CH 3-2: Empirical and Molecular Formulas: Determination from Experimental Data

Chapter 3: Molecules, Compounds, and Nomenclature

Overview

This section covers the determination of chemical formulas from experimental data, focusing on empirical and molecular formulas, and the use of CHNS (Carbon, Hydrogen, Nitrogen, Sulfur) analysis. These concepts are fundamental in General Chemistry for understanding the composition and structure of compounds.

Empirical and Molecular Formulas

Definitions

• Empirical Formula: The simplest whole-number ratio of atoms of each element in a compound. For example, the empirical formula of glucose (C6H12O6) is CH2O.

• Molecular Formula: The actual number of atoms of each element in a molecule of the compound. For glucose, the molecular formula is C6H12O6.

Examples of Compounds with Empirical Formula CH2O

Name	Molecular Formula	Whole Number Multiple	Molar Mass (g/mol)	Use/Function
Formaldehyde	CH2O	1	30.03	Disinfectant, preservative
Lactic Acid	C3H6O3	3	90.08	Forms in muscle during exercise
Glucose	C6H12O6	6	180.16	Major nutrient for energy in cells

Determining Empirical Formulas from Experimental Data

General Procedure

1. Assume a 100 g sample to convert mass percent to grams for each element.

2. Convert the mass of each element to moles using atomic weights.

3. Divide each mole value by the smallest number of moles to get the simplest ratio.

4. If necessary, multiply all ratios by a whole number to obtain whole-number subscripts.

Worked Example

Given a 100 g sample with 40.92 g C, 4.58 g H, and 54.50 g O:

1. Find moles of each element:

   • C: $\frac{40.92\ \text{g}}{12.01\ \text{g mol}^{-1}} = 3.407\ \text{mol}$

   • H: $\frac{4.58\ \text{g}}{1.008\ \text{g mol}^{-1}} = 4.54\ \text{mol}$

   • O: $\frac{54.50\ \text{g}}{16.00\ \text{g mol}^{-1}} = 3.406\ \text{mol}$

2. Divide by the smallest number of moles (3.406):

   • C: $\frac{3.407}{3.406} = 1$

   • H: $\frac{4.54}{3.406} = 1.33$

   • O: $\frac{3.406}{3.406} = 1$

3. Since H is not a whole number, multiply all ratios by 3:

   • C: $1 \times 3 = 3$

   • H: $1.33 \times 3 = 4$

   • O: $1 \times 3 = 3$

4. Empirical formula: C3H4O3

Table: Multiplying Decimal Subscripts to Whole Numbers

Decimal Subscript	Equivalent Fraction	Multiply by
0.20	1/5	5
0.25	1/4	4
0.33	1/3	3
0.50	1/2	2
0.66	2/3	3
0.80	4/5	5

CHNS Analysis

Overview of CHNS Analysis

CHNS analysis is a technique used to determine the elemental composition of organic compounds by combusting a sample in oxygen at high temperature. The products are analyzed to determine the amounts of carbon, hydrogen, nitrogen, and sulfur present.

• All carbon in the sample is converted to CO2.

• All hydrogen is converted to H2O.

• All nitrogen is converted to N2 (or NO/NO2 and then N2).

• All sulfur is converted to SO2.

Example: Determining Empirical Formula from CHNS Data

A hydrocarbon of unknown formula CxHy is analyzed. The products of combustion are measured to determine the empirical formula.

• Given: 0.6003 g H2O and 2.445 g CO2 produced.

• Calculate moles of H2O: $\frac{0.6003\ \text{g}}{18.02\ \text{g mol}^{-1}} = 0.03331\ \text{mol}$

• Calculate moles of H: $0.03331\ \text{mol H}_2\text{O} \times \frac{2\ \text{mol H}}{1\ \text{mol H}_2\text{O}} = 0.06663\ \text{mol H}$

• Calculate mass of H: $0.06663\ \text{mol H} \times 1.008\ \text{g mol}^{-1} = 0.06716\ \text{g}$

• Calculate moles of CO2: $\frac{2.445\ \text{g}}{44.01\ \text{g mol}^{-1}} = 0.05556\ \text{mol}$

• Calculate moles of C: $0.05556\ \text{mol CO}_2 \times \frac{1\ \text{mol C}}{1\ \text{mol CO}_2} = 0.05556\ \text{mol C}$

• Calculate mass of C: $0.05556\ \text{mol C} \times 12.01\ \text{g mol}^{-1} = 0.6673\ \text{g}$

• Calculate mass of O by difference: $\text{mass O} = \text{sample mass} - \text{mass C} - \text{mass H}$

• Empirical formula is determined from the mole ratios.

Calculating Molecular Formulas for Compounds

Relationship Between Empirical and Molecular Formulas

• The molecular formula is a whole-number multiple of the empirical formula.

• Mathematical relationship: $\text{Molecular Formula Mass} = n \times \text{Empirical Formula Mass}$ where $n$ is an integer.

• To find $n$: $n = \frac{\text{Molar Mass (from experiment)}}{\text{Empirical Formula Molar Mass}}$

Example

• Empirical formula: C3H4O3, empirical formula mass = 88.06 g/mol

• Molar mass of compound (from experiment): 177.5 g/mol

• $n = \frac{177.5}{88.06} = 2.016 \approx 2$

• Molecular formula: C6H8O6

Additional Example: Aniline (C6H7N)

• Given: Combustion of 9.71 g aniline yields 6.63 g H2O and 14.6 g CO2.

• Molar mass of aniline: 93.12 g/mol

• Empirical formula determined from combustion data: C6H7N

• Molecular formula matches empirical formula in this case.

Summary Table: Steps for Determining Empirical and Molecular Formulas

Step	Description
1	Convert mass percent to grams (assume 100 g sample)
2	Convert grams to moles using atomic masses
3	Divide by smallest number of moles to get ratios
4	Multiply to obtain whole numbers if necessary
5	Write empirical formula
6	Calculate empirical formula mass
7	Divide experimental molar mass by empirical formula mass to find n
8	Multiply subscripts in empirical formula by n to get molecular formula

Key Points

• Empirical formulas give the simplest ratio of elements; molecular formulas give the actual number of atoms.

• CHNS analysis is a standard method for determining elemental composition in organic compounds.

• Experimental data (mass of combustion products) can be used to determine empirical and molecular formulas.