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
In other words ….
m(CH4)= 32g m(O2)= ?
M (CH4) = M(C) + 4*M(H) = 12 + 4*1 = 16 g/mol
n = m = 32g = 2 mol M 16 g/mol
m n M
m = mass
M = Mr = Molar mass n = number of moles
Predicting masses
Example 2:
How many grams of water (H
2O) are formed in the complete combustion of methane (CH
4) with 8 grams of oxygen (O
2)
Given: m(O
2) = 8 g Needed: m(H
2O) = ?
Calculation: M(O
2) = 2*M(O) = 2*(16 g/mol) = 32 g/mol n(O
2) = m(O
2)/M(O
2) = 8 g / (32 g/mol) = 0,25 mol M(H
2O) = 2*M(H) + M(O) = 2*(1 g/mol) + 16 g/mol =
18 g/mol
From chemical equation: n(H
2O) = n(O
2) n(H
2O) = 0,25 mol
m(H
2O) = n(H
2O)* M(H
2O) = 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
m
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