Question

What is the molecular formula of each compound? (a) Empirical formula \(\mathrm{CH}(\mathscr{M}=78.11 \mathrm{~g} / \mathrm{mol})\) (b) Empirical formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2}(\mathscr{M}=74.08 \mathrm{~g} / \mathrm{mol})\) (c) Empirical formula \(\mathrm{HgCl}(\mathscr{M}=472.1 \mathrm{~g} / \mathrm{mol})\) (d) Empirical formula \(\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{2}(\mathscr{A}=240.20 \mathrm{~g} / \mathrm{mol})\)

Short answer

(a): \(\mathrm{C}_{6} \mathrm{H}_{6}\), (b): \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2}\), (c): \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\), (d): \(\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{2} \)

Step-by-step solution

Understanding the Empirical and Molecular Formula Relationship

The molecular formula of a compound is a whole number multiple of its empirical formula. The molecular mass (or molar mass, \(\textsl{\( \mathscr{M} \)}\)) helps determine this multiple.

Calculating the Empirical Formula Mass

Determine the empirical formula mass (EFM) by summing the atomic masses of all atoms in the empirical formula. Use the periodic table for atomic masses.

Determining the Multiplication Factor

The multiplication factor is given by -Step 3.1- Calculating theoretical molecular mass of empirical formula by adding the individual atomic masses.-Step 3.2- Determining the ratio of the molar mass to the theoretical molecular mass\[(n = \frac{{\mathscr{M}}}{{\text{{EFM}}}})\] we get a ratio which needs to be multiplied with empirical formula to get Molecular formula.

Step 4a: Solve Part (a)

Given empirical formula \( \mathrm{CH} \) and \( \mathscr{M} = 78.11 \mathrm{~g} / \mathrm{mol} \):\( \text{EFM} = 12.01 \mathrm{~g} / \mathrm{mol} + 1.01 \mathrm{~g} / \mathrm{mol} = 13.02 \mathrm{~g} / \mathrm{mol} \)\( n = \frac{{78.11}}{{13.02}} \approx 6 \)Molecular formula = \((\mathrm{CH})_6 = \mathrm{C}_{6} \mathrm{H}_{6} \)

Step 5b: Solve Part (b)

Given empirical formula \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2} \) and \( \mathscr{M} = 74.08 \mathrm{~g} / \mathrm{mol} \):\( \text{EFM} = 3(12.01) + 6(1.01) + 2(16.00) \mathrm{~g} / \mathrm{mol} = 74.08 \mathrm{~g} / \mathrm{mol} \)\( n = \frac{{74.08}}{{74.08}} = 1 \)Molecular formula = \( \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2} \)

Step 6c: Solve Part (c)

Given empirical formula \( \mathrm{HgCl} \) and \( \mathscr{M} = 472.1 \mathrm{~g} / \mathrm{mol} \):\( \text{EFM} = 200.59 \mathrm{~g} / \mathrm{mol} + 35.45 \mathrm{~g} / \mathrm{mol} = 236.04 \mathrm{~g} / \mathrm{mol} \)\( n = \frac{{472.1}}{{236.04}} \approx 2 \)Molecular formula = \((\mathrm{HgCl})_2 = \mathrm{Hg}_{2} \mathrm{Cl}_{2} \)

Step 7d: Solve Part (d)

Given empirical formula \( \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{2} \) and \( \mathscr{A} = 240.20 \mathrm{~g} / \mathrm{mol} \):\( \text{EFM} = 7(12.01) + 4(1.01) + 2(16.00) \mathrm{~g} / \mathrm{mol} = 120.10 + 4.04 + 32.00 = 156.14 \mathrm{~g} / \mathrm{mol} \)\( n = \frac{{240.20}}{{156.14}} \approx 1.54 \)The multiplication factor results in a fractional number close to 1.54, which typically indicates reevaluation, ignoring experimental/sample errors molecular formula remains as the empirical formula.

Key concepts

Empirical Formula

The empirical formula represents the simplest ratio of elements in a compound. This formula only considers the relative number of each type of atom. For instance, in the empirical formula \(\text{CH}\), there is one carbon atom for each hydrogen atom, but it does not tell us the exact number of atoms in a molecule.

\[ \text{Empirical Formula Mass (EFM)} = \sum_{i} (\text{Number of atoms of element}_i \times \text{Atomic mass of element}_i) \] This means you sum up all the atomic masses of each element based on the empirical formula to get the EFM.

