Lactose crystallization: Looking beyond lactose content

When discussing ice cream formulation, few ingredients generate as much concern as lactose.

Almost every ice cream maker has heard the same warning:

Too much lactose will make your ice cream sandy.

Although this statement is not wrong, it is incomplete.

Lactose crystallization is not controlled simply by the amount of lactose in a recipe. It depends on a dynamic balance involving temperature, water availability, supersaturation, molecular mobility, and time.

This balance explains why two recipes containing exactly the same amount of lactose can behave very differently during storage.

Lactose is an unusual sugar

Compared with sucrose and the monosaccharides commonly used in ice cream, lactose has limited solubility in water. Its equilibrium solubility also depends strongly on temperature and on the other solutes present. Early measurements found that sucrose can reduce lactose solubility relative to pure water, with a larger effect as sucrose concentration rises (Nickerson and Moore, 1926).

This matters during freezing. The amount of lactose remains constant while the amount of liquid water decreases. Eventually, the remaining liquid phase may no longer keep all lactose dissolved.

At that point, the solution becomes supersaturated. Thermodynamic conditions permit crystallization, but they do not determine when a detectable crystal will appear.

The recipe is not what lactose sees

When formulating ice cream, we usually think in terms of ingredients:

• milk;
• cream;
• sugars;
• stabilizers;
• proteins.

From the point of view of a lactose molecule, one question dominates:

How much liquid water is available to dissolve me?

Two recipes may contain exactly the same percentage of lactose while presenting different concentration states because they contain different amounts of water.

Consider two simplified recipes.

Recipe	Water	Lactose
A	68%	5%
B	56%	5%

Both contain 5% lactose. Recipe B nevertheless places each gram of water under a larger lactose load.

Try keeping lactose at 5%, then reduce water from 68% to 56%. Compare grams of lactose per gram of water with mass fraction in the lactose-water subsystem.

Lactose in available water

Reset

Water

68% of mix

Lactose

5.0% of mix

Lactose per gram of liquid water0.074 g/g

Aqueous-phase mass fraction6.8%

Recipe A

68% water, 5% lactose0.074 g/g water

Recipe B

56% water, 5% lactose0.089 g/g water

A first approximation

A useful first estimate expresses lactose as a fraction of the lactose-water subsystem:

$$C=\frac{L}{L+W}$$

where:

• $L$ is lactose content, expressed as a mass fraction of the mix;
• $W$ is water content on the same mass basis;
• $C$ is lactose mass fraction in this simplified aqueous subsystem.

For Recipe A:

$$C_A=\frac{0.05}{0.05+0.68}=0.0685$$

For Recipe B:

$$C_B=\frac{0.05}{0.05+0.56}=0.0820$$

Recipe B therefore has an aqueous lactose mass fraction of 8.2%, compared with 6.85% for Recipe A.

This result supports comparison of formulations. It cannot establish solubility in a complete ice cream mix or predict crystal appearance because it assumes that all recipe water remains liquid and available.

Water is not just water

Ice cream is a multiphase system, not a simple sugar solution.

Water interacts with proteins, hydrocolloids, salts, and sugars. More importantly, an increasing fraction becomes ice as temperature falls. Frozen water is no longer part of the liquid solvent phase.

This removal of water concentrates every solute left in the unfrozen serum. The process is called cryoconcentration.

The amount of lactose does not increase. Its concentration does.

Connecting lactose crystallization to ice fraction

Our chapter on how much ice forms in ice cream derives this calculation. Explorer below uses Goff–Hartel equilibrium method: at each temperature, it solves for liquid water remaining while recalculating sucrose-equivalent and salt contributions against that shrinking water phase.

For lactose, only unfrozen water acts as solvent. Define $f_{\mathrm{ice}}(T)$ as the fraction of total recipe water frozen at temperature $T$. If total water content is $W$, liquid water remaining is:

$$W_{\mathrm{liquid}}=W\left(1-f_{\mathrm{ice}}(T)\right)$$

Substituting liquid water into the first approximation gives:

$$C(T)=\frac{L}{L+W\left(1-f_{\mathrm{ice}}(T)\right)}$$

As temperature decreases:

1. fraction of water frozen increases;
2. liquid water decreases;
3. lactose concentration in the remaining aqueous phase increases.

