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Chap 1 Quantitative Chemistry
1.2 Formulas
1.2.1
1.2.2
1.2.3
1.2.4
1.2.5
1.2.6
3 hours
Assessment statement
Obj Teacher’s notes
Define the terms relative
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atomic mass (Ar) and relative
molecular mass (Mr).
Calculate the mass of one mole 2 The term molar mass (in g mol–1) will be used.
of a species from its formula.
Solve problems involving the
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relationship between the
amount of substance in moles,
mass and molar mass.
Distinguish between the terms
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empirical formula and
molecular formula.
Determine the empirical
3 Aim 7: Virtual experiments can be used to demonstrate this.
formula from the percentage
composition or from other
experimental data.
Determine the molecular
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formula when given both the
empirical formula and
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experimental data.
Chemical Formulas
• Chemical formulas show the ratio of the number of
atoms present in a compound or ion.
• Magnesium chloride MgCl2
– 2 chloride ions for every 1 magnesium ion (Ionic compound)
• Acetic Acid CH3COOH
– 2 atoms carbon, 2 atoms oxygen, 4 atoms hydrogen in one
molecule of acetic acid (molecular compound)
• Ionic compound consist of a metal (positive ion) and nonmetal (negative ion).
• Molecular (covalent) compounds are made of non-metals
only.
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Relative Atomic Mass (Ar)
• The mass of an atom is compared with that of an atom
of carbon-12. (Carbon-12 standard)
• The relative atomic mass of carbon-12 is taken to be 12.
• Relative masses are based on elements having isotopes
(same identity, different “stuff” within core)
• For example, naturally occurring chlorine consists of atoms of
relative isotopic masses 35 (75%) and 37 (25%).
– Its relative atomic mass (Ar) is 35.5.
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Relative Molecular Mass (Mr)
• The MASS of a MOLECULE
• The mass of a molecule is compared with that of an
atom of carbon-12. (Carbon-12 standard)
• A relative molecular mass can be calculated easily by
adding together the relative atomic masses of the
constituent atoms that have units g mol-1
• e. g., ethanol: CH3CH2OH, has a Mr of 46.08 g mol-1
(C2+ H6+ O = 24.02 + 6.06 + 16.00 = 46.08)
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Example: Molecular Mass
• Find the molecular mass of Glucose C6H12O6.
• Solving Process:
– Add the atomic masses of all the atoms in the C6H12O6
• 6 C atoms
6 x 12.0 = 72.0 g mol-1
• 12 H atoms
12 x 1.0 = 12.0 g mol-1
• 6 O atoms
6 x 16.0 = 96.0 g mol-1
Formula mass of glucose
= 180.0 g mol-1
– This mass represents one mole of glucose.
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Example: Formula Mass
• Find the formula mass of Copper (II) phosphate,
Cu3(PO4)2.
• Solving Process:
– Add the atomic masses of all the atoms in the Cu3(PO4)2 formula.
Remember the subscript applies the entire polyatomic ion.
• 3 Cu atoms
3 x 63.5 = 190.5 g mol-1
• 2 P atoms
2 x 30.9 = 61.8 g mol-1
• 8 O atoms
8 x 16.0 = 128.0 g mol-1
Formula mass of Cu3(PO4)2
= 380.3 g mol-1
– This mass represents one mole of Copper (II) phosphate.
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Chap 1 Quantitative Chemistry
1.1 The mole concept and Avogadro’s constant
Assessment statement
1.1.1 Apply the mole concept to
substances.
2 hours
Obj Teacher’s notes
2 The mole concept applies to all kinds of particles: atoms,
molecules, ions, electrons, formula units, and so on. The
amount of substance is measured in moles (mol). The
approximate value of Avogadro’s constant (L),
6.02 × 1023 mol–1, should be known.
TOK: Chemistry deals with enormous differences in
scale. The magnitude of Avogadro’s constant is beyond
the scale of our everyday experience.
1.1.2 Determine the number of
particles and the amount of
substance (in moles).
3 Convert between the amount of substance (in moles)
and the number of atoms, molecules, ions, electrons
and formula units.
