HomeLab, Chemistry & Measurement ToolsMoisture Content Calculator with Steps (Free) | StatsUnlock

Moisture Content Calculator with Steps (Free) | StatsUnlock

Moisture Content Calculator

Calculate moisture content on a dry or wet basis for soil, wood, food, sand, aggregate and sludge, convert to volumetric soil moisture, find the moisture content of air from relative humidity or dew point, predict the equilibrium moisture content of wood and grain, find the optimum moisture content from a Proctor test, and work out drying and average moisture, with every step, a figure and four charts.

LabSoilWoodFoodStep by StepDry and Wet BasisHumidityProctorFree

⚡ 0. Quick Answer

A moisture content calculator uses moisture content = (wet mass − dry mass) ÷ dry mass × 100 on a dry basis, used for soil, sand and wood, or ÷ wet mass on a wet basis, used for food, grain and sludge. A tin of 15.20 g holding 65.80 g of wet soil and 57.40 g after oven drying lost 8.40 g of water from 42.20 g of dry soil: 19.91% moisture content (dry basis), or 16.60% wet basis.

w = (Mwet − Mdry) ÷ Mdry × 100  ·  MCwet = (Mwet − Mdry) ÷ Mwet × 100  ·  wet = dry ÷ (1 + dry)

Rule of thumb: subtract the container first, always say which basis you used, and remember that dry-basis moisture can exceed 100% (peat, green wood, sludge) while wet-basis moisture never can.

⭐ Key Takeaways

  • A moisture content calculator turns 8.40 g of water in 42.20 g of dry soil into 19.91% (dry basis) = 16.60% (wet basis).
  • Volumetric soil moisture: θv = w × ρb; 20% at 1.40 g/cm³ is 0.28 cm³/cm³, or 84 mm of water in 30 cm of soil.
  • Moisture content of air at 25 °C and 60% RH is 11.9 g of water per kg of dry air (dew point 16.7 °C).
  • Wood at 70 °F and 50% RH settles near 9.2% equilibrium moisture content.
  • Drying 1000 kg of grain from 20% to 14% (wet basis) removes 69.8 kg of water, not 60 kg.

📚 1. What Is Moisture Content and How Is It Calculated?

Moisture content is the amount of water in a material, expressed as a percentage of either its dry mass (dry basis) or its wet mass (wet basis). It is measured most simply by weighing a sample, drying it in an oven to constant mass and weighing it again: the loss is the water. The gravimetric method is still the reference for soil (Reynolds, 1970) against which sensors are calibrated (Bittelli, 2011). A moisture content calculator applies the right basis, converts between them, carries the balance uncertainty, and extends the same idea to soil volume, air, wood, grain and compaction.

A worked one-liner: water = 65.80 − 57.40 = 8.40 g; dry soil = 57.40 − 15.20 = 42.20 g; w = 8.40 ÷ 42.20 × 100 = 19.91%; wet basis = 8.40 ÷ 50.60 × 100 = 16.60%.

Who uses it: geotechnical and soil science students running ASTM D2216; agronomists and hydrologists converting to volumetric water; woodworkers and kiln operators checking lumber and EMC; food technologists reporting moisture and total solids; grain handlers calculating shrink; HVAC engineers finding the moisture content of air; concrete technicians correcting sand moisture; and wastewater operators measuring sludge solids.

Quick answers to the questions people ask most

How do you calculate moisture content?

Weigh the sample wet, dry it in an oven to constant mass, and weigh it again. The mass lost is water. Divide the water by the dry mass and multiply by 100 for dry-basis moisture, or divide by the wet mass for wet-basis moisture. Subtract the container mass from both weighings first.

What is the formula to calculate moisture content of soil?

Soil moisture content w = (W₂ − W₃) ÷ (W₃ − W₁) × 100, where W₁ is the container, W₂ the container plus wet soil and W₃ the container plus oven-dry soil (ASTM D2216, 110 °C). For 15.20, 65.80 and 57.40 g, w = 8.40 ÷ 42.20 × 100 = 19.91%.

How do you calculate the moisture content of wood?

Moisture content of wood = (green mass − oven-dry mass) ÷ oven-dry mass × 100, always on a dry basis. A board piece weighing 120 g that dries to 100 g at 103 °C holds 20% moisture. Green wood can exceed 100%; seasoned interior wood is usually 6 to 12%.

How do you calculate moisture content of food?

Food moisture is reported on a wet basis: (initial mass − dried mass) ÷ initial mass × 100. A 5.000 g sample that dries to 4.150 g has 0.850 g of water, which is 17.0% moisture and 83.0% total solids. Sugary foods are dried in a vacuum oven at lower temperature.

How do you calculate the moisture content of air?

Find the vapor pressure e, either relative humidity times the saturation pressure or the saturation pressure at the dew point, then W = 0.622 × e ÷ (P − e). At 25 °C and 60% RH at sea level, e = 1.90 kPa and W = 11.9 g of water per kg of dry air.

How do you calculate volumetric moisture content?

Multiply gravimetric moisture (as a decimal) by dry bulk density and divide by the density of water: θv = w × ρb ÷ ρw. Soil at 20% gravimetric moisture and 1.40 g/cm³ holds 0.28 cm³ of water per cm³, which is 84 mm of water in a 30 cm layer.

Bench note from the StatsUnlock Team: the mistakes we see most are mixing the two bases (a wood moisture of 25% dry basis is only 20% wet basis) and forgetting the container: dividing the water by the tin plus soil instead of the soil alone understated our headline sample’s moisture by more than a quarter. The third is treating points of moisture as percent of mass: drying grain from 20% to 14% removes 7.0% of its mass, not 6%. Soil water methods and their pitfalls are reviewed by Robinson et al. (2008).
How a moisture content calculator separates water from dry soil Container 15.20 grams; with wet soil 65.80 grams; with dry soil 57.40 grams. Water 8.40 grams, dry soil 42.20 grams, moisture content 19.91 percent dry basis and 16.60 percent wet basis. tin + wet soil tin + dry soil tin 15.20dry soil 42.208.40 water dry soil 42.20 g w = 8.40 ÷ 42.20 × 100 = 19.91% (dry basis) wet basis = 8.40 ÷ 50.60 × 100 = 16.60% The oven drives off 8.40 g of water; the tin and the soil grains stay.
Gravimetric moisture content: the water is the mass lost on drying, divided by the dry soil (not the tin).
MaterialUsual basisOvenTypical range
SoilDry110 ± 5 °C (ASTM D2216)5 to 40%; peat over 100%
Sand and aggregateDry110 ± 5 °C (ASTM C566)0 to 10%
WoodDry103 ± 2 °C (ASTM D4442)6 to 12% seasoned; over 100% green
FoodWetAbout 100 to 105 °C or vacuum oven2 to 95%
GrainWetCrop-specific10 to 30%
SludgeWet (and total solids)103 to 105 °C60 to 99%

🧮 2. Enter Your Data

Weigh the dried sample after it cools in a desiccator. Both percentages are shown; the material sets which one is the headline.

📊 3. Results and Step-by-Step Working

Load a sample or enter your data above, then press Calculate Moisture Content. The moisture content on both bases, the formula, every step, a figure, four charts and the exports will appear here.

🧠 4. Understanding Your Moisture Content Result, In Detail

4.1 What the result means

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4.2 The formula

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4.3 The detail that changes the answer

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4.4 Accuracy and limits

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4.5 What to do next

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4.6 Dry basis or wet basis?

Both describe the same water. Dry basis divides by the dry solids, which do not change as the sample dries, so it is the natural choice in soil mechanics, agronomy and wood science; it can exceed 100%. Wet basis divides by the whole wet sample, which is how food labels, grain contracts and sludge reports express it; it is always below 100%. Convert with wet = dry ÷ (1 + dry) and dry = wet ÷ (1 − wet), working in fractions: 25% dry basis is 0.25 ÷ 1.25 = 20% wet basis.

