Question

How many grams of F are in 205 g CaF2 ?

Answer 1

Answer:

We know that,

Atomic mass of calcium = 40g

Atomic mass of flourine = 19g

So,

Molecular mass of

$CaF_2 = 40 + 2(19) \\ = 40 + 38 \\ = 78g$

It give that,

in 78g of CaF2

38g of F is present.

So,

In 1g of CaF2 , presense of F =

$\frac{38}{78}$

So,

In 205g of CaF2 ,

$F \: is \: present = \dfrac{38}{78} \times 205 \\ = \frac{7790}{78} \\ =99.87g$

Answer 2

Answer:

The molecular mass of CaF2

= 40 +19* 2

= 78 gm.

Now, given mass of CaF2 is 205 gm

Therefore moles of CaF2

= (given mass)/(molar mass)

= 205 /78

= 2.628 moles.

Now if you look at the compound CaF2, you can see that

1 mole of CaF2 has 2 moles of Fluoride.

1 mole of CaF2 has 2 moles of Fluoride.as per this equation:

CaF2 => Ca+2 + 2F-

Therefore moles of Fluoride present

= 2 * moles of CaF2

= 2 * 2.628

= 5.25 moles of Fluoride.

Mass of fluorine present

= mole of fluorine * molar mass

= 5.25 * 19

= 99.87 gm

The answer is 99.87 gm of fluorine is present in 205 gm of CaF2.