Skip to content
Brent$117.6Gas TTF$1,192USD453.87 ₸
MunaiHub
Reservoir engineeringFluid composition and phase behaviour Basic 15 min read

Composition of reservoir fluids

Reservoir fluid composition is the list of the substances that make up the oil or gas of an accumulation, with the share of each. The hydrocarbon part (the part made of carbon and hydrogen) is a mixture of hundreds of compounds, from methane to heavy resins and asphaltenes. It is mixed with nitrogen, carbon dioxide, hydrogen sulfide and other substances that are not hydrocarbons. For engineering calculations we write the composition in mole fractions (the share of the molecules of each substance). We list the light components one by one and group the heavy ones in one or several fractions, for example C7+ (everything heavier than hexane).

Why it matters

The composition decides whether the fluid in the reservoir is oil or gas. It also decides how much gas and liquid a well will deliver at the surface, what the density and viscosity of the oil are, and at what pressure gas will come out of the oil. The content of hydrogen sulfide and carbon dioxide decides the choice of materials, the processing of the produced fluids and the safety measures. For the sour fields of the Pre-Caspian Basin (fields with hydrogen sulfide) this is one of the main design conditions.

Quick refresher

  • Oil and natural gas are mixtures of hydrocarbons. The more small molecules a mixture holds, the “lighter” it is. At one end is dry gas, which is almost entirely methane. At the other end is black oil (ordinary dark oil), in which the C7+ fraction makes up almost half of the moles and most of the mass.
  • The main classes of hydrocarbons are alkanes CnH2n+2\mathrm{C}_n\mathrm{H}_{2n+2} (normal and branched), cycloalkanes (naphthenes) CnH2n\mathrm{C}_n\mathrm{H}_{2n}, and aromatics, which contain a benzene ring. The heaviest are the resins and asphaltenes, with sulfur, nitrogen and oxygen.
  • The non-hydrocarbon components are nitrogen, carbon dioxide, hydrogen sulfide, water with salts, and traces of mercury and helium. The acid gases H2S\mathrm{H_2S} and CO2\mathrm{CO_2} cause corrosion (they attack steel), and hydrogen sulfide is deadly poisonous.
  • For a gas, the mole fraction and the volume fraction are the same, but the mass fraction is not. A heavy component with a small mole fraction can still make up a noticeable share of the mass.
  • We group everything heavier than hexane into the C7+ fraction and describe it by two numbers: its molar mass and its specific gravity. For equation-of-state calculations (calculations with a formula that links pressure, volume, temperature and composition) we split it into several pseudo-components.
  • We often express the density of stock-tank oil (oil at the surface, after its gas has come out) in degrees API. Water is 10 °API, and the lighter the oil, the more degrees it has.
Key relationships
M=∑iziMiM = \sum_i z_i M_i

molar mass of a mixture from the mole fractions

wi=ziMi∑jzjMjw_i = \dfrac{z_i M_i}{\sum_j z_j M_j}

mass fraction of a component

γg=Mg28.97,pi=yi p\gamma_g = \dfrac{M_g}{28.97},\qquad p_i = y_i\,p

specific gravity of a gas relative to air; partial pressure of a gas component (the pressure is absolute)

∘API=141.5γo−131.5{}^{\circ}\mathrm{API} = \dfrac{141.5}{\gamma_o} - 131.5

γo\gamma_o is the specific gravity of the oil relative to water at 60 °F (15.6 °C)

Common pitfalls
  • Confusing mole percentages with mass percentages: 62 mol% of methane in the gas of the lesson example is only 40% of its mass.
  • Assuming that more °API means a heavier oil. The scale runs the other way.
  • Taking the C7+ fraction for “heptane”. It is hundreds of compounds, and we measure its properties for every sample.
  • Calculating the partial pressure from the gauge pressure (the pressure above atmospheric). Dalton's law uses absolute pressure.
  • Hoping to detect hydrogen sulfide by its smell. At dangerous concentrations the sense of smell stops registering it.

After this lesson you can

  • list the main classes of hydrocarbons — alkanes (chain molecules), cycloalkanes (ring molecules), aromatics (molecules with a benzene ring), resins and asphaltenes (the heaviest, tar-like molecules) — and explain why each matters for production
  • name the non-hydrocarbon components of reservoir fluids and explain why hydrogen sulfide and carbon dioxide are dangerous
  • convert a composition from mole fractions to mass fractions; calculate the molar mass (the mass of one mole) and the specific gravity (the density relative to air) of a gas; and calculate the partial pressure of a component (the part of the total pressure that belongs to it)
  • explain how a composition is represented for calculations: individual components, the C7+ fraction with its molar mass and density, and pseudo-components (groups of compounds treated as one)
  • convert oil density to degrees API (an oilfield scale of density) and back, and judge from the composition which type of fluid it probably is

Oil and gas are mixtures of hydrocarbons

Oil and natural gas are natural mixtures in which hydrocarbons dominate. Hydrocarbons are compounds of carbon and hydrogen. By mass, oil is usually 83–87% carbon and 10–14% hydrogen. The rest is sulfur, nitrogen, oxygen and traces of metals. Oil contains hundreds and thousands of individual compounds. An engineer does not need each of them. What matters is how the mixture as a whole behaves when pressure and temperature change.

