[Hjsk]

Suggest a possible hydrocarbon formula if at a pressure of 0.15MPa and a temperature of 87 ° C, it has a density of 3.51 g/l and 1 liter of this substance contains 4.52 * 10^23 atoms
Heeeeelp

[chenbeier]

From the amount of atoms you can calculate the moles.

From the moles you can calculate the volume by using ideal gas law.

From the volume and the density you can calculate the mass.

From mass and moles you can calculate the molar mass.

[AWK]

Rearrange Ideal Gas Law and calculate the molecular mass in one step.

[Borek]

From the amount of atoms you can calculate the moles.

Number of moles of what?

[Babcock_Hall]

@OP,

According to the Forum Rules (see red link above), you must show your attempt before we can help you.  With the hints you have already been given, this should not be a problem.

[chenbeier]

From the amount of atoms you can calculate the moles.

Number of moles of what?

moles of the substace what is looking for

[Borek]

From the amount of atoms you can calculate the moles.

Number of moles of what?

moles of the substace what is looking for

How?

[chenbeier]

1 mol = 6,023 * 10²³  => x mol = 4,52 *10²³ x = 0,75 mol.

[AWK]

And how many atoms does the molecule contain?

In this task, the number of atoms in 1 liter of gas is a piece of superfluous information.
With hydrocarbon masses below 100, the composition of the hydrocarbon can be unequivocally determined from its weight alone.
eg.
For MW 100 we can fit C₇H₁₆ or C₈H₄.

[Borek]

1 mol = 6,023 * 10²³  => x mol = 4,52 *10²³ x = 0,75 mol.

No. See what AWK wrote and read the question once again.

[chenbeier]

pV = nRT , n =m/M

pV = mRT/M

M = mRT/pV

Rho = m/V

M = RhoRT/p


Given T = 360,15 K ,p = 150000 N/m², spec. Gravity = 3.51 g/l

M = 3.51kg/m³ *8.314 Nm/molK*360.15 K/150000 N/m² = 0,07 kg/mol = 70 g/ mol.

Formula  C₅H₁₀ = 5 *12 + 10 * 1 = 70

n = m/M   n= 3,51 g/70 g/mol = 0,05 mol in 1 l

4,51 *10²³ Atoms divided by 15 = 3 *10²² molecules C₅H₁₀

This devided by 6.023*10²³ gives also 0.05 mol

[AWK]

The whole discussion is due to the imprecision of Chenbeier hints.
"From the number of atoms you can calculate the moles."
Moles must be combined with microscopic individuals (atoms, molecules, charges, etc.).

From the available data, you can directly calculate the mole of atoms from the given number of atoms (0.75 mol of atoms) or the number of moles of molecules from the ideal gas equation using a volume of 1 liter contained in density (which gives 0.05 moles of molecules). From this number, the mass of the molecule can be counted in the next step (M = 3.51/0.05 = 70.2).
By rearrangement the ideal gas equation for M, this can be counted in one step (M = dRT/p where d = density of gas). This gives the same value for M.
In the next step, for M = 70, only one C₅H₁₀ hydrocarbon formula can be deduced by fitting (70 divided by 12 equals 5 with the remainder of 10).

[chenbeier]

That is what I did in last approach.

[Babcock_Hall]

The OP did not do any work.