Download Empirical Formula

Empirical Formula
Objectives
• To be able to calculate the empirical
formula.
• To be able to balance equations.
• To be able to use equations to calculate
the mass of reactants and products.
Remember
• What is the Relative Atomic Mass of
Sodium?
• What is the Relative Formula Mass of
Sodium Chloride?
Remember
• What is the Relative Atomic Mass of Sodium?
Answer: 23
• What is the Relative Formula Mass of Sodium
Chloride?
Answer: 23 + 35.5 = 58.5
Empirical Formula
If we know the percentage composition
of a compound we can work out the ratio
of the numbers of atoms in the
compound.
This is called the Empirical Formula.
Example
If 9g of aluminium reacts with 35.5g of
chlorine, what is the empirical formula of the
compound formed?
Step 1: What is the ratio of atoms?
9/27 = 0.3 moles of aluminium atoms
35.5/35.5 = 1 mole of chlorine atoms
Finally
So:
0.3 moles of aluminium atoms
1 mole of chlorine atoms
Therefore, we need 3 Cl for every 1 Al.
So, formula is AlCl3
Try these…
1. A compound contains 16g of sulfur and
24g of oxygen. What is its empirical
formula? (Ar Values: S = 32, O = 16)
2. A compound contains 44.8g of iron and
19.2g of oxygen. What is its empirical
formula? (Ar Values: Fe = 56, O = 16)
A compound contains 16g of sulfur and
24g of oxygen. What is its empirical
formula? (Ar Values: S = 32, O = 16)
Step 1: What is the ratio of atoms?
16/32 = 0.5 moles of sulfur atoms
24/16 = 1.5 mole of chlorine atoms
So 1 S needs 2 Cl, so formula is SCl2
Try this one….
A compound contains 44.8g of iron and
19.2g of oxygen. What is its empirical
formula? (Masses: Fe = 56, O = 16)
A compound contains 44.8g of iron and
19.2g of oxygen. What is its empirical
formula? (Masses: Fe = 56, O = 16)
Step 1: What is the ratio of atoms?
44.8/56 = 0.8 moles of iron atoms
19.2/16 = 1.2 mole of oxygen atoms
So 2 Fe needs 3 O, so formula is Fe2O3
Step 1: What is the ratio of atoms?
44.8/56 = 0.8 moles of iron atoms
19.2/16 = 1.2 mole of oxygen atoms
If you multiply by 10, you get 8 Fe to 12 O
If you divide by 4, you get 2 Fe to 3 O.
So 2 Fe needs 3 O, so formula is Fe2O3
Balancing
Hydrogen reacts with oxygen to produce
water.
2
H2 + O2
 2 H20
Can you balance this equation?
Using equations
2H2 + O2
 2H20
This means that 2 moles of Hydrogen
reacts with 1 mole of oxygen.
This reaction produced 2 moles of water.
Using equations
2H2 + O2
 2H20
Ar of H = 1, so mass of 1 mole of H2 = 2 x 1 = 2g.
Equation tells us we have 2 moles so we need
4g.
Ar of O = 16, so mass of 1 mole of O2 = 2 x 16 =
32g. Equation tells us we only need 1 mole so
we need 32g.
Mr of H2O = 16 + 2 = 18, so mass of 1 mole of
H2O = 18g. Equation tells us we produce 2
moles so we produce 32g.
Using equations
2NaOH + Cl2  NaOCl + NaCl + H20
If we have a solution containing 100g of
NaOH, how much chlorine gas should we
pass through the solution to make
bleach (NaOCl)?
Using equations
2NaOH + Cl2  NaOCl + NaCl + H20
Ar of H = 1
Ar of O = 16
Ar of Na = 23
Ar of Cl = 35.5
Using equations
2NaOH + Cl2  NaOCl + NaCl + H20
1 mole of NaOH = 23 + !6 + 1 = 40g
1 mole of Cl2 = 35.5 x 2 = 70g
So, 100g of NaOH is 100/40 = 2.5 moles
Using equations
2NaOH + Cl2  NaOCl + NaCl + H20
Equation states that for every 2 moles of
NaOH we need only 1 mole of Chlorine.
So we need 2.5/2 = 1.25 moles of chlorine.
If 1 mole of Chlorine has a mass of 71g, then
1.25 x 71 = 88.75g of chlorine to react with
100g of NaOH.
Objectives
• To be able to calculate the empirical
formula.
• To be able to balance equations.
• To be able to use equations to calculate
the mass of reactants and products.
Try your best to…
•Write down the
steps to balancing
equations!
