Problem 6 - Chemistry in 30 days

Question

An oxide of iodine \((1=127)\) contains \(25.4 \mathrm{~g}\) of iodine for \(8 \mathrm{~g}\) of oxygen. lis formula could be (a) \(10_{23}\) (b) \(1.0\) (c) \(_{2} \mathrm{O}_{5}\) (d) \(1_{2} \mathrm{O}_{7}\)

Step-by-Step Solution

Answer

The formula of the iodine oxide is \( I_2O_5 \).

Step 1: Determine moles of iodine and oxygen

First, we calculate the moles of iodine and oxygen. The atomic weight of iodine is given as 127, so moles of iodine is \( \frac{25.4}{127} \approx 0.2 \). The atomic weight of oxygen is 16, so moles of oxygen is \( \frac{8}{16} = 0.5 \).

Step 2: Calculate the mole ratio

To find the simplest whole number ratio, divide the moles of each element by the smallest number of moles calculated. Divide moles of iodine and oxygen by 0.2. This gives iodine: \( \frac{0.2}{0.2} = 1 \) and oxygen: \( \frac{0.5}{0.2} = 2.5 \).

Step 3: Convert to whole numbers

The ratio of iodine to oxygen is 1:2.5. To convert this to whole numbers, multiply both numbers by 2 to get iodine:oxygen = 2:5.

Step 4: Determine the empirical formula

The whole number ratio is Iodine:Oxygen = 2:5, giving an empirical formula of \( I_2O_5 \).

Key Concepts

Understanding the Mole Concept

The mole concept is a fundamental idea in chemistry used to relate the mass of substances to the number of entities, such as atoms or molecules, they contain. It's essentially a bridge between the macroscopic and the atomic scales. One mole of a substance contains approximately 6.022 x 10\(^23\) entities (Avogadro's number), whether they're atoms, molecules, ions, or electrons.
To calculate the number of moles from a given mass, the formula is: \[\text{Number of moles} = \frac{\text{mass in grams}}{\text{molar mass in g/mol}}\]For the given problem, the atomic weight (which acts as molar mass for elements) was used. Here's how you can think about it:

• For iodine, the atomic weight is 127 g/mol, hence the moles of iodine were calculated as \( \frac{25.4}{127} \approx 0.2 \text{ moles} \).
• For oxygen, with an atomic weight of 16 g/mol, it resulted in \( \frac{8}{16} = 0.5 \text{ moles} \).

This mole calculation is essential because it gives us a way to compare substances by entirely different masses and allows us to think about chemical reactions on a particle level.

Determining the Empirical Formula

The empirical formula represents the simplest whole number ratio of atoms in a compound. It doesn't necessarily represent the actual number of atoms in a molecule, but it shows the proportion of elements relative to each other.
To find this, compare the moles of each element calculated using the mole concept. In practice:

• Divide each element's mole count by the smallest number of moles to normalize the ratio.
• For the problem, iodine and oxygen's moles were divided by 0.2.
• This resulted in a preliminary ratio of 1:2.5 for iodine to oxygen.

Since an empirical formula must have whole numbers, the ratio 1:2.5 was multiplied by 2 to yield 2:5. Thus, the empirical formula was determined to be \(I_2O_5\).
This step is crucial because it eliminates fractions in a formula and provides the simplest representation of the compound's elemental composition.

Molar Mass Calculation in Practice

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a critical concept because it's used to convert between the mass of a substance and the amount in moles, as was necessary in the original exercise.
For individual elements, the atomic weights from the periodic table serve as molar masses. For compounds, the molar mass is the sum of the atomic masses of all elements in the formula, each multiplied by its respective subscript in the chemical formula.
In your specific example:

• The molar mass of iodine was used as a premise to calculate that 25.4 g is approximately 0.2 moles.
• Similarly, for oxygen, the typical molar mass calculation allowed finding that 8 g equals 0.5 moles.

Molar mass aids in

• balancing chemical equations,
• determining the proportions in a mixture, and
• applying the principles of stoichiometry in reactions.

In essence, understanding molar mass is essential not only for solving textbook problems but for practical chemistry applications in laboratories and industries.