In the lab, students decomposed a sample of calcium carbonate by heating it over a Bunsen burner and collected carbon dioxide according to the following equation. $$ \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ (a) How many \(\mathrm{mL}\) of carbon dioxide gas were generated by the decomposition of \(6.24 \mathrm{~g}\) of calcium carbonate at STP? (b) If \(52.6 \mathrm{~L}\) of carbon dioxide at STP were needed, how many moles of calcium carbonate would be required?

Understanding the molar mass of a compound is crucial when working with chemical reactions. The molar mass is essentially the weight of one mole (6.02 \(\times\) 10^23 particles) of a substance and is expressed in grams per mole (g/mol). It's obtained by summing the atomic masses of all the atoms in the molecular formula.
For calcium carbonate (CaCO₃), we calculate its molar mass by adding the molar masses of one calcium atom (Ca), one carbon atom (C), and three oxygen atoms (O). With the atomic weights from the periodic table, 40.08 g/mol for calcium, 12.01 g/mol for carbon, and 16.00 g/mol for each oxygen, the molar mass of CaCO₃ comes out to 100.09 g/mol.

Molar Mass of Calcium Carbonate Calculation:

• Molar mass of Ca: 40.08 g/mol \(\times\) 1 = 40.08 g/mol
• Molar mass of C: 12.01 g/mol \(\times\) 1 = 12.01 g/mol
• Molar mass of O: 16.00 g/mol \(\times\) 3 = 48.00 g/mol
• Total molar mass of CaCO₃: 40.08 + 12.01 + 48.00 = 100.09 g/mol

This information is not only necessary for the understanding of the compound but is fundamentally important when converting between the mass of a substance and the number of moles. It is the first step in any stoichiometric calculation.

Stoichiometry is the area of chemistry that involves calculating the amounts of reactants or products in a chemical reaction. It is based on the conservation of mass and the concept of the mole. When dealing with stoichiometry, it's essential to use a balanced chemical equation, which gives the molar ratios of reactants and products.
In the case of calcium carbonate decomposition, the balanced equation is CaCO₃(s) \(\rightarrow\) CaO(s) + CO₂(g). Each mole of calcium carbonate yields one mole of carbon dioxide. This 1:1 molar ratio is fundamental to determine how much product is formed from a particular amount of reactant or vice versa.

Applying Stoichiometry:

• Determine the number of moles of the reactant, based on its mass and molar mass.
• Use the mole ratio from the balanced equation to find the moles of product formed.
• Convert the moles of product into desired units: mass, volume, or particles depending on the requirement.

The stoichiometric calculations can be extended to volumes when dealing with gases, keeping in mind the gas laws and conditions such as STP.

Standard Temperature and Pressure (STP) is a common reference for gases, defined as 0 degrees Celsius (273.15 Kelvin) and 1 atmosphere of pressure. When a gas is at STP, one mole of the gas occupies 22.4 liters.
This constant volume is known as the molar volume of a gas at STP. It's vital for converting between the volume of a gas and the amount in moles. The molar volume applies to any ideal gas, meaning it's an excellent approximation for most gases under STP conditions.

Volume Calculation at STP:

• Identify the volume of one mole of any gas at STP: 22.4 L.
• For a known amount in moles, multiply it by this molar volume to find the volume the gas would occupy at STP.
• Remember, this is a simplified approach, and real gases may deviate slightly from this value due to non-ideal behavior.

In practical applications like the decomposition of calcium carbonate, we can predict the volume of carbon dioxide produced when we have the number of moles of calcium carbonate reacting because the gas follows the molar volume at STP. This connection is an essential part of stoichiometry and helps relate the abstract concept of moles to a more understandable and measurable quantity, such as liters or milliliters of gas.