What is the empirical formula of a compound composed of 27.9 g of potassium (K) and 5.71g of oxygen (O)?

To find the empirical formula, we need to follow these steps:

### 1. **Convert the mass of each element to moles:**
We use the molar mass of each element to convert grams to moles.

- **Potassium (K):**
  \[
  \text{Moles of K} = \frac{\text{Mass of K}}{\text{Molar mass of K}} = \frac{27.9\ \text{g}}{39.10\ \text{g/mol}} = 0.713\ \text{mol}
  \]

- **Oxygen (O):**
  \[
  \text{Moles of O} = \frac{\text{Mass of O}}{\text{Molar mass of O}} = \frac{5.71\ \text{g}}{16.00\ \text{g/mol}} = 0.357\ \text{mol}
  \]

### 2. **Find the simplest whole-number ratio of moles:**
To do this, we divide both values by the smallest number of moles.

\[
\frac{0.713\ \text{mol K}}{0.357\ \text{mol}} = 2
\]
\[
\frac{0.357\ \text{mol O}}{0.357\ \text{mol}} = 1
\]

### 3. **Write the empirical formula:**
The ratio of potassium to oxygen is 2:1. Therefore, the empirical formula is:
\[
K_2O
\]

### Final Answer:
The empirical formula of the compound is **K₂O**.

Would you like more details or have any questions? 

Here are 5 related questions:
1. How does the molar mass affect the calculation of the empirical formula?
2. What is the difference between the empirical and molecular formulas?
3. How can you find the molecular formula from the empirical formula?
4. How would you calculate the empirical formula if a third element was involved?
5. Can you determine the empirical formula if only percentages of elements are given?

**Tip:** When calculating empirical formulas, always divide by the smallest mole value to get a simple whole-number ratio.

What is the empirical formula of a compound composed of 27.9 g of potassium (K) and 5.71 g of oxygen (O)?

Learn how to calculate the empirical formula for a compound made of 27.9 g of potassium (K) and 5.71 g of oxygen (O). This solution covers stoichiometry concepts, converting mass to moles, and finding the simplest whole-number ratio, ideal for high school chemistry students.

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