Ethanol Rocket

Introduction

Theory

Ethanol 3d model

Atom positions
-1.1712	0.2997	0.0000	O
-0.0463	-0.5665	0.0000	C
1.2175	0.2668	0.0000	C
-0.0958	-1.2120	0.8819	H
-0.0952	-1.1938	-0.8946	H
2.1050	-0.3720	-0.0177	H
1.2426	0.9307	-0.8704	H
1.2616	0.9052	0.8886	H
-1.1291	0.8364	0.8099	H

HCH bonds are assumed to be undistorded tetraherdal angle 109.5 degrees. Actually the electrons repeal each other. The HOC bond is 104.5 deg because. . .

Valence shell electron-pair repulsion theory (VSEPR theory). . .

Van der Waals radius
Atom	Radius
C	170 pm	1.42
H	120 pm	1.00
O	152 pm	1.27

Use CPK coloring convention, white (hydrogen), black (carbon) and red (oxygen).

Ethanol reaction with Oxygen and Air

Ethanol reaction with oxygen

${\textrm {C}}_{2}{\textrm {H}}_{6}{\textrm {O}}+3{\textrm {O}}_{2}\to 2{\textrm {CO}}_{2}+3{\textrm {H}}_{2}{\textrm {O}}$

The molecular weight of ethanol is $46.069$ g/mol, and the molar weight of oxygen is 32 g/mol. The oxygen--ethanol fuel ratio is $3\times 32/46=2.1$ . We need $2.1$ kg of oxygen to $1$ kg of ethanol. The air consists of 23.2 mass-% of oxygen, thus the air--ethanol ratio is $2.1/0.232=9$ .

The volume-% of oxygen in air is 20.9%. The volume of the bottle is $V=500$ ml which gives the amount of oxygen to be $0.209V=0.209\times 500=104.5$ ml $=0.1045\times 1.429=0.149$ g of oxygen, which gives the amount of ethanol $0.126/2.1=0.072$ g $=0.072/0.789=0.091$ ml. OR directly using air--ethanol ratio we have $\rho V/9=1.2041\times 0.5/9=0.066$ gram. That amount equals to $V/\rho _{\textrm {Ethanol}}=0.066g/(789g/l)=0.084$ ml. Almost the same result. See the attached spreadsheet for detailed calculations (and mistakes). The small difference in amounts is due to the density of oxygen, which is at STP (0 degrees).