Chapter 2, Problem 2.73P

Interpretation Introduction

Interpretation:

Fertilizers given in the problem statement has to be arranged in decreasing order of mass percentage of nitrogen.

Concept Introduction:

Relative amount of the elements present in a particular compound is same always.  It does not matter where the compound is obtained or prepared.  Mass percentage can be calculated by simple calculation using the mass of the element present in the compound and the total mass of the compound.

Explanation of Solution

(a)

Mass percentage of nitrogen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$:

Molecular formula of fertilizer is ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$.  Atomic mass of $\text{N}$ is $\text{14}\text{.0067}\text{g/mol}$.  Atomic mass of $\text{H}$ is $\text{1}\text{.00794}\text{g/mol}$.  Atomic mass of $\text{O}$ is $\text{15}\text{.9994}\text{g/mol}$.  Molecular mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is calculated as shown below.

  $\begin{array}{c}\text{Molecular}\text{mass}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}=2(\text{atomic}\text{mass}\text{of}\text{N)}+4(\text{atomic}\text{mass}\text{of}\text{H)}\\ +3(\text{atomic}\text{mass}\text{of}\text{O)}\\ =2(14.0067\text{g/mol) + 4(1}\text{.00794}\text{g/mol)}\\ \text{ + 3(15}\text{.9994}\text{g/mol)}\\ =\text{28}\text{.0134}\text{g/mol + 4}\text{.03176}\text{g/mol + 47}\text{.9982}\text{g/mol}\\ =\text{80}\text{.043}\text{g/mol}\end{array}$

Therefore, molecular mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $\text{80}\text{.043 g/mol}$.

Mass percentage of nitrogen:

Considering the atomic mass of nitrogen, and the molecular mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$, the mass percentage of nitrogen can be calculated as shown below.

    $\begin{array}{c}\text{Mass}\text{Percentage}\text{of}\text{Nitrogen}=\frac{\text{Atomic mass}\text{of}\text{nitrogen}}{\text{Molecular mass}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\times 100\\ =\frac{2(14.0067\text{g/mol)}}{\text{80}\text{.043}\text{g/mol}}\times 100\\ =34.98\%\end{array}$

Therefore, mass percentage of nitrogen in ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $34.98\%$.

(b)

Mass percentage of nitrogen in ${\text{NH}}_{\text{3}}$:

Molecular formula of fertilizer is ${\text{NH}}_{\text{3}}$.  Atomic mass of $\text{N}$ is $\text{14}\text{.0067}\text{g/mol}$.  Atomic mass of $\text{H}$ is $\text{1}\text{.00794}\text{g/mol}$.  Molecular mass of ${\text{NH}}_{\text{3}}$ is calculated as shown below.

  $\begin{array}{c}\text{Molecular}\text{mass}\text{of}{\text{NH}}_{\text{3}}=1(\text{atomic}\text{mass}\text{of}\text{N)}+3(\text{atomic}\text{mass}\text{of}\text{H)}\\ =1(14.0067\text{g/mol) + 3(1}\text{.00794}\text{g/mol)}\\ =\text{14}\text{.0067}\text{g/mol + 3}\text{.02382}\text{g/mol}\\ =\text{17}\text{.031}\text{g/mol}\end{array}$

Therefore, molecular mass of ${\text{NH}}_{\text{3}}$ is $\text{17}\text{.031 g/mol}$.

Mass percentage of nitrogen:

Considering the atomic mass of nitrogen, and the molecular mass of ${\text{NH}}_{\text{3}}$, the mass percentage of nitrogen can be calculated as shown below.

    $\begin{array}{c}\text{Mass}\text{Percentage}\text{of}\text{Nitrogen}=\frac{\text{Atomic mass}\text{of}\text{nitrogen}}{\text{Molecular mass}\text{of}{\text{NH}}_{\text{3}}}\times 100\\ =\frac{14.0067\text{g/mol}}{\text{17}\text{.031}\text{g/mol}}\times 100\\ =82.24\%\end{array}$

Therefore, mass percentage of nitrogen in ${\text{NH}}_{\text{3}}$ is $82.24\%$.

(c)

Mass percentage of nitrogen in ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$:

Molecular formula of fertilizer is ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$.  Atomic mass of $\text{N}$ is $\text{14}\text{.0067}\text{g/mol}$.  Atomic mass of $\text{H}$ is $\text{1}\text{.00794}\text{g/mol}$.  Atomic mass of $\text{O}$ is $\text{15}\text{.9994}\text{g/mol}$.  Atomic mass of sulfur is $\text{32}\text{.065}\text{g/mol}$.  Molecular mass of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated as shown below.

