Why ice cream never freezes completely

Two ice creams are stored overnight in the same freezer at −18 °C.

The first is hard enough to bend a spoon.

The second can be scooped almost immediately.

Their temperature is identical, yet their texture is completely different.

Why?

The usual answer is that one recipe contains more sugar. That is directionally correct, but incomplete. The amount of sugar matters, the type of sugar matters, the quantity of available water matters, and the salts naturally present in dairy ingredients matter as well.

All of these variables influence one central property:

Freezing point depression determines how much water can remain liquid at a given temperature.

Ice cream does not cross a single temperature and suddenly become solid. It freezes progressively. As temperature falls, part of water crystallizes while the rest remains trapped in an increasingly concentrated liquid phase.

Relationship between temperature and frozen water is called freezing curve.

Understanding that curve connects formulation decisions to practical outcomes:

• hardness at storage temperature;
• scoopability;
• resistance to heat shock;
• melting behaviour;
• flavour release;
• ice-crystal growth;
• perceived creaminess.

This article develops that relationship step by step. We begin with simplest possible system: sugar dissolved in water. We then move toward a complete ice cream mix containing several carbohydrates and milk salts. Finally, we compare calculation strategies and examine why they do not always predict same freezing curve.

Start with an experiment

Consider three solutions, each prepared with:

• 1000 g of water;
• 100 g of sugar.

Only sugar differs:

1. sucrose;
2. glucose;
3. fructose.

All three contain same mass of sugar. Should they freeze at same temperature?

At first glance, that seems reasonable. If mass concentration is identical, perhaps effect should also be identical.

It is not.

Glucose and fructose solutions experience approximately twice as much freezing point depression as sucrose solution.

Reason is not that glucose is intrinsically more powerful than sucrose. Freezing point depression depends primarily on number of dissolved particles, not directly on their mass.

Sucrose has molar mass of approximately:

$$M_{\mathrm{sucrose}} = 342.30\ \mathrm{g\,mol^{-1}}$$

Glucose and fructose have molar masses close to:

$$M_{\mathrm{glucose}} \approx M_{\mathrm{fructose}} \approx 180.16\ \mathrm{g\,mol^{-1}}$$

For 100 g of each sugar:

$$n_{\mathrm{sucrose}} = \frac{100}{342.30} = 0.292\ \mathrm{mol}$$ $$n_{\mathrm{glucose}} = \frac{100}{180.16} = 0.555\ \mathrm{mol}$$

Glucose solution therefore contains almost twice as many dissolved molecules. More molecules disturb equilibrium between liquid water and ice more strongly. Consequently, glucose solution must be cooled further before ice and liquid water can coexist.

Equal mass or equal particle count?

Equal massEqual moles

Water mass

1000 g

Sugar mass

100 g

Sugar	Mass	Molar mass	Moles	Molality	ΔTf
Sucrose	100.0 g	342.30 g/mol	0.292	0.292 mol/kg	0.54°C
Glucose	100.0 g	180.16 g/mol	0.555	0.555 mol/kg	1.03°C
Fructose	100.0 g	180.16 g/mol	0.555	0.555 mol/kg	1.03°C

Keep sugar mass constant, then switch to constant molar amount. Distinction is essential: at equal mass, glucose produces larger effect than sucrose; at equal number of moles, their ideal colligative effect is approximately same.

The simplest freezing-point equation

For an ideal dilute solution, freezing point depression can be estimated with:

$$\Delta T_f = iK_fm$$

where:

• $\Delta T_f$ is freezing point depression in degrees Celsius;
• $i$ is van ’t Hoff factor;
• $K_f$ is cryoscopic constant of solvent;
• $m$ is molality of dissolved solute.

For water:

$$K_f \approx 1.86\ ^\circ\mathrm{C\,kg\,mol^{-1}}$$

Molality is number of moles of solute per kilogram of solvent:

$$m = \frac{n_{\mathrm{solute}}}{m_{\mathrm{water,kg}}}$$

For sugars such as sucrose, glucose and fructose, no dissociation into ions is normally assumed in this elementary calculation, so:

$$i \approx 1$$

For solution containing 100 g of sucrose in 1 kg of water:

$$m_{\mathrm{sucrose}} = \frac{100/342.30}{1} = 0.292\ \mathrm{mol\,kg^{-1}}$$

Therefore:

$$\Delta T_{f,\mathrm{sucrose}} = 1 \times 1.86 \times 0.292 \approx 0.54^\circ\mathrm{C}$$

Predicted initial freezing temperature is approximately:

$$T_f \approx -0.54^\circ\mathrm{C}$$

For 100 g of glucose:

$$m_{\mathrm{glucose}} = \frac{100/180.16}{1} = 0.555\ \mathrm{mol\,kg^{-1}}$$ $$\Delta T_{f,\mathrm{glucose}} = 1 \times 1.86 \times 0.555 \approx 1.03^\circ\mathrm{C}$$

Glucose solution therefore begins to freeze near:

$$T_f \approx -1.03^\circ\mathrm{C}$$

Ideal model captures first practical lesson:

At equal mass, smaller molecules generally produce larger freezing point depression because they provide more dissolved particles.

