 \, \text{g (lost water)} = 2.125 \, \text{g} + 0.509 \, \text{g} = 2.634 \, \text{g}\]

Calculate Mass Percent of \(\mathrm{CuCl}_2 \cdot 2\mathrm{H}_2\mathrm{O}\)

Finally, to calculate the mass percent of \(\mathrm{CuCl}_2 \cdot 2\mathrm{H}_2\mathrm{O}\) in the original sample, use the formula for mass percent:\[\text{Mass Percent} = \left(\frac{\text{Mass of } \mathrm{CuCl}_2 \cdot 2\mathrm{H}_2\mathrm{O}}{\text{Mass of original sample}}\right) \times 100\%\]Substitute the known values:\[\text{Mass Percent} = \left(\frac{2.634 \, \text{g}}{2.634 \, \text{g}}\right) \times 100\% = 100\%\]

Key concepts

Hydrated Salts and Water of Crystallization

Hydrated salts are fascinating compounds that incorporate water molecules within their crystalline structures. This bound water is not merely trapped but is an integral part of the crystal's structure and stability. This water is termed as "water of crystallization." It doesn't act as free water. Instead, it participates in forming the crystal lattice, ensuring the proper shape and integrity of the salt.
For instance, in the compound \(\mathrm{CuCl}_{2} \cdot 2\mathrm{H}_{2}\mathrm{O}\), each formula unit is associated with two water molecules as its water of crystallization.

When hydrated salts like \(\mathrm{CuCl}_{2} \cdot 2\mathrm{H}_{2}\mathrm{O}\) are heated, the water molecules are released, often resulting in a visible change in the physical appearance like color change or powdery texture. This is primarily because the compound transitions from a hydrated to an anhydrous state, where the crystal structure modifies due to the absence of water molecules.

Chemical Composition Analysis

Understanding chemical composition analysis is crucial for determining the proportions of elements or compounds in a substance. This analysis becomes significant when dealing with compounds that lose part of their mass during processes like heating, which often happens with hydrated salts.
In our exercise, identifying the chemical composition involved inspecting the initial and final masses of the sample after the water was driven off. The chemical composition was examined by measuring:

• The initial mass of the sample (2.634 g, containing \(\mathrm{CuCl}_{2} \cdot 2\mathrm{H}_{2}\mathrm{O}\)) which includes both the \(\mathrm{CuCl}_{2}\) and the water molecules.


• The final mass (2.125 g), representing the sample after heating, thereby consisting mostly of anhydrous \(\mathrm{CuCl}_{2}\).

By subtracting the final from the initial mass, one calculates the mass of water lost, giving insight into how much of the hydrates constituted the original sample.

Stoichiometry in Chemical Calculations

Stoichiometry is a core pillar in chemistry that allows us to calculate reactant and product quantities in chemical reactions. It often involves balancing equations and using mole ratios, but in this exercise, the focus was on mass changes.
The mass percent calculation performed in the exercise is a perfect example of stoichiometry in action. We used stoichiometric principles to understand how much of the initial compound was present in its full, hydrated form. To find the mass percent of \(\mathrm{CuCl}_{2} \cdot 2\mathrm{H}_{2}\mathrm{O}\) in the original sample:

• The total mass of the hydrate in the original sample was divided by the initial total mass of the sample.
• This ratio was then multiplied by 100 to convert it into a percentage, providing the mass percent.

Stoichiometry allowed us to quantitatively analyze the chemical composition and verify that the entirety of the sample's initial mass was attributed to \(\mathrm{CuCl}_{2} \cdot 2\mathrm{H}_{2}\mathrm{O}\), resulting in a 100% mass purity. Understanding such calculations hones our capability to evaluate chemical purities effectively.