Final answer:
The formula for lead (II) chromite is A) PbCrO4. This is deduced by balancing the charges using the crisscross method, where the lead ion (Pb2+) and the chromate ion (CrO42-) both have charges that cancel each other out, resulting in a one-to-one ratio without any need to crisscross. The correct chemical formula is A) PbCrO4.
Step-by-step explanation:
The question relates to determining the correct formula for an ionic compound, specifically lead (II) chromite. To approach this question, understanding the crisscross method of writing chemical formulas is essential. The crisscross method is a technique used to write the formula of ionic compounds by balancing the charges of the ions.
In the case of lead (II) chromite, lead (II) has a +2 charge, denoted by the Roman numeral II, and the chromate ion (CrO4) has a -2 charge. Thus, when using the crisscross method, the charges balance out, and there's no need to crisscross the numbers as they are the same. The correct chemical formula is simply PbCrO4, where there is one lead ion for every chromate ion.
To answer the student's specific problem using the given data, where 207 g of lead combines with oxygen to form 239 g of lead oxide, we first find the mass of oxygen by subtracting the mass of lead from the mass of lead oxide: 239 g - 207 g = 32 g of oxygen. This corresponds to 2 moles of oxygen (32 g / 16 g/mol = 2 mol), given that the atomic mass of oxygen is 16.0 u. Since the starting mass of lead corresponds to 1 mole (207 g / 207 u/mol = 1 mol), the mole ratio of Pb to O is 1:2, suggesting a lead oxide formula of PbO2. This calculation is separate from chromite but is relevant information to help the student understand the mole concept and stoichiometry.