91Ó°ÊÓ

Stoichiometry is a fundamental concept in chemistry that revolves around the quantitative relationships between the reactants and products in a chemical reaction. When a reaction occurs, stoichiometry allows us to analyze and predict the amount of products that would be generated from a certain amount of reactants. This concept requires a balanced chemical equation, where the number of atoms for each element in the reactants equals the number in the products.

In our original exercise, the stoichiometry involved in the chemical reaction of the decomposition of calcium carbonate to calcium oxide and carbon dioxide is crucial. The balanced equation is:

• \[ \text{CaCO}_3(s) \longrightarrow \text{CaO}(s) + \text{CO}_2(g) \]

The balanced equation tells us that one mole of calcium carbonate decomposes to produce one mole of calcium oxide and one mole of carbon dioxide. This mole-to-mole ratio indicates the stoichiometry between the reactants and the products.

By using the stoichiometric relationship from the balanced equation, it’s possible to determine that the moles of carbon dioxide are directly related to the moles of calcium carbonate decomposed.

Molar mass is an important concept that aids in converting between the mass of a substance and the amount in moles. It is defined as the mass of one mole of a substance and is usually expressed in grams per mole (g/mol). Calculating molar mass involves summing the atomic masses of each element in a compound formula.

In the context of our exercise, the molar masses for calcium carbonate (\(\text{CaCO}_3\)) and carbon dioxide (\(\text{CO}_2\)) are crucial for further calculations:

• The molar mass of \(\text{CaCO}_3\) is calculated as:
  \[ 40.08 + 12.01 + 3 \times 16.00 = 100.09 \, \text{g/mol} \]
• The molar mass of \(\text{CO}_2\) is calculated as:
  \[ 12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol} \]

These calculations serve as the backbone for converting between mass and moles, such as when determining the moles of carbon dioxide produced from a known mass, or when calculating how much calcium carbonate was present.

Percent composition is a component of analytical chemistry that expresses the percentage by mass of each element in a compound. It is also an effective way to describe the makeup of a mixture or a sample, as shown in the exercise involving sandstone.

To calculate percent composition, we use the formula:

• \[ \text{Percent composition} = \left( \frac{\text{mass of component}}{\text{total mass of mixture}} \right) \times 100\% \]

In the exercise, after calculating the mass of calcium carbonate decomposed, we determined the mass of silica (\(\text{SiO}_2\)) remaining in the rock. Thus, the percent composition of silica in the sandstone sample is:

• \[ \frac{9.72 \, \text{mg of SiO}_2}{18.7 \, \text{mg total sample}} \times 100\% \approx 51.98\% \]

This percentage indicates how much of the original sandstone sample was made up of silica, providing valuable insight into the composition of geological samples. Understanding this concept is essential for many branches of science, including chemistry, geology, and environmental science.