Commonly, to get these atomic masses, one refers to the periodic table, where the atomic mass of each element is listed.

For example, for the empirical formula \(\text{CH}\):

• \(\text{Carbon (C)}: 12.01 \, g/mol \)
• \(\text{Hydrogen (H)}: 1.01 \, g/mol \)

Therefore, the empirical formula mass for \(\text{CH}\) would be calculated as follows:
\[ \text{EFM} = 12.01 \times 1 + 1.01 \times 1 = 13.02 \, g/mol \]
This is the basis for determining the molecular formula, as it shows the simplest integer ratio of elements in the compound.

Molar Mass

The molar mass, denoted by \( \mathscr{M} \), is the mass of one mole of a given substance (a chemical element or chemical compound). It is essential for converting between grams and moles, which is pivotal for stoichiometric calculations. The unit for molar mass is grams per mole (g/mol).

For instance, the molar mass of water (H2O) is calculated by summing up the atomic masses of hydrogen and oxygen, each multiplied by their respective number of atoms in the formula:
\[ \mathscr{M}(\text{H}_2\text{O}) = 2(1.01 \text{ g/mol}) + 16.00 \text{ g/mol} = 18.02 \text{ g/mol} \] When dealing with a compound's molecular formula, the molar mass is an integral part in conjunction with the empirical formula to find the accurate molecular formula.

Consider an empirical formula \( \text{CH} \) with a molar mass \( \mathscr{M} = 78.11 \, g/mol \). Here, \( \mathscr{M} \) would help establish the exact molecular formula by comparing it to the empirical formula mass.

Atomic Mass

Atomic mass, in the context of chemistry, is the mass of a single atom, typically expressed in atomic mass units (amu). This concept is crucial because the atomic masses are used to derive the empirical and molecular formulas of compounds.

In the periodic table, each element is represented along with its atomic mass. For example:

• Carbon (C): 12.01 amu
• Hydrogen (H): 1.01 amu
• Oxygen (O): 16.00 amu

To find the empirical formula mass (EFM) of a compound, you sum the atomic masses of all constituent elements in the empirical formula. Let's take \( \text{HgCl} \) as an example:

• Mercury (Hg): 200.59 amu
• Chlorine (Cl): 35.45 amu

Hence, the EFM for \( \text{HgCl} \) is:
\[ \text{EFM} = 200.59 + 35.45 = 236.04 \text{ amu} \]
Understanding atomic masses thus helps in calculating formula masses which are further used to identify the molecular compositions.

Multiplication Factor

The multiplication factor helps in deriving the molecular formula from the empirical formula. It is a ratio that represents how many times the empirical unit needs to be multiplied to get the molecular formula.

This factor is computed as the ratio of the molar mass \(\mathscr{M}\) to the empirical formula mass (EFM):
\[ n = \frac{ \mathscr{M}}{ \text{EFM}} \] where:

• \(\mathscr{M}\) is the molar mass of the compound.
• EFM is the empirical formula mass.

The resulting number \( n \) dictates how many times the empirical formula must be multiplied to arrive at the molecular formula.

In the scenario where the empirical formula is \( \text{C}_3\text{H}_6 \text{O}_2 \), and given \( \mathscr{M} = 74.08 \text{ g/mol} \), we have:
\[ n = \frac{74.08}{74.08} = 1 \] Thus, the empirical formula is indeed the molecular formula: \( \text{C}_3 \text{H}_6 \text{O}_2 \).
Understanding how to calculate \( n \) is essential when dealing with molecular formulas.

Periodic Table

The periodic table is a chart where all the known elements are organized based on their atomic number, electron configuration, and recurring chemical properties. It is a fundamental tool in chemistry to determine atomic masses and understand how elements relate to one another.

Each element in the table is represented by its chemical symbol, atomic number, and atomic mass. For instance, Carbon is written as:

• C: \( 12.01 \text{ amu} \)

The periodic table is divided into:

• Groups: Vertical columns with elements that have similar properties.
• Periods: Horizontal rows where elements have the same number of electron shells.

Using the periodic table, one can accurately gather the atomic masses needed to compute empirical and molecular formulas.

For example, if calculating the empirical formula mass of \( \text{C}_7 \text{H}_4 \text{O}_2 \):

• Carbon (C): \( 12.01 \text{ g/mol} \)
• Hydrogen (H): \( 1.01 \text{ g/mol} \)
• Oxygen (O): \( 16.00 \text{ g/mol} \)

The periodic table thus serves as a critical reference for many chemical calculations.