Chapter 10, Problem 215A

Interpretation Introduction

Interpretation:

The molecular formula for hydrazine should be predicted.

Concept introduction:

The representation of atoms of a compound in simple whole number ratio is known as empirical formula.

Chemical formula or molecular formula represents the type of atoms and number of atoms present in a molecule using numerical subscripts and atomic symbol.

The rules for determining empirical formula is given as follows:

• Determine the mass of elements in given compound.
• Calculate number of moles using molar mass of each compound or element.
• Divide each number of moles by smallest number of mole value calculated in second step.
• Round the value calculated in step 3 to nearest whole number.

Answer to Problem 215A

The molecular formula for hydrazine is ${\text{N}}_2{\text{H}}_4$.

Explanation of Solution

Assuming the percentage is given by mass that means 87.45 g, and 12.55 g of nitrogen, and hydrogen are present in the vitamin D₃.

1. Mass of nitrogen = 87.45 g

Mass of hydrogen = 12.55 g

Now, calculating the moles of nitrogen, and hydrogen using their molar mass as follows:

$\begin{array}{c}\text{Moles}\text{of}\text{nitrogen}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{87.45\text{g}}{14.0067\text{g/mol}}\\ {\text{n}}_{\text{N}}=6.24\text{mol}\\ \text{Moles}\text{of}\text{hydrogen}=\frac{\text{12}\text{.55}\text{g}}{\text{1}\text{.008}\text{g/mol}}\\ {\text{n}}_{\text{H}}=12.45\text{mol}\end{array}$

2. Now, dividing each mole value by smallest mole value as follows:

$\begin{array}{c}{\text{n}}_{\text{N}}=\frac{6.24}{6.24}\\ =1\\ {\text{n}}_{\text{H}}=\frac{12.45}{6.24}\\ =2\end{array}$

3. Thus, the empirical formula of hydrazine is ${\text{NH}}_{\text{2}}$.

Now, n that is the ratio of given molar mass and empirical formula mass is calculated as follows:

$\text{n}=\frac{\text{Molecular}\text{formula}\text{mass}}{\text{Empirical}\text{formula}\text{mass}}$

The empirical formula mass is calculated as follows:

$\begin{array}{c}\text{Empirical}\text{}\text{formula}\text{mass}=\left(1\times \text{N}+2\times \text{H}\right)\text{g/mol}\\ =\left(1\times 14.0067+2\times 1.008\right)\text{g/mol}\\ \text{Empirical}\text{}\text{formula}\text{mass}=16.0227\text{g/mol}\end{array}$

Calculation of n as follows:

$\begin{array}{c}\text{n}=\frac{\text{Molecular}\text{formula}\text{mass}}{\text{Empirical}\text{formula}\text{mass}}\\ =\frac{\text{32}\text{.04}\text{g/mol}}{\text{16}\text{.0227}\text{g/mol}}\\ =1.99\\ \text{n}=2\end{array}$

Now, the number obtained will be multiplied to empirical formula as follows:

$\begin{array}{c}\text{Molecular}\text{formula}={\text{NH}}_2\times 2\\ ={\text{N}}_2{\text{H}}_4\end{array}$