What is the specific gravity?
Type a density, or the mass of an object in air and its apparent mass in water. The specific gravity calculator divides by the density of water at 4 °C (or your own reference liquid) and gives the density in g/cm³, kg/m³ and lb/ft³.
- Specific gravity
- 19.32
The specific gravity is 19.32, a density of 19.32 g/cm³.
- Density (g/cm³)
- 19.32
- Density (kg/m³)
- 19,320
- Density (lb/ft³)
- 1,206.11
Specific gravity: 19.32. The specific gravity is 19.32, a density of 19.32 g/cm³.
How to calculate
Computes specific gravity, the density of a substance divided by the density of water, from its density or by weighing an object in air and in water, with its density in g/cm³, kg/m³ and lb/ft³.
Example with the default inputs (Method From density, Density of the substance 19.32 g/cm³, Density of water (reference) 1 g/cm³): The specific gravity is 19.32, a density of 19.32 g/cm³.
Method: Specific gravity = ρ ÷ ρ_water. By weighing: specific gravity = m_air ÷ (m_air − m_water), since m_air − m_water is the mass of water displaced; density = specific gravity × ρ_water.
- The reference is water at 4 °C, 1.000 g/cm³ (1,000 kg/m³), unless you type another reference density.
- When weighing, the object is fully under water, holds no air, and does not absorb water; the apparent mass in water is the scale reading then.
- Units convert exactly: 1 g/cm³ = 1,000 kg/m³; 1 lb = 0.45359237 kg; 1 ft³ = 0.028316846592 m³; 1 gal = 0.003785411784 m³.
Worked examples
Each example is checked against the calculator on every build.
- Method From density, Density of the substance 19.32 g/cm³, Density of water (reference) 1 g/cm³ gives Specific gravity 19.32, Density (g/cm³) 19.32, Density (kg/m³) 19,320, Density (lb/ft³) 1,206.108198.Source: OpenStax, College Physics 2e, §11.2 Density, Table 11.1 (water at 4 °C 1.000, gold 19.32, mercury 13.6, in 10³ kg/m³), https://openstax.org/books/college-physics-2e/pages/11-2-density (retrieved 2026-10-18); OpenStax, College Physics 2e, §11.7 Archimedes’ Principle (specific gravity = ρ̄ ÷ ρw, the density of water at 4.00 °C; apparent mass loss = mass of fluid displaced; Example 11.10: 8.630 g in air, 7.800 g in water, 0.830 cm³, 10.4 g/cm³), https://openstax.org/books/college-physics-2e/pages/11-7-archimedes-principle (retrieved 2026-10-18)
- Method By weighing, Mass in air 8.63 g, Apparent mass in water 7.8 g, Density of water (reference) 1 g/cm³ gives Specific gravity 10.39759, Density (kg/m³) 10,397.590361, Mass of water displaced (g) 0.83, Volume of the object (cm³) 0.83.Source: OpenStax, College Physics 2e, §11.7 Archimedes’ Principle (specific gravity = ρ̄ ÷ ρw, the density of water at 4.00 °C; apparent mass loss = mass of fluid displaced; Example 11.10: 8.630 g in air, 7.800 g in water, 0.830 cm³, 10.4 g/cm³), https://openstax.org/books/college-physics-2e/pages/11-7-archimedes-principle (retrieved 2026-10-18)
- Method From density, Density of the substance 0.9996 g/cm³, Density of water (reference) 1 g/cm³ gives Specific gravity 0.999552, Density (lb/ft³) 62.4.Source: OpenStax, College Physics 2e, §11.7 Archimedes’ Principle (specific gravity = ρ̄ ÷ ρw, the density of water at 4.00 °C; apparent mass loss = mass of fluid displaced; Example 11.10: 8.630 g in air, 7.800 g in water, 0.830 cm³, 10.4 g/cm³), https://openstax.org/books/college-physics-2e/pages/11-7-archimedes-principle (retrieved 2026-10-18); NIST SP 811, Appendix B.8 (1 lb = 0.45359237 kg, 1 ft³ = 0.028316846592 m³)
- Method From density, Density of the substance 13.6 g/cm³, Density of water (reference) 1 g/cm³ gives Specific gravity 13.6, Density (g/cm³) 13.6.Source: OpenStax, College Physics 2e, §11.2 Density, Table 11.1 (water at 4 °C 1.000, gold 19.32, mercury 13.6, in 10³ kg/m³), https://openstax.org/books/college-physics-2e/pages/11-2-density (retrieved 2026-10-18)
How it works
From density. Specific gravity is the density of the substance divided by the density of water (OpenStax, Eq. 11.40):
specific gravity = ρ ÷ ρ_water
The reference is water at 4 °C, 1.000 g/cm³ = 1,000 kg/m³, unless you type another reference density.
