8.4

In previous sections, the focus was on determining the empirical formula, which reflects the simplest whole-number ratio of elements within a compound. This foundational understanding is critical for grasping how compounds are structured on a molecular level.
Section 8.4 focuses on determining the molecular formula, which delineates the actual number of each type of atom present in a molecule of the compound. Unlike the empirical formula, the molecular formula does not reduce ratios and provides insight into the composition of the actual molecule.
To solve problems for both the empirical formula and the molecular formula, an essential piece of information is required: the molar mass of the molecular compound. This measurement is vital because it allows for the conversion between the empirical formula and the molecular formula by determining how many times the empirical formula can fit into the molecular compound based on mass.
It is imperative that the empirical formula must always be determined first before the molecular formula can be accurately found. This ensures a systematic approach towards understanding the molecular structure of the substance in question and allows for accurate representation of its chemical properties.

Procedure for Determining Molecular Formulas

Step 1: Calculate the molar mass of the empirical formula.
- To achieve this, add the molar masses of each atom present in the empirical formula. This calculation is based on the sum of their individual atomic weights, which can be obtained from the periodic table. It is vital to use precise atomic weights to ensure accuracy in the final calculation.
Step 2: Obtain the whole-number multiplier (n).
- This is done by dividing the molar mass of the molecular compound (which is provided in the problem) by the calculated molar mass of the empirical formula. This step requires careful attention to detail to ensure accurate division.
- $\frac{\text{Molar Mass of Molecular Formula}}{\text{Molar Mass of Empirical Formula}} = n$
Step 3: Calculate the molecular formula.
- Multiply the subscripts of the empirical formula by the whole number $n$ found in Step 2. This mathematical operation represents the actual formula of the compound and reflects its true molecular identity. It is this formula that chemists utilize when discussing the compound in more advanced chemistry concepts and applications.

Example 1: Transitioning from Empirical to Molecular Formula

Problem: What is the molecular formula for a compound whose empirical formula is $C_5H_4$ if the molar mass of the compound is $128.16\,g/mol$ ?
Step 1: Sum the molar mass of the empirical formula ( $C_5H_4$ ).
- Carbon ( $C$ ): $12.01 \times 5 = 60.05\,g/mol$
- Hydrogen ( $H$ ): $1.01 \times 4 = 4.04\,g/mol$
- Total empirical molar mass: $60.05 + 4.04 = 64.09\,g/mol$
Step 2: Divide the given molecular molar mass by the empirical molar mass.
- $\frac{128.16\,g/mol}{64.08\,g/mol} = 2$
Step 3: Multiply the subscripts.
- $C_5H_4$ multiplied by 2 becomes $C_{10}H_8$ , representing the molecular formula for this compound, which provides further insight into its chemical behavior.

Example 2: Boron-Hydrogen Compound

Problem: What is the molecular formula for a compound whose empirical formula is $BH_3$ if the molar mass is $27.7\,g/mol$ ?
Step 1: Sum the molar mass of the empirical formula ( $BH_3$ ).
- Boron ( $B$ ): $10.81\,g/mol$
- Hydrogen ( $H$ ): $1.01 \times 3 = 3.03\,g/mol$
- Total empirical molar mass: $10.81 + 3.03 = 13.84\,g/mol$
Step 2: Divide the given molecular molar mass by the empirical molar mass.
- $\frac{27.7\,g/mol}{13.84\,g/mol} = 2$
Step 3: Multiply the subscripts.
- $BH_3$ multiplied by 2 becomes $B_2H_6$ , indicating a compound where two boron atoms are present with six hydrogen atoms, revealing its complete molecular identity.

Practice Problem 1: Hydrogen and Oxygen

Question: What is the molecular formula of a compound that has a molar mass of $34\,g/mol$ and an empirical formula of $HO$ ?
Step 1: Find the molar mass of the empirical formula ( $HO$ ).
- Hydrogen ( $H$ ): $1.01\,g/mol$
- Oxygen ( $O$ ): $16.00\,g/mol$
- Total: $1.01 + 16.00 = 17.01\,g/mol$
Step 2: Divide the molecular molar mass by the empirical molar mass.
- $\frac{34\,g/mol}{17.01\,g/mol} = 2$
Step 3: Multiply the subscripts.
- $HO$ multiplied by 2 becomes $H_2O_2$ , which is a well-known chemical known as hydrogen peroxide, revealing its complete molecular structure.

Practice Problem 2: Full Determination From Percent Composition

Question: A compound is composed of $54.52\%\,C$ , $9.17\%\,H$ , and $36.31\%\,O$ . The molar mass of the molecular formula is $88.1\,g/mol$ . Determine the empirical and molecular formulas.
Phase 1: Determine the Empirical Formula.
- Assume a 100g sample: This allows the conversion of percentages directly to grams, simplifying the calculations.
- $54.52\,g\,C$
- $9.17\,g\,H$
- $36.31\,g\,O$
- Convert grams to moles:
  - Carbon: $54.52\,g\,C \times \frac{1\,mole\,C}{12.01\,g\,C} = 4.540\,mole\,C$
  - Hydrogen: $9.17\,g\,H \times \frac{1\,mole\,H}{1.01\,g\,H} = 9.08\,mole\,H$
  - Oxygen: $36.31\,g\,O \times \frac{1\,mole\,O}{16.00\,g\,O} = 2.269\,mole\,O$
- Divide by the smallest mole value to find the ratio:
  - $C = \frac{4.540}{2.269} \approx 2$
  - $H = \frac{9.08}{2.269} \approx 4$
  - $O = \frac{2.269}{2.269} = 1$
- Empirical Formula: $C_2H_4O$ , showcasing a more simplified representation of the compound's constituents.
Phase 2: Determine the Molecular Formula.
- Sum the molar mass of the empirical formula ( $C_2H_4O$ ):
- $C: 12.01 \times 2 = 24.02\,g/mol$
- $H: 1.01 \times 4 = 4.04\,g/mol$
- $O: 16.00 \times 1 = 16.00\,g/mol$
- Total: $24.02 + 4.04 + 16.00 = 44.06\,g/mol$
- Divide the molecular molar mass by the empirical molar mass:
- $\frac{88.1\,g/mol}{44.06\,g/mol} = 2$ , indicating how many empirical formula units fit into the molecular formula.
- Multiply the subscripts:
- $C_2H_4O$ becomes $C_4H_8O_2$ , reinforcing our understanding of its complete chemical identity and how this compound behaves significantly in various chemical settings and reactions.