Question

How many fluorine atoms are present in 5.85 g of c2f4?

Answer 1

M C2F4 = 100 g/mol

In 100 g of C2H4 we have 4 × 6,02 × 10^23 atoms of fluorine.

In 5,85 g of C2F4 we have x atoms of fluorine.

100 g ---- 2,408 × 10^24 atoms
5,85 g --- x atoms

x = 5,85 × 2,408 × 10^24 / 100 = 1,408 × 10^23

5,85 g of C2F4 contains 1,408 × 10^23 atoms of fluorine.

:-) ;-)

Answer 2

Answer:

$1.41x10^{23} atomsF$

Explanation:

Hello,

At first, molecular mass of dicarbon tetrafluoride is:

$M_{C_2F_4}=12*2+19*4=100g/mol$

Now, by applying the mole-mass relationship with the Avogradro's number one finds that:

$5.85gC_2F_4*\frac{1molC_2F_4}{100gC_2F_4}*\frac{4molF}{1molC_2F_4} *\frac{6.022x10^{23}atomsF }{1molF} =1.41x10^{23} atomsF$

Best regards.