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.
By StatsUnlock Team · Formulas from ASTM D2216, D4442 and D698, the Magnus vapor-pressure equation and the Wood Handbook EMC model, checked against published soil, wood and atmospheric science · Last updated:
⚡ 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.
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.
| Material | Usual basis | Oven | Typical range |
|---|---|---|---|
| Soil | Dry | 110 ± 5 °C (ASTM D2216) | 5 to 40%; peat over 100% |
| Sand and aggregate | Dry | 110 ± 5 °C (ASTM C566) | 0 to 10% |
| Wood | Dry | 103 ± 2 °C (ASTM D4442) | 6 to 12% seasoned; over 100% green |
| Food | Wet | About 100 to 105 °C or vacuum oven | 2 to 95% |
| Grain | Wet | Crop-specific | 10 to 30% |
| Sludge | Wet (and total solids) | 103 to 105 °C | 60 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.
Henderson form: 1 − RH = exp[−K (T + C) MN], M as a decimal dry basis, T in °C. Take crop constants from ASABE D245 or a published fit; sample 10 uses a wheat fit with T in kelvin (C = 273.15).
One sample per line: wet mass, dry mass, optional container mass and optional label, separated by commas or tabs. A header row is skipped. Results include the mean and SD of the dry-basis moisture.
📊 3. Results and Step-by-Step Working
Working (Given, Required, Equation, Substitute, Solve)
Uncertainty budget
Step by step
Full results table
Your sample, drawn from the numbers
Charts
Four plots that change with the mode: the two bases, what the sample is made of, and how the result responds to the inputs.
1.
2.
3.
4.
🧠 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
| Rule | Write | Avoid |
|---|---|---|
| State the basis | 19.9% (dry basis) | 19.9% moisture |
| Name the method | oven-dried at 110 °C (ASTM D2216) | moisture was measured |
| Use the field’s symbol | w = 19.9%; θv = 0.28 cm³ cm³ | “MC 0.28%” |
| Points, not percent, for differences | dried by 6 percentage points | dried by 6% |
| Match decimals to the balance | 19.91 ± 0.04% | 19.905213% |
| Give air moisture with its unit | 11.9 g kg⁻¹ dry air | moisture 11.9 |
5.2 Seven ready-to-copy write-ups
∑ 6. What Is the Moisture Content Calculation Formula? All Formulas Used
🔀 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.
📝 8. How to Use This Moisture Content Calculator
- 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.
- 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.
- Choose the material. Soil, aggregate and wood report a dry basis; food, grain and sludge a wet basis. Both are always shown.
- 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.
- Add the balance resolution. Optional: the balance tolerance gives an expanded uncertainty and a budget showing which weighing matters most.
- 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.
- Press Calculate Moisture Content. Press the button or Enter. The result, a Verified badge, the formula and every step appear in Section 3.
- 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.
- 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.
- 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.
=(B2-C2)/(C2-A2)*100. It returns 19.91.9.1 The formulas you need
| You want | Excel formula | Note |
|---|---|---|
| Moisture content, dry basis | =(B2-C2)/(C2-A2)*100 | Soil, wood, aggregate |
| Moisture content, wet basis | =(B2-C2)/(B2-A2)*100 | Food, grain, sludge |
| Dry to wet basis | =F2/(100+F2)*100 | 19.91 → 16.60 |
| Wet to dry basis | =G2/(100-G2)*100 | 16.60 → 19.91 |
| Volumetric moisture | =F2/100*H2 | H2 = 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.
A2 15.20, B2 65.80, C2 57.40.
2 Water and dry solids
In D2 type =B2-C2; in E2 type =C2-A2.
D2 shows 8.40; E2 shows 42.20.
3 Moisture content
In F2 type =D2/E2*100.
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
| Situation | What happens | Fix |
|---|---|---|
| Rounded intermediate values | Rounding water to 8.4 before dividing changes the last digit | Round only the final percentage |
| ROUND | Excel rounds half away from zero; R and Python round half to even | Format 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.