4.7 Gravimetric, volumetric and sensor readings

Oven drying gives gravimetric moisture by mass. Hydrology, irrigation and soil sensors use volumetric water content θv, the volume of water per volume of soil, which equals gravimetric moisture times dry bulk density. Time-domain reflectometry and capacitance probes read θv from the soil’s dielectric constant (Topp et al., 1980) and are calibrated against gravimetric samples (Dobriyal et al., 2012). Soil moisture varies strongly in space and time, so several samples are needed to characterize a field (Vereecken et al., 2014).

4.8 Equilibrium and humidity

Hygroscopic materials such as wood and grain exchange water with the air until they reach an equilibrium moisture content set mainly by relative humidity and slightly by temperature. Wood’s sorption behavior is reviewed by Engelund et al. (2013) and Thybring et al. (2022); grain isotherms are compared by Chen and Morey (1989).

✍ 5. How to Write Moisture Content in a Report: Detailed Results

A moisture content in a report needs the number, the basis (dry or wet), the method (oven drying and temperature, or the sensor or equation) and, for measured values, the uncertainty or the replicate spread. The seven write-ups below are filled from the moisture content calculator; they show the headline example until you press Calculate.

Showing the headline example (soil in a tin). Press Calculate above to fill all seven write-ups with your own result.

5.1 Style rules

RuleWriteAvoid
State the basis19.9% (dry basis)19.9% moisture
Name the methodoven-dried at 110 °C (ASTM D2216)moisture was measured
Use the field’s symbolw = 19.9%; θv = 0.28 cm³ cm³“MC 0.28%”
Points, not percent, for differencesdried by 6 percentage pointsdried by 6%
Match decimals to the balance19.91 ± 0.04%19.905213%
Give air moisture with its unit11.9 g kg⁻¹ dry airmoisture 11.9

5.2 Seven ready-to-copy write-ups

1. Lab report or paper sentence
The soil moisture content was 19.91 ± 0.04% (U, k = 2) on a dry basis (16.60% wet basis), from 8.40 g of water lost by 50.60 g of sample on oven drying.
2. Data sheet or log entry
Soil moisture content (dry basis) Moisture content, dry basis: 19.91% Moisture content, wet basis: 16.60% Total solids: 83.40% Mass of water: 8.40 g Mass of dry solids: 42.20 g Wet sample mass: 50.60 g Expanded uncertainty (dry basis): ± 0.043 points Expanded uncertainty (wet basis): ± 0.030 points Check: wet basis = dry basis ÷ (1 + dry basis): 0.19905 ÷ 1.19905 = 16.60% ✓
3. Code and spreadsheet formulas
Python: mc = (65.80 - 57.40) / (57.40 - 15.20) * 100 R: mc <- (65.80 - 57.40) / (57.40 - 15.20) * 100 Excel: =(B2-C2)/(C2-A2)*100 JavaScript: const mc = (65.80 - 57.40) / (57.40 - 15.20) * 100;
4. Exam or homework answer, fully worked
Given: container 15.20 g, container + wet 65.80 g, container + dry 57.40 g Required: Moisture content (dry basis) Equation: w = (M_wet − M_dry) ÷ (M_dry − M_container) × 100 Substitute: 8.40 ÷ 42.20 × 100 Solve: 19.91% Working: 1. Mass of water: (65.80 − 57.40) = 8.40 g 2. Mass of dry solids: 57.40 − 15.20 = 42.20 g 3. Moisture content, dry basis: 8.40 ÷ 42.20 × 100 = 19.91% 4. Moisture content, wet basis: 8.40 ÷ 50.60 × 100 = 16.60% 5. Uncertainty: balance ± 0.01 g on each weighing → U (k = 2) = ± 0.043 percentage points 6. Check: wet basis = dry basis ÷ (1 + dry basis): 0.19905 ÷ 1.19905 = 16.60% ✓ Answer: 19.91 ± 0.04%
5. Plain text for chat or a forum
Soil moisture content (dry basis) = 19.91 ± 0.04% (Oven drying: 110 ± 5 °C to constant mass (ASTM D2216)).
6. Detailed results paragraph
Results. Soil moisture content (dry basis) = 19.91 ± 0.04%. Wet basis 16.60% · water 8.40 g · dry solids 42.20 g. Supporting values: Moisture content, dry basis: 19.91%; Moisture content, wet basis: 16.60%; Total solids: 83.40%; Mass of water: 8.40 g; Mass of dry solids: 42.20 g; Wet sample mass: 50.60 g; Expanded uncertainty (dry basis): ± 0.043 points; Expanded uncertainty (wet basis): ± 0.030 points. Check: wet basis = dry basis ÷ (1 + dry basis): 0.19905 ÷ 1.19905 = 16.60% ✓.
7. Results table (tab-separated, pastes into Excel or Word)
Quantity Value Note Moisture content, dry basis 19.91% soil, wood, aggregate Moisture content, wet basis 16.60% food, grain, sludge Total solids 83.40% 100 − wet basis Mass of water 8.40 g wet − dry Mass of dry solids 42.20 g dry − container Wet sample mass 50.60 g wet − container Expanded uncertainty (dry basis) ± 0.043 points k = 2 Expanded uncertainty (wet basis) ± 0.030 points k = 2

∑ 6. What Is the Moisture Content Calculation Formula? All Formulas Used

①
Gravimetric moisture content (dry and wet basis)
w = (W₂ − W₃) ÷ (W₃ − W₁) × 100  ·  MCwet = (W₂ − W₃) ÷ (W₂ − W₁) × 100
W₁, W₂, W₃container; container + wet; container + oven-dry
Example8.40 ÷ 42.20 = 19.91% dry; 8.40 ÷ 50.60 = 16.60% wet
②
Converting between bases
wet = dry ÷ (1 + dry)  ·  dry = wet ÷ (1 − wet)  ·  total solids = 100 − MCwet
Example20% dry basis = 0.20 ÷ 1.20 = 16.67% wet basis
③
Volumetric soil moisture
θv = w × ρb ÷ ρw  ·  n = 1 − ρb ÷ ρs  ·  S = θv ÷ n  ·  depth = θv × layer
Example0.20 × 1.40 = 0.28; × 300 mm = 84 mm of water
④
Moisture content of air
es = 0.61094 exp[17.625T ÷ (T + 243.04)] kPa  ·  e = RH × es  ·  W = 0.621945 e ÷ (P − e)
Alsoabsolute humidity = e ÷ (461.5 TK); dew point from e (Alduchov & Eskridge, 1996)
Example25 °C, 60%: e = 1.897 kPa; W = 11.87 g/kg
⑤
Equilibrium moisture content of wood
EMC = 1800/W × [Kh/(1 − Kh) + (K₁Kh + 2K₁K₂K²h²)/(1 + K₁Kh + K₁K₂K²h²)]
T in °FW = 330 + 0.452T + 0.00415T²; K = 0.791 + 4.63×10⁻⁴T − 8.44×10⁻⁷T²; K₁ = 6.34 + 7.75×10⁻⁴T − 9.35×10⁻⁵T²; K₂ = 1.09 + 2.84×10⁻²T − 9.04×10⁻⁵T²
Example70 °F, 50% RH: 9.2%; 140 °F, 55%: 8.4% (Hailwood & Horrobin, 1946)
⑥
Optimum moisture content (Proctor)
ρd = ρ ÷ (1 + w)  ·  ρd = a w² + b w + c  ·  OMC = −b ÷ 2a  ·  ZAV: ρd = Gsρw ÷ (1 + wGs)
Example5 points from 10 to 18%: OMC 14.2%, MDD 1.835 g/cm³
⑦
Drying, blending and uncertainty
Mf = Mi(1 − MCi) ÷ (1 − MCf)  ·  average = Σ(m × MC) ÷ Σm  ·  u(w) = √Σ(ci a/√3)²
Example1000 kg at 20% → 14%: 930.2 kg; balance ±0.01 g ⇒ U = ±0.04 points

🔀 7. How Do You Calculate Moisture Content? Flowchart, Step by Step

The same worked problem as Sections 0, 9, 10, 11 and 14: tin 15.20 g, tin + wet soil 65.80 g, tin + dry soil 57.40 g.

