Chapter 2, Problem 2.124P

(a)

Interpretation Introduction

Interpretation:

The simplest whole-number ratio of $\text{N}$ to $\text{O}$ in three different nitrogen oxides is to be determined.

Concept introduction:

The formula to find an amount(mol) is:

  $\text{Amount}\left(\text{mol}\right)=\frac{\text{mass}}{\text{molar mass}}$        (1)

Following are the steps to determine the simplest ratio of $\text{N}$ to $\text{O}$ of a compound.

Step 1: Convert the mass percentage of each atom into its mass.

Step 2: Convert the mass of each atom into moles.

Step C: Divide each subscript by the smallest subscript and after that multiply with the proper number to make the whole number.

Answer to Problem 2.124P

The simplest whole-number ratio of $\text{N}$ to $\text{O}$ for sample I is $1:1$, for sample II is $2:3$ and for sample III is $2:5$.

Explanation of Solution

Sample I:

The expression to calculate the percentage of $\%\text{ O}$ is as follows:

  $\%\text{ O}=\left(100\text{ \%}\right)-\left(\text{Percentage of nitrogen}\right)$                                     (3).

Substitute $46.69\text{ \%}$ for the percentage of nitrogen in the equation (3).

  $\begin{array}{c}\%\text{ O}=\left(100\text{ \%}\right)-\left(46.69\text{ \%}\right)\\ =53.31\%\end{array}$

Consider $100\text{ g}$ of each sample.

Calculate the mass of $\text{N}$ from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{ of N}=\left(100\text{ g of sample}\right)\left(\frac{46.69\%\text{ N}}{100\text{ \% sample}}\right)\\ =46.69\text{ g}\end{array}$

Substitute $46.69\text{ g}$ for mass and $14.01\text{ g/mol}$ for molecular mass in equation (1) to calculate moles of $\text{N}$.

  $\begin{array}{c}\text{Moles of N}=\frac{46.69\text{ g}}{14.01\text{ g/mol}}\\ =3.3326\text{ mol}\end{array}$

Calculate the mass of oxygen from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{of O}=\left(100\text{ g of sample}\right)\left(\frac{52.31\%\text{ O}}{100\text{ g of sample}}\right)\\ =52.31\text{ g }\end{array}$

Substitute $52.31\text{ g}$ for mass and $16.00\text{ g/mol}$ for molecular mass in equation (1) to calculate moles of oxygen.

  $\begin{array}{c}\text{Moles of O}=\frac{52.31\text{ g}}{16.00\text{ g/mol}}\\ =3.3319\text{ mol}\end{array}$

Construct the preliminary formula and use the values $3.3326\text{ mol}$ and $3.3319\text{ mol}$ directly in the preliminary formula as subscripts of nitrogen and oxygen element symbols respectively.

    $\text{Preliminary formula}={\text{N}}_{3.3326\text{ mol}}{\text{O}}_{3.3319\text{ mol}}$

Divide each subscript by the smallest subscript and after that multiply to make the whole number. Now, construct the simplest ratio of $\text{N}$ and $\text{O}$.

  $\begin{array}{c}\text{Simplest ratio of }\left(\text{NO}\right)={\text{N}}_{\frac{3.3326\text{ mol}}{3.3319\text{ mol}}}{\text{O}}_{\frac{3.3319\text{ mol}}{3.3319\text{ mol}}}\\ ={\text{N}}_{1.0002}{\text{O}}_{1.000}\\ ={\text{N}}_{\text{1}}{\text{O}}_{\text{1}}\end{array}$

The simplest whole number ratio is $1:1$. Hence, $\text{N:O}$ is present in the ratio of $1:1$ in sample I.

Sample II:

Substitute $36.85\text{ \%}$ for the percentage of nitrogen in the equation (3).

  $\begin{array}{c}\%\text{ O}=\left(100\text{ \%}\right)-\left(36.85\text{ \%}\right)\\ =63.15\%\end{array}$

Calculate the mass of $\text{N}$ from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{ of N}=\left(100\text{ g of sample}\right)\left(\frac{36.85\%\text{ N}}{100\text{ \% sample}}\right)\\ =36.85\text{ g}\end{array}$

Substitute $36.85\text{ g}$ for mass and $14.01\text{ g/mol}$ for molecular mass in equation (1) to calculate moles of $\text{N}$.

  $\begin{array}{c}\text{Moles of N}=\frac{36.85\text{ g}}{14.01\text{ g/mol}}\\ =2.6303\text{ mol}\end{array}$

Calculate the mass of oxygen from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{of O}=\left(100\text{ g of nitrogen oxide}\right)\left(\frac{63.15\%\text{ O}}{100\text{ \% nitrogen oxide}}\right)\\ =63.15\text{ g }\end{array}$

Substitute $63.15\text{ g}$ for mass and $16.00\text{ g/mol}$ for molecular mass in equation (1) to calculate moles of oxygen.

  $\begin{array}{c}\text{Moles of O}=\frac{63.15\text{ g}}{16.00\text{ g/mol}}\\ =3.9469\text{ mol}\end{array}$

Construct the preliminary formula and use the values $2.6303\text{ mol}$ and $3.9469\text{ mol}$ directly in the preliminary formula as subscripts of nitrogen and oxygen element symbols respectively.

  $\text{Preliminary formula}={\text{N}}_{2.6303\text{ mol}}{\text{O}}_{3.9469\text{ mol}}$

Divide each subscript by the smallest subscript and after that multiply to make the whole number. Now, construct the simplest ratio of $\text{N}$ and $\text{O}$.