The state of a mixture depends first of all on the size of its molecules. A mixture of small molecules such as methane and ethane is a gas under ordinary conditions. A mixture of larger molecules is a liquid. The heaviest components — resins, asphaltenes and high-molecular-weight paraffins — are, in pure form, solid or as thick as bitumen (tar). In the reservoir, at high pressure and temperature, the mixture may be single-phase (all in one state). At the surface it may separate into gas and liquid. The topic Phase behaviour: pure substances and binary mixtures explains how this happens.

The pores of a reservoir always contain water as well as hydrocarbons. This is irreducible water, which the oil and gas could not push out when they filled the trap (the rock structure that holds oil and gas). Capillary forces (the forces of surface tension in narrow pores) hold it in place (see Fluid saturation). Formation water is also a reservoir fluid, but we describe its composition differently: by the dissolved salts.

Classes of hydrocarbons

A carbon atom forms four chemical bonds. We tell hydrocarbons apart by two features. The first is how the carbon atoms are joined: in an open chain or in a ring. The second is what bonds there are between them: only single bonds (such compounds are called saturated), or also double and triple bonds (unsaturated).

Molecular structure of hydrocarbons of different classesalkanes: isomerscycloalkanearomaticn-butaneC₄H₁₀isobutaneC₄H₁₀cyclohexaneC₆H₁₂benzeneC₆H₆

Four molecular diagrams. Normal butane is a zigzag chain of four carbon atoms. Isobutane is a central carbon atom with three neighbours. Cyclohexane is a hexagonal ring of six carbon atoms with single bonds. Benzene is a hexagonal ring in which single and double bonds alternate.

SchematicMolecular structure of the different classes. Normal butane (n-butane) and isobutane are isomers (same formula, different structure) with the formula C4H10\mathrm{C_4H_{10}}. Cyclohexane is a cycloalkane and benzene is an aromatic. Dots are carbon atoms and double lines are double bonds. Hydrogen atoms are not shown.
Classes of hydrocarbons in oil and gas
Class
Alkanes (paraffins)
Formula
CnH2n+2\mathrm{C}_n\mathrm{H}_{2n+2}
Structure
open chain, single bonds; normal and branched
Why it matters for production
the main part of most oils and almost all gas; heavy paraffins precipitate on cooling
Class
Cycloalkanes (naphthenes)
Formula
CnH2n\mathrm{C}_n\mathrm{H}_{2n}
Structure
rings of 5–6 carbon atoms, single bonds
Why it matters for production
similar to the alkanes in their properties, but denser
Class
Aromatics (arenes)
Formula
CnH2n−6\mathrm{C}_n\mathrm{H}_{2n-6} for benzene homologues
Structure
benzene ring
Why it matters for production
the densest for the same number of carbon atoms; they dissolve asphaltenes
Class
Resins and asphaltenes
Formula
—
Structure
large multi-ring molecules with S, N, O and metals
Why it matters for production
raise density and viscosity; asphaltenes can precipitate

Hydrocarbons with double and triple bonds (alkenes, alkynes) are chemically reactive and practically absent from reservoir fluids: they form during oil refining.

Alkanes, or paraffins, are the most widespread class. Methane, ethane, propane and the butanes are gases under normal conditions. From pentane up to about hexadecane C16\mathrm{C_{16}} they are liquids. Normal (straight-chain) alkanes from about C17\mathrm{C_{17}} upward are solid at room temperature: this is paraffin wax. Dissolved in oil, the wax does no harm. But when the oil cools in the well and in pipelines, the wax crystallises and deposits on the walls (Retrograde condensation, wax, asphaltenes and high-pressure fluids).

From butane onwards, the same atoms can be joined in different ways. The result is isomers: substances with the same composition but a different structure and different properties. Normal butane boils at −0.5 °C, and branched isobutane at −11.7 °C. The number of isomers grows quickly with the number of carbon atoms. Pentane has 3, heptane has 9, decane has 75, and the alkane C20\mathrm{C_{20}} has more than 366 thousand. This is why we cannot describe the heavy part of oil compound by compound.

Cycloalkanes, called naphthenes in petroleum practice, have their carbon atoms closed into rings of five or six: cyclopentane, cyclohexane and their derivatives with side chains. Aromatics contain a benzene ring. The simplest are benzene, toluene and the xylenes. For the same number of carbon atoms, aromatics are noticeably denser than alkanes. The specific gravity of liquid n-hexane is about 0.66, of cyclohexane 0.78, and of benzene 0.88.

The heaviest part of oil is the resins and asphaltenes. These are large molecules made of several fused rings, with atoms of sulfur, nitrogen, oxygen and metals (vanadium and nickel). Their share decides how dark, dense and viscous the oil is. Asphaltenes stay in the oil in a colloidal state (as tiny suspended particles), surrounded by resins. They may precipitate (settle out as a solid) when the pressure falls or when the oil is mixed with injected gas or carbon dioxide.