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/cm310.3 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
| Block | Purpose |
|---|---|
| moisture() | Checks the masses, then water ÷ dry solids and water ÷ wet mass |
| Uncertainty | Sensitivity coefficients for the three weighings, a/√3 each, U = 2u |
| Volumetric and air | θv = w ρb; Magnus vapor pressure and W = 0.621945 e/(P − e) |
| Middle blocks | Volumetric moisture, moisture content of air, wood EMC and a quadratic Proctor fit |
| Last block | Proctor 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.
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/cm311.3 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
| Block | Purpose |
|---|---|
| moisture() | Checks the masses, then water ÷ dry solids and water ÷ wet mass |
| Uncertainty | Sensitivity coefficients for the three weighings, a/√3 each, U = 2u |
| Volumetric and air | θv = w ρb; Magnus vapor pressure and W = 0.621945 e/(P − e) |
| Middle blocks | Volumetric moisture, moisture content of air, wood EMC and a quadratic Proctor fit |
| Last block | Proctor 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 (%) |
|---|---|---|
| 0 | 0.00 | 100.00 |
| 5 | 4.76 | 95.24 |
| 10 | 9.09 | 90.91 |
| 15 | 13.04 | 86.96 |
| 20 | 16.67 | 83.33 |
| 25 | 20.00 | 80.00 |
| 30 | 23.08 | 76.92 |
| 40 | 28.57 | 71.43 |
| 50 | 33.33 | 66.67 |
| 60 | 37.50 | 62.50 |
| 80 | 44.44 | 55.56 |
| 100 | 50.00 | 50.00 |
| 150 | 60.00 | 40.00 |
| 200 | 66.67 | 33.33 |
| 300 | 75.00 | 25.00 |
wet = dry ÷ (100 + dry) × 100. Exact. Last checked: October 2026.
12.2 Moisture content of saturated air (101.325 kPa)
| °C | °F | Saturation vapor pressure (kPa) | Moisture content at 100% RH (g/kg) | At 50% RH (g/kg) | Absolute humidity at 100% (g/m³) |
|---|---|---|---|---|---|
| −10 | 14.0 | 0.287 | 1.77 | 0.88 | 2.36 |
| −5 | 23.0 | 0.422 | 2.60 | 1.30 | 3.41 |
| 0 | 32.0 | 0.611 | 3.77 | 1.88 | 4.85 |
| 5 | 41.0 | 0.872 | 5.40 | 2.69 | 6.79 |
| 10 | 50.0 | 1.226 | 7.62 | 3.79 | 9.38 |
| 15 | 59.0 | 1.702 | 10.63 | 5.27 | 12.80 |
| 20 | 68.0 | 2.333 | 14.66 | 7.24 | 17.25 |
| 25 | 77.0 | 3.162 | 20.03 | 9.86 | 22.98 |
| 30 | 86.0 | 4.237 | 27.14 | 13.28 | 30.28 |
| 35 | 95.0 | 5.618 | 36.51 | 17.73 | 39.50 |
| 40 | 104.0 | 7.375 | 48.82 | 23.49 | 51.03 |
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% RH | 30% RH | 40% RH | 50% RH | 60% RH | 70% RH | 80% RH | 90% RH |
|---|---|---|---|---|---|---|---|---|
| 30 (−1.1) | 4.6 | 6.3 | 7.9 | 9.5 | 11.3 | 13.6 | 16.5 | 21.0 |
| 50 (10.0) | 4.6 | 6.3 | 7.9 | 9.5 | 11.2 | 13.4 | 16.4 | 20.9 |
| 70 (21.1) | 4.5 | 6.2 | 7.7 | 9.2 | 11.0 | 13.1 | 16.0 | 20.5 |
| 90 (32.2) | 4.3 | 5.9 | 7.4 | 8.9 | 10.6 | 12.6 | 15.4 | 19.9 |
| 110 (43.3) | 4.0 | 5.6 | 7.0 | 8.5 | 10.0 | 12.0 | 14.7 | 19.1 |
| 130 (54.4) | 3.7 | 5.2 | 6.6 | 7.9 | 9.5 | 11.3 | 14.0 | 18.2 |
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
| Material | Temperature | Standard or practice | Basis reported |
|---|---|---|---|
| Soil | 110 ± 5 °C | ASTM D2216 | Dry |
| Organic or gypsum soil | about 60 °C | ASTM D2216 note | Dry |
| Aggregate and sand | 110 ± 5 °C | ASTM C566 | Dry |
| Wood | 103 ± 2 °C | ASTM D4442 | Dry |
| Food | 100 to 105 °C (air) or about 70 °C (vacuum) | AOAC-type methods | Wet |
| Sludge and biosolids | 103 to 105 °C | Total solids method | Wet (total solids) |
Food methods summarized by Bradley (2010). Last checked: October 2026.
