Chapter 3, Problem 3.42QP

(a)

Interpretation Introduction

Interpretation: Based on mass percentage the richest source of nitrogen has to be identified from the given fertilizers.

Concept introduction: The mass percentage of each elements present in a compound is known as the percent composition by mass of the compound.

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{an}\text{element}\text{=}\frac{\text{n}\text{×atomic}\text{mass}\text{of}\text{element}}{\text{molecular}\text{or}\text{formula}\text{mass}\text{of}\text{compound}}\text{×100\%}\\ \text{n}\text{-}\text{The}\text{number}\text{of}\text{elements present}\text{in}\text{a}\text{molecule}\text{or}\text{formula}\text{unit}\text{of}\text{the}\text{compound}\end{array}$

To determine: The percentage mass of nitrogen in urea.

Answer to Problem 3.42QP

The richest source of nitrogen by mass is ammonia $\left({\text{NH}}_{\text{3}}\right)$

$\text{percentage}\text{mass}\text{of}\text{N}\text{in}\text{ammonia}\text{=}\text{82}\text{.2447}\text{\%}$

Explanation of Solution

Molecular mass of urea is calculated as follows:

$\text{Molecular}\text{formula}\text{of}\text{urea}\text{is}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\text{CO}$

$\begin{array}{l}\text{Molecular mass of urea}\text{=}\left(\left(\text{2×mass}\text{of}\text{N}\right)\text{+}\left(\text{4×mass}\text{of}\text{H}\right)\text{+}\text{mass}\text{of}\text{C}\text{+}\text{mass}\text{of}\text{O}\right)\text{amu}\\ \text{=}\left(\left(\text{2×14}\text{.0067}\right)\text{+}\left(\text{4×1}\text{.00794}\right)\text{+}\text{12}\text{.0107}\text{+}\text{15}\text{.9994}\right)\text{amu}\\ \text{=}\left(\text{28}\text{.0134}\text{+}\text{4}\text{.03176}\text{+}\text{12}\text{.0107}\text{+}\text{15}\text{.9994}\right)\text{amu}\\ \text{=}\text{60}\text{.05526}\text{amu}\end{array}$ The molecular mass of urea was calculated by the sum atomic masses nitrogen, hydrogen, carbon and oxygen atoms.

The percentage mass of nitrogen in ammonia is calculated as follows:

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{N}\text{in}\text{Urea}\text{=}\frac{\text{2×atomic}\text{mass}\text{of}\text{N}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}\text{Urea}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{2×14}\text{.0067}\text{amu}}{\text{60}\text{.05526}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{28}\text{.0134}\text{amu}}{\text{60}\text{.05526}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{46}\text{.6460}\text{\%}\end{array}$

$\text{percentage}\text{mass}\text{of}\text{N}\text{in}\text{Urea}\text{=}\text{46}\text{.646}\text{\%}$

The percentage mass of an element in its compound is calculated by using the atomic mass of that element and molecular mass of the compound.

(b)

Interpretation Introduction

Interpretation: Based on mass percentage the richest source of nitrogen has to be identified from the given fertilizers.

Concept introduction: The mass percentage of each elements present in a compound is known as the percent composition by mass of the compound.

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{an}\text{element}\text{=}\frac{\text{n}\text{×atomic}\text{mass}\text{of}\text{element}}{\text{molecular}\text{or}\text{formula}\text{mass}\text{of}\text{compound}}\text{×100\%}\\ \text{n}\text{-}\text{The}\text{number}\text{of}\text{elements present}\text{in}\text{a}\text{molecule}\text{or}\text{formula}\text{unit}\text{of}\text{the}\text{compound}\end{array}$

To determine: The percentage mass of nitrogen in ammonium nitrate.

Explanation of Solution

The molecular mass of ammonium nitrate is calculated as follows:

$\text{molecular formula of ammonium nitrate is}\left({\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\right)$

$\begin{array}{l}\text{Molecular mass of Ammonium nitrate}\text{=}\left(\left(\text{2×mass}\text{of}\text{N}\right)\text{+}\left(\text{4×mass}\text{of}\text{H}\right)\text{+}\left(\text{3×mass}\text{of}\text{O}\right)\right)\text{amu}\\ \text{=}\left(\left(\text{2×14}\text{.0067}\right)\text{+}\left(\text{4×1}\text{.00794}\right)\text{+}\left(\text{3×}\text{15}\text{.9994}\right)\right)\text{amu}\\ \text{=}\left(\text{28}\text{.0134}\text{+}\text{4}\text{.03176}\text{+}\text{47}\text{.9982}\right)\text{amu}\\ \text{=}\text{80}\text{.0434}\text{amu}\end{array}$

The molecular mass of ammonium nitrate was calculated by the sum of atomic masses of nitrogen, hydrogen and oxygen atoms.

The percentage mass of nitrogen in ammonium nitrate is calculated as follows:

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{N}\text{in}\text{Ammonium nitrate}\text{=}\frac{\text{2×atomic}\text{mass}\text{of}\text{N}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}\text{Ammonium nitrate}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{2×14}\text{.0067}\text{amu}}{\text{80}\text{.4336}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{28}\text{.0134}\text{amu}}{\text{80}\text{.4336}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{34}\text{.8280}\text{\%}\end{array}$

The percentage mass of an element in its compound is calculated by using the atomic mass of that element and molecular mass of the compound.

