Calculate Density and Specific Gravity

Question 1:

Water, at its maximum density (temperature of 3.98^o C, or approximately 4^o C), weighs 62.428 lb/ft³. Based on this reference density, determine the specific gravities of the following liquids:

Sea water: D = 63.99 lb/ft³
Mercury: D = 849 lb/ft³
Olive oil: D = 57.3 lb/ft³
Castor oil: D = 60.5 lb/ft³
Kerosene: D = 51.2 lb/ft³
Benzene: D = 56.1 lb/ft³
Acetone: D = 49.4 lb/ft³
Glycerin: D = 78.6 lb/ft³
Methanol (methyl alcohol): D = 49.41 lb/ft³
Ethanol (ethyl alcohol): D = 49.27 lb/ft³
Ethylene glycol (ethanediol): D = 69.22 lb/ft³
Gasoline: D = 41 lb/ft3 to 43 lb/ft³

Answer:

Sea water: 1.025
Mercury: 13.6
Olive oil: 0.918
Castor oil: 0.969
Kerosene: 0.820
Benzene: 0.899
Acetone: 0.791
Glycerin: 1.26
Methanol: 0.7915
Ethanol: 0.7892
Ethylene glycol: 1.109
Gasoline: 0.657 to 0.689

Question 2:

Toluene has a density of 0.8669 g/cm³ at 20^o C. Calculate its density in units of pounds per cubic feet and its specific gravity (unitless).

Answer:

Specific gravity = 0.8669 (same as density in units of g/cm³ )

The units of grams and cubic centimeters are defined in such a way that their density quotient in relation to pure water is 1. This is similar to the Celsius temperature scale, similarly defined at the 0^o and 100^o points by the freezing and boiling points of pure water, respectively.

To calculate the density of toluene in units of pounds per cubic feet, simply multiply the density of water (62.428 lb/ft³ ) by the specific gravity of toluene (0.8669):

(62.428 lb/ft³ )(0.8669) = 54.12 lb/ft³

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Credits: Tony R. Kuphaldt