12.5 Volumetric water content θv (cm³/cm³)
| w (%) | ρb 1.1 | ρb 1.2 | ρb 1.3 | ρb 1.4 | ρb 1.5 | ρb 1.6 | ρb 1.7 |
|---|---|---|---|---|---|---|---|
| 5 | 0.055 | 0.060 | 0.065 | 0.070 | 0.075 | 0.080 | 0.085 |
| 10 | 0.110 | 0.120 | 0.130 | 0.140 | 0.150 | 0.160 | 0.170 |
| 15 | 0.165 | 0.180 | 0.195 | 0.210 | 0.225 | 0.240 | 0.255 |
| 20 | 0.220 | 0.240 | 0.260 | 0.280 | 0.300 | 0.320 | 0.340 |
| 25 | 0.275 | 0.300 | 0.325 | 0.350 | 0.375 | 0.400 | 0.425 |
| 30 | 0.330 | 0.360 | 0.390 | 0.420 | 0.450 | 0.480 | 0.510 |
| 35 | 0.385 | 0.420 | 0.455 | 0.490 | 0.525 | 0.560 | 0.595 |
| 40 | 0.440 | 0.480 | 0.520 | 0.560 | 0.600 | 0.640 | 0.680 |
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 1000 | Water removed per 1000 | Shrink (%) |
|---|---|---|---|
| 16 | 976.7 | 23.3 | 2.33 |
| 17 | 965.1 | 34.9 | 3.49 |
| 18 | 953.5 | 46.5 | 4.65 |
| 19 | 941.9 | 58.1 | 5.81 |
| 20 | 930.2 | 69.8 | 6.98 |
| 22 | 907.0 | 93.0 | 9.30 |
| 24 | 883.7 | 116.3 | 11.63 |
| 26 | 860.5 | 139.5 | 13.95 |
| 28 | 837.2 | 162.8 | 16.28 |
| 30 | 814.0 | 186.0 | 18.60 |
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.
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.
| Quantity | Value | Note |
|---|---|---|
| Water | 12.7 g | 85.3 − 72.6 |
| Dry soil | 52.1 g | 72.6 − 20.5 |
| Moisture content | 24.38% (dry basis) | 19.60% wet basis |
What it means: Always subtract the can before dividing.
How to write it: "w = 24.38% (ASTM D2216, 110 °C)."
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.
| Quantity | Value | Note |
|---|---|---|
| Moisture content | 15.56% | 70 ÷ 450 |
| Wet basis | 13.46% | 70 ÷ 520 |
| Verdict | too wet for furniture | target about 6 to 9% |
What it means: Kiln-dry or acclimatize before use.
How to write it: "Moisture content 15.6% (oven-dry method, ASTM D4442)."
Moisture in cheese
Wet basis and total solids.
A 3.000 g cheese sample dries to 1.920 g in a vacuum oven.
| Quantity | Value | Note |
|---|---|---|
| Moisture (wet basis) | 36.0% | 1.080 ÷ 3.000 |
| Total solids | 64.0% | 100 − 36.0 |
| Dry basis | 56.25% | 1.080 ÷ 1.920 |
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%."
