Problem 163 - Chemistry Matter and Change

Question

Express the composition of each compound as the mass percent of its elements (percent composition). \begin{equation} \begin{array}{l}{\text { a. sucrose }\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) \quad \text { c. magnetite }\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right)} \\ {\text { b. aluminum sulfate }\left(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\right)}\end{array} \end{equation}

Step-by-Step Solution

Answer

Sucrose: C: 42.06%, H: 6.48%, O: 51.46%; Aluminum sulfate: Al: 15.77%, S: 28.12%, O: 56.10%; Magnetite: Fe: 72.36%, O: 27.64%.

Step 1: Determine Molar Mass of Sucrose

Calculate the molar mass of sucrose \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \) by finding the sum of the atomic masses of all the atoms: \( 12 \times 12.01 \text{ (for C)} + 22 \times 1.01 \text{ (for H)} + 11 \times 16.00 \text{ (for O)} = 342.30 \text{ g/mol} \).

Step 2: Calculate Percent Composition of Sucrose

Calculate the percent by mass of each element in sucrose. For carbon: \( \left(\frac{12 \times 12.01}{342.30}\right) \times 100 = 42.06\% \). For hydrogen: \( \left(\frac{22 \times 1.01}{342.30}\right) \times 100 = 6.48\% \). For oxygen: \( \left(\frac{11 \times 16.00}{342.30}\right) \times 100 = 51.46\% \).

Step 3: Determine Molar Mass of Aluminum Sulfate

Calculate the molar mass of aluminum sulfate \( \text{Al}_{2}(\text{SO}_{4})_{3} \). This is done by adding \( 2 \times 26.98 \text{ (for Al)} + 3 \times 32.07 \text{ (for S)} + 12 \times 16.00 \text{ (for O)} = 342.15 \text{ g/mol} \).

Step 4: Calculate Percent Composition of Aluminum Sulfate

Calculate the mass percent of each element. For aluminum: \( \left(\frac{2 \times 26.98}{342.15}\right) \times 100 = 15.77\% \). For sulfur: \( \left(\frac{3 \times 32.07}{342.15}\right) \times 100 = 28.12\% \). For oxygen: \( \left(\frac{12 \times 16.00}{342.15}\right) \times 100 = 56.10\% \).

Step 5: Determine Molar Mass of Magnetite

Calculate the molar mass of magnetite \( \text{Fe}_{3}\text{O}_{4} \) by summing the atomic masses: \( 3 \times 55.85 \text{ (for Fe)} + 4 \times 16.00 \text{ (for O)} = 231.55 \text{ g/mol} \).

Step 6: Calculate Percent Composition of Magnetite

For magnetite, calculate the mass percent by each element. For iron: \( \left(\frac{3 \times 55.85}{231.55}\right) \times 100 = 72.36\% \). For oxygen: \( \left(\frac{4 \times 16.00}{231.55}\right) \times 100 = 27.64\% \).

Key Concepts

Molar Mass Calculation

Calculating molar mass is a fundamental concept in chemistry that helps us understand the composition of chemical compounds. The molar mass of a compound is determined by finding the sum of the atomic masses of all the atoms present in the compound. This calculation is crucial when we want to understand the mass percent of each element in a compound to establish its formula units.

• Take sucrose \ \(\text{C}_{12}\text{H}_{22}\text{O}_{11}\ \) as an example. To determine its molar mass, we add up the atomic masses of all carbon, hydrogen, and oxygen atoms present in the formula:
• For carbon (C), we multiply 12 moles by the atomic mass of carbon (12.01 g/mol).
• For hydrogen (H), it's 22 moles times 1.01 g/mol.
• And for oxygen (O), it's 11 moles times 16.00 g/mol.
• This gives us a total molar mass of 342.30 g/mol for sucrose.

By breaking down the compound into sections by element, it ensures that each atomic mass contributes correctly to the total molar mass.

Chemical Compounds

Chemical compounds are substances composed of two or more different elements chemically bonded together. They play a pivotal role in chemistry as they form the molecules and materials we encounter around us. Understanding their composition is key to grasping chemical reactions and properties.Chemical compounds are represented by chemical formulas, where symbols of elements denote the constituent atoms. These formulas also indicate the ratios in which elements combine. For instance:

• Sucrose is composed of carbon, hydrogen, and oxygen, denoted as \(\text{C}_{12}\text{H}_{22}\text{O}_{11}\).
• Magnetite, a different compound, has a formula \(\text{Fe}_{3}\text{O}_{4}\), distinct in elements and proportions.
• Aluminum sulfate, \(\text{Al}_{2}(\text{SO}_{4})_{3}\), contains aluminum, sulfur, and oxygen in a specific ratio.

This formulaic representation not only demonstrates the makeup but also allows for calculations like molar mass and percent composition, crucial for practical applications in science.

Mass Percent

Mass percent, or mass composition, is a useful way to express the concentration of an element in a compound. It is defined as the ratio of the mass of a particular element to the total mass of the compound, multiplied by 100%. This percent composition provides insights into the proportion of each element in the compound.To calculate the mass percent, follow these steps:

• First, calculate the molar mass of the entire compound, as we’ve seen with sucrose (342.30 g/mol).
• Next, determine the mass contribution of each element. Say, for oxygen in sucrose, it's \(11 \times 16.00\) g for all oxygen atoms.
• Finally, divide this by the molar mass of sucrose and multiply by 100% to find the oxygen mass percent, which results in 51.46%.

Determining mass percent is essential for analyzing chemical formulas and predicting reaction yields, ensuring precise measurements in lab and industry settings.

Atomic Mass

Atomic mass is the weighted average mass of an element's atoms, measured in atomic mass units (amu). It is a crucial property used in calculating the molar mass of compounds, as it reflects the mass of different isotopes and their abundances in nature. A few key points:

• Atomic masses are listed on the periodic table and provide a way to calculate the molar mass of compounds.
• For instance, carbon has an atomic mass of approximately 12.01 amu, and this value is multiplied by the number of carbon atoms present in a compound to find its total mass contribution.
• Precision in atomic mass is what allows chemists to accurately gauge how much of each element is in a compound, vital for mass percent calculations.

Atomic mass forms the backbone of any molar mass calculation and is a vital reference in the study and application of chemistry.