Chapter 12, Problem 19E

To determine

The percentage of nitrogen by mass in each of the given compound has to be calculated.

Answer to Problem 19E

Percentage mass of nitrogen in urea is 47 percent.

Percentage mass of nitrogen in Ammonium nitrate is 35 percent.

Percentage mass of nitrogen in Ammonium sulfate is 21 percent.

Explanation of Solution

Mole of any elements is defined as “The amount of the element whose, mass in grams is equal to the atomic mass which is expressed in atomic mass units ($\text{u}$).  This definition means that mole of any elements will contain the same number of atoms as mole of other element.  This constant value is known as Avogadro’s number.

Avogadro’s number ($N_0$) is the number of atoms that is present in per mole of any element or number of formula units present in per mole of any substance.  This is equal to $6.02\times {10}^{23}\text{atoms/mol}$.  The total number of hydrogen molecules and oxygen molecules in one mole will be equal to the Avogadro’s number.

The product of total number of moles and the atomic mass of the substance gives the total mass of the substance.

Formula mass of urea (${\text{CO(NH}}_{\text{2}}{\text{)}}_{\text{2}}$) can be calculated as shown below,

$\begin{array}{l}\text{Total number of nitrogen atoms}\text{=}\text{2}\\ \text{Mass of one Nitrogen atom}\text{=}\text{14}\text{.00}\text{u}\\ \therefore \text{Mass of two Nitrogen atoms}\text{=}\text{2}\times \text{14}\text{.00}\text{u}\\ \text{=}\text{28}\text{.00}\text{u}\\ \text{Total number of Oxygen atoms}\text{=}\text{1}\\ \text{Mass of one Oxygen atom}\text{=}\text{16}\text{.00}\text{u}\\ \text{Total number of Carbon atoms}\text{=}\text{1}\\ \text{Mass of one Carbon atom}\text{=}\text{12}\text{.00}\text{u}\\ \text{Total number of hydrogen atoms}\text{=}\text{4}\\ \text{Mass of one Hydrogen atom}\text{=}\text{1}\text{.008}\text{u}\\ \therefore \text{Mass of four Hydrogen atoms}\text{=}\text{4}\times \text{1}\text{.008}\text{u}\\ \text{=}\text{4}\text{.032}\text{u}\end{array}$

Formula mass of Urea can be calculated as,

$1m_C+1m_N+1m_O+4m_H=60.03\text{g/mol}$

Therefore,

$\begin{array}{l}\text{Percentage mass of Nitrogen}\text{=}\frac{\text{28}\text{g}}{\text{60}\text{.03}\text{g/mol}}\text{×}\text{100}\\ \text{=}\text{47}\text{percent}\end{array}$

The percentage mass of nitrogen in urea is 47 percent.

Formula mass of ammonium nitrate (${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$) can be calculated as shown below,

$\begin{array}{l}\text{Total number of nitrogen atoms}\text{=}\text{2}\\ \text{Mass of one Nitrogen atom}\text{=}\text{14}\text{.00}\text{u}\\ \therefore \text{Mass of two Nitrogen atoms}\text{=}\text{2}\times \text{14}\text{.00}\text{u}\\ \text{=}\text{28}\text{.00}\text{u}\\ \text{Total number of Oxygen atoms}\text{=}\text{3}\\ \text{Mass of one Oxygen atom}\text{=}\text{16}\text{.00}\text{u}\\ \therefore \text{Mass of two Oxygen atoms}\text{=}\text{3}\times \text{16}\text{.00}\text{u}\\ \text{=}\text{48}\text{.00}\text{u}\\ \text{Total number of hydrogen atoms}\text{=}\text{4}\\ \text{Mass of one Hydrogen atom}\text{=}\text{1}\text{.008}\text{u}\\ \therefore \text{Mass of four Hydrogen atoms}\text{=}\text{4}\times \text{1}\text{.008}\text{u}\\ \text{=}\text{4}\text{.032}\text{u}\end{array}$

Formula mass of Ammonium nitrate can be calculated as,

$2m_N+3m_O+4m_H=80.03\text{g/mol}$

Therefore,

$\begin{array}{l}\text{Percentage mass of Nitrogen}\text{=}\frac{\text{28}\text{g}}{\text{80}\text{.03}\text{g/mol}}\text{×}\text{100}\\ \text{=}\text{35 percent}\end{array}$

The percentage mass of nitrogen in ammonium nitrate is 35 percent.

Formula mass of ammonium sulfate (${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$) can be calculated as shown below,

$\begin{array}{l}\text{Total number of nitrogen atoms}\text{=}\text{2}\\ \text{Mass of one Nitrogen atom}\text{=}\text{14}\text{.00}\text{u}\\ \therefore \text{Mass of two Nitrogen atoms}\text{=}\text{2}\times \text{14}\text{.00}\text{u}\\ \text{=}\text{28}\text{.00}\text{u}\\ \text{Total number of Oxygen atoms}\text{=}\text{3}\\ \text{Mass of one Oxygen atom}\text{=}\text{16}\text{.00}\text{u}\\ \therefore \text{Mass of four Oxygen atoms}\text{=}\text{4}\times \text{16}\text{.00}\text{u}\\ \text{=}\text{64}\text{.00}\text{u}\\ \text{Total number of hydrogen atoms}\text{=}\text{8}\\ \text{Mass of one Hydrogen atom}\text{=}\text{1}\text{.008}\text{u}\\ \therefore \text{Mass of eight Hydrogen atoms}\text{=}8\times \text{1}\text{.008}\text{u}\\ \text{=}8.\text{064}\text{u}\\ \text{Total number of sulfur atoms}\text{=}\text{1}\\ \text{Mass of one Sulfur atom}\text{=}\text{32}\text{.56}\text{u}\end{array}$

Formula mass of Ammonium sulfate can be calculated as,

$2m_N+4m_O+8m_H+1m_S=132.62\text{g/mol}$

Therefore,

$\begin{array}{l}\text{Percentage mass of Nitrogen}\text{=}\frac{\text{28}\text{g}}{\text{132}\text{.62}\text{g/mol}}\text{×}\text{100}\\ \text{=}\text{21 percent}\end{array}$

The percentage mass of nitrogen in ammonium sulfate is 21 percent.

Conclusion:

The percentage of nitrogen by mass in each of the given compound was calculated.