Non-hydrocarbon components

Besides hydrocarbons, reservoir fluids almost always contain non-hydrocarbon components. There are usually few of them. But they decide equipment corrosion, safety, the processing of the produced fluids and their price, and sometimes the phase behaviour of the fluid (whether it is gas, liquid or both).

Non-hydrocarbon components and why they matter
Component
Nitrogen N2\mathrm{N_2}
Where found
gases, gas caps
Why it matters
inert (does not react); lowers the heating value of the gas; raises the bubble point pressure of oil
Component
Carbon dioxide CO2\mathrm{CO_2}
Where found
gases and gas condensates, less often oils
Why it matters
forms carbonic acid with water — corrosion of steel; dissolves readily in oil
Component
Hydrogen sulfide H2S\mathrm{H_2S}
Where found
gases and oils, especially in carbonate reservoirs
Why it matters
deadly poisonous; corrosion and sulfide stress cracking of steel (steel cracks in the presence of H2S\mathrm{H_2S})
Component
Mercaptans (strong-smelling sulfur compounds) and other sulfur compounds
Where found
oils and condensates
Why it matters
odour, corrosion; sulfur lowers the price of the oil
Component
Water with dissolved salts
Where found
always
Why it matters
scale deposits (mineral crusts); compatibility with injected water; forms hydrates (ice-like solids) with gas
Component
Mercury, helium
Where found
some gases
Why it matters
mercury destroys aluminium heat exchangers; helium is extracted as a valuable product

Hydrogen sulfide and carbon dioxide are called acid gases. A gas with a noticeable hydrogen sulfide content is called sour gas, and a gas without it is called “sweet” gas. Do not confuse sour gas with sulfur dioxide, SO2\mathrm{SO_2}, which Russian-language practice calls “sulfurous gas”.

Carbon dioxide is not aggressive in itself. But in water it forms carbonic acid, and this causes carbon dioxide corrosion of carbon steel. It dissolves readily in oil, lowers its viscosity and increases its volume. This is the basis of CO2\mathrm{CO_2} injection for enhanced oil recovery. At the same time, carbon dioxide can cause asphaltenes to precipitate.

Hydrogen sulfide is dangerous in three ways. It is poisonous, it is flammable, and in the presence of water it causes corrosion and sulfide stress cracking. This is the brittle failure of loaded steel (it breaks suddenly, with no stretching first), especially high-strength steel. We choose materials for hydrogen sulfide environments according to a dedicated international standard (ISO 15156, also known as NACE MR0175). The measure of how aggressive the gas is the partial pressure of hydrogen sulfide: the product of its mole fraction in the gas and the absolute pressure of the gas. For carbon and low-alloy steels, we must confirm resistance to sulfide stress cracking from as little as about 0.3 kPa (0.05 psi).

Mole, mass and volume fractions

We can write the composition of a mixture in three ways: in mole fractions, mass fractions or volume fractions. The numbers come out different.

Mole fraction ziz_i
The share of the moles (that is, the molecules) of component ii among all the moles of the mixture. The mole fractions add up to one. We write yiy_i for the gas phase, xix_i for the liquid phase and ziz_i for the mixture as a whole. In percent, it is mole percent, mol%.
Mass fraction wiw_i
The share of the mass of the mixture that belongs to the component (wt%). We use mass fractions, for example, to report the results of distilling the heavy part of oil.
Volume fraction
The share of the volume that the component would occupy on its own at the same conditions (vol%). For ideal gases it equals the mole fraction: equal volumes of gases at the same pressure and temperature contain the same number of molecules. For liquids it does not. Equal volumes of different liquids contain different numbers of molecules, and, besides, volumes do not add up exactly on mixing.

We do phase-equilibrium and gas-law calculations in mole fractions. The pressure of a gas, solubility and the distribution of components between phases depend on the number of molecules, not on their mass. That is why a gas analysis is reported in mole percent (for a gas, the same as volume percent). To get to mass, we need the molar masses of the components.

Molar mass of a mixture
M=∑iziMiM = \sum_i z_i M_i
where:
  • MMmolar (average) mass of the mixture, g/mol = kg/kmol
  • ziz_imole fraction of component ii, fraction
  • MiM_imolar mass of the component; for the C7+ fraction, its measured molar mass, g/mol

Units: g/mol (kg/kmol)

Holds when
  • The mole fractions of all the components, including the non-hydrocarbon ones and the C7+ fraction, add up to one.

It is the mass of one mole of the mixture. We use it to find the gas density, to convert mole fractions to mass fractions, and to convert an amount in moles to a mass.

Conversion from mole fractions to mass fractions
wi=ziMi∑jzjMjw_i = \frac{z_i M_i}{\sum_j z_j M_j}
where:
  • wiw_imass fraction of component ii, fraction
  • zi, Miz_i,\ M_imole fraction and molar mass of the component, —, g/mol

Units: fraction

Holds when
  • The composition is given completely, including the C7+ fraction with its molar mass.

