Which of the following contains the maximum number of atoms and why?

4g of $H_{2}$

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16g of $O_{2}$

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28g of $N_{2}$

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18g of $H_{2} O$

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Solution

The correct option is A
4g of $H_{2}$

Explanation for correct option:
The given elements can be compared only in terms of moles.
The element with maximum number of moles will have the maximum number of atoms.
Step 1: Given data
Mass of $H_{2}$ = 4g
Mass of $O_{2}$ = 16g
Mass of $N_{2}$ = 28g
Mass of $H_{2} O$ = 18g
Step 2: Formula used
A mole is defined as the mass of the substance which consists of an equal quantity of basic units.
The formula for the number of moles formula is expressed as: $N u m b e r o f m o l e s = \frac{G i v e n m a s s o f s u b s \tan c e}{M a s s o f 1 m o l e o f s u b s \tan c e}$
Step 3: Calculation of number of moles.
Number of moles of $H_{2}$
Mass of one mole of $H_{2}$ = 2g
Applying the formula for number of moles,
$N u m b e r o f m o l e s o f H_{2} = \frac{4}{2} = 2$
Number of moles of $O_{2}$
Mass of one mole of $O_{2}$ = 32g
Applying the formula for the number of moles,
$N u m b e r o f m o l e s o f O_{2} = \frac{16}{32} = 0.5$
Number of moles of $N_{2}$
Mass of one mole of $N_{2}$ = 28g
Applying the formula for the number of moles,
$N u m b e r o f m o l e s o f N_{2} = \frac{28}{28} = 1$
Number of moles of $H_{2} O$
Mass of one mole of $H_{2} O$ = 18g
Applying the formula for the number of moles,
$N u m b e r o f m o l e s o f H_{2} O = \frac{18}{18} = 1$
As 4g $H_{2}$ has the maximum number of moles, therefore, it has the maximum number of atoms.
Explanation for the incorrect options:
Option B: 16g of $O_{2}$
According to calculations, 16g of $O_{2}$ contains 0.5 moles.
Hence, it does not have the maximum number of atoms.
Therefore, is option is incorrect.
Option C: 28g of $N_{2}$
According to calculations, 28g of $N_{2}$ contains 1 mole.
Hence, it does not have the maximum number of atoms.
Therefore, is option is incorrect.
Option C: 18g of $H_{2} O$
According to calculations, 18g of $H_{2} O$ contains 1 mole.
Hence, it does not have the maximum number of atoms.
Therefore, is option is incorrect.
Hence, option A 4g of $H_{2}$ is the correct option.

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