Freezing point is not frozen-water fraction

Previous calculation predicts temperature at which freezing begins. It does not tell us how much water is frozen at −5, −10 or −18 °C.

This distinction is fundamental.

Initial freezing point answers:

At what temperature can first ice crystals appear?

Freezing curve answers:

At a given temperature, what fraction of available water has crystallized?

For ice cream formulation, second question is usually more useful.

Imagine a mix initially containing:

• 600 g of water;
• 150 g of dissolved sugars;
• 250 g of fat, proteins and other solids.

When first ice appears, nearly all 600 g of water is still liquid. As cooling continues, perhaps 100 g crystallizes. Dissolved sugars remain in liquid phase. They are now dissolved in approximately 500 g of water instead of 600 g.

If another 100 g freezes, same sugars become concentrated in only 400 g of liquid water. Molality therefore rises continuously:

$$m = \frac{n_{\mathrm{solute}}}{m_{\mathrm{unfrozen\ water}}}$$

As $m_{\mathrm{unfrozen\ water}}$ decreases, molality increases. Freezing point of remaining serum becomes lower:

$$\Delta T_f = iK_fm$$

This creates characteristic progressive freezing process:

$$\text{cooling} \rightarrow \text{ice formation} \rightarrow \text{serum concentration} \rightarrow \text{larger FPD} \rightarrow \text{more cooling required}$$

Process is self-limiting. Every increment of ice formation makes remaining water harder to freeze.

Follow freeze concentration

Temperature

-18°C

Water

600 g

Sucrose

100 g

Dextrose

50 g

Ice mass

541.1 g

Unfrozen water

58.9 g

Water frozen

90.2%

Serum sugars

254.8%

Molality

9.68 mol/kg

Residual

3.6e-15°C

Inspect curve

Temperature -18.00°CWater frozen, ideal: 90.2%

At every temperature, simulation exposes current ice mass, remaining liquid water, serum sugar concentration, molality and residual equilibrium error. Numerical feedback loop matters more than animation of ice crystals.

From one sugar to a real formulation

Ideal equation is easy to apply to a single sugar in water. Real ice cream mix is more complicated.

It may contain:

• sucrose;
• dextrose;
• fructose;
• lactose from milk;
• glucose syrup containing molecules of several sizes;
• salts from milk and whey;
• sodium chloride;
• alcohol or polyols;
• proteins;
• stabilizers;
• fat.

Not all components affect freezing in same way.

Small dissolved molecules have strong colligative effect because many particles are present per unit mass. Large molecules contribute fewer particles per unit mass. Insoluble fat does not participate directly in aqueous freezing equilibrium.

Proteins and hydrocolloids can strongly influence viscosity, water mobility and ice-crystal growth, but their direct molar contribution to freezing point depression is comparatively small because of high molecular mass.

Component	Direct contribution to FPD	Other important effects
Sucrose	High	Sweetness, solids
Dextrose	Very high per unit mass	Sweetness, softness
Fructose	Very high per unit mass	High sweetness
Lactose	Moderate	Crystallization risk
Glucose syrup	Depends on molecular-weight distribution	Body, viscosity, solids
Milk salts	Significant	Ionic contribution
Fat	Negligible direct colligative effect	Structure, lubrication, flavour
Proteins	Small direct effect	Emulsification, water binding, body
Stabilizers	Small direct effect	Viscosity, recrystallization control

Important phrase is direct contribution.

A stabilizer may transform sensory and storage behaviour without materially changing calculated FPD. Conversely, a small quantity of low-molecular-mass solute may alter FPD significantly without adding much body.

FPD is important, but it is not a complete texture model.

