Copper (2) fluoride contains 37.42% F by mass. calculate the mass of fluorine (in grams) contained in 35.6 g of copper(2) fluoride.
answered above.
To calculate the mass of fluorine contained in 35.6 g of copper(II) fluoride, we need to use the given percentage of F by mass.
First, we need to convert the percentage into a decimal form by dividing it by 100:
37.42% ÷ 100 = 0.3742
Next, we will use the molar masses of copper(II) fluoride (CuF2) and fluorine (F) to find the mass of fluorine in one mole of copper(II) fluoride.
The molar mass of copper (Cu) is 63.55 g/mol, and the molar mass of fluorine (F) is 18.998 g/mol. Since there are two fluorine atoms in one mole of copper(II) fluoride, we multiply the molar mass of fluorine by 2:
2 * 18.998 g/mol = 37.996 g/mol
This means that in one mole of copper(II) fluoride, the mass of fluorine is 37.996 grams.
To calculate the mass of fluorine in 35.6 g of copper(II) fluoride, we can use a proportion:
Mass of fluorine / Mass of copper(II) fluoride = Percentage of F by mass / 100
Let's plug in the values:
Mass of fluorine / 35.6 g = 0.3742 / 100
Now, we can solve for the mass of fluorine:
Mass of fluorine = 35.6 g * (0.3742 / 100)
Mass of fluorine = 0.133 g
Therefore, the mass of fluorine contained in 35.6 g of copper(II) fluoride is approximately 0.133 grams.