Question

Copper(II) fluoride contains 37.42% F by mass. Calculate the mass of fluorine (in g) in 55.5 g of copper(II) fluoride.

Answer 1

Answer:

There are 20.8 g of fluorine in 55.5 g of copper (II) fluoride

Explanation:

x % by mass of a species in a specimen means there are x g of the species in total 100 g of a specimen

37.42 % F by mass means 100 g of copper (II) fluoride contains 37.42 g of F.

So, 100 g of copper (II) fluoride contains 37.42 g of F

55.5 g of copper (II) fluoride contains $\frac{37.42\times 55.5}{100}$g of F or 20.8 g of F

Hence there are 20.8 g of fluorine in 55.5 g of copper (II) fluoride.

Answer 2

The mass of fluorine present in 55.5 g of copper is 20.08 grams.

What is copper fluoride?

Copper fluoride is an inorganic compound whose chemical formula is $CuF_2$. It is a white crystalline solid.

Given, the mass of F is 37.42%

To find the mass of fluorine when the of copper is 55.5 g

% means there is that gram of the elements is present in total 100 g.

That means 37.42% means 37.42 grams of F is present in 100 grams of $CuF_2$

So, in 55.5 g of $CuF_2$

$\dfrac{37.42 \times 55.5 }{100} g = 20.08\; grams \;of\;F$

Thus, the mass of fluorine present in 55.5 g of copper is 20.08 grams.

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