The reverse conversion is zi=wi/Mi∑jwj/Mjz_i = \dfrac{w_i/M_i}{\sum_j w_j/M_j}. A heavy component with a small mole fraction can make up a noticeable share of the mass. A light component does the opposite.

Specific gravity of a gas relative to air
γg=MgMair=Mg28.97\gamma_g = \frac{M_g}{M_{\text{air}}} = \frac{M_g}{28.97}
where:
  • γg\gamma_ggas specific gravity — the ratio of its density to the density of air at the same pressure and temperature, —
  • MgM_gmolar mass of the gas, g/mol
  • MairM_{\text{air}}molar mass of dry air, 28.97 g/mol, g/mol

Units: dimensionless

Holds when
  • We compare the gas and air at the same pressure and temperature, as ideal gases. At standard conditions this is accurate to better than 1%.

Pure methane has γg=0.554\gamma_g = 0.554. Dry natural gas usually has 0.55–0.65. Associated gas (gas produced together with oil) that is rich in propane and butanes has 0.8 or more. The gas density at standard conditions is ρg=1.204 γg\rho_g = 1.204\,\gamma_g kg/m³ at 0.101325 MPa and 20 °C (at 15 °C, 1.225 γg1.225\,\gamma_g).

Partial pressure of a gas component (Dalton's law)
pi=yi pp_i = y_i\,p
where:
  • pip_ipartial pressure of component ii, MPa or kPa
  • yiy_imole fraction of the component in the gas, fraction
  • ppabsolute pressure of the gas, MPa or kPa

Units: any pressure units; the pressure is absolute

Holds when
  • The gas is close to ideal. At high pressures the partial pressure remains a convenient rule-of-thumb measure, but for phase-equilibrium calculations we use fugacity (an “effective pressure” for real fluids) instead (Vapour–liquid equilibrium).

It is the part of the gas pressure that belongs to the component. We use the partial pressures of H2S\mathrm{H_2S} and CO2\mathrm{CO_2} to measure how aggressive the gas is to steel.

Worked exampleSour associated gas: molar mass, mass fractions and the partial pressure of hydrogen sulfide
Given (assumed data)
Separator gas composition, mol%
N2\mathrm{N_2} 1.0; CO2\mathrm{CO_2} 3.0; H2S\mathrm{H_2S} 10.0; methane 62.0; ethane 12.0; propane 7.0; butanes 3.0; pentanes 2.0
Molar masses in the same order, g/mol
28.01; 44.01; 34.08; 16.04; 30.07; 44.10; 58.12; 72.15
Separator pressure
2.0 MPa (absolute)
Standard conditions
0.101325 MPa and 20 °C
Solution
  1. The molar mass of the mixture is the sum of the products ziMiz_i M_i over all components: M=24.84M = 24.84 g/mol. Methane makes the largest contribution (0.62⋅16.04=9.940.62\cdot 16.04 = 9.94 g/mol), and hydrogen sulfide adds 0.10⋅34.08=3.410.10\cdot 34.08 = 3.41 g/mol.
  2. Gas specific gravity: γg=24.84/28.97=0.857\gamma_g = 24.84/28.97 = 0.857.
  3. Molar volume of an ideal gas at standard conditions: Vm=RT/p=8.314⋅293.15/101.325=24.06V_m = RT/p = 8.314\cdot 293.15/101.325 = 24.06 m³/kmol, where R=8.314R = 8.314 kJ/(kmol·K) is the universal gas constant and the pressure is in kPa. Gas density: ρg=M/Vm=1.03\rho_g = M/V_m = 1.03 kg/m³.
  4. Mass fraction of hydrogen sulfide: w=3.41/24.84=0.137w = 3.41/24.84 = 0.137, that is, 13.7 wt%; of methane, 40 wt%, although by moles it is 62%.
  5. Partial pressure of hydrogen sulfide in the separator (the vessel that splits the well stream into gas, oil and water): pH2S=0.10⋅2.0=0.20p_{\mathrm{H_2S}} = 0.10\cdot 2.0 = 0.20 MPa, or 200 kPa. This is about 670 times the 0.3 kPa guideline.
  6. In 1000 m³ of gas (at standard conditions) there are 1000/24.06=41.61000/24.06 = 41.6 kmol of substance, of which hydrogen sulfide accounts for 4.164.16 kmol, or 4.16⋅34.08=1424.16\cdot 34.08 = 142 kg.
Answer: The molar mass of the gas is 24.84 g/mol, and γg=0.857\gamma_g = 0.857. Hydrogen sulfide is 10 mol% but 13.7 wt%; its partial pressure in the separator is 200 kPa, and every thousand cubic metres of gas contains about 142 kg of hydrogen sulfide.

What it means. The gas is heavier than ordinary natural gas because of the hydrogen sulfide, carbon dioxide and heavy hydrocarbons. The H2S\mathrm{H_2S} partial pressure is hundreds of times above the threshold from which we check steel for sulfide stress cracking, so all the equipment must resist hydrogen sulfide. We need mass fractions for the material balances of gas treatment plants: we use them to calculate how much sulfur we will get from the gas. The data in this example are assumed for illustration.

