Chapter ST, Problem 1PQ

Interpretation Introduction

Interpretation:

The simplest formula that contains $34.9\%$ of $\text{Na}$, $16.4\%$ of $\text{B}$ and $48.6\%$ of oxygen has to be determined.

Concept Introduction:

Mole is S.I. unit. Number of moles is calculated as ratio of mass of compound to molar mass of compound.

Molar mass is sum of total mass of all atoms that make up mole of particular molecule that is mass of 1 mole of compound. The S.I unit is $\text{g/mol}$.

The expression to relate number of moles, mass and molar mass of compound is as follows:

  $\text{Number of moles}=\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$

Empirical formula represents simplest positive integer ratio of atoms in the compound. It only gives proportions of elements in compound. Molecular formula comprises of chemical symbols for respective elements followed by numeric subscript that denotes number of atom of each element present in the molecule.

Answer to Problem 1PQ

Option (A) is the correct one.

Explanation of Solution

Reason for correct option:

Consider sample that contains $34.9\%$ of $\text{Na}$, $16.4\%$ of $\text{B}$ and $48.6\%$ of $\text{O}$ has a mass of $100\text{g}$. Then mass of $\text{Na}$, $\text{B}$ and $\text{O}$ is $34.9\text{g}$, $16.4\text{g}$ and $48.6\text{g}$ respectively.

The formula to calculate moles of $\text{Na}$ is as follows:

  $\text{Moles of Na}=\left(\frac{\text{Mass of Na}}{\text{molar mass of Na}}\right)$        (1)

Substitute $34.9\text{g}$ for mass of $\text{Na}$, $22.9897\text{ g/mol}$ for molar mass of $\text{Na}$ in equation (1).

  $\begin{array}{c}\text{Moles of Na}=\left(\frac{34.9\text{g}}{22.9897\text{g/mol}}\right)\\ =1.5180\text{mol}\end{array}$

The formula to calculate moles of $\text{B}$ is as follows:

  $\text{Moles of B}=\left(\frac{\text{Mass of B}}{\text{molar mass of B}}\right)$        (2)

Substitute $16.4\text{g}$ for mass of $\text{B}$, $10.811\text{ g/mol}$ for molar mass of $\text{B}$ in equation (2).

  $\begin{array}{c}\text{Moles of B}=\left(\frac{16.4\text{g}}{10.811\text{g/mol}}\right)\\ =1.5169\text{mol}\end{array}$

The formula to calculate moles of $\text{O}$ is as follows:

  $\text{Moles of O}=\left(\frac{\text{Mass of O}}{\text{molar mass of O}}\right)$        (3)

Substitute $48.6\text{g}$ for mass of $\text{O}$, $15.999\text{ g/mol}$ for molar mass of $\text{O}$ in equation (3).

  $\begin{array}{c}\text{Moles of O}=\left(\frac{48.6\text{g}}{15.999\text{g/mol}}\right)\\ =3.0376\text{mol}\end{array}$

Preliminary formula for sample is formed with moles of $\text{Na}$, $\text{O}$ and $\text{B}$ written in subscripts. Therefore it can be written as follows:

  $\text{Preliminary formula}={\text{Na}}_{1.5180}{\text{B}}_{1.5169}{\text{O}}_{3.0376}$

Each of subscripts of $\text{Na}$, $\text{O}$ and $\text{B}$ is divided by smallest value to determine empirical formula of compound. The smallest value is 1.5169. Therefore empirical formula of given compound is as follows:

  $\begin{array}{c}\text{Empirical formula}={\text{Na}}_{\frac{1.5180}{1.5169}}{\text{B}}_{\frac{1.5169}{1.5169}}{\text{O}}_{\frac{3.0376}{1.5169}}\\ ={\text{Na}}_{1.000}{\text{B}}_1{\text{O}}_{2.1804}\\ \approx {\text{NaBO}}_2\end{array}$

Hence option (A) is correct.

Reason for incorrect option:

Empirical formula of given compound does not match with ${\text{NaBO}}_3$, ${\text{Na}}_2{\text{B}}_4{\text{O}}_7$ and ${\text{Na}}_3{\text{BO}}_3$. Hence option (B), (C) and (D) are incorrect.

Conclusion

The simplest formula that contains $34.9\%$ of $\text{Na}$, $16.4\%$ of $\text{B}$ and $48.6\%$ of oxygen is ${\text{NaBO}}_2$. Hence, correct option is (A).