Copper(II) fluoride contains 37.42% F by mass. Use this % to calculate the mass of fluorine in grams contained 28.5 g of copper (ii) fluoride

10.6647g Flourine in 28.5g

28.5 x 0.3742 = ?

10.7g Fluorine (Sig Figs) contained in 28.5g Copper (II) fluoride

11.1

Well, well, looks like we've got some chemistry here! Let me sprinkle a bit of humor into it for you.

To find the mass of fluorine in 28.5 g of copper(II) fluoride, we'll use the percentage of fluorine in the compound. But before we dive in, let me ask you something: Why do chemists like nitrates so much? Because they're cheaper than day rates!

Now, let's get back to business. The percentage of fluorine in copper(II) fluoride is 37.42%. So, if you have 28.5 g of copper(II) fluoride, the mass of fluorine can be calculated as follows:

Mass of fluorine = (% of fluorine / 100) x mass of copper(II) fluoride
Mass of fluorine = (37.42 / 100) x 28.5 g

Calculating this, we get:
Mass of fluorine = 0.3742 x 28.5 g

And the answer is... *drumroll please*...
Approximately 10.68 grams of fluorine! Now that's a heavy dose of humor!

To solve this problem, we can follow the steps below:

1. Determine the molar mass of copper(II) fluoride (CuF2):
- The atomic mass of copper (Cu) is 63.55 g/mol.
- The atomic mass of fluorine (F) is 18.9984 g/mol.
- Since there are two fluorine atoms in one formula unit of copper(II) fluoride, we multiply the atomic mass of fluorine by 2.
- The molar mass of copper(II) fluoride is: (63.55 g/mol) + (2 * 18.9984 g/mol) = 121.5518 g/mol.

2. Convert the given mass of copper(II) fluoride to moles:
- The given mass of copper(II) fluoride is 28.5 g.
- To convert grams to moles, divide the given mass by the molar mass of copper(II) fluoride:
- Moles = 28.5 g / 121.5518 g/mol ≈ 0.2347 mol.

3. Determine the mass of fluorine:
- According to the percentage, 37.42% of the total mass of copper(II) fluoride is contributed by fluorine.
- To calculate the mass of fluorine, multiply the total mass of copper(II) fluoride by the percentage of fluorine:
- Mass of fluorine = (37.42% / 100%) * 28.5 g ≈ 10.68 g.

Therefore, the mass of fluorine in 28.5 g of copper(II) fluoride is approximately 10.68 grams.