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

3.1.0 – Learning objectives

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

Differentiate density and specific gravity.
Utilize density and specific gravity to interchangeably find the 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 the 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.

$Density=\rho$

$Specific{\text{ }}gravity={\frac {{\rho }_{sample}}{{\rho }_{H_{2}O}}}$

3.1.2 – Example 1

Natural gas is volumetrically made up of 94.44% Methane ( $CH_{4}$ ), 4.24% Ethane ( $C_{2}H_{6}$ ), 0.22% Propane ( $C_{3}H_{8}$ ), 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:

$1{\text{ }}L\cdot 0.9444=0.9444{\text{ }}L{\text{ }}[CH_{4}]$

$1{\text{ }}L\cdot 0.0424=0.0424{\text{ }}L{\text{ }}[C_{2}H_{6}]$

$1{\text{ }}L\cdot 0.0022=0.0022{\text{ }}L{\text{ }}[C_{3}H_{8}]$

$1{\text{ }}L\cdot 0.0078=0.0078{\text{ }}L{\text{ }}[N_{2}]$

$1{\text{ }}L\cdot 0.0032=0.0032{\text{ }}L{\text{ }}[CO_{2}]$

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:

$944.4{\text{ }}L{\text{ }}[CH_{4}]\cdot {\frac {0.72g}{L}}=679.97{\text{ }}g$

$42.4{\text{ }}L{\text{ }}[C_{2}H_{6}]\cdot {\frac {1.34g}{L}}=0.05682{\text{ }}g$

$2.2{\text{ }}L{\text{ }}[C_{3}H_{8}]\cdot {\frac {1.97g}{L}}=0.004334{\text{ }}g$

$7.8{\text{ }}L{\text{ }}[N_{2}]\cdot {\frac {1.251g}{L}}=0.009758{\text{ }}g$

$3.2{\text{ }}L{\text{ }}[CO_{2}]\cdot {\frac {1.977g}{L}}=0.006326{\text{ }}g$

The total mass of the 1 litre mixture would be the sum of these masses, which comes to $680.05{\text{ }}g$ . The mass fractions of the components then are:

${\frac {679.97}{680.05}}{\text{ }}[CH_{4}]=0.99988$

${\frac {0.05682}{680.05}}{\text{ }}[C_{2}H_{6}]=0.00008355$

${\frac {0.004334}{680.05}}{\text{ }}[C_{3}H_{8}]=0.00000637$

${\frac {0.009758}{680.05}}{\text{ }}[N_{2}]=0.00001435$

${\frac {0.006326}{680.05}}{\text{ }}[CO_{2}]=0.0000093$

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

${\overline {\rho }}_{natural{\text{ }}gas}=\sum _{i=1}^{n}{x_{i}}{\rho }_{i}=0.99988\cdot {\frac {0.72{\text{ }}g}{L}}=0.72{\text{ }}{\frac {g}{L}}$

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 $CCl_{4}$ ( $\rho =1.595{\text{ }}g/cm^{3}$ ) in a pipe is 100.0 cm $^{3}$ /min. What is the mass flow rate of the $CCl_{4}$ ?

${\dot {m}}_{CCl_{4}}=100.0{\text{ }}{\frac {cm^{3}}{min}}\times 1.595{\text{ }}{\frac {g}{cm^{3}}}=159.5{\text{ }}{\frac {g}{min}}$