Hydrates
% H2O in a Hydrate
Certain substances have water molecules associated with their structure. Many salts
fall into this category and we refer to them as hydrated salts. Different salts have
different numbers of of water molecules - BaCl2•2H2O; CuSO4•5H2O.
PROBLEM: Calculate the % by mass of water in one mole of BaCl2•2H2O
%H2O = (mass of H2O in one Mol/mass of BaCl2•2H2O) x 100
%H2O = (36 grams/243 grams) x 100 = 14.8% H2O
How do you make Rice Krispy
Treats?
• No really how do you
make them? Write down
a recipe in the form of a
formula!
Reaction Types
 You should be able to identify the
following types of reactions:
 Combination or Synthesis Reaction
 Decomposition or Analysis Reaction
 Single Replacement Reaction
 Double Replacement Reaction
 Combustion Reactions
 Complete
 Incomplete
Combination (Synthesis)
 In a combination reaction
usually two elements combine
to form a compound.
2Na(s) + Cl2(g) = 2NaCl(s)
Combination (Synthesis)
 Complete the following combination reactions by
balancing them:
N2(g) + H2(g) = NH3(g)
Al(s) + O2(g) = Al2O3(s)
S(s) + O2(g) = SO3(g)
Decomposition (Analysis)
 More complex compounds
breakdown into simpler
substances or elements.
CaCO3(s) = CaO(s) + CO2(g)
 Notice in this case the equation is already
balanced, this will not always be the case.
Decomposition (Analysis)
 Balance the following decomposition
reactions:
HI(g) = H2(g) + I2(g)
N2O5(g) = N2(g) + O2(g)
KClO3(s) = KCl(s) + O2(g)
Single Replacement
 In a single replacement reaction one of the
species in a formula is replaced by
another species:
2NaBr(aq) + Cl2(g) = 2NaCl(aq) + Br2(l)
Single replacement
Mg(s) + 2HCl(aq) = MgCl2(aq) + H2(g)
Single replacement
Single Replacement
 Balance the following single replacement
reactions:
Cu(s) + AgNO3(aq) = Cu(NO3)2(aq) + Ag(s)
Zn(s) + HCl(aq) = ZnCl2(aq) + H2(g)
Zn(s) + H2SO4(aq) = ZnSO4(aq) + H2(g)
Double Replacement
 In a double replacement reaction the
species simply exchange places:
AgNO3(aq) + NaCl(aq) = AgCl(s) + NaNO3(aq)
In this case, when we check both sides of the equation we find that there is
no need to adjust the coefficients - it is balanced
Double Replacement
 Balance the following double replacement
reactions:
Ba(NO3)2(aq) + K2SO4(aq) = BaSO4(s) + KNO3(aq)
Al(NO3)3(aq) + KOH(aq) = Al(OH)3(s) + KNO3(aq)
Cd(NO3)2(aq) + H2S(g) = CdS(s) + HNO3(aq)
Combustion Reactions
 Complete combustion generally involves the
burning of an organic compound in oxygen
to form carbon (IV) oxide and water as
products:
CH4(g) + 2O2(g) = CO2(g) + 2H2O(g)
Combustion Reactions
 Incomplete combustion reactions generally
involve the formation of carbon (II) oxide
and water as products:
2CH4(g) + 3O2(g) = 2CO(g) + 4H2O(g)
Here are your 2 options
• Create a visual
representation of
2 of the three
chemical
reactions! Be
creative!
– Ie, make a food
item, shoes, car,
• Write a story
about all the
three types of
chemical
reactions.
Poster Includes:
• Details notes, Important
Vocabulary, Create two
questions with “twelve words”
word stems, with and test
question numbers on the test.
Seminar
Zapata, Oscar
Alston, Eric
Thompson, Rasheed
Callands, Carlos
Stewart, Joi
Connor, Patrick
Ramirez-Rodriguez, Lucimar
Lesane, Virgil
Lockett, Janell
Eliab, Donovan
Cruz, Ariadna
Brown, Lakelle
Washington, Naomi
Pickard, Jasmine
McCain, Franklin
Stamey, Jania
Javed, Hamzah
Kidd, Christopher
McCall, Brayla
Robinson, Ashleigh
Little, Isaiah
McCain, Franklin
Ramirez, Luis
Atchabao, Ibn Khassir
Santiago Boria, Orangellys
Battle, Rickey
Boyd, Corey
Hopkins, Joshua
Muhammad-Brown, Infinity
Washington, Naomi
Brunson, Joshua
Atchabao, Ibn Khassir
Test Questions
1.1.1 Questions 1-7
1.1.2 Questions 8-12
1.2.1 Questions 13-19
1.2.2 Questions 20-25