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 $=0.1045\times 1.314=0.137$ g of oxygen, which gives the amount of ethanol $0.137/2.1=0.0659$ g $=0.0659/0.789=0.0834$ 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 using the methods. See the attached spreadsheet for detailed calculations.