# Technical Forum

## Alcohol determination

### Introduction

This method is not an officially recognised method for the determination of alcohol in wine. There are many methods used today the standard ones being, distillation/hydrometry, distillation/densitometry, by near infrared spectroscopy (NIR) and by gas chromatography (GC). This is however, a method able to be used by the Enthusiast wine maker with minimal equipment. Needed is a volumetric flask and a digital balance capable of reading to at least one decimal place and a beaker.

The concept involves the following:

- Tare a volumetric flask. (a container that has an know accurate volume i.e 250.0 mL)
- Fill the flask with wine to the mark and then weigh the wine.
- Add this wine quantitatively to a beaker.
- Gently heat the wine (keep the temperature below 80°C) allowing the alcohol and any water to evaporate from the wine.
- When all the alcohol has evaporated allow the beaker to cool then quantitatively transfer the liquid back into the tared volumetric flask and make up to volume.
- Reweigh the flask.

The initial weight of wine will reflect the weight of water and dissolved wine ingredients plus the weight of contained alcohol. This weight will likely be less than 250 g as the alcohol has a density of 0.79 (i.e. less than water).

The volume of alcohol | = | m_{2} - m_{1} / 0.21 |

Where:

m_{1} is the initial weight of the wine

m_{2} is the weight of wine afterwards

### Example

Assume the wine weight (m_{1}) is 253.7 g.

After evaporating the wine and adding it to the flask then topping the flask up with water the weight will be heavier than before. The greater the amount of alcohol in the wine the greater will be the new weight. This is because the lighter alcohol is now replaced by the heavier water.

Assume the weight (m_{2}) is now 260.0 g

The volume of alcohol = m_{2} - m_{1} / 0.21

The volume of alcohol becomes = (260.0 - 253.7) / 0.21

The volume of alcohol is = 30 mL

or as a percentage

(30 x 100) / 250 = 12% v/v

### Proof

m_{2} - m_{1} = V_{2}ρ_{2} - V_{1}ρ_{1}

where:

m_{1} is the weight of wine

m_{2} is the weight with the lost alcohol replaced with water

V_{1} is the volume alcohol in the wine

V_{2} is the volume of the lost alcohol replaced by water

ρ_{1} is the density of alcohol, nominally 0.79

ρ_{2} is the density of water, nominally 1.00

now:

m_{2} - m_{1} represents the difference between the weight of the volume of alcohol and the same volume of water therefore:

m_{2} - m_{1} = V_{2}ρ_{2} - V_{1}ρ_{1}

of course: V_{1} = V_{2}

m_{2} - m_{1} = V_{1}ρ_{2} - V_{1}ρ_{1}

m_{2} - m_{1} = V_{1}(ρ_{2} - ρ_{1})

V_{1} = m_{2} - m_{1} / (ρ_{2} - ρ_{1})

V_{1} = m_{2} - m_{1} / (1.00 - 0.79)

V_{1} = m_{2} - m_{1} / 0.21

The volume of alcohol | = | m_{2} - m_{1} / 0.21 |

The method works regardless of the size of the volumetric flask used, however, 250 mL is a good compromise with regard to the accuracy of the method and the amount of wine wasted in the test. To calculate the percent alcohol you need to know this volume.

% v/v alcohol | = | (The volume of Alcohol X 100) / volumetric flask size in mL |

Large amounts of volatile acids if present may interfere with the test. Their affect can be minimised by the addition of a known volume of sodium hydroxide solution. This volume needs to be accounted for when calculating the % alcohol.