Question

Question: How many atoms of nitrogen are present in 5.00 g of each of the following?

a. glycine,${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{O}}_{\mathbf{2}}\mathbf{N}$

b. magnesium nitride

c. calcium nitrate

d. dinitrogen tetroxide

Short answer

The atoms of nitrogen present:

a. $0.4011\times {10}^{23}\mathrm{Natoms}$

b. $0.596\times {10}^{23}\mathrm{Natoms}$

c. $0.367\times {10}^{23}\mathrm{Natoms}$

d. $0.654\times {10}^{23}\mathrm{Natoms}$

Step-by-step solution

Definition of mole

A mole is theamount of substance that contains many elementary entities of atoms in 0.012 kilogram of carbon 12.

a. glycine, C2H5O2N

The glycine mole is given as:

$$\begin{gathered} =\frac{5.00}{75.07{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}} \\ =0.0666\mathrm{moles} \end{gathered}$$

Now,

$$\begin{gathered} =0.0666\times 6.023\times {10}^{23}\mathrm{Natoms} \\ =0.4011\times {10}^{23}\mathrm{Natoms} \end{gathered}$$

b. magnesium nitride

The magnesium mole is given as:

$$\begin{gathered} =\frac{5.00\mathrm{g}}{100.95{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}} \\ =0.0495\mathrm{moles} \end{gathered}$$

Now,

$$\begin{gathered} =0.0990\times 6.023\times {10}^{23}\mathrm{Natoms} \\ =0.596\times {10}^{23}\mathrm{Natoms} \end{gathered}$$

c. calcium nitrate

The calcium mole is given as:

$$\begin{gathered} =\frac{5.00\mathrm{g}}{164.1{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}} \\ =0.0304\mathrm{moles} \end{gathered}$$

Now,

$$\begin{gathered} =0.0609\times 6.023\times {10}^{23}\mathrm{Natoms} \\ =0.367\times {10}^{23}\mathrm{Natoms} \end{gathered}$$

d. dinitrogen tetroxide

Thenitrogen mole is given as:

$$\begin{gathered} =\frac{5.00\mathrm{g}}{92.02{\displaystyle \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}} \\ =0.0543\mathrm{moles} \end{gathered}$$

Now,

$$\begin{gathered} =2\times 0.0543\times 6.023\times {10}^{23}\mathrm{Natoms} \\ =0.654\times {10}^{23}\mathrm{Natoms} \end{gathered}$$