1. Starter/Repeat
How many helium atoms are there in 7g of helium gas?

2. Mass ratios and the molar volume
How many hydrogen atoms fit into that balloon?

3. Today’s topics Repeating a bit of stoichiometry
Applying masses to chemical equations
Finding out about the connection between amount of substance and volume of a gas

4. How many atoms react to form…
Stoichiometry is the calculation of reactants and products in a chemical reaction by balancing the equation
How is this equation balanced?

5. Predicting masses
We can use balanced molecular and chemical equations to predict the masses of reactants and products as well as the other way around
Example: How many grams of oxygen (O2) react with 32 grams of methane (CH4) in a complete combustion (see right)?
Given: m(CH4) = 32 g Needed: m(O2) = ?
Calculation: M(CH4)= M(C) + 4*M(H) = 12 g/mol + 4*(1 g/mol) = 16 g/mol
M(O2) = 2*M(O) = 2*(16 g/mol) = 32 g/mol
n(CH4) = m(CH4)/M(CH4) = 32 g / (16 g/mol) = 2 mol
From chemical equation: n(O2) = 2*n(CH4)
n(O2) = 2*(2 mol) = 4 mol
m(O2) = n(O2)* M(O2) = 4 mol * 32 g/mol = 128 g

6. In other words …. m n M m(CH4)= 32g m(O2)= ?
M (CH4) = M(C) + 4*M(H) = *1 = 16 g/mol
n = m = 32g = 2 mol
M g/mol m
n M
m = mass
M = Mr = Molar mass
n = number of moles

7. Predicting masses
Example 2: How many grams of water (H2O) are formed in the complete combustion of methane (CH4) with 8 grams of oxygen (O2)
Given: m(O2) = 8 g Needed: m(H2O) = ?
Calculation: M(O2) = 2*M(O) = 2*(16 g/mol) = 32 g/mol
n(O2) = m(O2)/M(O2) = 8 g / (32 g/mol) = 0,25 mol
M(H2O) = 2*M(H) + M(O) = 2*(1 g/mol) + 16 g/mol = g/mol
From chemical equation: n(H2O) = n(O2)
n(H2O) = 0,25 mol
m(H2O) = n(H2O)* M(H2O) = 0,25 mol * 18 g/mol = 4,5 g

8. Work through the tasks on worksheet 21 (front)

9. Let‘s jump back to Amadeo Avogadro
Avogadro’s discoveries around substance amounts were based around observations on gases.
He observed, that 24 litres of hydrogen gas and 12 of oxygen gas react to produce 9 grams of liquid water.
At the same time, 48 grams of magnesium burn with 48 litres of oxygen to produce magnesium oxide (MgO)
Calculate the substance amount of the gases involved

10. The molar volume
The volume of gaseous substances is directly tied to the amount of molecules in the gas
The volume of one mole of molecules of a gas is called molar volume Vm
The molar volume for all gases is approximately the same!
Vm(at 0°C) = 22,4 l/mol
Vm(at 25°C/room temperature) ≈ 24 l/mol
Why does the molar volume change with temperature?

11. Using the molar volume
The molar volume is directly connected to the amount of substance and the volume of the gas.
𝑉 𝑚 = 𝑉 𝑛 𝑜𝑟

12. Work through the tasks on worksheet 21 (reverse)

13. In conclusion
A hot air balloon holds approximately litres of air
Calculate the approximate mass of litres of air at room temperature (80 % of volume nitrogen, 20% oxygen)
Modern zeppelins are filled with helium gas. Calculate the mass of litres of helium at room temperature.
The mass of the air in a flying hot air balloon is a lot smaller than the one we calculated. Explain!

14. EXTENSION: What is the molarity of water?
Water is H2O, water has a molar mass of 18 𝑔 𝑚𝑜𝑙
There are 1,000 g of water in a liter
How many moles of water are there in one liter?

15. EXTENSION 2: The Avogadro constant’s unit
If you take a look at german year 9 chemistry books, things can get a bit confusing. The number 6,02 x is referred to as „Loschmidtzahl“ NL
The Avogadro constant NA on the other hand is used to convert a number of atoms or molecules (Teilchenzahl N) to substance amounts n via the equation 𝒏= 𝑵 𝑵 𝑨
Because of this the actual Avogadro constant has a unit. Can you figure out what it is?