The alcohol degree (°) is expressed as a volume percentage, i.e. number of ml of absolute ethanol in 100 ml of mixture (v / v). This volumetric expression does not correspond to the percentage by mass since ethanol has a lower density than water (approx. 0.79). Alcohol degree calculations can be performed using formulas and a conversion table. An example is elaborated below and the table can be found in the attached PDF.
It is important to realize that if the pharmacy uses 'Ethanol 96%', it often does not contain 96% (v / v). Potable ethanol (according to the Ph.Eur.) Contains only 96% (v / v) ethanol due to the presence of water drawn from the air (azeotrope mixture). Denaturans are still added in most raw materials. This is usually 3% (v / v) ether in Belgium. As a result, denatured '96% ethanol' may contain only 90% ethanol (v / v). The exact content can always be found in the analysis certificate.
Example: Preparation of a 40 ° alcohol solution from 96.2 ° alcohol
The table shows that:
alcohol 40 ° contains 33.27g alcohol per 100g or 31.57g alcohol per 100ml. (interpolation)
alcohol 96.2 ° contains 94.10g alcohol per 100g or 76.03g alcohol per 100ml.
1) by weight: prepare 100 g alcohol at 40 °
100g alcohol at 40 ° contains 33.27g absolute alcohol
100g of alcohol at 96.2 ° contains 94.10g of absolute alcohol
- 33.27 / 94.10% = 35.36 (35.36g alcohol of 96.2 ° contains 33.27g absolute alcohol)
- Dissolve 35.36g alcohol from 96.2 ° with water up to 100g
2) based on volume: prepare 100 ml of alcohol at 40 °
a) by weight
100ml 40 ° alcohol contains 31.57g absolute alcohol
100ml alcohol from 96.2 ° contains 76.03g absolute alcohol / 100mL or 94.1g / 100g
- 31.57 g / 94.1% (m / m) = 33.549 g of 96.2% ethanol (v / v) (33.549 g of 96.2 ° alcohol contains 31.57 g of absolute alcohol)
- Dissolve 33.549 g of alcohol at 96.2 ° with water up to 100ml
b) by volume
100ml x 40 ° = X ml x 96.2 °
Dissolve X ml alcohol from 96.2 ° with water to 100ml.
Dissolve 41.58ml alcohol from 96.2 ° with water to 100ml.