How to Calculate Acid and Alcohol Amounts for Target-MW Polyester Polyols?

1 Polycondensation Degree

The first step is to understand the degree of polycondensation, represented by n. Polyester polyols consist of a hydroxyl group at both ends and repeating unit segments in the middle. During the reaction, water is released, and its removal drives the reaction forward.

The theoretical degree of polymerization n is calculated as:

n = (Polyester molecular weight – Average molecular weight of alcohol) ÷ Average molecular weight of repeating unit

Example

To synthesize polyethylene adipate glycol with a number-average molecular weight of 3000, and where diethylene glycol accounts for 40% of the total alcohol by mass, calculate the polycondensation degree.

Given:

Relative molecular weight of adipic acid = 146

According to formula (11-53):

Average molecular weight of mixed alcohol = (Mnj × Mnk)/[Mnk × A + Mnj × (1−a)]

= (62 × 106)/(106 × 0.6 + 62 × 0.4) = 74.34

According to formula (11-51):

M(unit) = M(acid) + M(alcohol) − 2 × M(water) = 146 + 74.34 − 2 × 18 = 184.34

(For details on formulas (11-53) and (11-51), please refer to the article How to calculate the molecular weight of polyester polyol repeating units?)

Thus, the theoretical polycondensation degree n for the M3000 polyester polyol is:

n = (3000 − 74.34) ÷ 184.34 = 15.87 ≈ 16

2 Actual Amount of Mixed Acid or Alcohol

When synthesizing a polyester polyol with a specific molecular weight, based on formulas (11-53) and (11-51), we calculate the theoretical value of n. For polyester diols with two hydroxyl ends, 1 mol of polyester diol requires (n+1) mol of alcohol and n mol of acid, so the molar ratio of alcohol to acid is:

(n+1)/n = 1 + (1/n)

The total theoretical amount of alcohol = Average molecular weight × (n+1)

If the alcohol mixture includes A, B, and C, then the amount of each alcohol = Mole fraction × Molecular weight

Example:

To synthesize 4 kg of polyethylene adipate glycol with a molecular weight of 4000, and the mass fraction of ethylene glycol in the mixed alcohol is 60%:

According to formula (11-53):

(62 × 106)/(106 × 0.6 + 62 × 0.4) = 74.34

According to formula (11-51):

146 + 74.34 − 36 = 184.34

Polycondensation degree n = (4000 − 74.34) ÷ 184.34 = 21.3 ≈ 21

Then:

Adipic acid required = 146 × 21 = 3066 g

Ethylene glycol required = 74.34 × 21 × 60% = 936.68 g

Diethylene glycol required = 74.34 × 21 × 40% = 624.46 g

3 Theoretical Volatile Substances

During production or lab-scale synthesis, water must be removed at high temperatures to continue polycondensation. Water distillation may also carry away some alcohols. Even with reflux devices, alcohol loss varies by equipment and operation.

Moreover, when acid value drops below 20–30 mgKOH/g, vacuum dehydration is needed to reduce the acid value to below 1 mgKOH/g. During this process, alcohol loss increases.

Therefore, the actual alcohol input usually exceeds the theoretical amount, depending on the efficiency and stability of the reflux system. Better reflux systems result in less alcohol loss and lower costs.

Volatile substances include:

1.Theoretical water generated for achieving the target molecular weight

2.Excess alcohol above the theoretical requirement