Question

A sample of an ionic compound that is often used as a dough conditioner is analyzed and found to contain 7.591g of potassium, 15.514g of bromine, and 9.319g of oxygen. what is the empirical formula for this compound?

Answer 1

To find the empirical formula, the masses of potassium, bromine, and oxygen in the compound are converted to moles and then divided by the smallest number of moles to get the simplest whole number ratio. The ratio reveals the empirical formula to be KBrO3.

Step-by-step explanation:

Empirical Formula Calculation

First, we need to convert the masses of potassium, bromine, and oxygen in the compound to moles. We do this by dividing the mass of each element by its respective molar mass: potassium (39.10 g/mol), bromine (79.90 g/mol), and oxygen (16.00 g/mol).

• Potassium: 7.591 g / 39.10 g/mol = 0.194 moles of potassium
• Bromine: 15.514 g / 79.90 g/mol = 0.194 moles of bromine
• Oxygen: 9.319 g / 16.00 g/mol = 0.582 moles of oxygen

Next, we find the simplest whole number ratio of moles of each element by dividing all by the smallest number of moles, which in this case is 0.194 moles:

• Potassium: 0.194 / 0.194 = 1
• Bromine: 0.194 / 0.194 = 1
• Oxygen: 0.582 / 0.194 = 3

Thus, the empirical formula for the compound is KBrO3.