Question

Which of the following contains the largest number of atoms?
A. $4{\text{g}}$ of ${{\text{H}}_2}$
B. $16{\text{g}}$ of ${{\text{O}}_2}$
C. $28{\text{g}}$ of ${{\text{N}}_2}$
D. $18{\text{g}}$ of ${{\text{H}}_2}{\text{O}}$

Answer 1

Hint:
Stoichiometry is the study of the quantitative aspects of chemical reactions. Chemical equations are concise representations of chemical reactions. Mole is defined as the quantity of a substance that contains the same number of ultimate particles as are present in $12{\text{g}}$ of carbon$ - 12$.

Complete step by step answer:
Stoichiometry deals with the numerical relationships of elements and compounds and the mathematical proportions of reactants and products in chemical transformations. Concentration is the amount of solute dissolved in a given amount of solution. There are different types of concentration units. Formula and molecular mass deal with individual atoms and molecules. Mole is the unit that relates the number of particles and mass.
One mole of an element contains $6.022 \times {10^{23}}$ particles. This absolute number is called Avogadro’s number. Mass of one mole of substance is called molar mass. Or molar mass of an element is equal to the molecular weight.

A. Molar mass of ${{\text{H}}_2}$, ${{\text{M}}_{{{\text{H}}_2}}} = 2{\text{g}}$
Mass of ${{\text{H}}_2}$, ${{\text{m}}_{{{\text{H}}_2}}} = 4{\text{g}}$
Number of moles of ${{\text{H}}_2}$, ${{\text{n}}_{{{\text{H}}_2}}} = \dfrac{{{{\text{m}}_{{{\text{H}}_2}}}}}{{{{\text{M}}_{{{\text{H}}_2}}}}} = \dfrac{{4{\text{g}}}}{{2{\text{g}}{\text{.mo}}{{\text{l}}^{ - 1}}}} = 2{\text{mol}}$
$2{\text{mol}}$ of ${{\text{H}}_2}$ contains $2 \times \left( {6.022 \times {{10}^{23}}} \right)$ atoms.

B. Molar mass of ${{\text{O}}_2}$, ${{\text{M}}_{{{\text{O}}_2}}} = 32{\text{g}}$
Mass of ${{\text{O}}_2}$, ${{\text{m}}_{{{\text{O}}_2}}} = 16{\text{g}}$
Number of moles of ${{\text{O}}_2}$, ${{\text{n}}_{{{\text{O}}_2}}} = \dfrac{{{{\text{m}}_{{{\text{O}}_2}}}}}{{{{\text{M}}_{{{\text{O}}_2}}}}} = \dfrac{{{\text{16g}}}}{{32{\text{g}}{\text{.mo}}{{\text{l}}^{ - 1}}}} = 0.5{\text{mol}}$
${\text{0}}{\text{.5mol}}$ of ${{\text{O}}_2}$ contains $0.5 \times \left( {6.022 \times {{10}^{23}}} \right)$ atoms.

C. Molar mass of ${{\text{N}}_2}$, ${{\text{M}}_{{{\text{N}}_2}}} = 28{\text{g}}$
Mass of ${{\text{N}}_2}$, \[{{\text{m}}_{{{\text{N}}_2}}} = 28{\text{g}}\]
Number of moles of ${{\text{N}}_2}$, ${{\text{n}}_{{{\text{N}}_2}}} = \dfrac{{{{\text{m}}_{{{\text{N}}_2}}}}}{{{{\text{M}}_{{{\text{N}}_2}}}}} = \dfrac{{{\text{28g}}}}{{{\text{28g}}{\text{.mo}}{{\text{l}}^{ - 1}}}} = 1{\text{mol}}$
${\text{1mol}}$ of ${{\text{N}}_2}$ contains $1 \times \left( {6.022 \times {{10}^{23}}} \right)$ atoms.

D. Molar mass of ${{\text{H}}_2}{\text{O}}$, ${{\text{M}}_{{{\text{H}}_2}{\text{O}}}} = 18{\text{g}}$
Mass of ${{\text{H}}_2}{\text{O}}$ , \[{{\text{m}}_{{{\text{H}}_2}{\text{O}}}} = 18{\text{g}}\]
Number of moles of ${{\text{H}}_2}{\text{O}}$, ${{\text{n}}_{{{\text{H}}_2}{\text{O}}}} = \dfrac{{{{\text{m}}_{{{\text{H}}_2}{\text{O}}}}}}{{{{\text{M}}_{{{\text{H}}_2}{\text{O}}}}}} = \dfrac{{{\text{18g}}}}{{{\text{18g}}{\text{.mo}}{{\text{l}}^{ - 1}}}} = 1{\text{mol}}$
${\text{1mol}}$ of ${{\text{H}}_2}{\text{O}}$ contains $1 \times \left( {6.022 \times {{10}^{23}}} \right)$ atoms.
Hence $4{\text{g}}$ of ${{\text{H}}_2}$ contains $2 \times \left( {6.022 \times {{10}^{23}}} \right)$ atoms which is larger than others.

Hence option A is correct.


Note:
Moles provide a bridge from molecular scale to real-world scale. One mole of molecules or formula units contain Avogadro number times the number of atoms or ions of each element in the compound. Each chemical equation provides information about the amount of reactants produced.