Free moisture in sand
Subtract absorption.
A 500 g sand sample dries to 480 g. Its absorption is 1.2%.
| Quantity | Value | Note |
|---|---|---|
| Total moisture | 4.17% | 20 ÷ 480 |
| Free moisture | 2.97% | total − absorption |
| Extra water per 1000 kg dry sand | 29.7 kg | reduce mix water by this |
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%."
Sludge moisture and total solids
Wet basis.
A 25.0 g sludge sample dries to 4.5 g at 105 °C.
| Quantity | Value | Note |
|---|---|---|
| Moisture (wet basis) | 82.0% | 20.5 ÷ 25.0 |
| Total solids | 18.0% | |
| Dry basis | 455.6% | over 100% is normal |
What it means: Report sludge as % total solids.
How to write it: "Total solids 18.0% (moisture 82.0%)."
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.
| Quantity | Value | Note |
|---|---|---|
| 20 °C, 40% RH | 5.78 g/kg | humidity ratio |
| 30 °C, 40% RH | 10.58 g/kg | |
| Ratio | 1.83 | same RH |
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)."
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.
| Quantity | Value | Note |
|---|---|---|
| Dry densities | 1.796, 1.864, 1.875, 1.825, 1.750 | g/cm³ |
| Optimum moisture content | 11.5% | −b ÷ 2a |
| Maximum dry density | 1.873 g/cm³ | 95% = 1.779 |
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."
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).
| Quantity | Value | Note |
|---|---|---|
| Wrong | 14.63% | tin included |
| Right | 19.91% | dry soil only |
| Error | −5.3 points | a quarter too low |
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
- Water. 65.80 − 57.40 = 8.40 g.
- Dry soil. 57.40 − 15.20 = 42.20 g.
- Dry basis. 8.40 ÷ 42.20 × 100 = 19.91%.
- Wet basis. 8.40 ÷ 50.60 × 100 = 16.60%.
- Uncertainty. ± 0.01 g balance: U = ± 0.04 points.
- Check. 0.1991 ÷ 1.1991 = 0.1660 ✓
14.2 On a calculator
| Step | Keys | Display |
|---|---|---|
| Dry basis | (65.80−57.40)÷(57.40−15.20)×100= | 19.905 |
| Wet basis | 8.40÷50.60×100= | 16.601 |
| Volumetric | 0.1991×1.40= | 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
- Weigh the clean, dry tin and lid (W1).
- Add a representative wet sample, close the lid and weigh at once (W2).
- Remove the lid and dry at 110 ± 5 °C to constant mass, usually overnight (12 to 16 h).
- Cool in a desiccator with the lid on, then weigh (W3).
- w = (W2 − W3) ÷ (W3 − W1) × 100. Run duplicates.
Record sheet
| Item | Value | Unit |
|---|---|---|
| Tin W1 | 15.20 | g |
| Tin + wet W2 | 65.80 | g |
| Tin + dry W3 | 57.40 | g |
| Water content w | 19.91 ± 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?
| Situation | Use | Calculator mode | Example |
|---|---|---|---|
| Soil, sand, aggregate lab test | Dry basis, oven drying | Wet and dry mass | 19.91% |
| Wood or lumber | Dry basis | Wet and dry mass | 120 → 100 g: 20% |
| Food, grain, sludge | Wet basis (and total solids) | Wet and dry mass | 5.000 → 4.150 g: 17.0% |
| Irrigation, hydrology, sensors | Volumetric θv | Volumetric | 0.28 cm³/cm³ |
| HVAC, drying air, storage rooms | Humidity ratio | Air | 11.9 g/kg |
| What wood or grain will settle at | EMC | Equilibrium | 9.2% at 70 °F, 50% |
| Earthworks compaction | Proctor OMC | Optimum moisture | 14.2% |
| Grain drying, blending lots | Mass balance | Drying | 69.8 kg removed |
- You need a defensible lab value.
