Converting Between Amount, Concentration, and Volume

Why Amount, Concentration, and Volume Must Agree

PBPK models may represent a compartment’s drug state as an amount, such as milligrams in an organ, or as a concentration, such as milligrams per litre of tissue. Either representation can be useful, but they are connected by the volume assigned to that compartment. Changing the state variable without changing the associated scaling can make an otherwise plausible model create or lose drug simply because its units were rewritten.

The conversion is conceptually simple; applying it consistently requires care. The volume must be the volume that corresponds to the modelled concentration, and its units must match the concentration units. When converting an ODE, the same relationship applies to rates of change, not just to the state and its initial value.

Definition: For compartment $i$, let $A_i$ be the drug amount, $C_i$ the compartment-average drug concentration, and $V_i$ the compartment volume. If $C_i$ is defined as amount per unit of that compartment’s volume, then $$ A_i=V_iC_i,\qquad C_i=\frac{A_i}{V_i}. $$ For example, if $A_i$ is measured in mg and $V_i$ in L, then $C_i$ is in mg/L. The relation assumes the amount represented by $A_i$ is associated with the same compartment volume used to define $C_i$.

The equation is a bookkeeping identity for the stated concentration and volume, not a claim that every concentration measured in a biological sample represents all drug in an organ. A tissue homogenate concentration, a plasma concentration, an unbound concentration, and an intracellular concentration can refer to different quantities. Use the conversion only when the model’s amount and concentration definitions correspond.

Convert the State and Its Rate Together

Suppose the compartment volume is constant. Differentiating $A_i=V_iC_i$ with respect to time gives the rate conversion. The constant-volume assumption matters: it allows $V_i$ to be taken outside the derivative.

\frac{dA_i}{dt}=V_i\frac{dC_i}{dt} \qquad\text{and}\qquad \frac{dC_i}{dt}=\frac{1}{V_i}\frac{dA_i}{dt}.

Thus, if an amount-based model specifies a net rate in mg/h, dividing that rate by a volume in L gives the corresponding concentration rate in mg/L/h. In the reverse direction, multiply a concentration rate by the volume to recover an amount rate. The state, initial condition, and every rate term must use the same state-variable convention.

Assumptions for the rate conversion: The compartment volume is constant over the time interval; the concentration is defined as $A_i/V_i$; and all amounts and rates refer to the same drug moiety and compartment. If volume varies with time, then $dA_i/dt$ is not simply $V_i\,dC_i/dt$.

For a time-varying volume, applying the product rule gives

\frac{dA_i}{dt} = V_i\frac{dC_i}{dt} + C_i\frac{dV_i}{dt}.

The additional term accounts for amount change associated with changing volume at the current concentration. Consequently, dividing an amount rate by a time-varying volume alone does not generally yield the concentration rate. The worked rate conversion below uses a fixed volume.

Worked Example: Converting a Tissue Concentration to Amount

A hypothetical tissue compartment has volume $0.750$ L and a modelled average concentration of $4.20$ mg/L. Calculate the amount in the compartment.

Use $A=VC$, with volume in L and concentration in mg/L:

A=(0.750\ \text{L})(4.20\ \text{mg/L})=3.15\ \text{mg}.

The litre units cancel, leaving mg. Therefore the compartment contains $3.15$ mg under the stated compartment-average concentration definition. If the volume were instead entered as $750$ mL, it would first need to be converted to $0.750$ L (or the concentration would need to be expressed per mL). Multiplying $750$ mL directly by $4.20$ mg/L mixes volume units and does not give a valid amount.

Changing an ODE from Amount to Concentration

Consider a fixed-volume compartment with an amount state $A$ satisfying an ODE. To express the model using concentration $C=A/V$, divide the entire amount-rate equation by $V$, then substitute $A=VC$. This transforms every term, not just the initial condition.

\frac{dA}{dt}=f(A,t) \quad\Longrightarrow\quad \frac{dC}{dt} =\frac{1}{V}f(VC,t), \qquad V\ \text{constant}.

The transformation follows directly: $dC/dt=(1/V)dA/dt$, and the amount appearing on the right-hand side can be replaced by $VC$. A useful check is that each term in the resulting concentration ODE has units of concentration per time.

Worked Example: Rewriting an Amount ODE as a Concentration ODE

A hypothetical compartment has fixed volume $2.00$ L. Its amount-based model is

\frac{dA}{dt}=12\ \text{mg/h}-(0.400\ \text{h}^{-1})A, \qquad A(0)=6.00\ \text{mg}.

Convert this model to a concentration state $C=A/V$. Divide the amount rate by $2.00$ L and substitute $A=VC$:

\frac{dC}{dt} = \frac{12\ \text{mg/h}}{2.00\ \text{L}} - (0.400\ \text{h}^{-1})C = 6.00\ \text{mg/L/h}-(0.400\ \text{h}^{-1})C.

The initial concentration is

C(0)=\frac{6.00\ \text{mg}}{2.00\ \text{L}} =3.00\ \text{mg/L}.

As a rate check, at $C=3.00$ mg/L the concentration-based equation gives $dC/dt=6.00-1.20=4.80$ mg/L/h. Multiplying by $2.00$ L gives $9.60$ mg/h. Substitution into the amount equation gives $12-(0.400)(6.00)=9.60$ mg/h as well. The two representations therefore give the same rate at the corresponding state.