  $\begin{array}{c}\text{Molecular}\text{mass}\text{of}{{\text{(NH}}_{\text{4}}}_{\text{)}}{\text{SO}}_{\text{4}}=2(\text{atomic}\text{mass}\text{of}\text{N)}+8(\text{atomic}\text{mass}\text{of}\text{H)}\\ +4(\text{atomic}\text{mass}\text{of}\text{O)}+1(\text{atomic}\text{mass}\text{of}\text{S)}\\ =2(14.0067\text{g/mol) + 8(1}\text{.00794}\text{g/mol)}\\ \text{ + 4(15}\text{.9994}\text{g/mol)+ 1(32}\text{.065}\text{g/mol)}\\ =\text{28}\text{.0134}\text{g/mol + 8}\text{.06352}\text{g/mol }\\ \text{ + 63}\text{.9976}\text{g/mol+ 32}\text{.065}\text{g/mol}\\ =\text{132}\text{.14 g/mol}\end{array}$

Therefore, molecular mass of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{132}\text{.14 g/mol}$.

Mass percentage of nitrogen:

Considering the atomic mass of nitrogen, and the molecular mass of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$, the mass percentage of nitrogen can be calculated as shown below.

    $\begin{array}{c}\text{Mass}\text{Percentage}\text{of}\text{Nitrogen}=\frac{\text{Atomic mass}\text{of}\text{nitrogen}}{\text{Molecular mass}\text{of}{{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}}\times 100\\ =\frac{2(14.0067\text{g/mol)}}{\text{132}\text{.14}\text{g/mol}}\times 100\\ =21.19\%\end{array}$

Therefore, mass percentage of nitrogen in ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $21.19\%$.

(d)

Mass percentage of nitrogen in ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{HPO}}_{\text{4}}$:

Molecular formula of fertilizer is ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{HPO}}_{\text{4}}$.  Atomic mass of $\text{N}$ is $\text{14}\text{.0067}\text{g/mol}$.  Atomic mass of $\text{H}$ is $\text{1}\text{.00794}\text{g/mol}$.  Atomic mass of $\text{O}$ is $\text{15}\text{.9994}\text{g/mol}$.  Atomic mass of phosphorus is $\text{30}\text{.973761}\text{g/mol}$.  Molecular mass of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is calculated as shown below.

  $\begin{array}{c}\text{Molecular}\text{mass}\text{of}{{\text{(NH}}_{\text{4}}}_{\text{)}}{\text{HPO}}_{\text{4}}=2(\text{atomic}\text{mass}\text{of}\text{N)}+9(\text{atomic}\text{mass}\text{of}\text{H)}\\ +4(\text{atomic}\text{mass}\text{of}\text{O)}+1(\text{atomic}\text{mass}\text{of}\text{P)}\\ =2(14.0067\text{g/mol) + 9(1}\text{.00794}\text{g/mol)}\\ \text{ + 4(15}\text{.9994}\text{g/mol)+ 1(30}\text{.973761}\text{g/mol)}\\ =\text{28}\text{.0134}\text{g/mol + 9}\text{.07146}\text{g/mol }\\ \text{ + 63}\text{.9976}\text{g/mol+ 30}\text{.973761}\text{g/mol}\\ =\text{132}\text{.06 g/mol}\end{array}$

Therefore, molecular mass of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is $\text{132}\text{.06 g/mol}$.

Mass percentage of nitrogen:

Considering the atomic mass of nitrogen, and the molecular mass of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{HPO}}_{\text{4}}$, the mass percentage of nitrogen can be calculated as shown below.

    $\begin{array}{c}\text{Mass}\text{Percentage}\text{of}\text{Nitrogen}=\frac{\text{Atomic mass}\text{of}\text{nitrogen}}{\text{Molecular mass}\text{of}{{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}}\times 100\\ =\frac{2(14.0067\text{g/mol)}}{\text{132}\text{.06}\text{g/mol}}\times 100\\ =21.21\%\end{array}$

Therefore, mass percentage of nitrogen in ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is $21.21\%$.

(e)

Mass percentage of nitrogen in ${\text{(NH}}_{\text{4}}{\text{)H}}_{\text{2}}{\text{PO}}_{\text{4}}$:

Molecular formula of fertilizer is ${\text{(NH}}_{\text{4}}{\text{)H}}_{\text{2}}{\text{PO}}_{\text{4}}$.  Atomic mass of $\text{N}$ is $\text{14}\text{.0067}\text{g/mol}$.  Atomic mass of $\text{H}$ is $\text{1}\text{.00794}\text{g/mol}$.  Atomic mass of $\text{O}$ is $\text{15}\text{.9994}\text{g/mol}$.  Atomic mass of phosphorus is $\text{30}\text{.973761}\text{g/mol}$.  Molecular mass of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{HPO}}_{\text{4}}$ is calculated as shown below.

