Question

How do you calculate the density of butane, ${C_4}{H_{10}}$ at STP .

Answer 1

$p = \dfrac{m}{V}$
or
density (p) is equal to mass (m) divided by volume (V).
(Density is defined as mass per unit volume.)
$\ L = d{m^3} \\\ mL = c{m^3} = d{m^3} = {10^{ - 3}}{m^3} \\\ \$
$Lm = {10^{ - 3}}Lc{m^3} = {10^{ - 6}}{m^2}$
Butane is a gas at room temperature and atmospheric pressure. Butane is highly flammable in nature. It is colorless and easily get liquefied and it can quickly vaporizes at room temperature
density=mass/volume so let's find a mass and a volume Butane
n-butane is an alkane with the formula ${C_4}{H_{10}}$
It is a colourless gas
It has gasoline-like or natural gas-like odour.

Complete step by step answer:
STP : STANDARD TEMPERATURE AND PRESSURE
We know that at STP ( ${0^0}C$ and $1atm$ ), $1$ mole of an ideal gas occupies $22.4d{m^3}$
This means mole of butane has a volume of $22.4d{m^3}$ at STP.
$\therefore V = 22.4d{m^3} = 22.4 \times {10^3}c{m^3} = 22.4 \times {10^{ - 3}}{m^3}$
$1$ mole of butane has the same mass in grams as relative molecular mass of butane. Molar mass
$\ = 4 \times 12 + 10 \times 1 = 58 \\\ \therefore m = 58g = 58 \times {10^{ - 3}}kg \\\ \$
Using $\rho = \dfrac{m}{V}$ where $\rho$ is density:
$\ \rho = \dfrac{{58}}{{22.4}} \times {10^3} \\\ \Rightarrow 2.59kg{m^{ - 3}} \\\ \$

Note: In order to calculate the density we must be careful towards the units as It is important that you keep all the quantities the same unit. As from the definition the density of gas is defined as the mass per unit volume of a substance. Since gases all occupy the same volume on a per mole basis, that’s why the density of a particular gas is dependent on its molar mass. A gas which has a small molar mass will also have a lower density than a gas which has a large molar mass. Gas densities are typically reported in grams per litre. In more simple words, Gas density can be calculated from molar mass and molar volume of gas.