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Chap 1.1 Mole Concept & Avogadro’s Constant
• Chemists do not deal with amounts of substance by
counting out individual atoms or molecules.
• Atomic Mass Units (amu) as a Very Small Number
• It is important to obtain a relationship between mass and
the number of particles in a compound.
• Molecular masses are in atomic mass units, which is a
very small unit.
• An atomic mass unit is only 1.66 x 10-24 grams.
• We need to use a larger unit for everyday use such as the
gram for laboratory use.
• We mass quantities of substances on balances.
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Avogadro’s Number
• Chemists have found that
6.022 x 1023 atoms of
Hydrogen have a mass of
1.0079 g.
• This number 6.022 x 1023 is
called Avogadro’s number.
• Commonly referred to as the
Mole.
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The Mole
• Avogadro’s number is an accepted SI standard.
• The symbol used to represent Avogadro's number is NA.
• This quantity can be expressed as 6.02205 x 1023 to be more
precise.
• This number of whatever we are questioning is called one
mole (mol) of things.
• Dozen: 12 of something. Whether feathers or dump trucks.
•
Same number, different masses
• Mole: just a name representing a number
(6.022 x1023) (just like dozen)
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The Mole (cont.)
• A mole of feathers and a mole of trucks are the same
number
– Are they the same mass?
• One mole of particles (atoms, ions, molecules,
electrons, formula units) has a mass (in grams)
– Happens to be equivalent in # to that of one particle in atomic mass units.
• If a mole of any particle has a mass of 4.02 g mol-1 ,
then a single particle has a mass of 4.02 amu (Atomic
Mass Units)
• If you have a single particle with a mass of 53.03 amu,
then a mole of the same particles will have a mass of
53.03 g.
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The Mole Road Map
Mass
6.022 x 1023
mole
mole
g
g
mole
Mole
mole
22.4 L
.
22.4 L
mole
“Stuff”
Stuff:
mole
6.022 x 1023
.
•Atoms
•Molecules
•Particles
•Electrons
•Ions
•Formula Units
For Gases only:
@ STP
(Standard Temp & Pressure)
(0 oC (273 K) & 1 atm)
Volume
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Comparison
• It is important to note that one mole of atoms contains
6.022 x 1023 atoms.
• One mole of molecules contains 6.022 x 1023 molecules.
• One mole of ions contains 6.022 x 1023 ions.
• NA, therefore can have any of these units (but the SAME
number).
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The Mole Road Map
Mass
6.022 x 1023
mole
mole
g
g
mole
Mole
mole
22.4 L
.
22.4 L
mole
“Stuff”
Stuff:
mole
6.022 x 1023
.
•Atoms
•Molecules
•Particles
•Electrons
•Ions
•Formula Units
For Gases only:
@ STP
(Standard Temp & Pressure)
(0 oC (273 K) & 1 atm)
Volume
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Example: Conversion to Moles
• 1.20 x 1025 molecules of NH3 will be how many moles?
• Solving Process:
– We have molecules and wish to covert to moles.
– One mole equals 6.022 x 1023 molecules.
– 1 mol/6.022 x 1023 molecules
– 1.20 x 1025 molecules NH3
1 mol NH3
.
6.022 x 1023 molecules NH3
= 19.9 mol NH3
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•
Try This!
Convert to the indicated units for each question.
1. 9.03 x 1023 atoms bromine to moles of bromine.
2. 3.5 mol of sodium to atoms of sodium
3. 2 mol of Calcium chloride (CaCl2) to:
a) Formula Units of calcium chloride
b) Number of calcium ions
c) Number of chloride ions
4. 4.75 mol of electrons to # of electrons.
5. 5.08 x 1023 molecules of oxygen to moles of oxygen
molecules. Moles of oxygen to atoms.*
*note: oxygen is a diatomic molecule, O2.
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Mole to Mole Calculations
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Chemical Calculations
• From the reaction involving nitrogen with hydrogen to
form ammonia
N2 + 3H2  2NH3
• The most important part of this BALANCED equation is
that 1 mole of nitrogen reacts with 3 moles of hydrogen
to form 2 moles of Ammonia.
