Chapter 9: Problem 98

Determine the empirical formula of the compound that is $43.2 \%$ by mass $\mathrm{K}, 39.1 \%$ by $\mathrm{mass} \mathrm{Cl}$, and also contains oxygen.

Short Answer

Expert verified

The empirical formula for the compound with $43.2\%$ K, $39.1\%$ Cl, and the rest being O is KClO.

Step by step solution

Convert mass percentages to grams

Assume we have 100 grams of the compound. This means we have 43.2 grams of K, 39.1 grams of Cl, and since this compound also contains oxygen, we can calculate the mass of oxygen by subtracting the mass of K and Cl from 100 grams: \[100 - 43.2 - 39.1 = 17.7 \text{ grams of O}\]

Convert grams to moles

Use the molar mass of each element to convert the mass to moles: For potassium (K): \[Moles \: K = \frac{43.2 \: g}{39.1 \: g/mol} = 1.104 \: moles\] For chlorine (Cl): \[Moles \: Cl = \frac{39.1 \: g}{35.5 \: g/mol} = 1.101 \: moles\] For oxygen (O): \[Moles \: O = \frac{17.7 \: g}{16.0 \: g/mol} = 1.106 \: moles\]

Finding the mole ratio

In order to find the mole ratio of each element in the compound, divide the moles of each element by the smallest moles value: \[ Ratio \: K = \frac{1.104}{1.101} = 1.003 \approx 1\] \[ Ratio \: Cl = \frac{1.101}{1.101} = 1\] \[ Ratio \: O = \frac{1.106}{1.101} = 1.005 \approx 1\] We see that, approximately, there's a 1:1:1 ratio between K, Cl, and O.