The molecule is electrically the neutral particle having all chemical properties inherent in this concrete substance. Including to gases: to oxygen, nitrogen, chlorine, etc. How it is possible to define quantity of molecules of gas?
Instruction
1. If you need to count, how much? molecules of oxygen contains in 320 grams of this gas under normal conditions, first of all, define what number of moles of oxygen is concluded in this quantity. According to Mendeleyev's table, it is possible to see that the rounded atomic mass of oxygen – 16 atomic units. As the oxygen molecule – diatomic, the mass of a molecule will make 32 atomic units. Therefore, number of moles 320/32 = 10.
2. Further you will be helped by the universal number of Avogadro called in honor scientist, who assumed that equal volumes of ideal gases under constant conditions contain identical quantities of molecules. It is designated by a N(A) symbol and is very big – approximately makes 6.022*10(23). Increase this number by the calculated number of moles of oxygen and you learn that required quantity of molecules in 320 grams of oxygen – 6.022*10(24).
3. And if you know the oxygen pressure and also the volume occupied by him and temperature? How to calculate quantity of its molecules at such data? And there is nothing difficult. It is only necessary to write down the universal equation of Mendeleyev-Klapeyrona for ideal gases: PV = RTM/m Where P is gas pressure in pascals, V – its volume in cubic meters, R – a universal gas constant, M – the mass of gas, and m – its molar weight.
4. Slightly transforming this equation, you receive: M = PVm/RT
5. As you have all necessary data (pressure, volume, temperature are set initially, R = 8.31, and the molar mass of oxygen = 32 grams/mol), you will elementary find the mass of gas at this volume, pressure and temperature. And further the problem is solved just as in the above example: N(A)M/m. Having made calculations, you learn how many molecules of oxygen contain under the set conditions.
6. It is possible to simplify even more the decision as in the received fraction of N(A) PVm/RTm molar masses is reduced, and remains: N(A)PV/RT. Having substituted sizes known to you in a formula, you receive the answer.