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Ch.3 - Molecules and Compounds
Tro - Chemistry: A Molecular Approach 5th Edition
Tro5th EditionChemistry: A Molecular ApproachISBN: 9780134874371Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 94

A 45.2-mg sample of phosphorus reacts with selenium to form 131.6 mg of the selenide. Determine the empirical formula of phosphorus selenide.

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1
Determine the mass of selenium in the compound by subtracting the mass of phosphorus from the total mass of the selenide: 131.6 \(\text{ mg}\) - 45.2 \(\text{ mg}\).
Convert the mass of phosphorus (45.2 mg) to moles using its molar mass (30.97 \(\text{ g/mol}\)).
Convert the mass of selenium (calculated in step 1) to moles using its molar mass (78.96 \(\text{ g/mol}\)).
Calculate the mole ratio of phosphorus to selenium by dividing the moles of each element by the smallest number of moles calculated in step 2 or 3.
Use the mole ratio to determine the empirical formula by expressing the ratio as the smallest whole numbers.

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Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of the elements present in that compound. It is determined by analyzing the relative amounts of each element in a sample. For example, if a compound contains 1 atom of phosphorus for every 2 atoms of selenium, its empirical formula would be PSe2.
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Molar Mass and Conversion

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To determine the empirical formula, it is essential to convert the mass of each element in the sample to moles using their respective molar masses. This conversion allows for the calculation of the ratio of the elements in the compound.
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Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It involves using balanced chemical equations to determine the amounts of substances consumed and produced. In this case, stoichiometry helps in calculating the moles of phosphorus and selenium to find their ratio in the empirical formula.
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