• Based on this information, we can write ratios that relate
moles of reactants to each other and to moles of
products.
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Mole – Mole Calculations
• How many moles of ammonia are produced when
0.60 mol of nitrogen reacts with hydrogen?
N2 + 3H2  2NH3
• The mole ratio of nitrogen to ammonia is 1:2
meaning for every 1 mole of Nitrogen, 2 moles of
ammonia are produced.
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VOLUME
VOLUME
MASS
x molar
mass
x 22.4 L
x molar
mass
x 22.4 L
 22.4 L
 molar
mass
 22.4 L
MASS
 molar
mass
Mole ratio
MOLES
MOLES
Mole ratio
 6.022 x 1023
x 6.022 x 1023
MOLECULES
 6.022 x 1023
x 6.022 x 1023
MOLECULES
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Mass – Mass Calculations
• Calculate the number of grams of NH3 produced by
the reaction of 5.40 g of hydrogen with an excess of
nitrogen.
• Solution:
– Balance the equation (if needed)
– Grams of H2 to moles of H2 to moles of NH3 to
grams of NH3
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Try This!
1. How many grams of acetylene are produced by adding
water to 5.00 g of CaC2?
Acetylene gas (C2H2) is produced by adding water to
calcium carbide.
CaC2 + 2H2O  C2H2 + Ca(OH)2
2. Using the same equation, determine how many moles
of CaC2 are needed to react completely with 49.0g of
water.
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Calculating Molecules of a Product
• How many molecules of oxygen are produced when
29.2 g of water is decomposed by electrolysis according
to this balanced equation?
2H2O (l) → 2H2 (g) + O2 (g)
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Try This!
1. How many molecules of oxygen are produced by the
decomposition of 6.54 g of potassium chlorate
(KClO3)?
2KClO3 (s) → 2KCl (s) + 3O2 (g)
2. The last step in the production of nitric acid is the
reaction of nitrogen dioxide with water. How many
grams of nitrogen dioxide must react with water to
produce 5.00 x 1022 molecules of nitrogen monoxide?
3NO2 (g) + H2O (l) → 2HNO3 (aq) + NO (g)
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Example: Conversion to Moles
• How many moles are represented by 21 g of ammonia,
NH3?
• Solving process:
– We have grams and want to convert to moles.
– Using the atomic mass from the periodic table, we can
calculate the atomic mass of NH3 (17 g mol-1 )
– 1 mole of NH3 has a mass of 17 g.
– We have a sample of 21 g of NH3
– Divide our sample by the mass of 1 mole of NH3
– We end up with 1.24 mol of Ammonia
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Example: Conversion to Mass
• What mass is 20 mol of NH3?
• Solving Process:
– Using your periodic table, determine the molecular
mass of NH3 (17.0 g mol-1)
– 17.0 g for 1 mol of NH3
– 20 mol NH3 x 17.0 g NH3
1 mol NH3
= 340 g of NH3 in 20 mol NH3
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Example: Conversion to Atoms
• How many atoms are in a 10.0 g sample of calcium
metal?
• Solving Process:
– 1 formula mass of calcium metal is 40.1 g mol-1.
– 10 g of calcium is 0.249 mol calcium
(10 g/40.1 g per mol)
– There are 6.02 x 1023 (NA) atoms in one mole
– NA(0.249 mol calcium) = # atoms in 10 g of calcium
metal
= 1.50 x 1023 atoms in 10 g of calcium
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Try This!
1.
2.
3.
4.
5.
6.
7.
8.
Make the following conversions:
7.92 formula units of Water (H2O) to moles.
2.65 mol of methane (CH4) to molecules.
207 g of sodium nitrite (NaNO3) to moles.
85.7 g of gold (III) sulfate (AuSO4) to formula units.
5.97 x 1023 formula units of magnesium bicarbonate
(Mg(HCO3)2)to grams
0.458 moles of sodium thiosulfate (Na2S2O3) to grams
1.99 x 1024 formula units of ammonium phosphate
((NH4)3PO4)to moles.
55 g of Mercury (Hg) to moles.
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