(c)

Interpretation Introduction

Interpretation: Based on mass percentage the richest source of nitrogen has to be identified from the given fertilizers.

Concept introduction: The mass percentage of each elements present in a compound is known as the percent composition by mass of the compound.

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{an}\text{element}\text{=}\frac{\text{n}\text{×atomic}\text{mass}\text{of}\text{element}}{\text{molecular}\text{or}\text{formula}\text{mass}\text{of}\text{compound}}\text{×100\%}\\ \text{n}\text{-}\text{The}\text{number}\text{of}\text{elements present}\text{in}\text{a}\text{molecule}\text{or}\text{formula}\text{unit}\text{of}\text{the}\text{compound}\end{array}$

To determine: The percentage mass of nitrogen in urea.

Explanation of Solution

Molecular mass of guanidine is calculated as follows:

$\text{Molecular}\text{formula}\text{of}\text{urea}\text{is }\left(\text{HNC}{\left({\text{NH}}_{\text{2}}\right)}_{\text{2}}\right)$

$\begin{array}{l}\text{Molecular mass of Guanidine}\text{=}\left(\left(\text{3×mass}\text{of}\text{N}\right)\text{+}\left(\text{5×mass}\text{of}\text{H}\right)\text{+}\left(\text{1×mass}\text{of}\text{C}\right)\right)\text{amu}\\ \text{=}\left(\left(\text{3×14}\text{.0067}\right)\text{+}\left(\text{5×1}\text{.00794}\right)\text{+}\text{12}\text{.0107}\right)\text{amu}\\ \text{=}\left(\text{42}\text{.0201}\text{+}\text{5}\text{.0397}\text{+}\text{12}\text{.0107}\right)\text{amu}\\ \text{=}\text{59}\text{.0705}\text{amu}\end{array}$

The molecular mass of guanidine was calculated by the sum atomic masses nitrogen, hydrogen, and carbon atoms.

The percentage mass of nitrogen in guanidine is calculated as follows:

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{N}\text{in}\text{Guanidine}\text{=}\frac{\text{3×atomic}\text{mass}\text{of}\text{N}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}\text{Guanidine}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{3×14}\text{.0067}\text{amu}}{\text{59}\text{.0705}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{42}\text{.0201}\text{amu}}{\text{59}\text{.0705}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{71}\text{.1355}\text{\%}\end{array}$

The percentage mass of an element in its compound is calculated by using the atomic mass of that element and molecular mass of the compound.

(d)

Interpretation Introduction

Interpretation: Based on mass percentage the richest source of nitrogen has to be identified from the given fertilizers.

Concept introduction: The mass percentage of each elements present in a compound is known as the percent composition by mass of the compound.

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{an}\text{element}\text{=}\frac{\text{n}\text{×atomic}\text{mass}\text{of}\text{element}}{\text{molecular}\text{or}\text{formula}\text{mass}\text{of}\text{compound}}\text{×100\%}\\ \text{n}\text{-}\text{The}\text{number}\text{of}\text{elements present}\text{in}\text{a}\text{molecule}\text{or}\text{formula}\text{unit}\text{of}\text{the}\text{compound}\end{array}$

To determine: The percentage mass of nitrogen in ammonium nitrate.

Explanation of Solution

The molecular mass of ammonia e is calculated as follows:

$\text{molecular formula of ammonia is}\left({\text{NH}}_{\text{3}}\right)$

$\begin{array}{l}\text{Molecular mass of Ammonia}\text{=}\left(\left(\text{1×mass}\text{of}\text{N}\right)\text{+}\left(\text{3×mass}\text{of}\text{H}\right)\right)\text{amu}\\ \text{=}\left(\text{14}\text{.0067}\text{+}\left(\text{3×1}\text{.00794}\right)\right)\text{amu}\\ \text{=}\left(\text{14}\text{.0067}\text{+}\text{3}\text{.02382}\right)\text{amu}\\ \text{=}\text{17}\text{.03052}\text{amu}\end{array}$

The molecular mass of ammonia was calculated by the sum atomic masses nitrogen, hydrogen atoms.

The percentage mass of nitrogen in ammonia is calculated as follows:

$\begin{array}{l}\text{Percent}\text{by}\text{mass}\text{of}\text{N}\text{in}\text{Ammonia}\text{=}\frac{\text{1×atomic}\text{mass}\text{of}\text{N}\text{in}\text{}\text{amu}}{\text{molecular}\text{mass}\text{of}\text{Ammonia}\text{in}\text{amu}}\text{×}\text{100\%}\\ \text{=}\frac{\text{14}\text{.0067}\text{amu}}{\text{17}\text{.03052}\text{amu}}\text{×}\text{100\%}\\ \text{=}\text{82}\text{.2447}\text{\%}\end{array}$

The percentage mass of an element in its compound is calculated by using the atomic mass of that element and molecular mass of the compound.