The C7+ fraction and pseudo-components

We do not get the composition of reservoir oil from one analysis of a sample under pressure. Usually we analyse the gas and the liquid separately: from the separator, or after the bottomhole sample (a sample taken at the bottom of the well) has been degassed in the laboratory. Then we recombine them, which means recalculating them to the reservoir mixture using the measured gas–oil ratio (Reservoir fluid sampling; a worked calculation is in the lesson on fluid types).

We analyse the gas by chromatography (a method that separates a mixture into its components). We measure each of the non-hydrocarbon gases and each hydrocarbon up to the pentanes separately, and the heavier ones in groups. We divide the liquid into fractions by carbon number (the number of carbon atoms in a molecule). The C8\mathrm{C_8} fraction includes all the compounds that boil between normal heptane and normal octane: alkanes, naphthenes and aromatics alike. Modern analysis goes up to C30\mathrm{C_{30}} and beyond. But we lump everything heavier than the chosen limit into a single plus fraction, C7+, or C10+, C20+ and so on.

Properties of reservoir fluid components
Component
Nitrogen N2\mathrm{N_2}
MM, g/mol
28.01
TbT_{\text{b}}, °C
−195.8
TcT_{\text{c}}, °C
−147.0
pcp_{\text{c}}, MPa
3.40
Component
Carbon dioxide CO2\mathrm{CO_2}
MM, g/mol
44.01
TbT_{\text{b}}, °C
−78.5*
TcT_{\text{c}}, °C
31.0
pcp_{\text{c}}, MPa
7.38
Component
Hydrogen sulfide H2S\mathrm{H_2S}
MM, g/mol
34.08
TbT_{\text{b}}, °C
−60.3
TcT_{\text{c}}, °C
100.0
pcp_{\text{c}}, MPa
9.0
Component
Methane C1\mathrm{C_1}
MM, g/mol
16.04
TbT_{\text{b}}, °C
−161.5
TcT_{\text{c}}, °C
−82.6
pcp_{\text{c}}, MPa
4.60
Component
Ethane C2\mathrm{C_2}
MM, g/mol
30.07
TbT_{\text{b}}, °C
−88.6
TcT_{\text{c}}, °C
32.2
pcp_{\text{c}}, MPa
4.87
Component
Propane C3\mathrm{C_3}
MM, g/mol
44.10
TbT_{\text{b}}, °C
−42.1
TcT_{\text{c}}, °C
96.7
pcp_{\text{c}}, MPa
4.25
Component
Isobutane
MM, g/mol
58.12
TbT_{\text{b}}, °C
−11.7
TcT_{\text{c}}, °C
134.7
pcp_{\text{c}}, MPa
3.64
Component
n-Butane
MM, g/mol
58.12
TbT_{\text{b}}, °C
−0.5
TcT_{\text{c}}, °C
152.0
pcp_{\text{c}}, MPa
3.80
Component
n-Pentane
MM, g/mol
72.15
TbT_{\text{b}}, °C
36.1
TcT_{\text{c}}, °C
196.6
pcp_{\text{c}}, MPa
3.37
Component
n-Hexane
MM, g/mol
86.18
TbT_{\text{b}}, °C
68.7
TcT_{\text{c}}, °C
234.5
pcp_{\text{c}}, MPa
3.03
Component
n-Heptane
MM, g/mol
100.20
TbT_{\text{b}}, °C
98.4
TcT_{\text{c}}, °C
267.1
pcp_{\text{c}}, MPa
2.74
Component
n-Decane
MM, g/mol
142.28
TbT_{\text{b}}, °C
174.1
TcT_{\text{c}}, °C
344.6
pcp_{\text{c}}, MPa
2.11
Component
Water H2O\mathrm{H_2O}
MM, g/mol
18.02
TbT_{\text{b}}, °C
100.0
TcT_{\text{c}}, °C
374.0
pcp_{\text{c}}, MPa
22.06

These are reference values. TbT_{\text{b}} is the normal boiling point (at 0.101325 MPa). TcT_{\text{c}} and pcp_{\text{c}} are the critical temperature and pressure (the point beyond which gas and liquid can no longer be told apart) (Phase behaviour: pure substances and binary mixtures). *Carbon dioxide does not boil at atmospheric pressure. It sublimes, which means it passes from solid straight into gas. It can be liquid only at pressures above 0.52 MPa.

We know the properties of the individual components precisely, but the C7+ fraction is different for every oil. Two measured numbers describe it: the molar mass M7+M_{7+} and the specific gravity γ7+\gamma_{7+} (relative to water at 60 °F). The molar mass reflects the size of the molecules, which means how volatile (how easily evaporating) the fraction is. At the same molar mass, the density tells us the type of hydrocarbons: an aromatic fraction is denser than a paraffinic one. For oils M7+M_{7+} is typically about 180–280 g/mol, and for gas condensates about 110–170 g/mol.

