Question

Which of the following contains the greatest number of atoms?
(Atomic masses: H = 1 u, N = 14 u, O = 16 u, C = 12 u, Ag = 108 u)

• A. 1.0 g butane $\left(C_4{H}_{10}\right)$
• B. 1.0 g nitrogen $\left(N_2\right)$
• C. 1.0 g silver $\left(Ag\right)$
• D. 1.0 g water $\left(H_{2}O\right)$

Answer 1

The correct option is A 1.0 g butane $\left(C_4{H}_{10}\right)$

(A) $1g$ of butane $C_4{H}_{10}$

$Molarmass=(at.mass.ofC\times 4)+(at.massofH\times 10)=(12\times 4)+(1\times 10)=58$

No. of moles $=\frac{Mass}{Molarmass}=\frac{1}{58}$

Total no. of atoms in one molecule of $C_4{H}_{10}$ = 4 carbon atoms + 10 hydrogen atoms = 14

No. of atoms in 1 g butane $=\frac{1}{58}\times 14\times N_A$, where $N_A=$ [Avogadro Number: $6.023\times {10}^{23}]$ $=0.24N_A$

(B) $1g$ of $N_2$

No. of moles $=\frac{Mass}{Molarmass}=\frac{1}{2\times 14}=\frac{1}{28}$

Total no. of atoms in one molecule of $N_2$ = 2

No. of atoms in 1 g $N_2$ $=2\times \frac{1}{28}\times N_A=\frac{1}{14}{N}_A=0.071N_A$

(C) $1g$ of $Ag$

No. of moles $=\frac{Mass}{Molarmass}=\frac{1}{108}$

Total no. of atoms in one molecule of $Ag$ = 1

No. of atoms in 1 g $Ag$ $=1\times \frac{1}{108}\times N_A$ $=0.0092N_A$

(D) $1g$ of $H_{2}O$

No. of moles $=\frac{Mass}{Molarmass}=\frac{1}{(2\times 1)+16}=\frac{1}{18}$

Total no. of atoms in one molecule of $H_{2}O$ = 2 atoms of hydrogen + 1 atom of oxygen = 3

No. of atoms in 1 g $H_{2}O$ $=3\times \frac{1}{18}\times N_A=0.1667N_A$

Hence, option A, butane has more number of atoms.