3.1 – Density and specific gravity (S.G.)

3.1.0 – Learning objectives

By the end of this notebook you should be able to:

  1. Differentiate density and specific gravity.
  2. Utilize density and specific gravity to interchangeably find mass and/or volumetric flow rates.

3.1.1 – Introduction

Density is the amount of mass per unit volume. Specific gravity (S.G.) is the ratio of density of the object to the density of a standard, usually water for a liquid or solid, and air for a gas. Both density and S.G. are common units in the determination of how much mass is in a chemical process based on the volumetric flow rate of the substance in the process.

\[Density = \rho\]
\[Specific \space gravity = \frac{{\rho}_{sample}}{{\rho}_{H_2O}}\]

3.1.2 – Example 1

Natural gas is volumetrically made up of 94.44% Methane (\(CH_4\)), 4.24% Ethane (\(C_2H_6\)), 0.22% Propane (\(C_3H_8\)), 0.78% Nitrogen (\(N_2\)), and 0.32% Carbon Dioxide (\(CO_2\)). What is the density of this natural gas mixture?

Let’s assume a total volume of 1 litre. This means there will be:

\[1 \space L \cdot 0.9444 = 0.9444 \space L \space [CH_4]\]
\[1 \space L \cdot 0.0424 = 0.0424 \space L \space [C_2H_6]\]
\[1 \space L \cdot 0.0022 = 0.0022 \space L \space [C_3H_8]\]
\[1 \space L \cdot 0.0078 = 0.0078 \space L \space [N_2]\]
\[1 \space L \cdot 0.0032 = 0.0032 \space L \space [CO_2]\]

Note we cannot use volumetric fractions directly to calculate the mixture density. Looking up the density of each component, the total mass and the mass fractions would be:

\[944.4 \space L \space [CH_4] \cdot \frac{0.72g}{L} = 679.97 \space g\]
\[42.4 \space L \space [C_2H_6] \cdot \frac{1.34g}{L} = 0.05682 \space g\]
\[2.2 \space L \space [C_3H_8] \cdot \frac{1.97g}{L} = 0.004334 \space g\]
\[7.8 \space L \space [N_2] \cdot \frac{1.251g}{L} = 0.009758 \space g\]
\[3.2 \space L \space [CO_2] \cdot \frac{1.977g}{L} = 0.006326 \space g\]

The total mass of the 1 litre mixture would be the sum of these masses, which comes to \(680.05 \space g\). The mass fractions of the components then are:

\[\frac{679.97}{680.05} \space [CH_4] = 0.99988\]
\[\frac{0.05682}{680.05} \space [C_2H_6] = 0.00008355\]
\[\frac{0.004334}{680.05} \space [C_3H_8] = 0.00000637\]
\[\frac{0.009758}{680.05} \space [N_2] = 0.00001435\]
\[\frac{0.006326}{680.05} \space [CO_2] = 0.0000093\]

The density of the mixture can be approximated by just the methane alone, the mass fraction is substantially more than the others.

\[\overline\rho_{natural \space gas} = \sum_{i=1}^{n} {x_i}{\rho}_i = 0.99988 \cdot \frac{0.72 \space g}{L} = 0.72 \space \frac{g}{L}\]

A common usage of densities and S.G. is the calculation of mass or volumetric flowrates, given one of the two factors, since:

\[\rho = \frac{\dot{m}}{\dot{V}} = \frac{m}{V}\]

Note: The dot above the variable means that the unit is the variable per unit time. (e.g. \(\dot{m}\) = \(mass/time\))

3.1.2 – Example 2

The volumetric flow rate of \(CCl_4\) ( \(\rho = 1.595 \space g/cm^3\) ) in a pipe is 100.0 cm\(^3\) /min. What is the mass flow rate of the \(CCl_4\)?

\[\dot{m}_{CCl_4} = 100.0 \space \frac{cm^3}{min} \times 1.595 \space \frac{g}{cm^3} = 159.5 \space \frac{g}{min}\]
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