For equation-of-state calculations (Cubic equations of state) we divide the C7+ fraction into parts. The mole fraction of the fractions falls in a regular way as the carbon number grows, roughly exponentially. This lets us extend the analysis beyond its limit. Then we lump neighbouring fractions together again into 3–5 pseudo-components: imaginary substances with average properties. A typical fluid model has 8–15 components. The calculation stays accurate enough and does not take too long.

For complex fluids, such as high-pressure, high-temperature gas condensates and oils prone to paraffin and asphaltene deposition, M7+M_{7+} and γ7+\gamma_{7+} alone are not enough. Then we find the group composition of the heavy fractions: the shares of paraffins, naphthenes and aromatics (PNA analysis, from paraffins, naphthenes, aromatics). To assess the stability of the asphaltenes, we find the shares of saturates, aromatics, resins and asphaltenes (SARA analysis, from saturates, aromatics, resins, asphaltenes).

Typical composition of reservoir fluids of different types, mol%
Fluid type
Dry gas
N2+CO2\mathrm{N_2} + \mathrm{CO_2}
1.5
C1\mathrm{C_1}
94.5
C2\mathrm{C_2}–C6\mathrm{C_6}
4.0
C7+
traces
Fluid type
Wet gas
N2+CO2\mathrm{N_2} + \mathrm{CO_2}
2.0
C1\mathrm{C_1}
87.0
C2\mathrm{C_2}–C6\mathrm{C_6}
10.0
C7+
1.0
Fluid type
Gas condensate
N2+CO2\mathrm{N_2} + \mathrm{CO_2}
2.5
C1\mathrm{C_1}
75.0
C2\mathrm{C_2}–C6\mathrm{C_6}
16.0
C7+
6.5
Fluid type
Volatile oil
N2+CO2\mathrm{N_2} + \mathrm{CO_2}
1.5
C1\mathrm{C_1}
58.0
C2\mathrm{C_2}–C6\mathrm{C_6}
24.5
C7+
16.0
Fluid type
Black oil
N2+CO2\mathrm{N_2} + \mathrm{CO_2}
1.0
C1\mathrm{C_1}
38.0
C2\mathrm{C_2}–C6\mathrm{C_6}
17.0
C7+
44.0

The data are assumed for illustration, because every real fluid has its own composition. The pattern, however, is reliable: the more C7+ in the mixture, the lower the gas–oil ratio and the “heavier” the fluid. The boundary between oils and gas condensates lies at about 12.5 mol% C7+ (Types of reservoir fluids).

Assumed composition of five types of reservoir fluid0255075100Mole fraction, %Dry gas94.5Wet gas87Gas condensate756.5Volatile oil5816Black oil3844N₂ + CO₂C₁C₂–C₆C₇₊

Five horizontal bars 100 percent long, divided into segments: nitrogen and carbon dioxide, methane, components from ethane to hexane, and the C7+ fraction. In dry gas almost the whole bar is methane, 94.5 percent. In wet gas and gas condensate the shares of ethane–hexane and C7+ grow. In volatile oil, C7+ takes up 16 percent. In black oil, methane is 38 percent and the C7+ fraction 44 percent of the bar.

CalculatedAssumed composition of the five types of reservoir fluid from the table above: from dry gas to black oil the share of methane falls and the share of the C7+ fraction grows.

Oil density and degrees API

The most readily available characteristic of produced oil is the density of the stock-tank oil. In Russian-language practice, we give it in kg/m³ at 20 °C (sometimes at 15 °C). In international practice, it is given as the specific gravity γo\gamma_o (the density relative to water) at 60 °F (15.6 °C), or in degrees API (after the American Petroleum Institute).

Oil density in degrees API
∘API=141.5γo−131.5{}^{\circ}\mathrm{API} = \frac{141.5}{\gamma_o} - 131.5
where:
  • ∘API{}^{\circ}\mathrm{API}density in degrees API, °API
  • γo\gamma_ospecific gravity of the stock-tank oil: ρo/ρw\rho_o/\rho_w, both densities at 60 °F (15.6 °C); the density of water at this temperature is 999.0 kg/m³, —

Units: °API is a conventional scale; γo\gamma_o is dimensionless

Holds when
  • The oil and water densities are at 60 °F and atmospheric pressure. We first bring a density measured at another temperature to 60 °F, using standard tables.
  • The formula applies to stock-tank oil and condensate, not to reservoir oil with dissolved gas.

The scale runs the other way: the lighter the oil, the more degrees. Water is 10 °API, and an oil heavier than water has less than 10 °API. The reverse conversion is γo=141.5131.5+∘API\gamma_o = \dfrac{141.5}{131.5 + {}^{\circ}\mathrm{API}}.

Classes of oils by density in degrees API
Class
Light
°API
above 31.1
Specific gravity γo\gamma_o
below 0.870
Class
Medium
°API
22.3–31.1
Specific gravity γo\gamma_o
0.870–0.920
Class
Heavy
°API
10–22.3
Specific gravity γo\gamma_o
0.920–1.000
Class
Extra-heavy
°API
below 10
Specific gravity γo\gamma_o
above 1.000

This is one of the common classifications. Other systems, including national standards, put the class boundaries elsewhere. The liquid of gas condensates and wet gases is usually lighter than oil: about 40–70 °API.

