Question

5. How many atoms and molecules of sulphur are present in 64.0 g of sulphur (S 8 )?

Answer 1

There are
`1.202* 10^(24)` atoms and
`1.502* 10^(23)` molecules in the compound.

Step-by-step explanation:

The molar mass of the sulphur is
`32.065\,(g)/(mol)`. The Avogradro's Law states that exists
`6.022* 10^(23)\,(atom)/(mol)`. The quantity of atoms in a quantity of mass is derived from dividing the mass by the molar mass and multiplying it by the Avogadro's Number. That is:

`n_(atom) = m_(S)\cdot (n_(A))/(M_(S))`

Where:

`m_(S)` - Mass of the sample, measured in grams.

`n_(A)` - Avogadro's Number, measured in atoms per mole.

`M_(S)` - Molar mass of the sulphur, measured in grams per mole.

If
`m_(S) = 64\,g`,
`n_(A) = 6.022* 10^(23)\,(atoms)/(mol)` and
`M_(S) = 32.065\,(g)/(mol)`, then:

`n_(atom) = (64\,g)\cdot \left((6.022* 10^(23)\,(atoms)/(mol) )/(32.065\,(g)/(mol) )\right)`

`n_(atom) = 1.202* 10^(24)\,atoms`

There are
`1.202* 10^(24)` atoms in the compound.

Now, the molecular weight of the compound is:

`M_{S_(8)} = 8\cdot \left(32.065\,(g)/(mol) \right)`

`M_{S_(8)} = 256.52\,(g)/(mol)`

The quantity of molecules in a quantity of mass is derived from dividing the mass by the molecular weight and multiplying it by the Avogadro's Number. That is:

`n_(molecule) = m_{S_(8)}\cdot \frac{n_(A)}{M_{S_(8)}}`

Where:

`m_{S_(8)}` - Mass of the sample, measured in grams.

`n_(A)` - Avogadro's Number, measured in atoms per mole.

`M_{S_(8)}` - Molecular weight of the compound (octosulphur), measured in grams per mole.

If
`m_{S_(8)} = 64\,g`,
`n_(A) = 6.022* 10^(23)\,(molecules)/(mol)` and
`M_{S_(8)} = 256.52\,(g)/(mol)`, then:

`n_(molecule) = (64\,g)\cdot \left((6.022* 10^(23)\,(molecules)/(mol) )/(256.52\,(g)/(mol) )\right)`

`n_(molecule) = 1.502* 10^(23)\,molecules`

There are
`1.502* 10^(23)` molecules in the compound.