  $\begin{array}{c}\text{Molecular}\text{mass}\text{of}{\text{(NH}}_{\text{4}}{\text{)H}}_{\text{2}}{\text{PO}}_{\text{4}}=1(\text{atomic}\text{mass}\text{of}\text{N)}+6(\text{atomic}\text{mass}\text{of}\text{H)}\\ +4(\text{atomic}\text{mass}\text{of}\text{O)}+1(\text{atomic}\text{mass}\text{of}\text{P)}\\ =1(14.0067\text{g/mol) + 6(1}\text{.00794}\text{g/mol)}\\ \text{ + 4(15}\text{.9994}\text{g/mol)+ 1(30}\text{.973761}\text{g/mol)}\\ =\text{14}\text{.0067}\text{g/mol + 6}\text{.04764}\text{g/mol }\\ \text{ + 63}\text{.9976}\text{g/mol+ 30}\text{.973761}\text{g/mol}\\ =\text{115}\text{.03 g/mol}\end{array}$

Therefore, molecular mass of ${\text{(NH}}_{\text{4}}{\text{)H}}_{\text{2}}{\text{PO}}_{\text{4}}$ is $\text{115}\text{.03 g/mol}$.

Mass percentage of nitrogen:

Considering the atomic mass of nitrogen, and the molecular mass of ${\text{(NH}}_{\text{4}}{\text{)H}}_{\text{2}}{\text{PO}}_{\text{4}}$, the mass percentage of nitrogen can be calculated as shown below.

    $\begin{array}{c}\text{Mass}\text{Percentage}\text{of}\text{Nitrogen}=\frac{\text{Atomic mass}\text{of}\text{nitrogen}}{\text{Molecular mass}\text{of}{{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}}\times 100\\ =\frac{1(14.0067\text{g/mol)}}{\text{115}\text{.03}\text{g/mol}}\times 100\\ =12.18\%\end{array}$

Therefore, mass percentage of nitrogen in ${\text{(NH}}_{\text{4}}{\text{)H}}_{\text{2}}{\text{PO}}_{\text{4}}$ is $12.18\%$.

(f)

Mass percentage of nitrogen in ${\text{KNO}}_{\text{3}}$:

Molecular formula of fertilizer is ${\text{KNO}}_{\text{3}}$.  Atomic mass of $\text{N}$ is $\text{14}\text{.0067}\text{g/mol}$.  Atomic mass of $\text{K}$ is $\text{39}\text{.0983}\text{g/mol}$.  Atomic mass of $\text{O}$ is $\text{15}\text{.9994}\text{g/mol}$.  Molecular mass of ${\text{KNO}}_{\text{3}}$ is calculated as shown below.

  $\begin{array}{c}\text{Molecular}\text{mass}\text{of}{\text{KNO}}_{\text{3}}=1(\text{atomic}\text{mass}\text{of}\text{N)}+1(\text{atomic}\text{mass}\text{of}\text{K)}\\ +3(\text{atomic}\text{mass}\text{of}\text{O)}\\ =1(14.0067\text{g/mol) + 1(39}\text{.0983}\text{g/mol)}\\ \text{ + 3(15}\text{.9994}\text{g/mol)}\\ =\text{14}\text{.0067}\text{g/mol + 39}\text{.0983}\text{g/mol + 47}\text{.9982}\text{g/mol}\\ =\text{101}\text{.10}\text{g/mol}\end{array}$

Therefore, molecular mass of ${\text{KNO}}_{\text{3}}$ is $\text{101}\text{.10 g/mol}$.

Mass percentage of nitrogen:

Considering the atomic mass of nitrogen, and the molecular mass of ${\text{KNO}}_{\text{3}}$, the mass percentage of nitrogen can be calculated as shown below.

    $\begin{array}{c}\text{Mass}\text{Percentage}\text{of}\text{Nitrogen}=\frac{\text{Atomic mass}\text{of}\text{nitrogen}}{\text{Molecular mass}\text{of}{\text{KNO}}_{\text{3}}}\times 100\\ =\frac{1(14.0067\text{g/mol)}}{\text{101}\text{.10}\text{g/mol}}\times 100\\ =13.85\%\end{array}$

Therefore, mass percentage of nitrogen in ${\text{KNO}}_{\text{3}}$ is $13.85\%$.

The order of decreasing mass percentage of nitrogen in the given fertilizer can be depicted as shown below.