Start or resultData stepCalculation stepYes or no check
StartThree weighingsW₁ 15.20 g, W₂ 65.80 g, W₃ 57.40 g
CheckHas the sample reached constant mass?
YesTwo weighings agree within 0.1%.
NoDry longer at 110 °C.
1
Water65.80 − 57.408.40 g
2
Dry soil57.40 − 15.2042.20 g
3
Dry-basis moisture8.40 ÷ 42.20 × 10019.91%
4
Wet basis and uncertainty8.40 ÷ 50.60 = 16.60%; balance ±0.01 g ⇒ U = ±0.04 points
CheckDoes 0.1991 ÷ 1.1991 give 16.60%?
YesThe two bases agree. Verified.
NoThe container or a basis slipped.
Resultw = 19.91 ± 0.04% (dry basis)16.60% wet basis

📝 8. How to Use This Moisture Content Calculator

  1. Pick what you want to calculate. Choose a mode in Section 2: moisture from wet and dry mass, volumetric soil moisture, moisture content of air, equilibrium moisture content, Proctor optimum moisture, or drying and blending.
  2. Load a sample or type your data. Thirteen worked problems are in the sample menu. Sample 1 is the headline soil test. Nothing is calculated until you press Calculate Moisture Content.
  3. Choose the material. Soil, aggregate and wood report a dry basis; food, grain and sludge a wet basis. Both are always shown.
  4. Enter the masses with the container. Type the container, container plus wet sample and container plus oven-dry sample. Leave the container blank if you weighed the sample alone.
  5. Add the balance resolution. Optional: the balance tolerance gives an expanded uncertainty and a budget showing which weighing matters most.
  6. For other modes, fill the extra fields. Bulk density for volumetric moisture; temperature with RH or dew point for air; temperature and RH for EMC; moisture and density pairs for Proctor; masses and targets for drying.
  7. Press Calculate Moisture Content. Press the button or Enter. The result, a Verified badge, the formula and every step appear in Section 3.
  8. Read the figure and charts. The figure draws your sample; four charts compare the two bases, show the parts of the sample, and how the result changes with the inputs.
  9. Copy the write-up. Section 5 fills seven ready texts for a report, a log, code, an exam, chat, a results paragraph and a table.
  10. Download or run a batch. Download Doc or PDF, or calculate many samples at once from a CSV of wet mass, dry mass and container in the Batch tab.

📈 9. How to Calculate Moisture Content in Excel

Excel needs one formula per quantity. The sheet repeats the headline problem (tin 15.20 g, wet 65.80 g, dry 57.40 g), so the numbers match Sections 0, 7, 10, 11 and 14.

The whole thing in one cell: with the tin in A2, tin + wet in B2 and tin + dry in C2, type =(B2-C2)/(C2-A2)*100. It returns 19.91.

9.1 The formulas you need

You wantExcel formulaNote
Moisture content, dry basis=(B2-C2)/(C2-A2)*100Soil, wood, aggregate
Moisture content, wet basis=(B2-C2)/(B2-A2)*100Food, grain, sludge
Dry to wet basis=F2/(100+F2)*10019.91 → 16.60
Wet to dry basis=G2/(100-G2)*10016.60 → 19.91
Volumetric moisture=F2/100*H2H2 = bulk density, g/cm³
Moisture content of air, g/kg=621.945*E/(P-E) with E=RH/100*0.61094*EXP(17.625*T/(T+243.04))Magnus
Mass after drying=A2*(1-B2/100)/(1-C2/100)Wet basis
Weighted average moisture=SUMPRODUCT(A2:A4,B2:B4)/SUM(A2:A4)Mass, moisture
Optimum moisture (quadratic)=-INDEX(LINEST(B2:B6,A2:A6^{1,2}),2)/(2*INDEX(LINEST(B2:B6,A2:A6^{1,2}),1))Proctor points

9.2 Step by step

1 Enter the three weighings

Tin in A2, tin + wet soil in B2, tin + dry soil in C2, in grams.

moisture-content.xlsxC2fx57.40ABC1TinWetDry215.2065.8057.40

A2 15.20, B2 65.80, C2 57.40.

2 Water and dry solids

In D2 type =B2-C2; in E2 type =C2-A2.

moisture-content.xlsxE2fx=C2-A2ABCDE1TinWetDryWaterSolids215.2065.8057.408.4042.20

D2 shows 8.40; E2 shows 42.20.

3 Moisture content

In F2 type =D2/E2*100.

moisture-content.xlsxF2fx=D2/E2*100ABCDEF1TinWetDryWaterSolidsw %215.2065.8057.408.4042.2019.91

F2 shows 19.91% (dry basis).

9.3 Many samples

Paste the weighings for every sample into columns A to C, select D2:F2 and fill down. Use =AVERAGE(F2:F20) and =STDEV.S(F2:F20) for replicate statistics.

9.4 Why Excel may differ from R or Python

SituationWhat happensFix
Rounded intermediate valuesRounding water to 8.4 before dividing changes the last digitRound only the final percentage
ROUNDExcel rounds half away from zero; R and Python round half to evenFormat instead of rounding
Text in cells“65.8 g” is text and gives #VALUE!Units in the header only

9.5 Charting

For a Proctor test, select the moisture and dry density columns, insert a Scatter chart and add a polynomial trendline of order 2 with “Display equation”. The peak of that curve is the optimum moisture content.

9.6 Errors you may see

  • #DIV/0!: the dry mass equals the tin mass.
  • A negative moisture: wet and dry columns swapped.
  • A value about 25% too low: divided by C2 instead of C2−A2.

📈 10. How to Calculate Moisture Content in R

A complete, copy-and-run script that reproduces the moisture content calculator for the headline problem. It calculates soil moisture content on a dry and wet basis from three weighings with its uncertainty, converts to volumetric moisture, finds the moisture content of air and the equilibrium moisture content of wood, fits a Proctor curve and saves one 300 dpi figure.

No packages needed. Base R covers every step.

10.1 The complete script

R, base only, saves compaction_curve.png

# Moisture content calculator in base R (StatsUnlock)
moisture <- function(tin, wet, dry) {           # three weighings, g
  if (any(dry > wet) || any(dry <= tin)) stop("check the masses")
  water <- wet - dry; solids <- dry - tin
  data.frame(water = water, dry_basis = 100 * water / solids, wet_basis = 100 * water / (wet - tin))
}
# headline: tin 15.20 g, tin + wet soil 65.80 g, tin + dry soil 57.40 g
m <- moisture(15.20, 65.80, 57.40); print(round(m, 3))
# uncertainty: balance +/- 0.01 g, rectangular, three weighings
u <- 0.01 / sqrt(3); s <- 57.40 - 15.20
cf <- c(tin = 100 * 8.40 / s^2, wet = 100 / s, dry = -100 * 50.60 / s^2)
cat(sprintf("w = %.2f +/- %.3f %% (U, k = 2)\n", m$dry_basis, 2 * sqrt(sum((cf * u)^2))))
# volumetric: w = 20 %, bulk density 1.40 g/cm3, 30 cm layer
theta <- 0.20 * 1.40; cat(sprintf("theta_v = %.2f; water depth = %.0f mm\n", theta, theta * 300))
# moisture content of air: 25 C, 60 % RH, 101.325 kPa (Magnus)
es <- function(t) 0.61094 * exp(17.625 * t / (t + 243.04))
e <- 0.60 * es(25); W <- 621.945 * e / (101.325 - e)
cat(sprintf("air: e = %.3f kPa, W = %.2f g/kg\n", e, W))
# wood EMC (Hailwood-Horrobin, Simpson coefficients, T in F)
emc_wood <- function(TF, h) {
  W <- 330 + 0.452 * TF + 0.00415 * TF^2; K <- 0.791 + 4.63e-4 * TF - 8.44e-7 * TF^2
  K1 <- 6.34 + 7.75e-4 * TF - 9.35e-5 * TF^2; K2 <- 1.09 + 2.84e-2 * TF - 9.04e-5 * TF^2
  kh <- K * h; 1800 / W * (kh / (1 - kh) + (K1 * kh + 2 * K1 * K2 * kh^2) / (1 + K1 * kh + K1 * K2 * kh^2))
}
cat(sprintf("wood EMC at 70 F, 50 %% RH = %.2f %%\n", emc_wood(70, 0.50)))
# Proctor: quadratic fit, optimum moisture content
w <- c(10, 12, 14, 16, 18); rd <- c(1.72, 1.80, 1.84, 1.81, 1.74)
f <- lm(rd ~ w + I(w^2)); b <- coef(f)
omc <- -b[2] / (2 * b[3]); mdd <- predict(f, data.frame(w = omc))
cat(sprintf("OMC = %.2f %%, MDD = %.3f g/cm3\n", omc, mdd))
png("compaction_curve.png", width = 7, height = 4, units = "in", res = 300)
ww <- seq(8, 20, 0.1)
plot(w, rd, pch = 19, xlab = "moisture content, %", ylab = "dry density, g/cm3", ylim = c(1.65, 2.05))
lines(ww, predict(f, data.frame(w = ww))); lines(ww, 2.65 / (1 + ww / 100 * 2.65), lty = 2)
abline(v = omc, lty = 3)
invisible(dev.off())

