Question

Determine the water of hydration for the following hydrates and write the chemical formula: (a) \(\mathrm{SrCl}_{2} \cdot \mathrm{XH}_{2} \mathrm{O}\) is found to contain \(18.5 \%\) water. (b) \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \cdot \mathrm{XH}_{2} \mathrm{O}\) is found to contain \(37.2 \%\) water. (c) \(\mathrm{CoSO}_{4} \cdot \mathrm{XH}_{2} \mathrm{O}\) is found to contain \(10.4 \%\) water. (d) \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} \cdot \mathrm{XH}_{2} \mathrm{O}\) is found to contain \(30.9 \%\) water.

Short answer

(a) \({\rm SrCl}_2\cdot6{\rm H}_2{\rm O}\), (b) \( {\rm Ni(NO}_3)_{2}{\cdot}6{\rm H}_2{\rm O} \), (c) \({\rm CoSO}_4{\cdot}1{\rm H}_2{\rm O} \), (d) \( {\rm Na}_2{\rm B}_4{\rm O}_7{\cdot}10{\rm H}_2{\rm O} \).

Step-by-step solution

Understand the Relationship

In a hydrate such as \( \, {\rm SrCl}_{2} \cdot {\rm XH}_{2}{\rm O} \, \), the percentage of water is given and is used to determine \( X \), which represents the number of water molecules (\( {\rm H}_{2}{\rm O} \)) bound with one formula unit of the compound. The formula is generally \( \text{hydrate} \cdot {\rm XH}_2{\rm O} \).

Calculate Molar Mass of Anhydrous Compound

Calculate the molar mass of the anhydrous part of each compound, ignoring the water portion.(a) \( {\rm SrCl}_2 \): \( 87.62 \,{\rm g/mol} + 2 \times 35.45 \,{\rm g/mol} = 158.52 \,{\rm g/mol} \). (b) \( {\rm Ni}{\rm (NO}_{3}{\rm )}_{2} \): \( 58.69 \,{\rm g/mol} + 2 \times (14.01 \,{\rm g/mol} + 3 \times 16.00 \,{\rm g/mol}) = 182.71 \,{\rm g/mol} \). (c) \( {\rm CoSO}_{4} \): \( 58.93 \,{\rm g/mol} + 32.07 \,{\rm g/mol} + 4 \times 16.00 \,{\rm g/mol} = 154.99 \,{\rm g/mol} \). (d) \( {\rm Na}_{2}{\rm B}_{4}{\rm O}_{7} \): \( 2 \times 22.99 \,{\rm g/mol} + 4 \times 10.81 \,{\rm g/mol} + 7 \times 16.00 \,{\rm g/mol} = 201.22 \,{\rm g/mol} \).

Calculate Mass of Water per Formula Unit

For each compound, calculate the mass of water per mole of hydrate:(a) \( 18.5\% \) of \( {\rm SrCl}_2 \cdot {\rm XH}_2{\rm O} = 0.185 \times {\rm M}_{\rm hydrate} \)(b) \( 37.2\% \) of \( {\rm Ni(N}{\rm O}_{3})_{2} \cdot {\rm XH}_2{\rm O} = 0.372 \times {\rm M}_{\rm hydrate} \)(c) \( 10.4\% \) of \( {\rm CoSO}_4 \cdot {\rm XH}_2{\rm O} = 0.104 \times {\rm M}_{\rm hydrate} \)(d) \( 30.9\% \) of \( {\rm Na}_2{\rm B}_4{\rm O}_7 \cdot {\rm XH}_2{\rm O} = 0.309 \times {\rm M}_{\rm hydrate} \)

Setup Equation and Solve for X

For each hydrate, solve for \( X \), the number of water molecules:(a) \( {\rm Y} = 18.5 \; \%, \; M_{\rm anhydrous} = 158.52 \,{\rm g/mol} \), eq.: \( \frac{{18.5 \cdot {M}_{{\rm hydrate}}}}{{18.015}} = X \). Solve for the hydrate mass: \( {M}_{{\rm hydrate}} = \frac{{M}_{{\rm anhydrous}}}{{1 - {\rm Y}}} \). Substitute in \( X = \frac{{0.185 \cdot {M}_{{\rm hydrate}}}}{18.015} \).(b) Similar method, for \( 37.2\% \): Solve for \( X \) using \( 0.372 = \frac{{18.015 \cdot X}}{{{M}_{\rm Ni(NO}_3)_{2}} + 18.015 \cdot X} \).(c) For \( 10.4\% \): \( 0.104 = \frac{{18.015 \cdot X}}{{{M}_{{\rm CoSO}_{4}} + 18.015 \cdot X}} \).(d) For \( 30.9\% \): \( 0.309 = \frac{{18.015 \cdot X}}{{{M}_{{\rm Na}_2{\rm B}_4{\rm O}_7} + 18.015 \cdot X}} \).

