Mole Road Map — Overview & Examples of Conversions

Before this, knowThe Mole — Definition & Avogadro's NumberMolar Mass — Definition & CalculationsConverting Moles to Volume — Overview & Examples

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Molar Road Map

To help you remember how to convert between moles, volume at STP, mass, and number of particles, we made this simple visual.

Image showing the conversions between moles to volume (22.L/1mol), moles and mass (molar mass/1mol) and moles and number of particles (6.02X10^23 particles/1mol)

Image source: By Caroline Monahan

Practice Problems!

Often, you'll be given moles and have to convert to volume. That's simple! Just use the conversion factor. Unfortunately, it's not always that easy. Sometimes, you are given a problem where you start with mass and need to get to volume or number of particles. In order to do this you must use 2 conversion factors. First, multiply the mass by the conversion to get to moles. Then from there, you can multiply the moles with another conversion factor to get to what you need!

To explore these common types of problems, try these two practice problems.

An analytical chemist has 0.345 mols of diatomic oxygen (O2O_{2}), what is the volume of O2O_{2} in STP?

Choose one answer

Read · 2 min

We are learning about the mole. We started by learning to calculate a molecule's molecular mass. We can look at the periodic table and find the atomic mass of each atom in our molecule. The sum of the molecule's atomic masses is the molecular mass. Next, we learned about the mole. In one mole of a substance, there are 6.022x10^23 particles. After learning about the molecular mass and the mole, we discovered the molar mass. The molar mass is the mass of one mole of a substance. Fortunately, we calculate molecular mass and molar mass in the same way. But, what if you have a gaseous molecule? It is conventional to talk about the volume of a gas. So, we learned how to convert a mole of gas to its volume.

The Three Major Conversions

From our previous lessons, we know three conversions. Remember, nn is the number of moles.

  • Moles to mass: n×(molar mass)n\times(molar\,mass)= X grams
  • Moles to particles: n×6.022×1023n\times6.022\times10^{23}= X particles
  • moles of gas to volume: ngas×22.4n_{gas}\times22.4= X litres

Example Problem

Let's look at an example problem. In 2001, Ian Ashpole floated to 11,000 ft above sea level using 600 toy balloons. He filled his balloons with helium. With 600 ballons, he burst his previous record of 10,000 ft. All the noble gases are essentially ideal gases. Why? Because they have a complete octet. So, they have very minimal electrostatic interactions. The only Van der Waal's interactions they have are dispersion forces.

With the ideal gas law, I calculated he filled each balloon with about 0.21 moles of helium. Let's do a few calculations to find out more about his balloons. As he floated into the sky, the temperature and pressure would change. So, he wouldn't be at atmospheric pressure. But, we'll assume that he was at standard temperature and pressure the whole time.

How many grams of helium did he use to fill each balloon? What about all 600 balloons?

  • Helium has an atomic mass of 4.002 u. Its molar mass is 4.00 gmol\frac{g}{mol}.
  • Moles to mass: n×(molar mass)n\times(molar\,mass)= 0.21mols×4.00gmol=0.84g0.21mols\times4.00\frac{g}{mol}= 0.84g.
  • Mass of all 600 Balloons: 0.84g×600=504.0g0.84g\times600= 504.0g.

How many helium atoms were in each balloon? How many atoms total?

  • Moles to particles: n×(6.022×1023)n\times(6.022\times10^{23})=0.21mols×(6.022×1023atomsmol)0.21mols\times(6.022\times10^{23}\frac{atoms}{mol})= 1.26×1023atoms1.26\times10^{23} atoms.
  • Atoms in all 600 Balloons: (1.26×1023atoms)×600(1.26\times10^{23}atoms)\times600= 7.56×1025atoms7.56\times10^{25} atoms. That's 7.56 ten-septillion atoms of helium!

What was the volume of each balloon at STP? The sum of all the balloons?

  • Moles of gas to volume: ngas×22.4n_{gas}\times22.4=0.21mols×22.4Lmol=4.7L0.21mols\times22.4\frac{L}{mol}= 4.7 L
  • The Volume of 600 Balloons: 4.7L×600=2820L4.7L\times600=2820 L

Just for fun, what is the radius of each balloon?

  • 4.7L→0.0047m34.7 L\rightarrow0.0047m^{3}
  • To calculate the radius we use the volume of a sphere. V=43\frac{4}{3}π\pir3r^{3} →\rightarrow r= .0047m3×34π3=0.10m\sqrt[3]{.0047m^{3}\times\frac{3}{4\pi}}= 0.10m

Video

(Video) Mole Highway Practice Problems

by Chem2Farr

In this video, Chem2Farr walks you through some Mole Roadmap practice problems. You'll learn to use the Mole Roadmap to guide your dimensional analysis. It will make the problems so much easier! If you need help with stoichiometry problems, this video can help! For more advanced stoichiometry, check out the extended stoichiometry roadmap.

Type a concept, like “slope” or “photosynthesis”.

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