Problem 40 - Chemistry The Science in Context

Question

How many moles of \(\mathrm{NaCl}\) are there in a crystal of halite that has a mass of \(6.82 \mathrm{g} ?\).

Step-by-Step Solution

Answer

Answer: There are approximately 0.1167 moles of NaCl in the crystal of halite.

Step 1: Calculate the molar mass of NaCl

To calculate the molar mass of NaCl, we need to find the molar mass of each element (Na and Cl) and add them together. From the periodic table, we have the molar mass of Na = 22.99 g/mol and Cl = 35.45 g/mol. The molar mass of NaCl is the sum of the molar masses of Na and Cl: M(NaCl) = M(Na) + M(Cl) = 22.99 + 35.45 = 58.44 \text{ g/mol}

Step 2: Calculate the number of moles of NaCl

Now we know the molar mass of NaCl, we can use the formula n = mass / molar mass to calculate the number of moles of NaCl in the crystal. We have the mass of the crystal = 6.82 g: n(NaCl) = \frac{6.82 \text{ g}}{58.44 \text{ g/mol}} = 0.1167 \text{ mol} So, there are approximately 0.1167 moles of NaCl in the crystal of halite.

Key Concepts

Molar Mass of NaCl

Understanding the molar mass of a compound is crucial for delving into the world of chemistry, and sodium chloride (NaCl) serves as a prime example. The molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). To find the molar mass of NaCl, simply add the molar mass of sodium (Na), which is 22.99 g/mol, to that of chlorine (Cl) with 35.45 g/mol.

This leads to a combined molar mass for NaCl:
\( M(\text{NaCl}) = M(\text{Na}) + M(\text{Cl}) = 22.99 + 35.45 = 58.44 \text{ g/mol} \).

Knowing this figure allows us to transition from the macroscopic world, with mass in grams, to the microscopic, dealing with moles, which represent a specific number of particles.

Stoichiometry

Stoichiometry lies at the heart of chemical reactions, representing the balance of mass and the proportions of reactants and products. It correlates the quantities in a chemical equation. To master stoichiometry, one must be adept at converting between grams and moles, leveraging the molar mass as a pivotal conversion factor.

Understanding the Mole Concept

In stoichiometry, the mole is a fundamental unit that measures the amount of substance. One mole of any substance contains the same number of entities, be they atoms, molecules, or ions, as there are in 12 grams of carbon-12. This connects to Avogadro's constant, another cornerstone concept.

Applying stoichiometry to the problem at hand, we calculate the number of moles (\(n\)) of NaCl by using the formula \(n = \frac{\text{mass}}{\text{molar mass}}\). This not only aids in determining the amount needed for a reaction but also helps predict the yield from given reactants.

Avogadro's Number

Embarking on the journey of understanding moles inevitably leads to Avogadro's number, which is a cornerstone in the land of chemistry. Avogadro's number, \(6.022 \times 10^{23}\), specifies the number of units (be it atoms, molecules, or ions) in one mole of a substance.

Avogadro's number isn't just a random figure; it was derived from meticulous measurements and enables chemists to directly relate macroscopic masses to the microscopic world of atoms and molecules. When we talk about the 0.1167 moles of NaCl from our previous example, we are actually referring to roughly \(0.1167 \times 6.022 \times 10^{23}\) individual sodium and chloride ions residing in the crystal.

Chemical Formula

A chemical formula reveals the types and numbers of atoms in the smallest unit of a substance. For NaCl, the chemical formula displays a 1:1 ratio of sodium ions to chloride ions. This formula is akin to a recipe, providing insight into the composition and stoichiometry of the compound.

In NaCl's formula, 'Na' signifies one sodium atom and 'Cl' stands for one chlorine atom, indicating that each pair combines to make one formula unit of the compound. This fundamental information is vital, as it lays the groundwork for conducting various calculations in chemistry, such as determining ratios in reactions and the calculation of the molar mass as seen previously.