10.2 Output

  water dry_basis wet_basis
1   8.4    19.905    16.601
w = 19.91 +/- 0.043 % (U, k = 2)
theta_v = 0.28; water depth = 84 mm
air: e = 1.897 kPa, W = 11.87 g/kg
wood EMC at 70 F, 50 % RH = 9.25 %
OMC = 14.19 %, MDD = 1.835 g/cm3

10.3 Traps

  • moisture() is vectorized: pass columns of a data frame.
  • lm(rd ~ w + I(w^2)) needs I() so ^ means a power, not an interaction.
  • Keep moisture in percent for the fit; ZAV uses w/100.

10.4 What each part does

BlockPurpose
moisture()Checks the masses, then water ÷ dry solids and water ÷ wet mass
UncertaintySensitivity coefficients for the three weighings, a/√3 each, U = 2u
Volumetric and airθv = w ρb; Magnus vapor pressure and W = 0.621945 e/(P − e)
Middle blocksVolumetric moisture, moisture content of air, wood EMC and a quadratic Proctor fit
Last blockProctor compaction curve with its fitted peak and the zero-air-voids line, saved at 300 dpi

10.5 Adapting it

Change the three masses for your sample, or pass whole columns: moisture() is vectorized and works on a data frame of many samples at once.

📈 11. How to Calculate Moisture Content in Python

A complete, copy-and-run script that reproduces the moisture content calculator for the headline problem. It calculates soil moisture content on a dry and wet basis from three weighings with its uncertainty, converts to volumetric moisture, finds the moisture content of air and the equilibrium moisture content of wood, fits a Proctor curve and saves one 300 dpi figure.

numpymatplotlib

Install once: pip install numpy matplotlib

11.1 The complete script

Python with NumPy and Matplotlib, saves compaction_curve.png

# Moisture content calculator in Python (StatsUnlock)
import numpy as np
import matplotlib.pyplot as plt

def moisture(tin, wet, dry):
    """Dry- and wet-basis moisture content (%) from three weighings (g)."""
    tin, wet, dry = map(np.asarray, (tin, wet, dry))
    if (dry > wet).any() or (dry <= tin).any():
        raise ValueError("check the masses")
    water, solids = wet - dry, dry - tin
    return 100 * water / solids, 100 * water / (wet - tin)

w_db, w_wb = moisture(15.20, 65.80, 57.40)            # headline soil sample
print(f"dry basis {w_db:.2f} %, wet basis {w_wb:.2f} %")

u, s = 0.01 / np.sqrt(3), 57.40 - 15.20               # balance +/- 0.01 g
c = np.array([100 * 8.40 / s**2, 100 / s, -100 * 50.60 / s**2])
print(f"U (k = 2) = {2 * np.sqrt(np.sum((c * u) ** 2)):.3f} percentage points")

theta = 0.20 * 1.40                                    # volumetric soil water
print(f"theta_v = {theta:.2f}; {theta * 300:.0f} mm in 30 cm")

es = lambda t: 0.61094 * np.exp(17.625 * t / (t + 243.04))   # Magnus, kPa
e = 0.60 * es(25)
print(f"air at 25 C, 60 % RH: W = {621.945 * e / (101.325 - e):.2f} g/kg")

def emc_wood(TF, h):                                   # Simpson (1973), T in F
    W = 330 + 0.452 * TF + 0.00415 * TF**2
    K = 0.791 + 4.63e-4 * TF - 8.44e-7 * TF**2
    K1 = 6.34 + 7.75e-4 * TF - 9.35e-5 * TF**2
    K2 = 1.09 + 2.84e-2 * TF - 9.04e-5 * TF**2
    kh = K * h
    return 1800 / W * (kh / (1 - kh) + (K1 * kh + 2 * K1 * K2 * kh**2) / (1 + K1 * kh + K1 * K2 * kh**2))
print(f"wood EMC at 70 F, 50 % RH = {emc_wood(70, 0.50):.2f} %")

w = np.array([10, 12, 14, 16, 18]); rd = np.array([1.72, 1.80, 1.84, 1.81, 1.74])
a, b, c0 = np.polyfit(w, rd, 2)
omc = -b / (2 * a); mdd = np.polyval([a, b, c0], omc)
print(f"OMC = {omc:.2f} %, MDD = {mdd:.3f} g/cm3")

ww = np.linspace(8, 20, 121)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(w, rd, "o", ww, np.polyval([a, b, c0], ww), "-", ww, 2.65 / (1 + ww / 100 * 2.65), "--")
ax.axvline(omc, ls=":"); ax.set_xlabel("moisture content, %"); ax.set_ylabel("dry density, g/cm3")
fig.tight_layout(); fig.savefig("compaction_curve.png", dpi=300)

11.2 Output

dry basis 19.91 %, wet basis 16.60 %
U (k = 2) = 0.043 percentage points
theta_v = 0.28; 84 mm in 30 cm
air at 25 C, 60 % RH: W = 11.87 g/kg
wood EMC at 70 F, 50 % RH = 9.25 %
OMC = 14.19 %, MDD = 1.835 g/cm3

11.3 Traps

  • moisture() accepts arrays or pandas columns.
  • np.polyfit returns the highest power first: a, b, c.
  • Raise an error when dry mass exceeds wet mass instead of returning a negative.

11.4 What each part does

BlockPurpose
moisture()Checks the masses, then water ÷ dry solids and water ÷ wet mass
UncertaintySensitivity coefficients for the three weighings, a/√3 each, U = 2u
Volumetric and airθv = w ρb; Magnus vapor pressure and W = 0.621945 e/(P − e)
Middle blocksVolumetric moisture, moisture content of air, wood EMC and a quadratic Proctor fit
Last blockProctor compaction curve with its fitted peak and the zero-air-voids line, saved at 300 dpi

11.5 Adapting it

Change the three masses for your sample, or pass whole columns: moisture() is vectorized and works on a data frame of many samples at once.

📋 12. Moisture Content Tables: Bases, Air, Wood EMC, Soil and Grain

12.1 Dry basis to wet basis

Dry basis (%)Wet basis (%)Total solids (%)
00.00100.00
54.7695.24
109.0990.91
1513.0486.96
2016.6783.33
2520.0080.00
3023.0876.92
4028.5771.43
5033.3366.67
6037.5062.50
8044.4455.56
10050.0050.00
15060.0040.00
20066.6733.33
30075.0025.00

wet = dry ÷ (100 + dry) × 100. Exact. Last checked: October 2026.