Worked exampleDegrees API and density
Given (assumed data)
Oil of field A
846 kg/m³ at 15.6 °C
Oil of field B
941 kg/m³ at 15.6 °C
Condensate
56 °API
Density of water at 15.6 °C
999.0 kg/m³
Solution
  1. Oil A: γo=846/999.0=0.847\gamma_o = 846/999.0 = 0.847; ∘API=141.5/0.847−131.5=35.6{}^{\circ}\mathrm{API} = 141.5/0.847 - 131.5 = 35.6.
  2. Oil B: γo=941/999.0=0.942\gamma_o = 941/999.0 = 0.942; ∘API=18.7{}^{\circ}\mathrm{API} = 18.7.
  3. Condensate: γo=141.5/(131.5+56)=0.755\gamma_o = 141.5/(131.5 + 56) = 0.755, and the density is 0.755⋅999.0=7540.755\cdot 999.0 = 754 kg/m³.
  4. The densities of oils A and B differ by 11%, and in degrees API by 17 units.
Answer: Oil A is 35.6 °API (light), oil B is 18.7 °API (heavy); the density of the condensate is about 754 kg/m³.

What it means. The API scale stretches out the range of light oils: a density difference of 11% becomes 17 degrees. From the density alone, already at the wellhead, we can guess what kind of fluid we are dealing with. A dark oil with a low °API is more likely “black”, and a pale liquid of 50–60 °API is more likely a condensate. But we establish the fluid type finally from the gas–oil ratio and laboratory studies (Types of reservoir fluids). The data in this example are assumed for illustration.

At the drilling rig, we also judge how light an oil is from indirect signs. We examine cuttings (rock chips brought up by the drilling fluid) and core under an ultraviolet lamp. Oil glows, and the colour of the glow changes from dull brown for heavy oils to yellow and bluish-white for light ones. The refractive index of oil (how strongly it bends light) also rises with its density. These are quick qualitative estimates, not a substitute for analysis.

Terms and symbols

Component
An individual substance (methane, CO2\mathrm{CO_2}) or a simplified group of substances (a fraction, a pseudo-component) whose proportion is stated in the composition of a mixture.
Mole fraction ziz_i (yiy_i, xix_i)
The share of the moles of a component in a mixture (in a gas, in a liquid). For an ideal gas it equals the volume fraction.
C7+ fraction
All the components heavier than hexane, lumped into one. We describe it by its molar mass M7+M_{7+} and specific gravity γ7+\gamma_{7+}.
Pseudo-component
An imaginary substance that replaces, in equation-of-state calculations, a group of components or fractions with similar properties.
Gas specific gravity γg\gamma_g
The ratio of the molar mass of the gas to the molar mass of air (28.97 g/mol).
Degrees API
A scale of stock-tank oil density that runs the other way: 141.5/γo−131.5141.5/\gamma_o - 131.5. Water is 10 °API.
Acid gases
Hydrogen sulfide and carbon dioxide. A gas containing hydrogen sulfide is called sour.
Partial pressure
The product of the mole fraction of a component in the gas and the absolute pressure of the gas. It measures how aggressive H2S\mathrm{H_2S} and CO2\mathrm{CO_2} are.

Common mistakes and misconceptions

Wrong

“The gas is 62% methane, so by mass it is more than half methane.”

In fact

The percentages in a gas analysis are mole (volume) percentages. In the gas of the lesson example the mass fraction of methane is about 40%, because heavy components with a small mole fraction weigh a lot.

Wrong

“An oil of 30 °API is heavier than an oil of 20 °API.”

In fact

The API scale runs the other way: 30 °API is γo≈0.876\gamma_o \approx 0.876, and 20 °API is γo≈0.934\gamma_o \approx 0.934. The 30 °API oil is lighter.

Wrong

“The C7+ fraction is heptane, and its properties are in the handbook.”

In fact

C7+ is all the compounds heavier than hexane, hundreds and thousands of substances. We measure its molar mass and density for every sample, and they differ from oil to oil.

Wrong

“There is only 1% hydrogen sulfide, so it is not dangerous.”

In fact

At a pressure of 10 MPa the partial pressure of such hydrogen sulfide is 100 kPa, hundreds of times above the threshold for sulfide stress cracking. And 1% is 10,000 ppm, so a leak of such gas is deadly dangerous.

Wrong

“The composition of the separator gas is the composition of the reservoir fluid.”

In fact

The gas and liquid of the separator are the products of splitting the reservoir mixture. We get the composition of the mixture by recombining the gas and liquid samples in the proportion that the gas–oil ratio gives.

