Problem 45 - Chemistry The Central Science

Question

Give the empirical formula of each of the following compounds if a sample contains (a) 0.0130 mol \(\mathrm{C}, 0.0390 \mathrm{mol}\) \(\mathrm{H},\) and \(0.0065 \mathrm{mol} \mathrm{O} ;\) (b) \(11.66 \mathrm{g}\) iron and 5.01 \(\mathrm{g}\) oxygen; (c) \(40.0 \% \mathrm{C}, 6.7 \% \mathrm{H},\) and 53.3\(\%\) o by mass.

Step-by-Step Solution

Answer

The empirical formulas for the given compounds are: (a) \(C_2H_6O\) (b) \(Fe_2O_3\) (c) \(CHO\)

Step 1: Compound (a)

Given moles of C, H, and O are 0.0130 mol, 0.0390 mol, and 0.0065 mol, respectively. Step 1: Calculate the ratio of moles: Since the given moles are already provided, we find the smallest value among (0.0130, 0.0390, 0.0065), which is 0.0065 mol. Step 2: Divide each element's moles by the smallest moles value: C: \(\frac{0.0130}{0.0065} = 2\) H: \(\frac{0.0390}{0.0065} = 6\) O: \(\frac{0.0065}{0.0065} = 1\) Empirical formula for Compound (a) is \(C_2H_6O\).

Step 2: Compound (b)

Given mass of iron (Fe) is 11.66 g and oxygen (O) is 5.01 g. Step 1: Convert mass to moles: Fe: \(\frac{11.66\,\text{g}}{55.85\,\text{g/mol}} = 0.2086\,\text{mol}\) O: \(\frac{5.01\,\text{g}}{16.00\,\text{g/mol}} = 0.3131\,\text{mol}\) Step 2: Divide each element's moles by the smallest moles value: Find the smallest value among (0.2086, 0.3131), which is 0.2086 mol. Fe: \(\frac{0.2086}{0.2086} = 1\) O: \(\frac{0.3131}{0.2086} ≈ 1.5\) Step 3: Multiply each ratio by a whole number (in this case, 2) to make it a whole number: Fe: \(1 × 2 = 2\) O: \(1.5 × 2 = 3\) Empirical formula for Compound (b) is \(Fe_2O_3\).

Step 3: Compound (c)

Given mass percentage of C, H, and O are 40.0%, 6.7%, and 53.3% respectively. Assume the total mass of the compound is 100 g. Step 1: Convert mass percentage to moles: C: \(\frac{40.0\,\text{g}}{12.01\,\text{g/mol}} = 3.331\,\text{mol}\) H: \(\frac{6.7\,\text{g}}{1.008\,\text{g/mol}} = 6.646\,\text{mol}\) O: \(\frac{53.3\,\text{g}}{16.00\,\text{g/mol}} = 3.331\,\text{mol}\) Step 2: Divide each element's moles by the smallest moles value: Find the smallest value among (3.331, 6.646, 3.331), which is 3.331 mol. C: \(\frac{3.331}{3.331} = 1\) H: \(\frac{6.646}{3.331} ≈ 2\) O: \(\frac{3.331}{3.331} = 1\) Empirical formula for Compound (c) is \(CHO_1\), or simply \(CHO\).

Key Concepts

Chemical Composition

Understanding chemical composition is essential for any student beginning their journey into the world of chemistry. It refers to the identification and quantification of the elements that make up a chemical compound. The chemical composition of a substance governs its properties, reactions, and behavior during chemical processes.

Empirical formulas play a critical role in presenting chemical composition in a simplified manner. They represent the simplest ratio of the elements within a compound without indicating the actual numbers or arrangement of atoms. For example, the empirical formula for glucose, C₆H₁₂O₆, is CH₂O, which shows the simplest whole number ratio between carbon, hydrogen, and oxygen atoms in the molecule.

To determine the empirical formula from an experimental data set, one needs to know the amount of each element in a compound, either in moles or in terms of mass. The steps involve calculating the relative number of moles of each element and then expressing these numbers in the smallest whole number ratio. This ratio reflects the proportional composition of the compound and is a fingerprint of its chemical identity.

Mole Concept

The mole concept is a foundational cornerstone of chemistry that allows chemists to count atoms, molecules, and ions in terms of quantity. One mole is Avogadro's number (\(6.022 \times 10^{23}\)) of particles, which can be atoms, molecules, or ions. It provides a bridge between the atomic and macroscopic worlds.

For instance, if we say we have 1 mole of carbon atoms, we mean that there are approximately \(6.022 \times 10^{23}\) carbon atoms present. The mole concept enables us to relate the mass of substances to the number of entities they contain and to participate in stoichiometry calculations which are central to chemical reactions and quantitative analysis.

Importance of the Mole Concept

The concept is critical in determining the chemical composition of a compound since it allows us to measure amounts of a substance in a scientifically consistent way. By using the molar masses of elements – the mass of one mole of a substance – chemists can calculate the number of moles present in a given mass of an element, and vice versa, which is the key process in determining empirical formulas.

Mass-to-Mole Conversion

Mass-to-mole conversion is an essential skill in chemistry that involves using the molar mass of an element or compound to convert between the mass of a substance and the number of moles. The molar mass serves as the conversion factor and is numerically equivalent to the atomic or molecular weight in grams per mole.

To convert mass to moles, the mass of the substance is divided by the molar mass. This step is crucial when determining the empirical formula of a compound from its mass composition, as seen in the above exercises. For example, if 5.01 grams of oxygen are present, to find the number of moles, you would divide by oxygen's molar mass (\(16.00\text{ g/mol}\)), resulting in approximately 0.3131 moles of oxygen.

Why Conversion Matters

This conversion is vital because reactions in chemistry typically involve combining mole ratios of substances, not mass ratios. Therefore, understanding how to convert mass to moles allows students to bridge the gap between a compound's tangible mass and its mole ratio as required for stoichiometric calculations, a foundation for material quantification and understanding the chemistry behind reactions.