Understanding the Impact of Molecular Weight on Percent Composition

In chemistry, percent composition refers to the mass percentage of each element in a compound. The molecular weight of a compound plays a crucial role in determining these percentages. In this article, we will explore three practical examples that illustrate how molecular weight impacts percent composition.

Example 1: Sodium Chloride (NaCl)

Sodium chloride, commonly known as table salt, is a prime example of how molecular weight influences percent composition. In the context of food chemistry, understanding the composition of sodium chloride is essential for nutrition labeling and food preservation.

To calculate the percent composition of sodium (Na) and chlorine (Cl) in sodium chloride, we first determine the molecular weight:

• Molecular Weight of NaCl:
  • Sodium (Na): 22.99 g/mol
  • Chlorine (Cl): 35.45 g/mol
  • Total: 22.99 + 35.45 = 58.44 g/mol

Next, we calculate the percent composition:

• Percent Na = (22.99 g/mol / 58.44 g/mol) x 100 = 39.3%
• Percent Cl = (35.45 g/mol / 58.44 g/mol) x 100 = 60.7%

This example illustrates how the molecular weight directly affects the mass percentages of individual elements in a compound, which is crucial for both nutritional information and chemical applications.

Example 2: Glucose (C6H12O6)

Glucose is a simple sugar that serves as a primary energy source for living organisms. It is also a common subject in biochemistry for studying metabolism and energy balance.

To understand the percent composition of glucose, we first calculate its molecular weight:

• Molecular Weight of C6H12O6:
  • Carbon (C): 12.01 g/mol (6 x 12.01 = 72.06 g/mol)
  • Hydrogen (H): 1.01 g/mol (12 x 1.01 = 12.12 g/mol)
  • Oxygen (O): 16.00 g/mol (6 x 16.00 = 96.00 g/mol)
  • Total: 72.06 + 12.12 + 96.00 = 180.18 g/mol

Now, we calculate the percent composition:

• Percent C = (72.06 g/mol / 180.18 g/mol) x 100 = 40.0%
• Percent H = (12.12 g/mol / 180.18 g/mol) x 100 = 6.7%
• Percent O = (96.00 g/mol / 180.18 g/mol) x 100 = 53.3%

This example demonstrates how molecular weight influences the percent composition of glucose, which is essential for understanding its role in metabolism and nutrition.

Example 3: Ethanol (C2H5OH)

Ethanol, commonly known as alcohol, is widely used in beverages, fuel, and as a solvent. Understanding its molecular weight and percent composition is vital for both regulatory purposes and consumer information.

To find ethanol’s percent composition, we first calculate its molecular weight:

• Molecular Weight of C2H5OH:
  • Carbon (C): 12.01 g/mol (2 x 12.01 = 24.02 g/mol)
  • Hydrogen (H): 1.01 g/mol (6 x 1.01 = 6.06 g/mol)
  • Oxygen (O): 16.00 g/mol (1 x 16.00 = 16.00 g/mol)
  • Total: 24.02 + 6.06 + 16.00 = 46.08 g/mol

Next, we calculate the percent composition:

• Percent C = (24.02 g/mol / 46.08 g/mol) x 100 = 52.1%
• Percent H = (6.06 g/mol / 46.08 g/mol) x 100 = 13.2%
• Percent O = (16.00 g/mol / 46.08 g/mol) x 100 = 34.7%

This case highlights how the molecular weight of ethanol affects its percent composition, which has implications for its production and use in various applications.

Conclusion

Understanding the impact of molecular weight on percent composition is crucial for various fields, including nutrition, biochemistry, and industrial applications. These examples provide a clear view of how molecular weight influences the mass percentages of elements in different compounds, enhancing our comprehension of their chemical properties and uses.