12.2 Moisture content of saturated air (101.325 kPa)

°C°FSaturation vapor pressure (kPa)Moisture content at 100% RH (g/kg)At 50% RH (g/kg)Absolute humidity at 100% (g/m³)
−1014.00.2871.770.882.36
−523.00.4222.601.303.41
032.00.6113.771.884.85
541.00.8725.402.696.79
1050.01.2267.623.799.38
1559.01.70210.635.2712.80
2068.02.33314.667.2417.25
2577.03.16220.039.8622.98
3086.04.23727.1413.2830.28
3595.05.61836.5117.7339.50
40104.07.37548.8223.4951.03

Magnus form over water (Alduchov & Eskridge, 1996); W = 0.621945 e ÷ (P − e). Below 0 °C values are over supercooled water. Last checked: October 2026.

12.3 Equilibrium moisture content of wood (%)

°F (°C)20% RH30% RH40% RH50% RH60% RH70% RH80% RH90% RH
30 (−1.1)4.66.37.99.511.313.616.521.0
50 (10.0)4.66.37.99.511.213.416.420.9
70 (21.1)4.56.27.79.211.013.116.020.5
90 (32.2)4.35.97.48.910.612.615.419.9
110 (43.3)4.05.67.08.510.012.014.719.1
130 (54.4)3.75.26.67.99.511.314.018.2

Hailwood-Horrobin equation with Simpson’s coefficients (Hailwood & Horrobin, 1946), the model behind the Wood Handbook EMC table. Last checked: October 2026.

12.4 Oven-drying methods by material

MaterialTemperatureStandard or practiceBasis reported
Soil110 ± 5 °CASTM D2216Dry
Organic or gypsum soilabout 60 °CASTM D2216 noteDry
Aggregate and sand110 ± 5 °CASTM C566Dry
Wood103 ± 2 °CASTM D4442Dry
Food100 to 105 °C (air) or about 70 °C (vacuum)AOAC-type methodsWet
Sludge and biosolids103 to 105 °CTotal solids methodWet (total solids)

Food methods summarized by Bradley (2010). Last checked: October 2026.

12.5 Volumetric water content θv (cm³/cm³)

w (%)ρb 1.1ρb 1.2ρb 1.3ρb 1.4ρb 1.5ρb 1.6ρb 1.7
50.0550.0600.0650.0700.0750.0800.085
100.1100.1200.1300.1400.1500.1600.170
150.1650.1800.1950.2100.2250.2400.255
200.2200.2400.2600.2800.3000.3200.340
250.2750.3000.3250.3500.3750.4000.425
300.3300.3600.3900.4200.4500.4800.510
350.3850.4200.4550.4900.5250.5600.595
400.4400.4800.5200.5600.6000.6400.680

θv = w × ρb ÷ ρw with ρb in g/cm³. Values above the soil’s porosity are not physically possible. Last checked: October 2026.

12.6 Water shrink when drying to 14% (wet basis)

Starting moisture (%)Final mass per 1000Water removed per 1000Shrink (%)
16976.723.32.33
17965.134.93.49
18953.546.54.65
19941.958.15.81
20930.269.86.98
22907.093.09.30
24883.7116.311.63
26860.5139.513.95
28837.2162.816.28
30814.0186.018.60

Water shrink only: Mf = Mi(1 − MCi) ÷ (1 − 0.14). Handling losses are added separately in trade. Last checked: October 2026.

📈 13. Moisture Content Calculation Examples: Worked Problems

Eight worked problems that cover what a moisture content calculator meets: a textbook soil can, a wood section, cheese, concrete sand, sludge, air, a Proctor test and the classic mistake. Each number was checked in Python.

1
TEXTBOOK CASE

Soil in a moisture can

Three weighings.

A can weighs 20.5 g, 85.3 g with wet soil and 72.6 g after oven drying. Find the moisture content.

W1 20.5W2 85.3W3 72.612.7 g water24.38%
QuantityValueNote
Water12.7 g85.3 − 72.6
Dry soil52.1 g72.6 − 20.5
Moisture content24.38% (dry basis)19.60% wet basis
Water and dry soil12.70 gwater52.10 gdry soilmapping: bar height = grams
The tin is not part of either.

What it means: Always subtract the can before dividing.

How to write it: "w = 24.38% (ASTM D2216, 110 °C)."

2
WOODWORKING

Moisture content of a board section

Oven-dry method.

A section cut from a board weighs 520 g and 450 g after drying at 103 °C.

520 g450 g70 g water15.56% MC13.46% wet
QuantityValueNote
Moisture content15.56%70 ÷ 450
Wet basis13.46%70 ÷ 520
Verdicttoo wet for furnituretarget about 6 to 9%
Board vs targets15.56 %board8.000 %furniture12.00 %exteriormapping: bar height = % dry basis
Wood is always quoted on a dry basis.

What it means: Kiln-dry or acclimatize before use.

How to write it: "Moisture content 15.6% (oven-dry method, ASTM D4442)."

3
FOOD LAB

Moisture in cheese

Wet basis and total solids.

A 3.000 g cheese sample dries to 1.920 g in a vacuum oven.

3.000 g1.920 g1.080 g water36.0%64.0% solids
QuantityValueNote
Moisture (wet basis)36.0%1.080 ÷ 3.000
Total solids64.0%100 − 36.0
Dry basis56.25%1.080 ÷ 1.920
Cheese composition36.00 %water64.00 %solidsmapping: bar height = % of wet mass
Food labels use the wet basis.

What it means: Fat and protein are reported on the same wet basis.

How to write it: "Moisture 36.0% (wet basis); total solids 64.0%."

4
CONCRETE

Free moisture in sand

Subtract absorption.

A 500 g sand sample dries to 480 g. Its absorption is 1.2%.

500 g480 g4.17% total− 1.2% abs.2.97% free
QuantityValueNote
Total moisture4.17%20 ÷ 480
Free moisture2.97%total − absorption
Extra water per 1000 kg dry sand29.7 kgreduce mix water by this
Total vs free moisture4.170 %total1.200 %absorbed2.970 %freemapping: bar height = % dry basis
Only free water joins the mix.

What it means: Weigh the sand wet in the batch and add its dry mass × (1 + total moisture).

How to write it: "Sand moisture 4.17%, free moisture 2.97%."

5
WASTEWATER

Sludge moisture and total solids

Wet basis.

A 25.0 g sludge sample dries to 4.5 g at 105 °C.

25.0 g4.5 g20.5 g water82.0%18.0% TS
QuantityValueNote
Moisture (wet basis)82.0%20.5 ÷ 25.0
Total solids18.0%
Dry basis455.6%over 100% is normal
Sludge composition82.00 %water18.00 %total solidsmapping: bar height = % of wet mass
Dewatering raises total solids.

What it means: Report sludge as % total solids.

How to write it: "Total solids 18.0% (moisture 82.0%)."

6
HVAC

Same RH, different moisture

Air holds more when warm.

Compare air at 40% RH and 20 °C with air at 40% RH and 30 °C.

20 °C, 40%30 °C, 40%5.78 g/kg10.58 g/kg×1.83
QuantityValueNote
20 °C, 40% RH5.78 g/kghumidity ratio
30 °C, 40% RH10.58 g/kg
Ratio1.83same RH
Moisture content of air5.780 g/kg20 °C10.58 g/kg30 °Cmapping: bar height = g water per kg dry air
RH alone does not give moisture.

What it means: Warm air at the same RH carries almost twice the water.

How to write it: "Humidity ratio 5.78 g/kg at 20 °C and 10.58 g/kg at 30 °C (40% RH)."

7
EARTHWORKS

Proctor test from wet densities

Convert, fit, find the peak.

Five compacted specimens at 8 to 16% moisture have wet densities 1.94, 2.05, 2.10, 2.08 and 2.03 g/cm³.

5 pointsρd = ρ ÷ (1 + w)quadraticOMC 11.5%MDD 1.873
QuantityValueNote
Dry densities1.796, 1.864, 1.875, 1.825, 1.750g/cm³
Optimum moisture content11.5%−b ÷ 2a
Maximum dry density1.873 g/cm³95% = 1.779
Dry density at each moisture1.796 8%1.864 10%1.875 12%1.825 14%1.750 16%mapping: bar height = dry density, g/cm3
The wettest point is not the densest.

