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Benzoyl peroxide is the active ingredient in an acne cream. Calculate the percent composition of benzoyl peroxide, \(\mathrm{C}_{14} \mathrm{H}_{10} \mathrm{O}_{4}\).

Short Answer

Expert verified
Carbon: 69.43%, Hydrogen: 4.16%, Oxygen: 26.41%.

Step by step solution

01

Calculate Molar Mass of Elements

First, determine the molar mass of each element in the compound. The atomic masses are approximately: Carbon (C) = 12.01 g/mol, Hydrogen (H) = 1.008 g/mol, and Oxygen (O) = 16.00 g/mol. The formula for benzoyl peroxide is \(\mathrm{C}_{14} \mathrm{H}_{10} \mathrm{O}_{4}\).
02

Calculate Molar Mass of Benzoyl Peroxide

Use the formula \(\mathrm{C}_{14} \mathrm{H}_{10} \mathrm{O}_{4}\) to find the total molar mass: \(14 \times 12.01 + 10 \times 1.008 + 4 \times 16.00\) which results in 168.14 + 10.08 + 64.00 = 242.22 g/mol.
03

Calculate Percent Composition

Calculate the percent composition of each element by dividing the total atomic mass of each element in the compound by the molar mass of benzoyl peroxide and then multiplying by 100%. For Carbon: \(\frac{168.14}{242.22} \times 100\% = 69.43\%\), for Hydrogen: \(\frac{10.08}{242.22} \times 100\% = 4.16\%\), and for Oxygen: \(\frac{64.00}{242.22} \times 100\% = 26.41\%\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
Understanding molar mass calculation is crucial for determining the composition of chemical compounds. The molar mass is the sum of the atomic masses of all the atoms in a molecule.
To calculate it, follow these simple steps:
  • Identify the atomic masses: Find the atomic mass of each element in the molecule from the periodic table.
  • Multiply atomic masses by their respective counts: Use the number of atoms of each element in the formula to calculate the contribution of each element. For example, in benzoyl peroxide (\(\mathrm{C}_{14} \mathrm{H}_{10} \mathrm{O}_{4}\)), Carbon has an atomic mass of 12.01 g/mol and appears 14 times, so its contribution is \(14 \times 12.01\).
  • Add the contributions: Sum all individual contributions to get the total molar mass of the compound.
This simple approach helps you find out the total weight of molecules and is the foundation for calculating percent composition.
Chemical Formula
The chemical formula of a compound reveals the types and numbers of atoms it contains. For benzoyl peroxide, the formula is \(\mathrm{C}_{14} \mathrm{H}_{10} \mathrm{O}_{4}\). This tells us:
  • There are 14 Carbon atoms.
  • There are 10 Hydrogen atoms.
  • There are 4 Oxygen atoms.
Each part of the formula is crucial for understanding the properties and behavior of the compound. Knowing how to interpret chemical formulas allows chemists to calculate molecular weights, understand reactions, and determine the percentages of each element within a compound.
The formula is the basis for many other calculations, such as stoichiometry, which involves the calculation of reactants and products in chemical reactions.
Stoichiometry
Stoichiometry is the field of chemistry that involves calculations based on the quantitative relationships of the reactants and products in chemical reactions. It allows chemists to predict the amounts of products that form in a reaction as well as the amounts of reactants needed.
In the context of calculating percent composition, stoichiometry begins with:
  • Determining molar mass: This is essential for finding the mass of all reactants and products. It ensures balanced and precise calculations.
  • Using balanced chemical equations: Equations must be balanced to correctly relate the quantities of substances involved.
  • Applying conversion factors: We use them to switch between units and elements, ensuring all calculations exhibit the law of conservation of mass.
By understanding and applying stoichiometry, you can explore the relationships between different chemical substances in detail, predict yields of reactions, and analyze efficiency. This forms the basis for further application in industries such as pharmaceuticals, environmental science, and material engineering.

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