# Proof of Alcohol Lab Report

The main purpose of this lab was to discover the proof of alcohol in the given unknown solution. The proof of alcohol in a solution is double its volume percent. Therefore if a bottle of grey goose vodka is 80 proof then its liquid solution has a volume of 40% alcohol. To find the proof in this experiment many steps were taken to separate the ethanol from the solution. Since ethanol is c in water it needs to be separated by adding diethyl ether and sodium acetate to cause a chemical reaction. The chemical reaction results in separation of ethanol form water. Since the ethanol and water is chemically separated the ethanol floats on top of the water, which allows them to be easily extracted separately using a separatory funnel. After using the seperatory funnel to physically separate the ethanol and unknown water solution, the volume of the unknown water solution and ethanol can be measured which will give the proof. Another purpose of the lab was to calculate the density of ethanol and water, which was done by taking a roughly 3 mL sample of each and finding the mass of them.

Procedure

The procedure was followed according to the lab manual. No changes were made to the procedure of this experiment. “Proof of Alcohol.” CHM1045L Online Manual. Spring 2009. 26 February 2009

Data

Part I

Mass of 10.0 mL graduated cylinder| 39.5763 g|

Volume of Ethanol| 2.95 mL|

Mass of Ethanol| 2.1367 g|

Density of Ethanol| .724 g/mL|

Part II

Mass of the 10.0 mL graduated Cylinder| 39.5763 g|

Volume of DI Water| 2.95 mL|

Mass of DI Water| 2.6766 g|

Density of DI Water| .907 g/mL|

Part III

Mass of 25 mL graduated cylinder| 60.9571 g|

Mass of graduated cylinder with solution| 84.0864 g|

Mass of Solution| 23.1293 g|

Volume of Solution| 25.0 mL|

Density of Solution| .925 g/mL|

Mass of Salt| 1.0700 g|

Mass of Water Layer| 19.6829 g|

Mass of Alcohol Layer| 4.5164 g|

Volume of Water Layer (using density in Part II)| 20.60 mL| Volume of Alcohol Layer (using density in Part I)| 4.40 mL| Density of Water Layer| .955 g/ml|

Percent Alcohol by Mass| 19.53 %|

Percent Alcohol based on efficiency| 22.459 %|

Percent Alcohol by Volume| 17.6 %|

Proof of Alcohol| 35.2|

Calculations

Mass of liquid:

Mass of Ethanol:

Mass of Graduated Cylinder 10 mL with Ethanol - Mass of Graduated Cylinder

41.713 g – 39.5763 g =2.1367g

Mass of Water:

Mass of Graduated Cylinder 10 mL with Water – Mass of Graduated Cylinder

42.2529 g - 39.5763 g = 2.6766 g

Mass of Solution:

Mass of Graduated Cylinder 25 mL with Solution – Mass of Graduated Cylinder

84.0864 g – 60.9571 g = 23.1293 g

Density of Liquid:

Density = Mass

Volume

Density of Ethanol:

= 2.1367g

2.95 mL

= 0.724 g/mL

Density of Water:

= 2.6766 g

2.95 mL

= 0.907 g/mL

Density of Solution:

= 23.1293 g

25.0 mL

= 0.925 g/mL

Mass of Water Layer:

Total Mass of Water Layer in Graduated Cylinder – Mass of Graduated Cylinder – Mass of salt

= 80.2640 g – 60.9571 g -1.0700 g

= 18.2369 g

Mass of Alcohol Layer:

Mass of Solution+ Mass of NaCl – Mass of Water Layer

= 23.1293 g+1.0700 g – 19.6829 g

= 4.5164 g

Volume of Alcohol Layer:

Unkknown volume – Water layer volume

=25 mL – 20.60 mL

= 4.40 mL

Percent of Alcohol Based on Mass:

% Alcohol = Mass of Alcohol After Extraction *100

Starting Mass of Solution

= 4.5164 g *100

23.1293 g

= 19.53 %

Percent of Alcohol Based on Volume:

% Alcohol = Volume or Mass of Alcohol After Extraction *100

Starting Mass or Volume of Solution

= 4.4 mL *100

25.0 mL

= 17.6 %

Percent of Alcohol Based on an 85% Efficiency:

% alcohol based on efficiency = % alcohol + (% alcohol x ( (100-% efficiency)/100))

= (19.53 %) + (19.53 % x ((100-85))

100

= (19.53 %) + (19.53 % x .015)

= 22.4595 %

Proof of...

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