Question

A good analytical balance (scale) can detect an increase in mass of 1 microgram. Suppose a film of gold is deposited on a 1-cm x 1-cm substrate. How thick would the film need to be before you could detect its mass? About how many atoms of gold would this correspond to?

Answer 1

Ans. # Density of gold metal = 19.32 g/ cm³

Given, minimum mass to be detected = 1.0 ug = 1.0 x 10^-6 g

Now,

Volume of 1.0 ug gold = Mass/ Density

                                    = (1.0 x 10^-6 g) / (19.32 g/ cm³)

                                    = 5.175983 x 10^-8 cm³

# Area of gold film = 1.0 cm x 1.0 cm = 1.0 cm²

Now,

            Thickness of gold film = Volume / area

                                    = (5.175983 x 10^-8 cm³) / (1.0 cm²)

                                    = 5.175983 x 10^-8 cm

# Atomic radius of gold atom = 1.46 angstrom = 1.46 x 10^-8 cm

Diameter of gold atom = 2 x radium = 2 x (1.46 x 10^-8 cm) = 2.92 x 10^-8 cm.

So, the thickness of one gold atom or one-atom thick sheet or layer of gold film is equal to 2.92 x 10^-8 cm = diameter of atom.

# Relative thickness of film = Required thickness of gold film / thickness of one atom

                                                = (5.175983 x 10^-8 cm) / (2.92 x 10^-8 cm/ atom)

                                                = 1.8 atom

                                                = 2 (nearest whole number)

Therefore, the film is around 2 atom thick.

# Calculating number of gold atom

            Moles of Au = Mass / molar mass

                                    = (1.0 x 10^-6 g) / (196.96657 g/ mol)

                                    = 5.0770 x 10^-9 mol

Number of atoms = Moles x Avogadro number

                                    = (5.0770 x 10^-9 mol) x (6.022 x 10²³ atoms)

                                    = 3.0574 x 10¹⁵ atoms