(b) How many moles of chloride ions are in 0.2550 g of aluminum chloride?

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Determine the molar mass of aluminum chloride (AlCl3). Aluminum (Al) has a molar mass of approximately 26.98 g/mol, and chlorine (Cl) has a molar mass of approximately 35.45 g/mol. The molar mass of AlCl3 is calculated as follows: Molar mass of AlCl3 = (1 \(\times\) 26.98) + (3 \(\times\) 35.45) g/mol.

Calculate the number of moles of aluminum chloride (AlCl3) in 0.2550 g using the formula: Number of moles = mass (g) / molar mass (g/mol).

Recognize that each formula unit of AlCl3 contains three chloride ions. Therefore, the number of moles of chloride ions is three times the number of moles of AlCl3.

Multiply the number of moles of AlCl3 by 3 to find the total number of moles of chloride ions.

The result from the previous step gives you the number of moles of chloride ions present in 0.2550 g of aluminum chloride.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a molecule. For aluminum chloride (AlCl3), the molar mass is approximately 133.33 g/mol, which is essential for converting grams of the compound to moles.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. In the case of aluminum chloride, each formula unit contains three chloride ions (Cl-). Understanding stoichiometry allows us to determine how many moles of chloride ions are present based on the moles of aluminum chloride.

Conversion of Grams to Moles

To find the number of moles from a given mass, the formula used is: moles = mass (g) / molar mass (g/mol). This conversion is crucial for solving the problem, as it allows us to calculate the moles of aluminum chloride first, which can then be used to find the moles of chloride ions through stoichiometric relationships.

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