Question

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

Answer 1

Answer:

There are $1.202\times 10^{24}$ atoms and $1.502\times 10^{23}$ molecules in the compound.

Explanation:

The molar mass of the sulphur is $32.065\,\frac{g}{mol}$. The Avogradro's Law states that exists $6.022\times 10^{23}\,\frac{atom}{mol}$. 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:

$n_{atom} = m_{S}\cdot \frac{n_{A}}{M_{S}}$

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 $m_{S} = 64\,g$, $n_{A} = 6.022\times 10^{23}\,\frac{atoms}{mol}$ and $M_{S} = 32.065\,\frac{g}{mol}$, then:

$n_{atom} = (64\,g)\cdot \left(\frac{6.022\times 10^{23}\,\frac{atoms}{mol} }{32.065\,\frac{g}{mol} }\right)$

$n_{atom} = 1.202\times 10^{24}\,atoms$

There are $1.202\times 10^{24}$ atoms in the compound.

Now, the molecular weight of the compound is:

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

$M_{S_{8}} = 256.52\,\frac{g}{mol}$

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 $m_{S_{8}} = 64\,g$, $n_{A} = 6.022\times 10^{23}\,\frac{molecules}{mol}$ and $M_{S_{8}} = 256.52\,\frac{g}{mol}$, then:

$n_{molecule} = (64\,g)\cdot \left(\frac{6.022\times 10^{23}\,\frac{molecules}{mol} }{256.52\,\frac{g}{mol} }\right)$

$n_{molecule} = 1.502\times 10^{23}\,molecules$

There are $1.502\times 10^{23}$ molecules in the compound.