Calculation of reacting masses, gas volumes and mole concentrations

Mole: The unit of amount of substance. One mole is roughly 6.02 x 1023

Number of moles = number of particles/number of particles in a mole

example: How many moles of Carbon is in 1.50 x 1024

number of moles = 1.50 x 1024 / 6.02 x 1023

Molar Mass: Mr, is the mass of one mole of something. The units are g mol -1

Number of moles = Mass of a substance / molar mass

example: How many moles of aluminium oxide are present in 5.1 g of Al203

Molar Mass = (2 x 27.0) + (3.16.0) = 120 g mol -1

Number of moles = 5.1 / 102 = 0.05 moles

Gas Volume for r.t.p

Room temperature and pressure = 298 K and 100 kPa

Molar gas volume = 24 dm3 mol -1 (or 24 000 cm3 mol-1)

number of moles = volume in dm3 / 24 or volume in cm3 / 24 000

Ideal Gas Equation

Ideal Gas Equation = pV = nRT or PressureVolume = NumberofmolesGasconstantTemperature

Gas constant = 8.314

Example: At what pressure would 0.400 moles of argon gas occupy 0.0100 m3 at 298 K?

p =nRT / V = 0.400 x 8.314 x 298 / 0.0100 = 99 100 pa

To convert:

• Pressure may be givin in kPa. To convert to Pa times by 1000
• Temperature may be given in C. To convert to K add 273
• Volume may be given in cm3. To covert from cm3 to m3 times by 10-6.
• Volume may be given in dm3. To convert from dm3 to m3 times by 10-3.

Example: What volume would 2.00 moles of argon gas occupy at 27.0 C and 100 kPA?

V = nRT/ P = 2.00 x 8.314 x 300 / 100 000 = 0.0499

Example: At a temperature tof 60.0 C and a pressure of 250 kPa, a gas occupied a volume of 110 cm3 and had a mass of 1.59 g. Find its relative molecular mas

You have been given temperature, pressure and volume. So you need to find the moles

n = pV / RT = 250 000 x (1.10/100) / 8.314 x 333 = 0.0993 moles

Now you have the number of moles you can work out the molar mass using the formular, molar mass = mass / number of moles = 1.59 / 0.0993 = 16.0 g mol -1

So the relative molecular mass is also 16.0

number of moles = concentration x volume (in dm3)

Concentration = Number of Moles / Volume in dm3

Volume in dm3 = Number of moles / Concentration

Examples: How many moles of lithium chloride are present in 25 cm3 of a 1.2 mol dm-3 solution of LiCl

Number of moles = concentration x volume = 1.2 x 0.025 = 0.030 moles

A solution of CaCl2 contains 0.2 moles of  calcium chloride in 0.4 dm3 water. What is the concentration of the solution?

Concentration = number of moles / volume = 0.2 / 0.4 = 0.5 mol dm -3

What mass of sodium hydroxide needs to be dissolved in 50 cm3 of water to make a 2.0 mol dm-3 solution

moles = concentration x volume in dm3 = 0.050 x 2.0 = 0.10 moles of NaOH

Mass = moles x Mr = 0.10 x (23 +16 +1) = 4.0

Mass = 4.0 g