Question

How many bromine atoms are present in 39.9 g of CH2Br2

Answer 1

Molar mass of CH2Br2: 173.83 g/mol

Given: 39.9 g / 173.83 g/mol = 0.230 mol

0.230 mol x 6.022x10^23 atoms/mol =

1.39 x 10^23 atoms

Answer 2

2.77 × 10²³ Br Atoms

Solution:

Data Given:

Mass of CH₂Br₂ = 39.9 g

M.Mass of CH₂Br₂ = 173.83 g.mol⁻¹

Step 1: Calculate Moles of CH₂Br₂ as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 39.9 g ÷ 173.83 g.mol⁻¹

Moles = 0.23 mol

Step 2: Calculate number of CH₂Br₂ Molecules,

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of CH₂Br₂ Molecules can be written as,

Moles = Number of CH₂Br₂ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of CH₂Br₂ Molecules,

Number of CH₂Br₂ Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

Number of CH₂Br₂ Molecules = 0.23 mol × 6.022 × 10²³ Atoms.mol⁻¹

Number of CH₂Br₂ Molecules = 1.38 × 10²³ CH₂Br₂ Molecules

Step 3: Calculate Number of Br Atoms:

As,

1 Molecule of CH₂Br₂ contains = 2 Atoms of Br

So,

1.38 × 10²³ Molecules of CH₂Br₂ will contain = X Atoms of Br

Solving for X,

X = (1.38 × 10²³ CH₂Br₂ × 2 Br) ÷ 1 CH₂Br₂

X = 2.77 × 10²³ Br Atoms