Question

How many grams of fluorine are contained in 8 molecules of boron trifluoride?

Answer 1

Each molecule of boron trifluoride (BF3) contains 3 fluorine atoms. Hence, 8 molecules BF3 would contain 24 fluorine atoms. Calculate the number of moles of fluorine,
     (24 fluorine atoms) x (1 mole/ 6.022 x10^23 atoms) = 3.985 x10^-23 moles fluorine

Multiply this number of moles by the molar mass of fluorine which is equal to 19 g/mol. The answer would be 4.78x10^-22 grams. 

Answer 2

Answer:

             7.57 × 10⁻²² g of F

Solution:

Data Given:

                 Number of Molecules  =  8

                 M.Mass of BF₃ =  67.82 g.mol⁻¹

                 Mass of Fluorine atoms  =  ?

Step 1: Calculate Moles of BF₃

           Moles  =  Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Putting value,

            Moles  =   8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

            Moles  =  1.33 × 10⁻²³ mol

Step 2: Calculate Mass of BF₃:

                   Moles  =  Mass ÷ M.Mass

Solving for Mass,

                   Mass  =  Moles × M.Mass

Putting values,

                   Mass  =  1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹

                   Mass  =  9.0 × 10⁻²² g

Step 3: Calculate Mass of Fluorine Atoms:

As,

                         67.82 g BF₃ contains  =  57 g of F

So,

                    9.0 × 10⁻²² g will contain  =  X g of F

Solving for X,

                       X =  (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g

                        X  =  7.57 × 10⁻²² g of F