A chromium-containing compound has the formula FexCryO4 and is 28.59% oxygen by mass. Find the values of x and y.

Verified step by step guidance

Start by assuming 100 g of the compound FexCryO4, which means you have 28.59 g of oxygen.

Calculate the moles of oxygen using its molar mass (16.00 g/mol): \( \text{moles of O} = \frac{28.59 \text{ g}}{16.00 \text{ g/mol}} \).

Let the molar mass of the compound be \( M = x \times \text{molar mass of Fe} + y \times \text{molar mass of Cr} + 4 \times \text{molar mass of O} \).

Set up the equation for the mass percentage of oxygen: \( \frac{4 \times 16.00}{M} \times 100 = 28.59 \).

Solve the equation for \( M \), and use the known molar masses of Fe and Cr to find the values of \( x \) and \( y \).

Key Concepts

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

Molar Mass and Percent Composition

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole. Percent composition refers to the percentage by mass of each element in a compound. To find the values of x and y in the formula FexCryO4, one must use the given mass percentage of oxygen to determine the total molar mass of the compound and subsequently the contributions of iron (Fe) and chromium (Cr).

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the conservation of mass. In this context, it involves using the molar ratios of the elements in the compound to relate the amounts of Fe, Cr, and O. By setting up equations based on the formula FexCryO4 and the known mass percentage of oxygen, one can solve for the unknown variables x and y.

Empirical and Molecular Formulas

An empirical formula represents the simplest whole-number ratio of elements in a compound, while a molecular formula shows the actual number of atoms of each element in a molecule. In this problem, determining the values of x and y will help establish the empirical formula of the chromium-containing compound. Understanding the relationship between these formulas is crucial for accurately interpreting the results of the calculations.

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Empirical vs Molecular Formula

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