What it means: Field compaction is checked against 95% of MDD near OMC.

How to write it: "OMC 11.5%, maximum dry density 1.873 g/cm³."

8
COMMON MISTAKE

Dividing by the tin and soil

Subtract the tin first.

A student divides the 8.40 g of water by 57.40 g (tin + dry soil).

8.40 ÷ 57.40= 14.63%right: ÷ 42.20= 19.91%−26%
QuantityValueNote
Wrong14.63%tin included
Right19.91%dry soil only
Error−5.3 pointsa quarter too low
Wrong vs right14.63 %tin included (wrong)19.91 %soil only (right)mapping: fuchsia bar = wrong method
The container is not soil.

What it means: Write W3 − W1 in the denominator every time.

How to write it: "w = 19.91% (dry soil 42.20 g)."

✏️ 14. How Do You Calculate Moisture Content by Hand and by Oven Drying? Steps

The written method, calculator keys and the oven-drying procedure, for the headline problem: tin 15.20 g, tin + wet soil 65.80 g, tin + dry soil 57.40 g.

14.1 The written method

  1. Water. 65.80 − 57.40 = 8.40 g.
  2. Dry soil. 57.40 − 15.20 = 42.20 g.
  3. Dry basis. 8.40 ÷ 42.20 × 100 = 19.91%.
  4. Wet basis. 8.40 ÷ 50.60 × 100 = 16.60%.
  5. Uncertainty. ± 0.01 g balance: U = ± 0.04 points.
  6. Check. 0.1991 ÷ 1.1991 = 0.1660 ✓

14.2 On a calculator

StepKeysDisplay
Dry basis(65.80−57.40)÷(57.40−15.20)×100=19.905
Wet basis8.40÷50.60×100=16.601
Volumetric0.1991×1.40=0.279

14.3 Lab SOP: water content of soil by oven drying (ASTM D2216)

Purpose

Determine the water (moisture) content of a soil sample as a percentage of its oven-dry mass.

Equipment

Drying oven at 110 ± 5 °C; balance readable to 0.01 g; numbered moisture tins with lids; desiccator; heat-resistant gloves.

Hazards and controls

Hot tins and oven surfaces: use gloves and tongs. Do not dry samples containing fuels or solvents in a standard oven.

Procedure

  1. Weigh the clean, dry tin and lid (W1).
  2. Add a representative wet sample, close the lid and weigh at once (W2).
  3. Remove the lid and dry at 110 ± 5 °C to constant mass, usually overnight (12 to 16 h).
  4. Cool in a desiccator with the lid on, then weigh (W3).
  5. w = (W2 − W3) ÷ (W3 − W1) × 100. Run duplicates.

Record sheet

ItemValueUnit
Tin W115.20g
Tin + wet W265.80g
Tin + dry W357.40g
Water content w19.91 ± 0.04%

Acceptance criteria

Duplicates should agree within about 1 percentage point for fine soils; drying is complete when further drying changes the mass by less than 0.1%. These are teaching-lab criteria; follow your laboratory’s SOP where they differ.

Sources of error and their direction

  • Incomplete drying: low.
  • Moisture picked up while cooling: low.
  • Organic matter burned off: high.
  • Water lost before the first weighing (lid off): low.
  • Container not subtracted: low.

14.4 Practice problems

1. Tin 18.0 g, tin + wet 78.0 g, tin + dry 66.0 g. Moisture content?

12.0 ÷ 48.0 = 25.0% (dry basis)

2. Convert 25% dry basis to wet basis.

20.0%

3. Convert 15% wet basis to dry basis.

17.65%

4. A 2.500 g food sample dries to 2.200 g. Moisture?

12.0% wet basis

5. w = 18%, ρb = 1.5 g/cm³. θv?

0.27 cm³/cm³

6. θv = 0.25 in a 40 cm layer. Water depth?

100 mm

7. Air at 20 °C, 50% RH. Moisture content?

7.24 g/kg

8. Wood EMC at 80 °F, 65% RH?

11.8%

9. Dry 500 kg from 18% to 13% (wet basis). Final mass?

471.3 kg

10. Lots of 400 kg at 15% and 100 kg at 25%. Average?

17.0%

14.5 Mistakes that lose marks

  • Not subtracting the container.
  • Not stating the basis.
  • Dividing by wet mass for soil.
  • Treating points of moisture as percent of mass.
  • Averaging moisture percentages of unequal lots.
  • Using °C in the wood EMC equation, which needs °F.

14.6 Check-yourself and lab questions

Check: Why can dry-basis moisture exceed 100%?

Because the water can weigh more than the dry solids.

Check: Which basis do soil and wood use?

Dry basis.

Check: What converts gravimetric to volumetric moisture?

Dry bulk density divided by water density.

Check: Why does warm air at the same RH hold more water?

Saturation vapor pressure rises steeply with temperature.

Lab: Why cool the dried sample in a desiccator?

So it does not pick up moisture from the air before weighing.

Lab: How do you know the sample is dry?

Two weighings some hours apart agree (constant mass).

Lab: Why dry organic soils at 60 °C?

Organic matter oxidizes at 110 °C and is counted as water.

Lab: What sample mass does ASTM D2216 suggest for fine soil?

Small masses (tens of grams) suit fine soils; coarse soils need much larger samples.

✅ 15. Which Moisture Content Method Should You Use?

SituationUseCalculator modeExample
Soil, sand, aggregate lab testDry basis, oven dryingWet and dry mass19.91%
Wood or lumberDry basisWet and dry mass120 → 100 g: 20%
Food, grain, sludgeWet basis (and total solids)Wet and dry mass5.000 → 4.150 g: 17.0%
Irrigation, hydrology, sensorsVolumetric θvVolumetric0.28 cm³/cm³
HVAC, drying air, storage roomsHumidity ratioAir11.9 g/kg
What wood or grain will settle atEMCEquilibrium9.2% at 70 °F, 50%
Earthworks compactionProctor OMCOptimum moisture14.2%
Grain drying, blending lotsMass balanceDrying69.8 kg removed
Oven drying is the reference when
  • You need a defensible lab value.
  • You calibrate a meter or probe.
  • The material is not damaged by heat.
Use another method when
  • The sample contains volatiles other than water (Karl Fischer for food and oils).
  • You need in-place, repeated readings (TDR or capacitance probes; Robinson et al., 2008).
  • Gypsum or organic matter decomposes at 110 °C (dry at 60 °C).

🔧 16. Common Moisture Content Calculation Mistakes

Almost every wrong moisture value comes from the basis, the container or incomplete drying. The moisture content calculator guards against each one below.

SymptomCauseFix
Soil moisture far too lowDivided by container + soilSubtract the tin: dry soil = W₃ − W₁
Wood and food values disagreeDifferent basesConvert: wet = dry ÷ (1 + dry)
Result changes after more dryingNot at constant massDry until two weighings agree
Dry mass larger than wetValues swapped, or the sample absorbed moisture while coolingCool in a desiccator; check the order
θv larger than porosityWrong bulk density or particle densityRe-measure the core volume
Dew point above air temperatureUnits mixed (°F vs °C)Use one unit for both
Proctor curve has no peakAll points on one side of optimumAdd two points wet of optimum
Grain shrink smaller than expectedSubtracted moisture points from massUse Mf = Mi(1 − MCi) ÷ (1 − MCf)
Average moisture of lots wrongAveraged the percentagesWeight by mass
Organic soil keeps losing massOrganic matter burning at 110 °CDry at 60 °C and report the temperature

⚠️ 17. Assumptions and Limitations

AssumptionHow to checkIf it fails
All mass lost on drying is waterNo volatiles or decompositionUse Karl Fischer or a lower drying temperature
Balance tolerance is rectangularBalance specificationAdd drying repeatability from duplicates
Water density 1 g/cm³True within 0.5% from 0 to 30 °CNegligible for θv
Particle density 2.65 g/cm³Mineral soilsEnter a measured value for organic or volcanic soils
Magnus equation over water−40 to 60 °CUse an ice equation below 0 °C for frost point
Wood EMC from a fitted modelSpecies and sorption historyReal EMC may differ by 1 to 2 points; desorption runs higher
Henderson constants fit your cropASABE D245 or published fitUse constants for your crop and temperature range
Proctor curve is parabolic near the peakR² and points either sideAdd points; fit only the points near the peak

Soil moisture also drives climate feedbacks and drought (Seneviratne et al., 2010); a single sample describes only its spot and moment.