By weighing. Weigh the object in air (m_air), then fully under water (m_water, the apparent mass). The apparent loss, m_air − m_water, is the mass of the water the object displaces, which has the object’s volume:
- Volume = (m_air − m_water) ÷ ρ_water
- Density = m_air ÷ volume = ρ_water × m_air ÷ (m_air − m_water)
- Specific gravity = m_air ÷ (m_air − m_water)
In both methods the page also shows the density, specific gravity × ρ_water, in g/cm³, kg/m³ and lb/ft³. By weighing it shows the mass of water displaced in grams and the object’s volume in cm³.
Units. 1 g/cm³ = 1 g/mL = 1 kg/L = 1,000 kg/m³; 1 lb/ft³ = 0.45359237 ÷ 0.028316846592 kg/m³ (about 16.0185); 1 lb/gal = 0.45359237 ÷ 0.003785411784 kg/m³ (about 119.826). Masses: 1 g = 0.001 kg, 1 mg = 10⁻⁶ kg, 1 oz = 0.028349523125 kg, 1 lb = 0.45359237 kg. Each typed value is read as the exact decimal you typed, in its unit, and the unit sizes are exact, so every result is an exact fraction until it is rounded for display.
Rules. Densities are from 0.000001 to 1,000,000 kg/m³, and the reference from 0.001 to 1,000,000 kg/m³. The mass in air is above 0 and the apparent mass in water is 0 or more, both at most 1,000,000 kg. When weighing, the apparent mass in water must be less than the mass in air; otherwise there is no answer (an object that floats needs a sinker, which this page does not handle).
Output format. Each value shows 6 significant figures.
Worked examples by hand
Gold (OpenStax Table 11.1). 19.32 g/cm³ ÷ 1.000 g/cm³ = 19.32. In lb/ft³: 19,320 ÷ 16.0184634 = 1,206.11 lb/ft³.
Is the coin silver? (OpenStax Example 11.10). m_air = 8.630 g, m_water = 7.800 g. Water displaced: 8.630 − 7.800 = 0.83 g, so the volume is 0.83 ÷ 1.000 = 0.83 cm³. Specific gravity = 8.630 ÷ 0.830 = 10.3976, a density of 10.3976 g/cm³ = 10,397.6 kg/m³ (the book: 10.4 g/cm³).
Water in US units. 62.4 lb/ft³ × 16.0184634 = 999.552 kg/m³; ÷ 1,000 = 0.999552.
Mercury (OpenStax Table 11.1). 13.6 g/cm³ ÷ 1.000 = 13.6.
Other questions people ask
What is specific gravity?
It is the density of a substance divided by the density of a reference liquid, usually water. Because the two densities share a unit, specific gravity has no unit. Gold, at 19.32 g/cm³, has a specific gravity of 19.32.
How do I calculate specific gravity from density?
Divide the density by the density of water at 4 °C, 1.000 g/cm³ or 1,000 kg/m³, in the same unit. Mercury at 13.6 g/cm³ has a specific gravity of 13.6 ÷ 1.000 = 13.6.
How do I find specific gravity by weighing in water?
Weigh the object in air, then hanging fully under water. The difference is the mass of water it pushes aside (Archimedes’ principle). Specific gravity = mass in air ÷ (mass in air − mass in water). A coin of 8.630 g that reads 7.800 g in water has 8.630 ÷ 0.830 = 10.4.
Is specific gravity the same as density?
Not quite. Specific gravity is a ratio with no unit, while density has units. Against water at 1 g/cm³, the specific gravity has the same number as the density in g/cm³, so 2.7 g/cm³ aluminum has a specific gravity of 2.7.
What does a specific gravity below 1 mean?
The substance is less dense than water, so a solid piece of it floats. Above 1 it sinks. Water itself has a specific gravity of 1.
Can I use a reference other than water?
Yes. Type the density of the reference liquid in place of water’s, for example water at the temperature you measured at. Water at 4 °C, 1.000 g/cm³, is the usual reference.