What is the empirical formula of a compound that contains 52.7% K and 47.3% Cl?
Understand the Problem
The question is asking to determine the empirical formula of a compound given the percentages of Potassium (K) and Chlorine (Cl) in the compound. To solve this question, we need to convert the percentages to grams (assuming a 100g sample), then to moles, and finally find the simplest whole number ratio of the moles of each element to determine the subscripts in the empirical formula.
Answer
$KCl$
Answer for screen readers
The empirical formula of the compound is $KCl$.
Steps to Solve
- Convert percentages to grams
Assume a 100g sample. This makes the percentages directly equivalent to grams. Potassium (K): 52.7% becomes 52.7 g Chlorine (Cl): 47.3% becomes 47.3 g
- Convert grams to moles
Divide the mass of each element by its molar mass. Molar mass of K = 39.10 g/mol Molar mass of Cl = 35.45 g/mol
Moles of K = $\frac{52.7 \text{ g}}{39.10 \text{ g/mol}} = 1.348 \text{ mol}$
Moles of Cl = $\frac{47.3 \text{ g}}{35.45 \text{ g/mol}} = 1.334 \text{ mol}$
- Find the simplest whole number ratio
Divide each mole value by the smallest mole value to find the simplest ratio. In this case, the smallest mole value is 1.334 mol.
Ratio of K = $\frac{1.348 \text{ mol}}{1.334 \text{ mol}} \approx 1.01 \approx 1$
Ratio of Cl = $\frac{1.334 \text{ mol}}{1.334 \text{ mol}} = 1$
- Write the empirical formula
The mole ratio of K to Cl is approximately 1:1. Therefore, the empirical formula is KCl.
The empirical formula of the compound is $KCl$.
More Information
Potassium chloride, with the empirical formula KCl, is an ionic compound composed of potassium and chloride ions. It is commonly known as a salt substitute for its salty taste.
Tips
A common mistake is not dividing by the smallest mole value to obtain the simplest ratio. Another common mistake is using the incorrect molar masses for the given elements. Also, rounding too early can lead to inaccurate ratios.
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