Find the mass of the C2H2 acetylene molecule and its density under normal conditions?

Normal conditions:
p = 101300 Pa.
T = 273 K.
Let’s calculate the molar mass of acetylene using the periodic table:
M (C2H2) = 12 * 2 + 1 * 2 = 26 g / mol = 26 * 10 ^ -3 kg / mol.
Mendeleev-Clapeyron equation
pV = (m / M) * R * T, where p is the gas pressure, V is the gas volume, m is the gas mass, M is the molar mass of the gas, R is the universal gas constant 8.31 J / (mol * K), T is the gas temperature.
let’s divide both parts by V:
p = m * R * T / (M * V) = (m / V) * (R * T / M).
We write down the formula for determining the mass, through the specific gravity:
m = ρ * V, hence:
ρ = m / V.
Let’s replace and display the density:
p = (m / V) * (R * T / M) = ρ * (R * T / M).
ρ = p * M / (R * T).
Substitute the numerical values:
ρ = p * M / (R * T) = 101300 * 26 * 10 ^ -3 / (8.31 * 273) = 1.16 kg / m³.

Let’s write an expression to determine the amount of a substance:
ν = m / M = N / Na, where m is the mass of the substance, M is the molar mass of the substance, N is the number of particles, Na is Avogadro’s constant Na = 6.023 * 10 ^ 23 1 / mol.
Molecule mass:
m0 = M / Na, substitute the numerical values ​​and find:
m0 = M / Na = 26 * 10 ^ -3 / 6.023 * 10 ^ 23 = 4.32 * 10 ^ -26 kg.

Answer: m0 = 4.32 * 10 ^ -26 kg, ρ = 1.16 kg / m³.