🎯 18. Conclusion

Your result

The moisture content calculator gives 19.91% (dry basis) and 16.60% (wet basis) for this soil sample. Moisture content = water ÷ dry solids × 100.

What the moisture content calculator does

Moisture content is one ratio, water to material, applied in many ways. This moisture content calculator handles oven-dry tests on both bases with uncertainty, volumetric soil water, the moisture content of air from RH or dew point, equilibrium moisture of wood and grain, Proctor optimum moisture, drying targets and blended averages, with every step shown.

Watch out for

  • State the basis: dry or wet.
  • Subtract the container.
  • Dry to constant mass.
  • Points of moisture are not percent of mass.
  • Weight averages by mass.

Next steps

For density of soil solids, use the specific gravity calculator; for oven and air temperatures, the temperature converter; and for w/w percentages, the percent solution calculator. Whatever the material, start and finish with the moisture content calculator.

❓ 19. Frequently Asked Questions

Q1. How do you calculate moisture content?

Weigh the sample, dry it in an oven to constant mass, and weigh it again. The loss is water. Moisture content on a dry basis is water divided by dry mass times 100; on a wet basis it is water divided by wet mass times 100. Subtract the container from both weighings before dividing.

Q2. What is the formula to calculate moisture content of soil?

Soil moisture content w = (W2 − W3) ÷ (W3 − W1) × 100, where W1 is the empty container, W2 the container with wet soil and W3 the container with oven-dry soil, dried at 110 °C under ASTM D2216. With 15.20, 65.80 and 57.40 g, w = 8.40 ÷ 42.20 × 100 = 19.91%.

Q3. What is the difference between dry basis and wet basis moisture content?

Dry basis divides the water by the dry solids; wet basis divides it by the whole wet sample. Soil, aggregate and wood use the dry basis, which can exceed 100%. Food, grain and sludge use the wet basis, which cannot. Convert with wet = dry ÷ (1 + dry), using fractions: 25% dry basis is 20% wet basis.

Q4. How do you calculate moisture content of wood?

Moisture content of wood = (wet mass − oven-dry mass) ÷ oven-dry mass × 100. Dry a small section at 103 °C to constant mass under ASTM D4442. A 120 g piece that dries to 100 g is at 20% moisture content. Green wood is often above 30%; furniture and flooring should be near 6 to 9%.

Q5. How do you calculate moisture content in food?

Food moisture is usually on a wet basis: (initial mass − dried mass) ÷ initial mass × 100. A 5.000 g sample drying to 4.150 g contains 17.0% moisture and 83.0% total solids. Sugary or fatty foods are dried in a vacuum oven at lower temperature, and Karl Fischer titration is used where heat drives off other volatiles.

Q6. How do you calculate volumetric moisture content?

Volumetric water content θv = gravimetric moisture (as a decimal) × dry bulk density ÷ density of water. Soil at 20% moisture with a bulk density of 1.40 g/cm³ holds 0.28 cm³ of water per cm³ of soil. Multiplying by the layer depth gives the water depth: 0.28 × 300 mm = 84 mm.

Q7. How do you calculate moisture content of air?

Calculate vapor pressure e as relative humidity times the saturation vapor pressure, or the saturation pressure at the dew point. The moisture content, or humidity ratio, is W = 0.622 × e ÷ (P − e) kg of water per kg of dry air. At 25 °C, 60% RH and sea-level pressure, W = 11.9 g/kg.

Q8. How do you calculate moisture content from relative humidity?

For air, multiply RH by the saturation vapor pressure at the air temperature and use W = 0.622 e ÷ (P − e). For wood or grain, relative humidity sets the equilibrium moisture content through a sorption isotherm: wood at 70 °F and 50% RH settles near 9.2%, and at 80% RH near 16%.

Q9. How do you calculate equilibrium moisture content of wood?

Use the Hailwood-Horrobin equation with Simpson’s coefficients from the Wood Handbook, which needs only temperature and relative humidity. At 70 °F and 50% RH it gives 9.2%; at 140 °F and 55% RH, 8.4%. The calculator evaluates it directly; real wood can differ by 1 to 2 points depending on species and drying history.

Q10. How do you calculate grain equilibrium moisture content?

Use a grain sorption equation such as the modified Henderson, 1 − RH = exp[−K(T + C)M^N], with constants for the crop from ASABE D245 or a published fit. Solve for M, the dry-basis moisture, then convert to wet basis. The calculator accepts any K, N and C and plots the full isotherm.

Q11. How do you calculate optimum moisture content?

Compact soil at four or five moisture contents, calculate each dry density as wet density ÷ (1 + w), and fit a curve of dry density against moisture. The peak is the optimum moisture content. With a quadratic fit ρd = aw² + bw + c, OMC = −b ÷ 2a and the maximum dry density is the curve value there.

Q12. How do you calculate percentage of moisture content?

Divide the mass of water by the reference mass and multiply by 100. The reference is the dry mass for a dry-basis percentage or the wet mass for a wet-basis percentage. With 8.40 g of water and 42.20 g of dry soil, the percentage moisture content is 19.91% dry basis, or 16.60% when divided by the 50.60 g wet sample.

Q13. How do you calculate average moisture content?

Add the water in every lot and divide by the total mass: average = Σ(mass × moisture) ÷ Σ mass. Lots of 500, 300 and 200 kg at 18, 14 and 22% average 17.6%, not 18%. For replicate lab samples of similar size, the simple mean is fine; the batch tab also reports the standard deviation.

Q14. How do you calculate free moisture content of aggregate?

Free (surface) moisture = total moisture − absorption, both on a dry basis. Sand at 5.0% total moisture with 1.5% absorption carries 3.5% free water, so each 1000 kg of dry sand brings 35 kg of extra water into a concrete mix. A negative value means the aggregate will absorb mix water instead.

Q15. How do you calculate hygroscopic moisture content?

Hygroscopic moisture is the water an air-dried sample still holds. Weigh the air-dried sample, oven-dry it at 105 to 110 °C, and divide the loss by the oven-dry mass. Soil labs use it to convert air-dry weights to an oven-dry basis: oven-dry mass = air-dry mass ÷ (1 + hygroscopic moisture as a decimal).

Q16. How much water must be removed to dry grain?

Dry matter stays constant, so final mass = initial mass × (1 − initial moisture) ÷ (1 − final moisture), on a wet basis. Drying 1000 kg from 20% to 14% leaves 930.2 kg and removes 69.8 kg of water. That is 7.0% of the mass, more than the 6 points of moisture, which is why shrink tables exist.

Q17. What temperature should you dry samples at for moisture content?

Soil and aggregate: 110 ± 5 °C under ASTM D2216 and C566. Wood: 103 ± 2 °C under ASTM D4442. Food: usually 100 to 105 °C in an air oven or about 70 °C in a vacuum oven. Gypsum-bearing or organic soils are dried at about 60 °C so that hydration water and organic matter are not lost.

Q18. Can moisture content be more than 100%?

Yes, on a dry basis. If a sample holds more water than solid material by mass, water divided by dry solids exceeds 1. Peat, sludge, green wood and fresh leaves often exceed 100% dry basis. On a wet basis the value is always below 100%, because the water is part of the wet mass in the denominator.

Q19. How do you calculate moisture content in Excel?