Compute the Value of X

Compute the number of water molecules for each compound.(a) For \( {\rm SrCl}_2{\cdot}9{\rm H}_2{\rm O} \): Approximately, \( X \approx 6 \).(b) For \( {\rm Ni(NO}_3)_{2}{\cdot}6{\rm H}_2{\rm O} \): \( X = 6 \).(c) For \( {\rm CoSO}_4{\cdot}7{\rm H}_2{\rm O} \): \( X \approx 1 \).(d) For \( {\rm Na}_2{\rm B}_4{\rm O}_7{\cdot}10{\rm H}_2{\rm O} \): \( X \approx 10 \).

Write the Chemical Formula with Hydrates

Write the chemical formula for each compound with the correct number of water molecules:(a) \({\rm SrCl}_2\cdot6{\rm H}_2{\rm O}\)(b) \( {\rm Ni(NO}_3)_{2}{\cdot}6{\rm H}_2{\rm O} \)(c) \({\rm CoSO}_4{\cdot}1{\rm H}_2{\rm O} \)(d) \( {\rm Na}_2{\rm B}_4{\rm O}_7{\cdot}10{\rm H}_2{\rm O} \)

Key concepts

Molar Mass Calculation

Understanding molar mass is fundamental in chemistry, especially when working with hydrates. The molar mass is the sum of the atomic masses of all atoms in a molecule. For instance, to determine the molar mass of an anhydrous compound like \[ \text{SrCl}_2 \], you add up the atomic masses of Strontium (Sr) and Chlorine (Cl).

• Sr has an atomic mass of approximately 87.62 g/mol.
• Each Cl has an atomic mass of about 35.45 g/mol, and since there are two Cl atoms, you multiply 35.45 g/mol by 2.

So, the total molar mass of \[ \text{SrCl}_2 \] would be \( 87.62 \, \text{g/mol} + 2 \times 35.45 \, \text{g/mol} = 158.52 \, \text{g/mol} \).
This method is applied similarly to other compounds like \[ \text{Ni(NO}_3)_2 \], where you calculate the anhydrous mass by including the atomic masses of Nickel (Ni), Nitrogen (N), and Oxygen (O). Understanding how to calculate molar mass is crucial for determining the chemical formula of hydrates accurately.

Chemical Formula Determination

After calculating the molar mass, determining the chemical formula of a hydrate involves knowing the proportion of water in the compound. When presented with a problem like \( \text{SrCl}_2 \cdot \text{XH}_2\text{O} \), where you know the percentage of water, you can set up an equation relating the anhydrous molar mass and the water percentage to find how many water molecules are associated per formula unit of the compound.
For example, if a compound is 18.5% water, you can find the molar mass of the whole hydrate (\( M_{\text{hydrate}} \)) using the anhydrous molar mass and solve for \( X \), representing the number of water molecules. This involves a little algebraic manipulation and the use of the water molar mass, which is 18.015 g/mol, to solve for \( X \).
Setting up these equations correctly is vital to determine the correct chemical formula of the hydrate, like \( \text{SrCl}_2\cdot6\text{H}_2\text{O} \). This indicates six water molecules per one formula unit of \( \text{SrCl}_2 \).

Water of Hydration

Water of hydration refers to water molecules that are chemically bound within a crystal structure of a compound. In hydrate chemistry, these water molecules are essential in maintaining the structure of the compound and are indicated in the chemical formula by a dot. For \( \text{SrCl}_2 \cdot \text{6H}_2\text{O} \), this means there are six water molecules associated with one unit of \( \text{SrCl}_2 \).

To determine how much water is present, chemists measure the percentage of water in the compound. Given that hydrates can easily lose their water content by heating, you can calculate the weight difference before and after heating to find how much water was lost, which helps in confirming the number of water molecules in the actual formula.

• This involves knowing the starting mass of the hydrate, % water content, and the total mass of the compound.
• With these factors in mind, you can establish the general formula used in calculations.

Understanding water of hydration is crucial since it directly affects the properties of the hydrate, such as stability, color, and solubility.