In depth: advanced detail

Watson characterisation factor

How can we judge the type of hydrocarbons in a fraction from two numbers, the boiling point and the density? At the same boiling point, paraffins are lighter than naphthenes, and naphthenes are lighter than aromatics. The Watson characterisation factor puts this into one number:

Watson characterisation factor
Kw=Tb1/3γK_w = \frac{T_b^{1/3}}{\gamma}
where:
  • KwK_wcharacterisation factor, —
  • TbT_bmean boiling point of the fraction in degrees Rankine: Tb [∘R]=1.8 Tb [K]T_b\,[{}^{\circ}\text{R}] = 1.8\,T_b\,[\text{K}], °R
  • γ\gammaspecific gravity of the fraction relative to water at 60 °F, —

Units: empirical: TbT_b only in degrees Rankine

Holds when
  • This is an empirical index. For narrow fractions we take their mean boiling point. For wide fractions we use a specially averaged boiling point of the whole fraction.

For normal alkanes KwK_w is about 12.5–13, for naphthenes about 11, and for aromatics about 10. For the heavy fractions of oils and condensates it is usually 11.5–12.5. Without a detailed composition, we use KwK_w to judge which type of hydrocarbon predominates, and to estimate other properties of the fraction from correlations (empirical formulas).

Worked exampleThree C6 hydrocarbons: close boiling points, different classes
Given (assumed data)
n-Hexane
boils at 68.7 °C, γ=0.664\gamma = 0.664
Cyclohexane
boils at 80.7 °C, γ=0.783\gamma = 0.783
Benzene
boils at 80.1 °C, γ=0.884\gamma = 0.884
Solution
  1. n-Hexane: Tb=1.8⋅(68.7+273.15)=615T_b = 1.8\cdot(68.7 + 273.15) = 615 °R; Kw=6151/3/0.664=12.8K_w = 615^{1/3}/0.664 = 12.8.
  2. Cyclohexane: Tb=637T_b = 637 °R; Kw=11.0K_w = 11.0.
  3. Benzene: Tb=636T_b = 636 °R; Kw=9.7K_w = 9.7.
Answer: KwK_w is 12.8 for n-hexane, 11.0 for cyclohexane and 9.7 for benzene.

What it means. The boiling points are close, but the densities and KwK_w differ. The Watson factor at once separates the paraffinic, naphthenic and aromatic character. For the C7+ fraction of a real oil we get an in-between value, which reflects a mixture of classes. The properties of these substances are reference values.

Connections to other topics

Check yourself

  1. 1

    Which class does the hydrocarbon C6H12\mathrm{C_6H_{12}} belong to, if all the bonds in its molecule are single?

    1. A.Aromatic
    2. B.Cycloalkane (naphthene)
    3. C.Alkane
    4. D.Alkene
    Show the answer

    Correct answer: B. Cycloalkane (naphthene)

    Only a closed ring has the formula CnH2n\mathrm{C}_n\mathrm{H}_{2n} with single bonds. Here it is cyclohexane or one of its isomers, for example methylcyclopentane. An alkene with the same formula would contain a double bond, and the alkane C6\mathrm{C_6} has the formula C6H14\mathrm{C_6H_{14}}.

  2. 2

    A gas contains 90 mol% methane and 10 mol% carbon dioxide. What is the mass fraction of CO2\mathrm{CO_2}?

    1. A.About 44%
    2. B.About 5%
    3. C.10%
    4. D.About 23%
    Show the answer

    Correct answer: D. About 23%

    M=0.9⋅16.04+0.1⋅44.01=18.84M = 0.9\cdot 16.04 + 0.1\cdot 44.01 = 18.84 g/mol; wCO2=4.40/18.84≈0.234w_{\mathrm{CO_2}} = 4.40/18.84 \approx 0.234. The heavy CO2\mathrm{CO_2} molecule makes up 10% of the moles but almost a quarter of the mass.

  3. 3

    A gas with 3 mol% hydrogen sulfide flows through a pipeline at an absolute pressure of 6 MPa. What is the partial pressure of H2S\mathrm{H_2S}? What does it mean for the choice of pipes?

    Show the answer

    pH2S=0.03⋅6=0.18p_{\mathrm{H_2S}} = 0.03\cdot 6 = 0.18 MPa, or 180 kPa, which is 600 times the 0.3 kPa guideline. The pipes, valves and welds must resist sulfide stress cracking. We select the materials according to the standard for hydrogen sulfide environments.

  4. 4

    The specific gravity of an oil is 0.876. How many degrees API is that?

    1. A.30 °API
    2. B.20 °API
    3. C.87.6 °API
    4. D.40 °API
    Show the answer

    Correct answer: A. 30 °API

    141.5/0.876−131.5=161.5−131.5=30.0141.5/0.876 - 131.5 = 161.5 - 131.5 = 30.0 °API — a medium oil by density.

  5. 5

    Why is the C7+ fraction described by its molar mass and density rather than by a list of substances?

    Show the answer

    The fraction contains thousands of compounds (the isomers of the alkane C20\mathrm{C_{20}} alone number more than 366 thousand), and every oil has its own set. For calculations, what matters is not the individual substances but their averaged properties. The molar mass tells us the size of the molecules and the volatility. The density at the same molar mass tells us the type of hydrocarbons. From these two numbers, correlations (empirical formulas) give the critical parameters and the other properties that the equation of state needs.

Updated: 30 September 2026