Question

How would you find out how many atoms are there in a compound?

Answer 1

Hint: The total mass occupied by Avogadro’s number of particles I.e.,$6.022\times {10}^{23}$ particles of a substance is known as its molar mass. It is generally provided in the units of$gram\left(gm\right)$ and is numerically equal to the molar mass of the molecule. $N_0$or Avogadro’s number is the very same number which represents the amount of particles present in $1$ mole of a substance irrespective of whether the particle is an atom, molecule, element or compound. So, one mole of carbon atoms is equal to $6.022\times {10}^{23}$ carbon atoms. To solve such questions, we need a proper knowledge of unitary methods.

Complete step-by-step answer:
The question is about how to calculate the total number of atoms of a particular element in specific or of all the elements altogether. We need to first know the number of moles of the compound. We will be using the following relation if we will be provided by the weight of the compound in units of grams.
$n={\displaystyle \frac{w}{M}}$
Where w is the given weight of the compound in units of grams
M is the molecular weight of the compound in units of $gmo{l}^{-1}$

After knowledge of the number of moles of the compound, we will be using the chemical formula of the compound to know the total number of moles of the atoms. For example, one mole of $NaCl$ contains one mole of sodium atoms and one mole of chloride atoms i.e., two moles of atoms in total. To calculate the number of atoms, we will be using the following concept. One mole of any element will contain $6.022\times {10}^{23}$ atoms. Thus, we can calculate the total number of the atoms in a compound by the following method.

Additional information:
$1$ Gram atomic mass or one-gram atom is the mass of ${\displaystyle \frac{1}{12}}th$ of the mass of $C-12$ atoms,
Molar mass is the mass of $6.022\times {10}^{23}$ atoms (or one mole atom) of an element expressed in grams.

Note: We must take note about the units of weight and molar mass in the question as this sometimes leads to grave calculative errors especially when weight is given in units other than the standard S.I. units of $grams\left(gm\right)$, commonly the weight is provided in units of microgram, milligram, nanogram etc. other than the standard units of gram.