Question

you tested the rocket with three mistures of fuel having different masses of oxygen and hydrogen. what is the relationship between the masses of oxygen and hydrogen in each of the tests?

Answer 1

Assuming your fuel is molecular hydrogen (H2) and the oxidizer is molecular oxygen (O2), the combustion stoichiometry is

2 H2 + O2 → 2 H2O.

That means 2 moles of H2 react with 1 mole of O2. Converting to masses:

- Molar masses: H2 = 2.016 g·mol⁻¹, O2 = 32.00 g·mol⁻¹.
- So the stoichiometric mass ratio (O2 mass : H2 mass) is 32.00 / (2 × 2.016) ≈ 7.94 : 1.

Thus, at stoichiometry:
- mass(O2) ≈ 7.94 × mass(H2),
- or mass(H2) ≈ 0.126 × mass(O2).

How to interpret your three tests:
- Compute mass(O2)/mass(H2) for each test.
- If ≈ 7.94 → stoichiometric (neither excess oxidizer nor fuel).
- If > 7.94 → excess oxygen (oxidizer-rich).
- If < 7.94 → excess hydrogen (fuel-rich).

Example: 1 g H2 needs ≈ 7.94 g O2; the stoichiometric mixture is ≈ 11.25% H2 and 88.75% O2 by mass.