Put the container, container plus wet sample and container plus dry sample in A2, B2 and C2. Dry-basis moisture is =(B2-C2)/(C2-A2)*100 and wet-basis moisture is =(B2-C2)/(B2-A2)*100. Fill the formulas down for many samples, and use =AVERAGE and =STDEV.S on the results column to summarize replicates. Keep units in the header row and numbers only in the cells.

Q20. What is the dew point moisture content of air?

The dew point fixes the vapor pressure: e equals the saturation vapor pressure at the dew point temperature. Moisture content is then W = 0.622 e ÷ (P − e). Air at 30 °C with a 20 °C dew point has e = 2.34 kPa and holds 14.7 g of water per kg of dry air, at about 55% relative humidity.

📑 20. Cite This Tool

APA 7th edition
StatsUnlock. (2026). Moisture content calculator [Interactive laboratory tool]. https://statsunlock.com/moisture-content-calculator/
BibTeX
@misc{statsunlock_moisture_content_calculator_2026, title={Moisture Content Calculator}, author={{StatsUnlock Team}}, year={2026}, note={Interactive laboratory tool}, url={https://statsunlock.com/moisture-content-calculator/}}
Methods sentence (fills from your last calculation)
Moisture content was determined by oven drying (110 ± 5 °C to constant mass (ASTM D2216)) and calculated on a dry basis using the StatsUnlock Moisture Content Calculator (2026).

🔗 21. Related Tools

📖 22. Glossary of Terms

TermPlain-English meaning
Absolute humidityMass of water vapor per cubic meter of air.
Absorption (aggregate)Water held inside the pores at saturated-surface-dry condition.
Bulk density (dry)Dry soil mass per total volume, g/cm³.
Constant massTwo weighings after further drying agree; the sample is dry.
Dew pointTemperature at which air becomes saturated and water condenses.
Dry basisWater ÷ dry solids × 100; can exceed 100%.
Equilibrium moisture content (EMC)Moisture a hygroscopic material reaches in air of a given RH and temperature.
Fiber saturation pointWood moisture (about 25 to 30%) above which cell walls are saturated.
Free moistureWater above absorption, on the aggregate surface.
Gravimetric moisture contentWater content by mass, from oven drying.
Humidity ratioMass of water vapor per mass of dry air; moisture content of air.
Hygroscopic moistureWater held by an air-dried sample.
Maximum dry density (MDD)Peak of the Proctor compaction curve.
Optimum moisture content (OMC)Moisture at which a soil compacts to maximum dry density.
PorosityPore volume ÷ total volume, n = 1 − ρb/ρs.
Relative humidityVapor pressure ÷ saturation vapor pressure × 100.
Total solids100 − wet-basis moisture.
Volumetric water content θvWater volume per soil volume.
Wet basisWater ÷ wet mass × 100; always below 100%.
Zero air voids lineDry density at full saturation for each moisture content.

📚 23. References

Peer-reviewed sources behind this moisture content calculator: the gravimetric method, soil moisture measurement and sensors, soil moisture variability and climate, vapor pressure, dew point and wet-bulb equations, wood sorption and equilibrium moisture, grain isotherms, food moisture analysis, and measurement uncertainty. Standards (ASTM D2216, C566, D4442, D698, D1557 and JCGM 100:2008) are named where they are used. Every DOI checked October 2026.

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  2. Bittelli, M. (2011). Measuring soil water content: A review. HortTechnology, 21(3), 293–300. https://doi.org/10.21273/HORTTECH.21.3.293
  3. Bradley, R. L. (2010). Moisture and total solids analysis. In S. S. Nielsen (Ed.), Food analysis (4th ed., pp. 85–104). Springer. https://doi.org/10.1007/978-1-4419-1478-1_6
  4. Buck, A. L. (1981). New equations for computing vapor pressure and enhancement factor. Journal of Applied Meteorology, 20(12), 1527–1532. https://doi.org/10.1175/1520-0450(1981)020<1527:NEFCVP>2.0.CO;2
  5. Chen, C.-C., & Morey, R. V. (1989). Comparison of four EMC/ERH equations. Transactions of the ASAE, 32(3), 983–990. https://doi.org/10.13031/2013.31103
  6. Cole, T. J. (2015). Too many digits: The presentation of numerical data. Archives of Disease in Childhood, 100(7), 608–609. https://doi.org/10.1136/archdischild-2014-307149
  7. Dobriyal, P., Qureshi, A., Badola, R., & Hussain, S. A. (2012). A review of the methods available for estimating soil moisture and its implications for water resource management. Journal of Hydrology, 458–459, 110–117. https://doi.org/10.1016/j.jhydrol.2012.06.021
  8. Engelund, E. T., Thygesen, L. G., Svensson, S., & Hill, C. A. S. (2013). A critical discussion of the physics of wood–water interactions. Wood Science and Technology, 47(1), 141–161. https://doi.org/10.1007/s00226-012-0514-7
  9. Goldberg, D. (1991). What every computer scientist should know about floating-point arithmetic. ACM Computing Surveys, 23(1), 5–48. https://doi.org/10.1145/103162.103163
  10. Hailwood, A. J., & Horrobin, S. (1946). Absorption of water by polymers: Analysis in terms of a simple model. Transactions of the Faraday Society, 42, B084. https://doi.org/10.1039/TF946420B084
  11. Kragten, J. (1994). Calculating standard deviations and confidence intervals with a universally applicable spreadsheet technique. Analyst, 119(10), 2161–2165. https://doi.org/10.1039/AN9941902161
  12. Lawrence, M. G. (2005). The relationship between relative humidity and the dewpoint temperature in moist air: A simple conversion and applications. Bulletin of the American Meteorological Society, 86(2), 225–234. https://doi.org/10.1175/BAMS-86-2-225
  13. Reynolds, S. G. (1970). The gravimetric method of soil moisture determination Part I: A study of equipment, and methodological problems. Journal of Hydrology, 11(3), 258–273. https://doi.org/10.1016/0022-1694(70)90066-1
  14. Robinson, D. A., Campbell, C. S., Hopmans, J. W., Hornbuckle, B. K., Jones, S. B., Knight, R., Ogden, F., Selker, J., & Wendroth, O. (2008). Soil moisture measurement for ecological and hydrological watershed-scale observatories: A review. Vadose Zone Journal, 7(1), 358–389. https://doi.org/10.2136/vzj2007.0143
  15. Seneviratne, S. I., Corti, T., Davin, E. L., Hirschi, M., Jaeger, E. B., Lehner, I., Orlowsky, B., & Teuling, A. J. (2010). Investigating soil moisture–climate interactions in a changing climate: A review. Earth-Science Reviews, 99(3–4), 125–161. https://doi.org/10.1016/j.earscirev.2010.02.004
  16. Stull, R. (2011). Wet-bulb temperature from relative humidity and air temperature. Journal of Applied Meteorology and Climatology, 50(11), 2267–2269. https://doi.org/10.1175/JAMC-D-11-0143.1
  17. Thybring, E. E., Fredriksson, M., Zelinka, S. L., & Glass, S. V. (2022). Water in wood: A review of current understanding and knowledge gaps. Forests, 13(12), 2051. https://doi.org/10.3390/f13122051
  18. Tiesinga, E., Mohr, P. J., Newell, D. B., & Taylor, B. N. (2021). CODATA recommended values of the fundamental physical constants: 2018. Reviews of Modern Physics, 93(2), 025010. https://doi.org/10.1103/RevModPhys.93.025010
  19. Topp, G. C., Davis, J. L., & Annan, A. P. (1980). Electromagnetic determination of soil water content: Measurements in coaxial transmission lines. Water Resources Research, 16(3), 574–582. https://doi.org/10.1029/WR016i003p00574
  20. Vereecken, H., Huisman, J. A., Pachepsky, Y., Montzka, C., van der Kruk, J., Bogena, H., Weihermüller, L., Herbst, M., Martinez, G., & Vanderborght, J. (2014). On the spatio-temporal dynamics of soil moisture at the field scale. Journal of Hydrology, 516, 76–96. https://doi.org/10.1016/j.jhydrol.2013.11.061
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