- You calibrate a meter or probe.
- The material is not damaged by heat.
- 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.
| Symptom | Cause | Fix |
|---|---|---|
| Soil moisture far too low | Divided by container + soil | Subtract the tin: dry soil = W₃ − W₁ |
| Wood and food values disagree | Different bases | Convert: wet = dry ÷ (1 + dry) |
| Result changes after more drying | Not at constant mass | Dry until two weighings agree |
| Dry mass larger than wet | Values swapped, or the sample absorbed moisture while cooling | Cool in a desiccator; check the order |
| θv larger than porosity | Wrong bulk density or particle density | Re-measure the core volume |
| Dew point above air temperature | Units mixed (°F vs °C) | Use one unit for both |
| Proctor curve has no peak | All points on one side of optimum | Add two points wet of optimum |
| Grain shrink smaller than expected | Subtracted moisture points from mass | Use Mf = Mi(1 − MCi) ÷ (1 − MCf) |
| Average moisture of lots wrong | Averaged the percentages | Weight by mass |
| Organic soil keeps losing mass | Organic matter burning at 110 °C | Dry at 60 °C and report the temperature |
⚠️ 17. Assumptions and Limitations
| Assumption | How to check | If it fails |
|---|---|---|
| All mass lost on drying is water | No volatiles or decomposition | Use Karl Fischer or a lower drying temperature |
| Balance tolerance is rectangular | Balance specification | Add drying repeatability from duplicates |
| Water density 1 g/cm³ | True within 0.5% from 0 to 30 °C | Negligible for θv |
| Particle density 2.65 g/cm³ | Mineral soils | Enter a measured value for organic or volcanic soils |
| Magnus equation over water | −40 to 60 °C | Use an ice equation below 0 °C for frost point |
| Wood EMC from a fitted model | Species and sorption history | Real EMC may differ by 1 to 2 points; desorption runs higher |
| Henderson constants fit your crop | ASABE D245 or published fit | Use constants for your crop and temperature range |
| Proctor curve is parabolic near the peak | R² and points either side | Add 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
StatsUnlock. (2026). Moisture content calculator [Interactive laboratory tool]. https://statsunlock.com/moisture-content-calculator/@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/}}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
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More lab, chemistry and measurement tools from StatsUnlock, with worked steps, code and references.
📖 22. Glossary of Terms
| Term | Plain-English meaning |
|---|---|
| Absolute humidity | Mass 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 mass | Two weighings after further drying agree; the sample is dry. |
| Dew point | Temperature at which air becomes saturated and water condenses. |
| Dry basis | Water ÷ 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 point | Wood moisture (about 25 to 30%) above which cell walls are saturated. |
| Free moisture | Water above absorption, on the aggregate surface. |
| Gravimetric moisture content | Water content by mass, from oven drying. |
| Humidity ratio | Mass of water vapor per mass of dry air; moisture content of air. |
| Hygroscopic moisture | Water 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. |
| Porosity | Pore volume ÷ total volume, n = 1 − ρb/ρs. |
| Relative humidity | Vapor pressure ÷ saturation vapor pressure × 100. |
| Total solids | 100 − wet-basis moisture. |
| Volumetric water content θv | Water volume per soil volume. |
| Wet basis | Water ÷ wet mass × 100; always below 100%. |
| Zero air voids line | Dry 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.
- Alduchov, O. A., & Eskridge, R. E. (1996). Improved Magnus form approximation of saturation vapor pressure. Journal of Applied Meteorology, 35(4), 601–609. https://doi.org/10.1175/1520-0450(1996)035<0601:IMFAOS>2.0.CO;2
- Bittelli, M. (2011). Measuring soil water content: A review. HortTechnology, 21(3), 293–300. https://doi.org/10.21273/HORTTECH.21.3.293
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
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