Additive colligative calculation

For a mixture of ideal, non-electrolyte solutes, total molality can be estimated by summing moles of each dissolved compound:

$$m_{\mathrm{total}} = \frac{\sum_j n_j}{m_{\mathrm{water,kg}}}$$

with:

$$n_j = \frac{m_j}{M_j}$$

Ideal freezing point depression becomes:

$$\Delta T_f = K_f \frac{\sum_j m_j/M_j}{m_{\mathrm{water,kg}}}$$

For a mixture containing sucrose and dextrose:

$$\Delta T_f = K_f\left( \frac{m_{\mathrm{sucrose}}/M_{\mathrm{sucrose}} + m_{\mathrm{dextrose}}/M_{\mathrm{dextrose}}}{m_{\mathrm{water,kg}}} \right)$$

Suppose 40 g of sucrose are replaced by 40 g of dextrose. Sugar mass remains constant, number of dissolved molecules increases, and therefore:

$$\Delta T_f \uparrow$$

At fixed storage temperature:

$$\text{unfrozen water} \uparrow$$

and usually:

$$\text{hardness} \downarrow$$

This is physical basis behind many practical sugar-balancing strategies. Locking total sugar mass, total sweetness, or total solids produces different substitutions; these constraints should not be conflated.

Replace sucrose with dextrose

Start with 150 g sucrose in 620 g water. Choose what must remain constant, then replace part of the sucrose.

Constraint

Keep sugar mass constantKeep sweetness constant

Sucrose replaced

50%

Quantity	Original	After substitution
Sucrose	150.0 g	75.0 g
Dextrose	0.0 g	75.0 g
Total sugar mass	150.0 g	150.0 g
Sweetness equivalent	150.0	127.5
Dissolved amount	0.438 mol	0.635 mol
Ideal initial FPD	1.31°C	1.91°C

The table keeps each constraint explicit. Replacing sugar gram for gram preserves solids, but not sweetness. Replacing it at equal sweetness may require a different total sugar mass. In both cases, the number of dissolved molecules, and therefore predicted FPD, changes.

Why ideal equation is not enough

Cryoscopic equation assumes a dilute ideal solution. An ice cream serum becomes neither dilute nor ideal during freezing.

As ice forms:

• solute concentrations increase sharply;
• interactions between molecules become more important;
• effective behaviour of glucose syrups depends on molecular distribution;
• salts dissociate into ions;
• some water may be associated with proteins or other solids;
• liquid phase can approach a highly concentrated, viscous state.

Equation $\Delta T_f = iK_fm$ remains an excellent conceptual model. It explains why particle count matters, predicts direction of ingredient substitutions, and provides useful estimates at modest concentrations.

But extrapolating it directly across complete freezing range can produce an unrealistic curve.

This leads to practical question:

How can we estimate FPD for a complete ice cream formulation without explicitly modelling every molecule?

One influential answer is sucrose-equivalent method described by Goff and Hartel.

The Goff–Hartel sucrose-equivalent approach

Method converts major carbohydrate sources in formulation into equivalent amount of sucrose. Resulting value is sucrose equivalent, or $SE$.

A published form of relationship is:

$$\begin{aligned} SE ={}& 0.545\,MSNF + 0.765\,WS + S \\ &+ 0.2\,(10DE\ CSS) + 0.6\,(36DE\ CSS) \\ &+ 0.8\,(42DE\ CSS) + 1.2\,(62DE\ CSS) \\ &+ 1.8\,HFCS + 1.9\,F \end{aligned}$$

where, depending on formulation convention:

• $MSNF$ represents milk solids-not-fat;
• $WS$ represents whey solids;
• $S$ represents sucrose;
• $CSS$ represents corn-syrup solids of specified dextrose equivalent;
• $HFCS$ represents high-fructose corn syrup;
• $F$ represents fructose or another pure monosaccharide such as dextrose.

Coefficients express approximate freezing-point contribution of each ingredient relative to sucrose. Sucrose has coefficient 1. Fructose has coefficient near 1.9. Low-DE glucose syrup has lower coefficient because average molecular mass is higher. Higher-DE syrup contains greater proportion of smaller saccharides and therefore has larger effect.

Method does not abandon molecular reasoning. It packages it into practical ingredient coefficients.

According to University of Guelph calculation procedure, $SE$ is first expressed in g per 100 g of mix, then converted to equivalent sucrose concentration in available water:

$$C_{SE} = 100\frac{SE}{W}$$

where $W$ is water content on same mass basis. Sugar contribution can then be approximated by polynomial used in published computational work:

$$FPD_{\mathrm{sugars}} = 0.00009\,C_{SE}^2 + 0.0612\,C_{SE}$$

Definition and basis of $SE$ must remain consistent. Dividing by water is essential during freeze concentration because available liquid water changes.

Follow one calculation from ingredients to FPD

Use the formulation developed below: 110 g sucrose, 40 g dextrose, 100 g milk solids-not-fat, and 620 g water.

