Question

Calculate the molar mass of each hydrated compound. Note that the water of hydration is included in the molar mass. (See page \(85 .\) ) (a) \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\)

Short answer

(a) 290.83 g/mol, (b) 249.72 g/mol.

Step-by-step solution

Understand the Formula

For both compounds, recognize that they are hydrates, which means they include water molecules in their structure. For example, \( \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \cdot 6 \mathrm{H}_{2O} \) includes 6 water molecules attached to \( \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \).

Calculate Molar Mass of Each Component

(a) For \( \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \cdot 6 \mathrm{H}_{2O} \), calculate:- \( \mathrm{Ni} = 58.69 \text{ g/mol} \)- \( \mathrm{N} = 14.01 \text{ g/mol} \), thus for \( 2 \: \mathrm{N} = 28.02 \text{ g/mol} \)- \( \mathrm{O} = 16.00 \text{ g/mol} \), thus for \( 6 \: \mathrm{O} = 96.00 \text{ g/mol} \) - \( \mathrm{H}_{2}O = 18.02 \text{ g/mol} \), thus for \( 6 \: \mathrm{H}_{2}O = 108.12 \text{ g/mol} \)(b) For \( \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2O} \), calculate:- \( \mathrm{Cu} = 63.55 \text{ g/mol} \)- \( \mathrm{S} = 32.07 \text{ g/mol} \)- \( \mathrm{O} = 16.00 \text{ g/mol} \), thus for \( 4 \: \mathrm{O} = 64.00 \text{ g/mol} \)- \( \mathrm{H}_{2}O = 18.02 \text{ g/mol} \), thus for \( 5 \: \mathrm{H}_{2}O = 90.10 \text{ g/mol} \)

Add the Molar Masses Together

(a) \( \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \cdot 6 \mathrm{H}_{2O} \):Add the molar masses from Step 2:\[58.69 + 28.02 + 96.00 + 108.12 = 290.83 \text{ g/mol}\](b) \( \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2O} \):Add the molar masses from Step 2:\[63.55 + 32.07 + 64.00 + 90.10 = 249.72 \text{ g/mol}\]

Conclusion

The molar masses of the hydrated compounds are as follows:(a) \( \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \cdot 6 \mathrm{H}_{2O} \) is approximately \( 290.83 \text{ g/mol} \).(b) \( \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2O} \) is approximately \( 249.72 \text{ g/mol} \).

Key concepts

Hydrated Compounds

Hydrated compounds are special types of chemical compounds that include water molecules as a part of their structure. These water molecules are known as water of hydration and are integral to the compound's overall chemical formula and properties. For instance, when looking at the compound \( \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \), the "6 \( \mathrm{H}_2 \mathrm{O} \)" indicates that six water molecules are associated with each formula unit of nickel(II) nitrate.

• Hydrated compounds often form crystalline structures, where the water molecules help stabilize the structure.
• Typically, the water of hydration can be removed by heating the compound, turning the hydrated compound into its anhydrous form.
• The presence of water significantly affects the molar mass, which must always be included in mass calculations.

Understanding the role of water in such compounds provides insight into how chemical structures can alter due to hydration, affecting both physical properties and chemical behavior.

Stoichiometry

Stoichiometry refers to the calculation of reactants and products in chemical reactions. It helps you know how much of each element is present and what amounts of substances are involved in reactions. In the context of molar mass calculations for hydrated compounds, stoichiometry is crucial as it dictates how each part of a compound comes together.The stoichiometry of a compound tells us the exact number of atoms of each element in a compound. For example, in \( \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_2\mathrm{O} \), stoichiometry tells us we have:

• One atom of Copper (Cu)
• One atom of Sulfur (S)
• Four atoms of Oxygen (O) in the sulfate ion, plus additional oxygen from the water molecules
• And, as indicated, five molecules of water as part of the compound's composition.

The stoichiometry of each element is determined by the subscripts in the chemical formula, and understanding these ratios is essential for accurate chemical calculations.

Chemical Formulas

Chemical formulas are symbolic representations of chemical compounds that show the elements present and their proportions. They provide specific information about the structures of compounds and help in determining their properties.

• The formula \( \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \) indicates that nickel nitrate is a hydrate with six water molecules per formula unit.
• Each element in a chemical formula is represented by a unique chemical symbol, with subscripts indicating the number of atoms.
• For instance, in \( \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_2\mathrm{O} \), \( \mathrm{Cu} \) stands for copper, \( \mathrm{SO_4} \) is the sulfate group, and \( \mathrm{H}_2\mathrm{O} \) counts for the water of hydration.

Understanding chemical formulas is essential for interpreting chemical reactions and processes, allowing for the accurate calculation of molar masses, reaction yields, and chemical behavior.

Molecular Weight

Molecular weight, or molar mass, is the weight of a given substance’s molecules measured in grams per mole (g/mol). It's a critical concept when studying chemical reactions and compound compositions.To calculate the molar mass of a compound, you sum the atomic weights of each element present in the molecular formula, multiplied by their respective stoichiometric coefficients. For example, when calculating the molar mass of \( \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} \cdot 6 \mathrm{H}_{2O}\), you would:

• Add up the atomic masses of nickel (Ni), nitrogen (N), and oxygen (O) as provided by the periodic table.
• Include the mass contributed by the water molecules \( \mathrm{H}_2\mathrm{O} \), multiplying by the number of water molecules present.

The accurate computation of molecular weight is fundamental in stoichiometric calculations, helping you figure out how much of each reactant is needed or product is formed in chemical reactions. This allows for effective and precise chemical experimentation and industry application.