Question

An unknown compound consists of \(82.98 \%\) potassium and \(17.02 \%\) oxygen. What is the empirical formula of the compound?

Short answer

The empirical formula of the unknown compound, which consists of $82.98 \%$ potassium and $17.02 \%$ oxygen, is K₂O.

Step-by-step solution

Convert mass percentages to grams

Assume we have 100 grams of the compound. With the given mass percentages, we can calculate the amounts of each element in grams: - Potassium: \( 82.98 \% \) of 100 grams = 82.98 grams - Oxygen: \( 17.02 \% \) of 100 grams = 17.02 grams

Convert grams to moles

To convert grams to moles, we need to divide the mass of each element by its corresponding molar mass. The molar masses of potassium (K) and oxygen (O) are approximately 39.10 g/mol and 16.00 g/mol, respectively. Moles of potassium (K): \[\frac{\text {grams of potassium}}{\text{molar mass of potassium}} = \frac{82.98 \text{g}}{39.1\,\text{g/mol}} \approx 2.122\,\text{moles}\] Moles of oxygen (O): \[\frac{\text{grams of oxygen}}{\text{molar mass of oxygen}} = \frac{17.02\, \text{g}}{16.00\, \text{g/mol}} \approx1.064\, \text{moles}\]

Find the mole ratio

To find the mole ratio, divide all moles by the smallest mole quantity: Mole ratio of potassium to oxygen: \[\frac{\text {moles of potassium}}{\text {moles of oxygen}} = \frac{2.122}{1.064} \approx 1.994\]

Round the numbers to the nearest whole number

Round the numbers from the previous step to the nearest whole number to find the empirical formula. Since the ratio is approximately \(2:1\), the empirical formula for the compound is K₂O.

Key concepts

Mass Percentages

Mass percentages refer to the proportion of each element in a compound expressed as a percentage of the total mass of the compound. This concept is crucial when you want to determine the composition of a compound. In our scenario, the compound is made up of \(82.98\%\) potassium and \(17.02\%\) oxygen.

Here's how we use mass percentages to find the empirical formula:

• First, assume a sample size, typically 100 grams, to make calculations straightforward.
• Then, apply the percentage to that mass to find the specific mass of each element. For potassium, it's \(82.98\text{ g}\) and for oxygen, it's \(17.02\text{ g}\).

Using mass percentages simplifies the process of finding the empirical formula by converting them directly into grams, allowing easy transition to the mole calculation.

Mole Ratio

Determining the mole ratio is a critical step in finding the empirical formula. It involves comparing the number of moles of each element in a compound.

To do this, convert the mass of each element to moles, then find the smallest whole number ratio between them. In this example, the moles are calculated as follows:

• For potassium: \(2.122\) moles
• For oxygen: \(1.064\) moles

Now, the mole ratio of potassium to oxygen is calculated by dividing by the smallest number, \(1.064\). This yields an approximate ratio of \(2:1\). This ratio is key to determining the simplest whole-number ratio of elements in the empirical formula, confirming that for every 2 atoms of potassium, there is 1 atom of oxygen.

Molar Mass

The molar mass of an element is the mass of one mole of that element, usually expressed in grams per mole (g/mol). This constant value allows the conversion between grams and moles, a crucial process when deriving the empirical formula from known masses.

For this particular exercise:

• Potassium (\(K\)): approximately \(39.10\, \text{g/mol}\)
• Oxygen (\(O\)): approximately \(16.00\, \text{g/mol}\)

These values serve as conversion factors. By dividing the grams of each element from the assumed 100g sample by their respective molar masses, you obtain the amount in moles, crucial for forming the mole ratio.

Potassium

Potassium is a chemical element with the symbol \(K\) and is known for its light weight and crucial role in biology. It's one of the most abundant elements in the Earth's crust.

In chemistry, potassium's reactivity and compounds, such as potassium oxide, are commonly studied. Here, potassium is converted from mass percentage into moles to help uncover the empirical formula of a compound.

When calculating the moles of potassium in our problem, the process is straightforward:

• Start with the mass: \(82.98\, \text{g}\)
• Convert to moles using molar mass: \(\frac{82.98\, \text{g}}{39.10\, \text{g/mol}} \approx 2.122\, \text{moles}\)

This calculation plays a key role in evaluating how much potassium contributes to the empirical formula and thus the compound's structure.

Oxygen

Oxygen is a fundamental life-supporting element represented by \(O\) on the periodic table. It is essential for combustion and is a key component of water and various organic compounds.

In our exercise, calculating the number of moles of oxygen is pivotal for determining the empirical formula. Given the mass of oxygen derived from the compound's mass percentage:

• The oxygen mass is \(17.02\text{ g}\)
• Convert this mass to moles with its molar mass: \(\frac{17.02\, \text{g}}{16.00\, \text{g/mol}} \approx 1.064\, \text{moles}\)

By understanding how oxygen's moles relate to potassium, outlined in the mole ratio, you help determine the accurate empirical formula. This consistency across these foundational calculations ensures precise results.