Recipe composition has not changed. Physical state has.

Turning ice fraction into a formulation limit

We can now answer a practical question: how much lactose could the remaining water hold before reaching an aqueous reference solubility?

A commonly reported correlation for total lactose solubility in water is (Huppertz and Gazi, 2016):

$$S(T)=10.9109\exp(0.02804T)$$

where $S(T)$ is expressed in grams of lactose per 100 g of water and $T$ in degrees Celsius. Applying this correlation below 0 °C is an extrapolation. Applying it to ice cream serum is another approximation because sucrose, salts, proteins, and hydrocolloids modify solvent environment.

Converting $S(T)$ to grams per gram of water, the lactose content corresponding to this aqueous reference is:

$$L_{\mathrm{ref}}(T)=\frac{S(T)}{100}W\left(1-f_{\mathrm{ice}}(T)\right)$$

A user-selected comparison margin $m$ can then define a more conservative target:

$$L_{\mathrm{target}}(T)=(1-m)L_{\mathrm{ref}}(T)$$

Consider Recipe A at −10 °C using Goff–Hartel calculation shown above:

• total water: 68% of mix;
• water frozen: 75.0% of total water;
• liquid water remaining: 16.99% of mix;
• extrapolated aqueous solubility: 0.0824 g lactose per gram of water;
• aqueous reference limit: 1.40% lactose in the mix;
• comparison target with a 20% margin: 1.12% lactose in the mix.

Recipe A contains 5% lactose. Its calculated saturation ratio is therefore:

$$R_{\mathrm{sat}}=\frac{5.00}{1.40}=3.57$$

This result does not mean Recipe A becomes sandy immediately. It means its unfrozen phase sits well above a pure-water equilibrium reference under the model assumptions. The large gap also explains why a universal lactose percentage cannot predict sandiness: many real ice creams operate in a supersaturated state while slow molecular mobility delays detectable crystal growth.

Move temperature from 0 °C toward storage conditions in explorer below. Watch frozen-water fraction, remaining liquid water, and lactose-to-water ratio change together. Change water, sucrose, dextrose, or MSNF and complete equilibrium curve is recalculated from composition.

Freeze concentration of lactose

Goff–Hartel equilibrium estimate

Reset

Temperature

-10.0°C

Water

680g

Sucrose

100g

Dextrose

30g

MSNF

90g

Lactose

5.0%

Safety margin

20%

Water frozen75.0%

Liquid water remaining17.0% of mix

Lactose per liquid water0.294 g/g

Reference aqueous solubility0.082 g/g water

Maximum before reference supersaturation1.40% of mix

Comparison target1.12% of mix

Saturation ratio3.57×

Reference zoneAbove aqueous reference

Illustrative frozen-phase view

Procedural frozen-phase view unavailable.

Frozen water

Pale regions show estimated frozen-water coverage.

Liquid serum

Brown-grey channels show remaining liquid serum.

Thermodynamic opportunity

Accent dots show lactose in liquid serum. Dot density follows calculated lactose per liquid water. Clusters indicate supersaturation visually, not predicted nuclei.

Warm points show thermodynamic opportunity, not predicted crystals, nucleation time, sandiness, or shelf life.

Freeze-concentration curve

How to use this result

• Below comparison target: formulation sits below chosen margin relative to aqueous reference. This is lowest thermodynamic-risk zone within model.
• Between target and reference: formulation approaches reference saturation. Small changes in ice fraction or composition can move it above reference.
• Above the aqueous reference: supersaturation is thermodynamically possible. Reduce lactose, preserve more liquid water, improve temperature stability, or validate storage life experimentally.

Compare formulations at same temperature, ice-fraction model, and margin. Lower saturation ratio means lower thermodynamic driving force under shared assumptions.

Do not read comparison target as guaranteed safe maximum. Model does not include solubility changes caused by sucrose, salts, proteins, hydrocolloids, lactose mutarotation, or glass-transition effects. It also cannot predict nucleation time, crystal size, or sensory detection.

Supersaturation does not mean immediate crystallization

Exceeding an equilibrium solubility limit creates a thermodynamic driving force. It does not produce a crystal immediately.

