## How to measure density?

**Density formula** of solid, liquid, and gas molecules is the measurement of mass per unit volume of material at a given temperature and pressure. In chemistry or physics, it is normally expressed by the symbol ρ or D. Mathematically density is measure by the formula, ρ = m/V, where m = mass of the substance and V = volume. A standard or normal density calculator uses for calculating densities at 0°C and 1 atmosphere pressure. The CGS and SI units of density are g/cm^{3} and kg/m^{3} and dimension [ML^{-3}]. The ideal gas law formula uses to calculate the molar mass and density of ideal gases but this equation is approximately obeyed under normal conditions. Therefore, the density measurement by the ideal gas equation, not an accurate one.

## Density of gases

The density of gases and vapour is the mass per unit volume at a given temperature and pressure. The standard or normal density of gases is the density at 0°C temperature and 1 atmosphere pressure. In the calculations of molecular weight, relative densities are used. It is the ratio of masses of an equal volume of gas and oxygen. The density of the gas is also related to Graham’s law of diffusion or effusion.

### Ideal gas density formula

The ideal gas equation for n-mole gases, PV = nRT. From the above equation P = nRT/V = gRT/MV = ρRT/M. ρ = symbol of the density of the gas, M = mass (g) per unit volume (cc). The molar mass of the gas can be experimentally determined by using this expression.

## Limiting density formula

Such density definition or calculation from the above gas equation measured only an approximate and not an accurate one for learning chemistry or chemical science. But the above equation accurately obeyed at very low pressure at P→0. So the molar mass value of this pressure gives accurate and the equation is represented as, M = (ρ/P)_{P→0} ×RT, where (ρ/P)_{P→0} = limiting the density of a gas.

### Pressure vs limiting density graph

But direct use of this relation poses some difficulty since, at P→0, ρ→0 and not possible to determine experimentally. Therefore, the above graphical measurement has taken advantage of the calculation of molar mass by density.

**Problem:** At 0°, the density of gaseous non-metallic oxide at 2 atm pressure is the same as that of oxygen at 5 atm. Find the molecular weight of the nonmetallic substance.

**Solution:** From the density formula, M = M_{O2} × (P_{O2}/P_{non-mettalic}) = 32 × (5/2) = 32 = 80 g/mol.

## What is vapour density?

Vapour density of a substance defined as the ratio of the density of the gaseous state of the substance and hydrogen under the same temperature and pressure. It is the quantity having no unit or dimensions and represented as, D_{0} = M_{0}/2, where M_{0} = formula molecular weight of the gasses.

### Abnormal vapour density

But for many chemical substances, like ammonium chloride, phosphorus pentachloride, hydrogen peroxide, iodine, etc have calculated vapour densities less than the theoretical measurement. Only at a high temp D_{0}≈D_{0}/2. This phenomenon is called abnormal vapour densities. The cause of this phenomenon due to the thermal bond dissociation of the above chemical substances.

NH_{4}Cl → NH_{3} + HCl |

PCl_{5} → PCl_{3} + Cl_{2} |

N_{2}O_{4} → 2NO_{2} |

I_{2} → 2I |

### Formula for Calculating Abnormal Density

At constant pressure due to the dissociation of the chemical bond, the volume of the gases increases with the increase of the mole number. Since mass remains the same and D_{0} decreases. The extent or fraction of the total number of molecules that suffer dissociation calculates the degree of dissociation. Let us take one molecule of substance A splits up into n molecules of B by the specific heat, A → nB.

At equilibrium for each g mole of A will be (1 – α) g moles of undissociated A and nα g moles of B. So the total number of g moles of gas present = (1-α) + nα or 1 + (n-1)α. When V = volume occupied per unit volume of gas or vapour and D_{0} define the density measurement absence of any dissociation. If D denotes the observed density when actual dissociation occurred and the total weight of the gas = W.

W = D_{0}V and W = D×V[1 + (n-1)α]

∴ D_{0}/D = 1 + (n-1)α

The above expression shows that, at high temperature, α = 1 (complete bond dissociation) and for NH_{4}Cl, and PCl_{5}, n=2. Thus, the density, D = D_{0}/2. The above formula is used to calculate the degree of dissociation.