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

Starter/Repeat

 How many helium atoms are there in 7g of

helium gas?

Mass ratios and the molar volume

How many hydrogen atoms fit into

that balloon?

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

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?

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

Predicting masses

 Example 2:

How many grams of water (H

O) are formed in the complete combustion of methane (CH

) with 8 grams of oxygen (O

)

 Given: m(O

) = 8 g Needed: m(H

O) = ?

 Calculation: M(O

) = 2*M(O) = 2*(16 g/mol) = 32 g/mol n(O

) = m(O

)/M(O

) = 8 g / (32 g/mol) = 0,25 mol M(H

O) = 2*M(H) + M(O) = 2*(1 g/mol) + 16 g/mol =

18 g/mol

From chemical equation: n(H

O) = n(O

) n(H

O) = 0,25 mol

m(H

O) = n(H

O)* M(H

O) = 0,25 mol * 18 g/mol = 4,5 g

 Work through the tasks on worksheet 21 (front)

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

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 V



The molar volume for all gases is approximately the same!

 Vm(at 0°C) = 22,4 l/mol

 V_m(at 25°C/room temperature) ≈ 24 l/mol



Why does the molar volume change with temperature?

Using the molar volume

 The molar volume is directly connected to the amount of substance and the volume of the gas.

  

 Work through the tasks on worksheet 21

(reverse)

In conclusion

 A hot air balloon holds approximately 3.000.000 litres of air

 Calculate the approximate mass of 3.000.000 litres of air at room temperature (80 % of

volume nitrogen, 20% oxygen)

 Modern zeppelins are filled with helium gas.

Calculate the mass of 3.000.000 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!

EXTENSION:

What is the molarity of water?

 Water is H

O, water has a molar mass of

 There are of water in a liter

 How many moles of

water are there in one liter?

  

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 10

is referred to as „Loschmidtzahl“ N

 The Avogadro constant N

on the other hand is used to convert a

number of atoms or molecules (Teilchenzahl N) to substance amounts n via the equation

 Because of this the actual Avogadro constant has a unit. Can you figure out what it is?

