AP Chemistry 1.3 Elemental Composition of Pure Substances Study Notes - New Syllabus 2024-2025
AP Chemistry 1.3 Elemental Composition of Pure Substances Study Notes- New syllabus
AP Chemistry 1.3 Elemental Composition of Pure Substances Study Notes – AP Chemistry – per latest AP Chemistry Syllabus.
LEARNING OBJECTIVE
- Explain the quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance.
Key Concepts:
- Elemental Composition of Pure Substances
Elemental Composition of Pure Substances
The elemental composition of a pure substance is the percentage by mass of each element present in the substance. It tells us what proportion of the compound’s mass comes from each element.
Types of Pure Substances
1. Molecular substances: Composed of individual molecules (e.g., \( \text{H}_2\text{O} \), \( \text{CO}_2 \)).
2. Ionic substances: Composed of ions held together in fixed proportions as described by a formula unit (e.g., \( \text{NaCl} \), \( \text{CaCO}_3 \)).
3. Atomic substances: Some pure elements exist as atoms (e.g., noble gases like Ar, Ne).
Law of Definite Proportions
This law states that the ratio of the masses of the constituent elements in any pure sample of a compound is always the same. In other words, water from any source will always contain hydrogen and oxygen in the same mass ratio of 1:8.
Empirical Formula
The empirical formula of a compound is the chemical formula that shows the lowest whole number ratio of atoms of each element present. For example:
- Glucose has the molecular formula \( \text{C}_6\text{H}_{12}\text{O}_6 \), but its empirical formula is \( \text{CH}_2\text{O} \).
- Hydrogen peroxide has the molecular formula \( \text{H}_2\text{O}_2 \), but its empirical formula is \( \text{HO} \).
Formula for Percent Composition
\( \% \text{Element} = \dfrac{(\text{mass of element in 1 mole of compound})}{(\text{molar mass of compound})} \times 100 \)
Conclusion
Pure substances can exist as molecules, ions, or atoms. According to the law of definite proportions, their elemental composition is fixed. Using percent composition data, chemists can calculate empirical formulas, which represent the simplest whole number ratio of atoms in a compound.
Example
Calculate the percent composition of oxygen in water (\( \text{H}_2\text{O} \)).
▶️ Answer/Explanation
Molar mass of \( \text{H}_2\text{O} = (2 \times 1.01) + 16.00 = 18.02 \, \text{g mol}^{-1} \)
Mass of oxygen = 16.00 g
\( \% \text{O} = \dfrac{16.00}{18.02} \times 100 = 88.8\% \)
Example
Find the percent composition of carbon in carbon dioxide (\( \text{CO}_2 \)).
▶️ Answer/Explanation
Molar mass of \( \text{CO}_2 = 12.01 + (2 \times 16.00) = 44.01 \, \text{g mol}^{-1} \)
Mass of carbon = 12.01 g
\( \% \text{C} = \dfrac{12.01}{44.01} \times 100 = 27.3\% \)
Example
Determine the empirical formula of a compound that contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
▶️ Answer/Explanation
Assume 100 g sample:
C = 40.0 g → \( \dfrac{40.0}{12.01} = 3.33 \, \text{mol} \)
H = 6.7 g → \( \dfrac{6.7}{1.01} = 6.63 \, \text{mol} \)
O = 53.3 g → \( \dfrac{53.3}{16.00} = 3.33 \, \text{mol} \)
Simplest whole number ratio: C:H:O = 3.33:6.63:3.33 ≈ 1:2:1
Empirical formula = \( \text{CH}_2\text{O} \)