1. Convert each source to sucrose equivalent.

$$SE = 110 + (1.9 \times 40) + (0.545 \times 100) = 240.5\ \mathrm{g}$$

The three visible contributions are 110 g from sucrose, 76 g from dextrose, and 54.5 g from lactose associated with milk solids-not-fat.

2. Express that equivalent amount relative to available water.

$$C_{SE} = 100\frac{240.5}{620} = 38.79$$

3. Calculate the sugar contribution.

$$FPD_{\mathrm{sugars}} = 0.00009(38.79)^2 + 0.0612(38.79) = 2.51^\circ\mathrm{C}$$

4. Calculate the separate milk-salt contribution.

$$FPD_{\mathrm{milk\ salts}} = 2.37\frac{100+0}{620} = 0.38^\circ\mathrm{C}$$

5. Add both contributions.

$$FPD_{\mathrm{total}} = 2.51 + 0.38 = 2.89^\circ\mathrm{C}$$

The estimated initial freezing point is therefore approximately $-2.89^\circ\mathrm{C}$. This sequence follows the same separation used in the FDS implementation: normalize sucrose equivalents by free water, calculate their FPD, then add the milk-salt term.

Contribution of milk salts

Sugars are not only dissolved particles in an ice cream mix. Milk ingredients introduce mineral salts that also depress freezing point.

Goff and Hartel give practical relationship for contribution associated with milk solids-not-fat and separately represented whey solids:

$$FPD_{\mathrm{milk\ salts}} = 2.37\left(\frac{MSNF+WS}{W}\right)$$

where $MSNF$, $WS$, and $W$ use consistent units. Set $WS=0$ when whey solids are not represented separately. Factor 2.37 represents empirical approximation based on average mineral composition and molecular behaviour of milk salts.

Total initial freezing point depression can then be estimated as:

$$FPD_{\mathrm{total}} = FPD_{\mathrm{sugars}} + FPD_{\mathrm{milk\ salts}}$$

and predicted initial freezing temperature is:

$$T_{f,0} = -FPD_{\mathrm{total}}$$

For formulations containing separately added sodium chloride, another explicit salt term may be introduced:

$$FPD_{\mathrm{NaCl}} = 0.581855\,C + 0.0034889\,C^2 + 0.0004314\,C^3$$

with:

$$C = 100\frac{m_{\mathrm{NaCl}}}{m_{\mathrm{water}}}$$

Complete expression becomes:

$$FPD_{\mathrm{total}} = FPD_{\mathrm{sugars}} + FPD_{\mathrm{milk\ salts}} + FPD_{\mathrm{NaCl}}$$

Separation is useful because milk salts and deliberately added sodium chloride need not be represented by same empirical term.

What initial FPD cannot answer

Initial freezing point tells us when ice can first appear. It does not determine how much water has frozen at a lower temperature. Once ice forms, remaining liquid water decreases and the serum becomes increasingly concentrated. The FPD calculation must then be applied against that changing liquid phase.

That turns the next calculation into an equilibrium problem: find the liquid-water mass for which serum FPD equals the magnitude of target temperature. Ice mass follows from difference between initial and remaining liquid water.

This chapter has established why that feedback exists. Next chapter compares two ways of quantifying it, defines each fraction denominator, and exposes numerical solution interactively.

Next: How much ice forms in ice cream?

References

• H. Douglas Goff, Freezing Point Depression of a Mix, University of Guelph.
• H. Douglas Goff and Richard W. Hartel, Ice Cream, Springer, 2013.
• Romain Bertin, FDS ice-cream calculation implementation, source code.
• A. Rinaldi et al., A Mathematical Modeling of Freezing Process in the Batch Production of Ice Cream, Foods 10(2), 2021.
• H. Douglas Goff, Low-temperature stability and the glassy state in frozen foods, Food Research International 25(4), 1992.
• C. Cogné, P. Laurent, J. Andrieu and J. Ferrand, Experimental Data and Modelling of Ice Cream Freezing, Chemical Engineering Research and Design 81(9), 2003.
• E. M. Drewett and R. W. Hartel, Ice crystallization in a scraped surface freezer, Journal of Food Engineering 78(3), 2007.
• K. Inoue et al., Modeling of the effect of freezer conditions on the hardness of ice cream using response surface methodology, Journal of Dairy Science 92(12), 2009.
• Q. Wang, G. Sala and E. Scholten, Functionality of sugars and sugar replacers in model frozen dessert systems, Current Research in Food Science, 2025.