Question

Calculate the mass of $6.022\text{ x }{{10}^{21}}$ molecules of nitrogen.

Answer 1

Hint: The number of moles of nitrogen is calculated by dividing the given number of molecules of nitrogen to the Avogadro's number. The mass of the given molecules can be calculated by multiplying the number of moles with the molecular mass of the nitrogen molecule.

Complete answer:
A mole of a compound is known as that amount of the substance which has a mass equal to gram formula mass or which contains Avogadro's number ($6.022\text{ x 1}{{\text{0}}^{23}}$) of formula units.
Avogadro’s number may be defined as the number of atoms present in one gram atom of the element or the number of molecules present in one gram molecule of the substance.
Now in the question the given number of nitrogen molecules = $6.022\text{ x }{{10}^{21}}$
To find the number of moles of nitrogen, we can divide the given number of molecules with the Avogadro’s number.
So, the formula is: $\text{moles = }\dfrac{\text{given number of molecules}}{\text{Avogadro }\!\!'\!\!\text{ s number}}$
The value of Avogadro’s number is $6.022\text{ x 1}{{\text{0}}^{23}}$.
So, putting the values in the formula, we get
$\text{number of moles = }\dfrac{6.022\text{ x 1}{{\text{0}}^{21}}}{6.022\text{ x 1}{{\text{0}}^{23}}}=0.01\text{ moles}$
We know that the number of moles is also equal to the value of the given mass of the atom or molecule divided by the molar mass of the atom or molecule.
The molar mass of nitrogen = \[28\text{ }g\text{ }/\text{ }mol\]
Therefore, the mass of the molecule is equal to the number of moles multiplied to the molar mass.
The formula is: $\text{mass = (moles x molar mass)}$
By putting the values, we get
$\text{mass = 0}\text{.01 x 28 = 0}\text{.28 g}$
Hence, the mass of nitrogen is 0.28 g.

Note: Avogadro's number and mole concept is used for calculating the actual mass of a single atom or a single molecule, the number of atoms present in a given mass of an element, number of molecules present in a given volume of gas, size of atoms, etc.