Question

A certain compound of bromine and fluorine is used to make UF6, which is an important chemical in the processing and reprocessing of nuclear fuel. The compound contains 58.37 mass percent bromine. What is its empirical formula

a. BrFb. BrF2c. Br2F3d. Br3Fe. BrF3

Answer 1

The empirical formula of the compound containing 58.37 mass percent bromine is determined by converting the mass percentages to moles and finding the simplest whole number ratio of the elements. The calculation reveals that the empirical formula is BrF3.

Step-by-step explanation:

To find the empirical formula of the bromine and fluorine compound containing 58.37 mass percent bromine, we will assume a 100 g sample of the compound. This assumption simplifies the calculation because the percentages can be directly treated as grams.

We have 58.37 g of bromine and 100 g - 58.37 g = 41.63 g of fluorine. Next, we calculate the moles of bromine and fluorine using their respective atomic masses (Br = 79.90 g/mol, F = 18.998 g/mol):

• Moles of Br = 58.37 g / 79.90 g/mol ≈ 0.730 moles
• Moles of F = 41.63 g / 18.998 g/mol ≈ 2.191 moles

We then find the simplest whole number ratio of moles of Br to F by dividing each by the smallest number of moles:

• Ratio of Br = 0.730 moles / 0.730 ≈ 1
• Ratio of F = 2.191 moles / 0.730 ≈ 3

So the empirical formula based on the smallest whole number mole ratio of bromine to fluorine is BrF3.

The correct answer to the student's question is e. BrF3.

Answer 2

Answer: The empirical formula for the given compound is
`BrF_3`

Step-by-step explanation:

We are given:

Percentage of Br = 58.37 %

Percentage of F = 41.63 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of Br = 58.37 g

Mass of F = 41.63 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Bromine =
`\frac{\text{Given mass of Bromine}}{\text{Molar mass of Bromine}}=(58.37g)/(79.90g/mole)=0.73moles`

Moles of Fluorine =
`\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=(41.63g)/(19g/mole)=2.19moles`

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.73 moles.

For Bromine =
`(0.73)/(0.73)=1`

For Fluorine =
`(2.19)/(0.73)=3`

• Step 3: Taking the mole ratio as their subscripts.

The ratio of Br : F = 1 : 3

Hence, the empirical formula for the given compound is
`BrF_3`