Question

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

Short answer

The empirical formulas for the given molecular formulas are: (a) \(\mathrm{AlBr}_{3}\) (b) \(\mathrm{CsH}_{10}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) (d) \(\mathrm{P}_{2} \mathrm{O}_{5}\) (e) \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\) (f) \(\mathrm{BNH}_{2}\)

Step-by-step solution

a) Al₂Br₆ Empirical Formula:

Divide each subscript by the greatest common divisor. In this case, the greatest common divisor for both Al and Br is 2. \(\mathrm{Al}_{2} \mathrm{Br}_{6}\) → \(\frac{\mathrm{Al}}{2} \frac{\mathrm{Br}}{3}\) → \(\mathrm{AlBr}_{3}\) The empirical formula for \(\mathrm{Al}_{2} \mathrm{Br}_{6}\) is \(\mathrm{AlBr}_{3}\).

b) CsH₁₀ Empirical Formula:

The greatest common divisor for Cs and H is 1. \(\mathrm{CsH}_{10}\) → \(\mathrm{CsH}_{10}\) The empirical formula for \(\mathrm{CsH}_{10}\) is the same as the molecular formula, \(\mathrm{CsH}_{10}\).

c) C₄H₈O₂ Empirical Formula:

Divide each subscript by the greatest common divisor, which is 2 for C, H, and O. \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\) → \(\frac{\mathrm{C}}{2} \frac{\mathrm{H}}{4} \frac{\mathrm{O}}{1}\) → \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) The empirical formula for \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\) is \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\).

d) P₄O₁₀ Empirical Formula:

Divide each subscript by the greatest common divisor, which is 2 for P and O. \(\mathrm{P}_{4} \mathrm{O}_{10}\) → \(\frac{\mathrm{P}}{2} \frac{\mathrm{O}}{5}\) → \(\mathrm{P}_{2} \mathrm{O}_{5}\) The empirical formula for \(\mathrm{P}_{4} \mathrm{O}_{10}\) is \(\mathrm{P}_{2} \mathrm{O}_{5}\).

e) C₆H₄Cl₂ Empirical Formula:

The greatest common divisor for C, H, and Cl is 1. \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\) → \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\) The empirical formula for \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\) is the same as the molecular formula, \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\).

f) B₃N₃H₆ Empirical Formula:

Divide each subscript by the greatest common divisor, which is 3 for B, N, and H. \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6}\) → \(\mathrm{B}_{1} \mathrm{N}_{1} \mathrm{H}_{2}\) → \(\mathrm{BNH}_{2}\) The empirical formula for \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6}\) is \(\mathrm{BNH}_{2}\).

Key concepts

Molecular Formula

The molecular formula of a compound provides the specific number of each type of atom present in a molecule. This formula is crucial in chemistry as it offers insight into the actual structure and composition of the compound. For example, the molecular formula for ethylene glycol is \( \mathrm{C}_4\mathrm{H}_8\mathrm{O}_2 \). This tells us that each molecule of ethylene glycol contains four carbon atoms, eight hydrogen atoms, and two oxygen atoms.

Understanding molecular formulas helps in predicting the types of chemical reactions the compound can participate in. It's like a detailed recipe that specifies the exact ingredients (atoms) and their quantities needed to create the compound.

• Specifies the kind and number of atoms in a molecule.
• Gives insight into the potential interactions and compound properties.
• Important for understanding chemical behavior and reactions.

Chemistry Problem-Solving

Approaching a chemistry problem requires a systematic method where each step builds upon the previous. Starting with analyzing the molecular formula is critical because it establishes a foundation to derive the empirical formula. Let’s consider how to solve such problems efficiently:

1. **Determine the Molecular Formula**: Identify given values for each element in the compound. This is the starting point, like knowing the ingredients before making a recipe.
2. **Find the Greatest Common Divisor (GCD)**: Determine the GCD of the subscripts in the molecular formula to identify how many numbers divide into each subscript evenly. This is akin to finding a common factor in math, easing simplification.
3. **Simplify using the GCD**: Divide each subscript by the GCD to reach the simplest ratio of atoms in the compound, which forms the empirical formula.

By applying these steps, students can consistently solve chemistry problems, developing problem-solving skills that improve with practice.

Chemical Subscripts Reduction

Reducing chemical subscripts is a key step in converting a molecular formula into an empirical formula. This involves simplifying the ratios of atoms to the smallest possible integers, providing a more straightforward representation of a compound.

Consider the example of \( \mathrm{C}_4\mathrm{H}_8\mathrm{O}_2 \):

• The molecular formula has subscripts 4, 8, and 2 for carbon, hydrogen, and oxygen, respectively.
• Here, the greatest common divisor is 2.
• Dividing each subscript by 2 gives \( \mathrm{C}_2\mathrm{H}_4\mathrm{O} \).

This reduction from \( \mathrm{C}_4\mathrm{H}_8\mathrm{O}_2 \) to \( \mathrm{C}_2\mathrm{H}_4\mathrm{O} \) provides the empirical formula, representing the simplest whole-number ratio of atoms in the compound.

This process may seem repetitive but is crucial for standardizing chemical formulas across different compounds. Whether you are working with complex biological molecules or simple inorganic compounds, mastering chemical subscripts reduction is always beneficial.