Problem 49 - Chemistry The Central Science

Question

Write the empirical formula corresponding to each of the following molecular formulas: (a) \(\mathrm{Al}_{2} \mathrm{Br}_{6},(\mathbf{b}) \mathrm{C}_{8} \mathrm{H}_{10}\) (c) \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2},(\mathbf{d}) \mathrm{P}_{4} \mathrm{O}_{10},(\mathbf{e}) \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2},(\mathbf{f}) \mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6} .\)

Step-by-Step Solution

Answer

The empirical formulas corresponding to the given molecular formulas are: (a) AlBr3 (b) C4H5 (c) C2H4O (d) P2O5 (e) C3H2Cl (f) BNH2

Step 1: a) Al2Br6

First, let's analyze the given molecular formula Al2Br6. The subscripts are 2 and 6. The greatest common divisor of 2 and 6 is 2. We will divide both subscripts by this common divisor: Al\(_{2\div2}\)Br\(_{6\div2}\) Which simplifies to: AlBr3 The empirical formula for Al2Br6 is AlBr3.

Step 2: b) C8H10

Now, let's analyze the given molecular formula C8H10. The subscripts are 8 and 10. The greatest common divisor of 8 and 10 is 2. We will divide both subscripts by this common divisor: C\(_{8\div2}\)H\(_{10\div2}\) Which simplifies to: C4H5 The empirical formula for C8H10 is C4H5.

Step 3: c) C4H8O2

Let's analyze the given molecular formula C4H8O2. The subscripts are 4, 8, and 2. The greatest common divisor of 4, 8, and 2 is 2. We will divide all subscripts by this common divisor: C\(_{4\div2}\)H\(_{8\div2}\)O\(_{2\div2}\) Which simplifies to: C2H4O The empirical formula for C4H8O2 is C2H4O.

Step 4: d) P4O10

Let's analyze the given molecular formula P4O10. The subscripts are 4 and 10. The greatest common divisor of 4 and 10 is 2. We will divide both subscripts by this common divisor: P\(_{4\div2}\)O\(_{10\div2}\) Which simplifies to: P2O5 The empirical formula for P4O10 is P2O5.

Step 5: e) C6H4Cl2

Let's analyze the given molecular formula C6H4Cl2. The subscripts are 6, 4, and 2. The greatest common divisor of 6, 4, and 2 is 2. We will divide all subscripts by this common divisor: C\(_{6\div2}\)H\(_{4\div2}\)Cl\(_{2\div2}\) Which simplifies to: C3H2Cl The empirical formula for C6H4Cl2 is C3H2Cl.

Step 6: f) B3N3H6

Let's analyze the given molecular formula B3N3H6. The subscripts are 3, 3, and 6. The greatest common divisor of 3, 3, and 6 is 3. We will divide all subscripts by this common divisor: B\(_{3\div3}\)N\(_{3\div3}\)H\(_{6\div3}\) Which simplifies to: BNH2 The empirical formula for B3N3H6 is BNH2.

Key Concepts

Molecular Formula Conversion

The process of determining an empirical formula from a molecular formula involves identifying the simplest integer ratio of atoms present within a compound. To achieve this, we look for the greatest common divisor (GCD) among the subscripts in the molecular formula and divide each subscript by this number.

For instance, consider the molecular formula \( C_8H_{10} \). With subscripts of 8 and 10, we find the GCD to be 2. By dividing each subscript by 2, we get a simplified empirical formula of \( C_4H_5 \). It's important to note that empirical formulas represent the ratio of elements in a compound, whereas the molecular formula details the exact number of atoms of each element in one molecule of the compound.

When converting molecular formulas to empirical formulas, always ensure to reduce the subscripts to the lowest possible whole numbers to accurately represent the molecule's basic chemical identity.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of the mole, allowing chemists to predict the amounts of substances consumed and formed in a chemical reaction.

In the context of empirical and molecular formulas, stoichiometry plays a crucial role. When given an empirical formula, one can utilize stoichiometry to determine the mass of each element in a given quantity of compound or to scale up to the molecular formula if the molar mass is known. For example, once the empirical formula of \( C_2H_4O \) is obtained from the molecular formula \( C_4H_8O_2 \) as shown in the textbook solution, one could use stoichiometry to find out how many grams of carbon, hydrogen, and oxygen are present in a \( 10 \text{g} \) sample of the compound.

Stoichiometry is fundamental in ensuring that chemical equations are balanced and obeys the law of conservation of mass, enabling precise calculations in chemical reactions and conversions.

Chemical Composition

Understanding chemical composition is pivotal in grasping the basic concepts of chemistry like empirical and molecular formulas. Chemical composition refers to the arrangement, type, and ratio of atoms in molecules of chemical substances. It determines the physical and chemical properties of a substance and dictates how it will react with other substances.

The empirical formula provides a simplistic view of a compound's chemical composition by denoting the ratio of the elements present. For instance, in the solution for \( C_6H_4Cl_2 \), by using the GCD method, we deduce the empirical formula \( C_3H_2Cl \), which tells us the simplest ratio of carbon to hydrogen to chlorine is 3:2:1. However, it doesn't give information about how atoms are bonded or its structure. The molecular formula, on the other hand, gives more detail on the actual number of atoms as it appears in a single molecule.

Understanding both empirical and molecular formulas thus provides a comprehensive picture of a compound's chemical makeup, allowing for predictive uses in reaction outcomes, product formation, and in understanding the molecular structure of compounds.