Water (Moisture) Content

w=MwMs×100w = \frac{M_w}{M_s}\times 100
sig figs

Worked example: 45 g of water on 250 g of solids → w = 18%. Press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Change any unit and the answer follows.

Where Water (Moisture) Content fits

Your answer feeds
In this toolkit

See also

Degree of Saturation (Se = wGs): Both use water content.Dry Unit Weight from Moist Unit Weight: Both use water content.Liquidity Index: Both use water content.
Learning zone

Water (Moisture) Content explained

MwMsw

The single most-run test in any soils laboratory: weigh the tin wet, dry it overnight at 110 ± 5 °C, weigh it again, and the difference is water. A sample losing 45 g from 295 g of wet soil leaves 250 g of solids, so w = 45 ÷ 250 × 100 = 18%. Everything downstream (dry unit weight, saturation, compaction control, Atterberg indices) is built on this number, which is why the tin is weighed to 0.01 g and why the balance is the first thing a technician calibrates.

The trap is in the denominator: water content is referred to the mass of dry solids, not to the total wet mass. Divide by the wet mass and you have computed something a mining engineer would call moisture and a soils engineer would call wrong, and the two differ by 15% at w = 18%. That definition is also why w can legitimately exceed 100%: a Norwegian quick clay or a Florida peat can hold two or three times its own dry weight in water, giving w = 250% with no arithmetic error anywhere. Do not "fix" such a number.

Water (Moisture) Content formula

w=MwMs×100w = \frac{M_w}{M_s}\times 100
Where
  • ww= Water content (%)
  • MwM_w= Mass of water (kg)
  • MsM_s= Mass of dry solids (kg)
Revised 31 October 2026
Cite this page

formula.expert. (2026, October 31). Water (Moisture) Content Formula & Calculator. https://formula.expert/solvers/water-content