Both dimethylhydrazine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NNH}_{2}\), and methylhydrazine, \(\mathrm{CH}_{3} \mathrm{NHNH}_{2}\), have been used as rocket fuels. When dinitrogen tetroxide $\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)$ is used as the oxidizer, the products are \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}_{2}\), and \(\mathrm{N}_{2}\). If the thrust of the rocket depends on the volume of the products produced, which of the substituted hydrazines produces a greater thrust per gram total mass of oxidizer plus fuel? (Assume that both fuels generate the same temperature and that \(\mathrm{H}_{2} \mathrm{O}(g)\) is formed.)

We need to have a balanced chemical equation for both the dimethylhydrazine and methylhydrazine reactions with dinitrogen tetroxide. For Dimethylhydrazine: \((CH_{3})_{2}NNH_{2} + N_{2}O_{4} \rightarrow H_{2}O + CO_{2} + N_{2}\) Balanced equation: \[2(CH_{3})_{2}NNH_{2} + 3N_{2}O_{4} \rightarrow 6H_{2}O + 2CO_{2} + 5N_{2}\] For Methylhydrazine: \(CH_{3}NHNH_{2} + N_{2}O_{4} \rightarrow H_{2}O + CO_{2} + N_{2}\) Balanced equation: \[4CH_{3}NHNH_{2} + 5N_{2}O_{4} \rightarrow 12H_{2}O + 4CO_{2} + 9N_{2}\]

Now, we will determine the molar mass of each reactant and product in the chemical equations: Dimethylhydrazine: \(2\times12.01 + 2\times1.01 + 14.01 + 2\times1.01 = 60.14 \, g/mol\) Methylhydrazine: \(12.01 + 3\times1.01 + 14.01 + 2\times1.01 = 46.08 \, g/mol\) Dinitrogen tetroxide: \(2\times14.01 + 4\times16.00 = 92.02 \, g/mol\) Water: \(2\times1.01 + 16.00 = 18.02 \, g/mol\) Carbon dioxide: \(12.01 + 2\times16.00 = 44.01 \, g/mol\) Dinitrogen: \(2\times14.01 = 28.02 \, g/mol\)

According to the balanced equations, we have the following relationships: Per mole of dimethylhydrazine: - Fuel: \(2(CH_{3})_{2}NNH_{2}\) - Oxidizer: \(3N_{2}O_{4}\) - Products: \(6H_{2}O + 2CO_{2} + 5N_{2}\) Total mass of reactants for dimethylhydrazine (fuel + oxidizer): \(2 \times 60.14 + 3 \times 92.02 = 396.34 \, g\) Total moles of products produced: \(6 + 2 + 5 = 13\, moles\) Thrust per gram for dimethylhydrazine: \(\frac{13 \, moles}{396.34 \, g} = 0.0328 \frac{moles}{g}\) Per mole of methylhydrazine: - Fuel: \(4CH_{3}NHNH_{2}\) - Oxidizer: \(5N_{2}O_{4}\) - Products: \(12H_{2}O + 4CO_{2} + 9N_{2}\) Total mass of reactants for methylhydrazine (fuel + oxidizer): \(4 \times 46.08 + 5 \times 92.02 = 560.34 \, g\) Total moles of products produced: \(12 + 4 + 9 = 25\, moles\) Thrust per gram for methylhydrazine: \(\frac{25 \, moles}{560.34 \, g} = 0.0446 \frac{moles}{g}\)

Now that we have the thrust per gram for both dimethylhydrazine (0.0328 moles/g) and methylhydrazine (0.0446 moles/g), we can see that methylhydrazine produces a greater thrust per gram total mass of oxidizer plus fuel.