Question

Find the empirical formula of each of the following compounds: (a) \(0.039 \mathrm{~mol}\) of iron atoms combined with \(0.052 \mathrm{~mol}\) of oxygen atoms; (b) \(0.903 \mathrm{~g}\) of phosphorus combined with \(6.99 \mathrm{~g}\) of bromine; (c) a hydrocarbon with 79.9 mass \% carbon

Short answer

(a) Fe₃O₄ (b) PBr₃ (c) CH₃

Step-by-step solution

Understand the Problem

We need to find the empirical formula for compounds given in parts a, b, and c. The empirical formula is the simplest whole-number ratio of atoms in a compound.

Analyze Part (a)

Given: 0.039 mol of iron (Fe) and 0.052 mol of oxygen (O).

Calculate the Mole Ratio for Part (a)

Divide each amount of moles by the smallest amount of moles to find the simplest ratio.For Fe: 0.039 mol / 0.039 mol = 1For O: 0.052 mol / 0.039 mol ≈ 1.33So the ratio becomes Fe:O as 1:1.33. Multiply by 3 to get whole numbers: 3:4.Empirical formula: Fe₃O₄

Analyze Part (b)

Given: 0.903 g of phosphorus (P) and 6.99 g of bromine (Br).

Convert Mass to Moles for Part (b)

Use molar masses to convert grams to moles.Molar mass of P: 30.97 g/molMolar mass of Br: 79.90 g/molMoles of P: 0.903 g / 30.97 g/mol ≈ 0.0292 molMoles of Br: 6.99 g / 79.90 g/mol ≈ 0.0875 mol

Calculate the Mole Ratio for Part (b)

Divide each amount of moles by the smallest amount of moles.For P: 0.0292 mol / 0.0292 mol = 1For Br: 0.0875 mol / 0.0292 mol ≈ 3Empirical formula: PBr₃

Analyze Part (c)

Given: Hydrocarbon with 79.9% carbon.Let the total mass be 100 g for simplicity.Mass of carbon (C) = 79.9 gMass of hydrogen (H) = 100 g - 79.9 g = 20.1 g

Convert Mass to Moles for Part (c)

Use molar masses to convert grams to moles.Molar mass of C: 12.01 g/molMolar mass of H: 1.01 g/molMoles of C: 79.9 g / 12.01 g/mol ≈ 6.66 molMoles of H: 20.1 g / 1.01 g/mol ≈ 19.9 mol

Calculate the Mole Ratio for Part (c)

Divide each amount of moles by the smallest amount of moles.For C: 6.66 mol / 6.66 mol = 1For H: 19.9 mol / 6.66 mol ≈ 3Empirical formula: CH₃

Key concepts

Mole Ratio

The mole ratio is crucial in finding the empirical formula. It represents the simplest whole-number ratio of moles of elements in a compound. To determine it, start by calculating the moles of each element present. Then, divide each mole value by the smallest mole quantity among them. This helps in simplifying the ratio.
If the ratios are not whole numbers, multiply by an appropriate factor to convert them into whole numbers. For example, in step (a) of the exercise, the mole ratio of iron to oxygen was found to be approximately 1:1.33. Multiplying both by 3 gave 3:4, leading to the empirical formula Fe₃O₄.

Molar Mass

Molar mass is the mass of one mole of a substance, usually in g/mol. It's essential for converting mass to moles. To find the molar mass of an element, refer to the periodic table. For a compound, sum the molar masses of its constituent elements.
For example, in step (b) of the exercise, the molar mass of phosphorus (P) is 30.97 g/mol and that of bromine (Br) is 79.90 g/mol. These values were used to convert the given masses (0.903 g P and 6.99 g Br) to moles, aiding the determination of the mole ratio and, subsequently, the empirical formula PBr₃.

Chemical Composition

Chemical composition refers to the identities and amounts of elements in a compound. It's expressed in terms of mass percentages, which can be converted to moles for empirical formula determination.
In part (c) of the exercise, a hydrocarbon with 79.9% carbon means, in a 100 g sample, there are 79.9 g of carbon and 20.1 g of hydrogen. Converting these masses to moles allows you to find the simplest ratio of C to H, resulting in the empirical formula CH₃.

Stoichiometry

Stoichiometry is the study of quantitative relationships in chemical reactions. It's used to determine how much of each element is involved in forming compounds.
When calculating empirical formulas, stoichiometry helps in balancing the moles of elements to form a simplified ratio.
For instance, in all parts of the exercise, stoichiometric principles are applied to convert masses to moles, determine their ratios, and express the empirical formulas accordingly. Mastering stoichiometry is fundamental for accurate chemical analysis and formula determination.