Chapter 10, Problem 216A

Interpretation Introduction

Interpretation:

The given table needs to be completed by calculating number of moles, mass in kg or molecules.

Concept introduction:

The relation between number of moles, mass and molar mass of a substance is as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Molar mass is mass of a substance in 1 mol. Also, in 1 mol of any substance, there are $6.022\times {10}^{23}$ formula units of that substance.

Answer to Problem 216A

The complete table is as follows:

Propellants	Molecular formula	Mass (kg)	Moles	Molecules
Hydrogen	H2	1.03 ×105	5.14×107	3.09×1031
Oxygen	O2	6.14 ×105	1.93×107	1.16×1031
Hydrazine	NH2NH2	493	1.54×104	9.27×1027
Monomethyl-hydrazine	CH3NHNH2	4909	1.04×105	6.26×1028
Dinitrogen tetroxide	N2O4	7.95×103	8.64×104	5.20×1028

Explanation of Solution

• Molecular formula = H₂
Number of moles = 5.14 ×10⁷

The calculation of molecules is shown below:

$\begin{array}{c}\text{molecules}\text{of}{\text{H}}_{\text{2}}\text{=5}\text{.14}\text{×}{\text{10}}^{\text{7}}\text{×}\text{6}\text{.022}{\text{×10}}^{\text{23}}\\ \text{=}\text{30}\text{.95}\text{×}{\text{10}}^{\text{30}}\\ \text{=}\text{3}\text{.09}\text{×}{\text{10}}^{31}\text{molecules}\end{array}$

The calculation of mass is shown below:

$\begin{array}{c}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{=}\text{moles}\text{×}\text{molar}\text{mass}\\ \text{=}\text{5}\text{.14}{\text{×10}}^{\text{7}}\text{mol}\text{×}{\text{2gmol}}^{\text{-1}}\\ \text{=}\text{10}\text{.28}{\text{×10}}^{\text{7}}\text{g}\\ \text{=}\text{10}\text{.28}{\text{×10}}^{\text{4}}\text{kg}\\ \text{=}\text{1}\text{.03}\text{×}{\text{10}}^5\text{kg}\end{array}$

• Molecular formula = O₂
Number of molecules = 1.16 × 10³¹

The calculation of moles is shown below:

$\begin{array}{c}\text{moles}\text{of}{\text{O}}_{\text{2}}\text{=}\frac{\text{molecules}}{{\text{N}}_{\text{A}}}\\ \text{=}\frac{\text{1}\text{.16}\text{×}{\text{10}}^{\text{31}}}{\text{6}\text{.022}{\text{×10}}^{\text{23}}}\\ \text{=}\text{0}\text{.1926}{\text{×10}}^{\text{8}}\\ \text{=1}\text{.93}\text{×}{\text{10}}^{\text{7}}\text{mol}\end{array}$

The calculation of mass is shown below:

$\begin{array}{c}\text{mass}\text{of}{\text{O}}_{\text{2}}\text{=}\text{moles}\text{×}\text{molar}\text{mass}\\ \text{=}\text{1}{\text{.93×10}}^{\text{7}}\text{mol}\text{}\text{×}{\text{32gmol}}^{\text{-1}}\\ \text{=}\text{61}\text{.44}\text{×}{\text{10}}^{\text{7}}\text{g}\\ \text{=}\text{6}\text{.14}{\text{×10}}^{\text{5}}\text{kg}\end{array}$

Molecular formula = N₂H₄

• Mass of N₂H₄ = 493 kg
The calculation of moles is shown below:

$\begin{array}{c}\text{moles}\text{of}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}}\\ \text{=}\frac{{\text{493×10}}^{\text{3}}\text{g}}{\text{32}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{1}\text{.54}{\text{×10}}^{\text{4}}\text{mol}\end{array}$

The calculation of molecules is shown below:

$\begin{array}{c}\text{molecules}\text{=}\text{moles}\text{×}{\text{N}}_{\text{A}}\\ \text{=}\text{1}\text{.54}{\text{×10}}^{\text{4}}\text{×}\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\\ \text{=}\text{9}\text{.27}\text{×}{\text{10}}^{\text{27}}\text{molecules}\end{array}$

• Molecular formula = CH₃NHNH₂
Mass of CH₃NHNH₂ = 4909 kg

The calculation of moles is shown below:

$\begin{array}{c}\text{moles}\text{of}{\text{CH}}_{\text{3}}{\text{NHNH}}_{\text{2}}\text{=}\frac{\text{mass}}{\text{molar}\text{masss}}\\ \text{=}\frac{\text{4909}\text{×}{\text{10}}^{\text{3}}\text{g}}{\text{47}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{104}\text{.44}\text{×}{\text{10}}^{\text{3}}\text{mol}\\ \text{=}\text{1}\text{.04}\text{×}{\text{10}}^{\text{5}}\text{mol}\end{array}$

The calculation of molecules is shown below:

$\begin{array}{c}\text{molecules}\text{=}\text{moles}\text{×}{\text{N}}_{\text{A}}\\ \text{=}\text{1}\text{.04}\text{×}{\text{10}}^{\text{5}}\text{×}\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\\ \text{=}\text{6}\text{.26}\text{×}\text{×}{\text{10}}^{\text{28}}\text{molecules}\end{array}$

• Molecular formula = N₂O₄
Moles of N₂O₄ = 8.64×10⁴ mol

The calculation of mass is shown below:

$\begin{array}{c}\text{mass}\text{of}{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{=}\text{moles}\text{×}\text{molar}\text{mass}\\ \text{=}\text{8}\text{.64}\text{×}{\text{10}}^{\text{4}}\text{×}\text{92}\\ \text{=}\text{794}\text{.88}\text{×}{\text{10}}^{\text{4}}\text{g}\\ \text{=}\text{7}\text{.95}\text{×}{\text{10}}^{\text{3}}\text{kg}\end{array}$

The calculation of molecules is shown below:

$\begin{array}{c}\text{molecules}\text{=}\text{moles}\text{×}{\text{N}}_{\text{A}}\\ \text{=}\text{8}\text{.64}\text{×}{\text{10}}^{\text{4}}\text{×}\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\\ \text{=}\text{5}\text{.20}\text{×}{\text{10}}^{\text{28}}\text{molecules}\end{array}$