The reverse conversion uses the same reasoning. Given a concentration ODE for a fixed volume, multiply the entire equation by $V$ and replace $C$ by $A/V$. This is especially important when an ODE includes terms expressed in different forms: each must be converted consistently before they are combined.

Worked Example: Converting a Concentration Rate to an Amount Rate

A hypothetical compartment has volume $0.250$ L. At one instant, its concentration-based ODE gives $dC/dt=2.00$ mg/L/h. Find the corresponding amount rate.

For a constant volume, $dA/dt=V(dC/dt)$. Substitution gives

\frac{dA}{dt} = (0.250\ \text{L})(2.00\ \text{mg/L/h}) = 0.500\ \text{mg/h}.

The result is $0.500$ mg/h. If the volume is instead written as $250$ mL, the matching concentration rate must be converted to mg/mL/h before multiplication: $2.00$ mg/L/h is $0.00200$ mg/mL/h, and $(250\ \text{mL})(0.00200\ \text{mg/mL/h})=0.500$ mg/h. Multiplying $250$ mL directly by $2.00$ mg/L/h would give a unit-inconsistent result.

Detecting Volume-Scaling Errors

A missed volume factor usually changes the magnitude of a term while leaving the equation’s overall form looking reasonable. A duplicated factor can cause the same problem in the opposite direction. A unit audit should therefore examine the state conversion, the initial condition, and each rate term independently.

Write down the state definition.
Specify whether the state is amount or concentration, its units, and the compartment volume to which it refers.

Check the state and initial condition.
Verify $A=VC$ numerically. If an initial amount is specified, divide by the matching volume to initialize a concentration state.

Check each rate conversion.
For a fixed volume, amount rates divide by volume to become concentration rates; concentration rates multiply by volume to become amount rates.

Reconcile the two representations.
At a corresponding state, multiply the concentration derivative by volume and compare it with the amount derivative. They should agree when the assumptions hold.

For example, an amount rate of $0.500$ mg/h in a $0.250$ L compartment corresponds to $2.00$ mg/L/h, not $0.125$ mg/L/h. The latter would result from multiplying the amount rate by the volume rather than dividing. Similarly, a volume entered as $250$ mL when the concentration is in mg/L creates a factor-of-1000 error unless the units are reconciled first.

A second diagnostic is to compare amounts across compartments at a shared concentration. At $5.00$ mg/L, a $0.200$ L compartment contains $1.00$ mg, whereas a $1.00$ L compartment contains $5.00$ mg. Equal concentrations do not imply equal amounts when volumes differ. Conversely, equal amounts in compartments of different volumes do not imply equal concentrations.

Choosing an Amount or Concentration State

An amount state makes the quantity of drug in each compartment explicit and can make dose bookkeeping and mass accounting more direct. A concentration state can be convenient when the model is naturally expressed in concentrations or when concentration is the quantity of interest. Neither convention is inherently more correct. The important requirement is that the state definition, compartment volume, initial condition, and ODE terms form one consistent representation.

A mixed model can also be valid: different compartments may use different state conventions. However, any term that connects states must be expressed in compatible units before it is added to an ODE. A concentration in one compartment cannot be added directly to an amount in another; a volume conversion or another explicitly defined relationship is needed.

Common pitfall: A modeller may reason, “The amount equation has units of mg/h, so to get a concentration rate I multiply by the compartment volume.” That is backwards: dividing mg/h by L gives mg/L/h. Multiplication by L converts a concentration rate to an amount rate. A related error is to convert an initial amount to concentration but leave the ODE’s rate terms in amount units.

These conversions do not determine which physiological volume is appropriate for a particular concentration. That choice depends on how the compartment and concentration are defined. Nor does the identity $A=VC$ resolve differences between total and unbound drug or between distinct tissue subcompartments. Such quantities must not be treated as interchangeable without a model definition connecting them.

A practical final check is to calculate amount from concentration and volume, then reconstruct concentration from that amount. The recovered value should match the original, apart from rounding. For ODEs, perform the corresponding check on rates. These checks are simple, but they catch many implementation errors before simulation results are interpreted.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

A hypothetical compartment has volume $0.400$ L and concentration $7.50$ mg/L. What is its amount?
An amount-based ODE gives $dA/dt=3.00$ mg/h in a fixed $1.50$ L compartment. What is $dC/dt$?
A concentration state is $2.00$ mg/L in a $0.100$ L compartment. What is the corresponding amount?
Why is $dA/dt=V\,dC/dt$ insufficient when compartment volume changes with time?
A modeller multiplies an amount rate in mg/h by compartment volume in L to obtain a concentration rate. Identify the error and give the correct operation.

Solutions. (1) $A=VC=(0.400\ \text{L})(7.50\ \text{mg/L})=3.00$ mg. (2) $dC/dt=(dA/dt)/V=(3.00\ \text{mg/h})/(1.50\ \text{L})=2.00$ mg/L/h. (3) $A=(0.100\ \text{L})(2.00\ \text{mg/L})=0.200$ mg. (4) When $V$ varies, the product rule gives $dA/dt=V\,dC/dt+C\,dV/dt$; the volume-change term cannot generally be omitted. (5) The operation is reversed. Divide an amount rate by volume to obtain a concentration rate: $(\text{mg/h})/\text{L}=\text{mg/L/h}$.