Before a crystal can grow:

• lactose molecules must move through the serum;
• a stable nucleus must form or an existing surface must act as a seed;
• molecules must attach to that nucleus;
• crystal faces must continue growing.

Controlled work on α-lactose monohydrate shows that nucleation induction time depends on both supersaturation and temperature (McLeod et al., 2011). Separate bulk-crystallization experiments also found that yield continued increasing over tens of hours under controlled aqueous conditions (Raghavan et al., 2001).

Ice cream adds further constraints. Its unfrozen phase becomes concentrated and viscous, and other macromolecules restrict mobility. Thermodynamic possibility and kinetic observability must therefore remain separate concepts.

Why temperature fluctuations matter

Temperature stability changes both ice structure and the environment surrounding lactose.

During warming, some ice melts and dilutes the serum. During cooling, water freezes again and reconcentrates it. Repeated cycles reorganize interfaces, change mobility, and create repeated opportunities for nucleation and growth.

Cooling-rate experiments in concentrated dairy streams show that cooling history affects lactose crystal yield and size distribution (Pandalaneni and Amamcharla, 2018). That system is not ice cream, but it demonstrates why crystallization cannot be separated from thermal history.

Increase cycle amplitude and cycle count below. The index is deliberately qualitative: it compares exposure to thermal disturbance but does not calculate crystal diameter or storage life.

Temperature-cycle explorer

Illustrative model

Reset

Cycle amplitude

±3.0°C

Number of cycles

4

Relative growth opportunity13 / 100

Stable storage does not guarantee that a supersaturated formulation will remain crystal-free forever. Unstable storage does not assign a universal crystallization time. It increases one important source of kinetic opportunity.

Can we predict lactose crystallization?

We can estimate:

• lactose content;
• liquid-water content under a stated ice-fraction model;
• lactose concentration in that liquid phase;
• equilibrium solubility for a defined model solution;
• degree of supersaturation when a suitable solubility model exists.

We cannot predict exactly when crystals will become detectable from those quantities alone.

Appearance time and sensory impact also depend on nucleation sites, lactose crystal form, mutarotation, viscosity, molecular mobility, thermal history, crystal growth, and detection threshold. Solubility curves should therefore be treated as thermodynamic references, not strict sandiness boundaries.

Farther and longer excursions into supersaturation generally increase crystallization opportunity. They do not convert a formulation calculation into a validated shelf-life prediction.

Final thoughts

Lactose does not crystallize simply because a recipe contains “too much lactose.”

Crystallization becomes possible when, at a given temperature and composition, the remaining liquid phase cannot keep all lactose dissolved. Whether crystals then appear depends on kinetic events unfolding over time.

This distinction changes formulation practice. Lactose percentage remains useful, but it must be interpreted together with total water, fraction of water frozen, serum composition, storage temperature, and thermal history.

Lactose crystallization is not a property of the recipe alone.

It is a property of the frozen system.

References

1. Nickerson, T. A., and Moore, E. E. “Lactose Solubility and Lactose Crystal Formation: I. Lactose Solubility.” Journal of Dairy Science 9, no. 6 (1926): 517–537. https://doi.org/10.3168/jds.S0022-0302(26)93924-6
2. McLeod, J., Paterson, A. H. J., Jones, J. R., and Bronlund, J. E. “Primary nucleation of α-lactose monohydrate: The effect of supersaturation and temperature.” International Dairy Journal 21, no. 7 (2011): 455–461. https://doi.org/10.1016/j.idairyj.2011.01.006
3. Raghavan, S. L., Ristic, R. I., Sheen, D. B., and Sherwood, J. N. “The bulk crystallization of alpha-lactose monohydrate from aqueous solution.” Journal of Pharmaceutical Sciences 90, no. 7 (2001): 823–832. https://doi.org/10.1002/jps.1036
4. Pandalaneni, K., and Amamcharla, J. K. “Evaluating the crystallization of lactose at different cooling rates from milk and whey permeates in terms of crystal yield and purity.” Journal of Dairy Science 101, no. 10 (2018): 8805–8814. https://doi.org/10.3168/jds.2018-14846
5. Huppertz, T., and Gazi, I. “Lactose in dairy ingredients: Effect on processing and storage stability.” Journal of Dairy Science 99, no. 8 (2016): 6842–6851. https://doi.org/10.3168/jds.2015-10033