  $\begin{array}{c}\text{Simplest ratio of }\left(\text{NO}\right)={\text{N}}_{\frac{2.6303\text{ mol}}{2.6303\text{ mol}}}{\text{O}}_{\frac{3.9469\text{ mol}}{2.6303\text{ mol}}}\\ ={\text{N}}_{\left(1.000\right)\left(2\right)}{\text{O}}_{\left(1.5001\right)\left(2\right)}\\ ={\text{N}}_2{\text{O}}_3\end{array}$

The simplest whole number ratio is $2:3$. Hence, $\text{N:O}$ is present in the ratio of $2:3$ in sample II.

Sample III:

Substitute $25.94\text{ \%}$ for the percentage of nitrogen in the equation (3).

  $\begin{array}{c}\%\text{ O}=\left(100\text{ \%}\right)-\left(25.94\text{ \%}\right)\\ =74.06\%\end{array}$

Calculate the mass of $\text{N}$ from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{ of N}=\left(100\text{ g of sample}\right)\left(\frac{25.94\%\text{ N}}{100\text{ \% sample}}\right)\\ =25.94\text{ g}\end{array}$

Substitute $25.94\text{ g}$ for mass and $14.01\text{ g/mol}$ for molecular mass in equation (1) to calculate moles of $\text{N}$.

  $\begin{array}{c}\text{Moles of N}=\frac{\text{ 25}.94\text{ g}}{14.01\text{ g/mol}}\\ =1.8515\text{ mol}\end{array}$

Calculate the mass of oxygen from the given mass percent as follows:

  $\begin{array}{c}\text{Mass}\left(\text{g}\right)\text{of O}=\left(100\text{ g of nitrogen oxide}\right)\left(\frac{74.06\%\text{ O}}{100\text{ \% nitrogen oxide}}\right)\\ =74.06\text{ g }\end{array}$

Substitute $74.06\text{ g}$ for mass and $16.00\text{ g/mol}$ for molecular mass in equation (1) to calculate moles of oxygen.

  $\begin{array}{c}\text{Moles of O}=\frac{74.06\text{ g}}{16.00\text{ g/mol}}\\ =4.6288\text{ mol}\end{array}$

Construct the preliminary formula and use the values $1.8515\text{ mol}$ and $4.6288\text{ mol}$ directly in the preliminary formula as subscripts of nitrogen and oxygen element symbols respectively.

  $\text{Preliminary formula}={\text{N}}_{1.8515\text{ mol}}{\text{O}}_{4.6288\text{ mol}}$

Divide each subscript by the smallest subscript and after that multiply to make the whole number. Now, construct the simplest ratio of $\text{N}$ and $\text{O}$.

  $\begin{array}{c}\text{Simplest ratio of }\left(\text{NO}\right)={\text{N}}_{\frac{1.8515\text{ mol}}{1.8515\text{ mol}}}{\text{O}}_{\frac{4.6288\text{ mol}}{1.8515\text{ mol}}}\\ ={\text{N}}_{\left(1.000\right)\left(2\right)}{\text{O}}_{\left(2.500\right)\left(2\right)}\\ ={\text{N}}_2{\text{O}}_5\end{array}$

The simplest whole number ratio is $2:5$. Hence, $\text{N:O}$ is present in the ratio of $2:5$ in sample III.

Conclusion

The mass percent of each element in a given compound determines the ratio in which the two elements are present in the compound.

(b)

Interpretation Introduction

Interpretation:

The number of grams of oxygen per $1.0\text{ g}$ of nitrogen is to be determined.

Concept introduction:

The formula to find an amount(mol) is:

  $\text{Amount}\left(\text{mol}\right)=\frac{\text{mass}}{\text{molar mass}}$

Answer to Problem 2.124P

The number of grams of oxygen for sample I is $1.14\text{ g}$, the number of grams of oxygen for sample II is $1.71\text{ g}$, and the number of grams of oxygen for sample III is $\text{2}\text{.86 g}$.

Explanation of Solution

The expression to calculate the number of grams of oxygen per $1.0\text{ g}$ of nitrogen is as follows:

  $\text{Number of grams of oxygen}=\frac{\text{mass of oxygen}}{\text{mass of nitrogen}}$        (6)

Sample I:

Substitute $53.31\text{ amu}$ for the mass of oxygen and $46.69\text{ amu}$ for the mass of nitrogen in equation (6).

  $\begin{array}{c}\text{Number of grams of oxygen}=\frac{53.31\text{ amu}}{46.69\text{ amu}}\\ =1.1418\text{ g}\\ \approx 1.14\text{ g}\end{array}$

Sample II:

Substitute $63.15\text{ amu}$ for the mass of oxygen and $36.85\text{ amu}$ for the mass of nitrogen in equation (6).

  $\begin{array}{c}\text{Number of grams of oxygen}=\frac{63.15\text{ amu}}{36.85\text{ amu}}\\ =1.7137\text{ g}\\ \approx 1.71\text{ g}\end{array}$

Sample III:

Substitute $74.06\text{ amu}$ for the mass of oxygen and $25.94\text{ amu}$ for the mass of nitrogen in equation (6).

  $\begin{array}{c}\text{Number of grams of oxygen}=\frac{74.06\text{ amu}}{25.94\text{ amu}}\\ =2.8550\text{ g}\\ \approx 2.86\text{ g}\end{array}$

Conclusion

As the mass percent of oxygen varies in different nitrogen oxide, the number of grams of oxygen per $1.00\text{ g}$ of nitrogen also varies.