Download Moles Level

The SI(metric) unit to measure
the amount of a substance.

Scientists set out to develop a basic unit of
measurement to convert from atoms to grams
to make obtaining samples easier and more
standard.
 Carbon-12
was used to set the standard, one
mole C = the number of atoms of C in 12 grams
of C-12.

Experiments show that there are 6.02 x 1023
carbon atoms in 12 grams of Carbon-12.

6.02 x 1023 things is a “mole” of any thing!
 Avogadro: scientist
who determined the
number of “things” in a mole.
Substance
Element
Representative Particle
Atom
Covalent Molecule
Molecule
Ionic Compound
Formula Unit
1 mole = molar mass (g)
1 mole = Avogadro’s number of anything else
or
1 mole = 6.02 x 1023 atoms/molecules/Formula units
 1st
choose which conversion factor you
need.
• 1mol = molar mass (g)
• 1 mol = 6.02 x 1023
 2nd
place the number in the problem over
one, and apply conversion factor so units
are diagonal.
 3rd multiply across the top & bottom, then
divide the top by the bottom.
 How
many atoms are in 74.2 moles MgO?
moles
atoms
moles
=


 How
many moles are in 9.352 x 1018
molecules of CO2?
molecules
moles
molecules
=

#1. How many moles are 1.20 x 1025 atoms of phosphorous?
19.9 mols P

#2. How many atoms are in 0.750 mol of Zn?
4.52 x 1023 atoms Zn

#3. How many molecules are in 0.400 mol N2O5?
2.41 x 1023 molecules N2O5

#4. How many moles are contained in 1.20 x 1024 molecules
CO2?
1.99 mols CO2
 Molar
mass is the mass in grams of one
mole of any pure substance.
 The
molar mass of any element is
numerically equivalent to its atomic mass
and has the units g/mol.
 The
molar mass of a compound can be
calculated from its chemical formula
 Determine how many atoms of each
element there are
 Multiply # of atoms by that element’s
atomic mass
 Add all atoms’ atomic masses together
 What
is the molar mass of Mg3(PO4)2?
Mg – 24.305 x 3 = 72.915
P – 30.974 x 2 = 61.948
O – 15.999 x 8 = 127.992
72.915 + 61.948 + 127.992 = 262.855 g/mol
 How
many moles are in 35.24 g NaCl?
grams
moles
grams
Molar mass of NaCl
goes here!
=

Molar mass of CH3Cl
goes here!

Molar mass of Ca3N2
goes here!
 What
is the mass of 0.256 moles of CO2?
Molar mass of CO2 goes
here!
moles
Grams
=
moles
#1. Find the mass in grams of 3.32 mol of K.
1.30 x 102 g K
#2. Find the mass in grams of 15.0 mol of H2SO4.
1470 g H2SO4
#3. Find the number of moles in 187 g of aluminum.
6.93 mol Al
#4. Find the number of moles in 11.0 g of methane (CH4).
0.686 mol CH4
#5. Find the mass in grams of 0.423 moles of calcium fluoride.
33.0 g CaF2
#6. Find the number of moles in 182.15 grams of lead (II) nitrate.
0.5500 mols Pb(NO3)2
 How
do you convert between moles and
particles?
• Use the conversion factor 1 mole = 6.02 x 1023
particles
 How
do you convert between moles and
mass?
• Use the conversion factor 1 mole = molar mass (g)
 How
would we convert between particles
and mass?
• Do a 2 step conversion using both conversion factors
 How
many grams are in 4.3 x 1024 formula
units of AlCl3?
 How
many atoms are in 76.5 g FeO?
 How
many molecules are in 62.21 grams
of ammonia, NH3?
 How
many grams are in 2.81 x 1021
formula unit of magnesium oxide?




#1. Calculate the number of molecules present in 4.29
g of nitrogen dioxide.
5.62 x 1022 molecules NO2
#2. Calculate the number of moles of sulfur atoms
present in 2.01g of sodium sulfide.
0.0258 moles S atoms
#3. Calculate the mass in grams of 2.49 x 1020 carbon
dioxide molecules.
0.0182 g CO2
#4. Calculate the grams of carbon in 12.2 mol of
sucrose, C12H22O11.
1760 g C
 The volume of a gas is usually measured
at 0oC and 1 atmosphere of pressure.
This is called standard temperature and
pressure (STP).
 At STP, one mole of any gas has a volume
of 22.4 L.
• 22.4 L is called the molar volume of a gas.
 Example
1: How many moles are in 4.13
L of sulfur dioxide gas?
 Example
2: How many liters of CO2 are
found in 0.1569 moles?
#1. What is the volume (liters) at STP of
0.960 mol of methane, CH4?
#2. At STP, how many moles are in 0.542 mL
of neon gas?
 The
percent each element makes up, by
mass, in a compound
• percent = (part divided by total) x 100
• Can be used as the number of grams of the
element per 100 grams of the compound
C2H6
Step 1: Mass of Compound
Determine molar mass of the
compound
Step 2: Mass of Elements
Find the individual masses of
each element in the compound
Step 3: Divide
Divide the mass of each element
by the total mass of compound.
Multiply by 100%.
12.011 g x 2 = 24.022 g
1.0079 g x 6 = 6.0474 g
30.069 g
24.022 g of C in molecule
6.0474 g of H in molecule
CH3Cl
Step 1: Mass of Compound
Determine molar mass of the
compound
Step 2: Mass of Elements
Find the individual masses of
each element in the compound
Step 3: Divide
Divide the mass of each element
by the total mass of compound.
Multiply by 100%.
12.011 g x 1 = 12.011 g
1.0079 g x 3 = 3.0237 g
35.453 g x 1 = 35.453 g
50.488 g
12.011 g C
3.0237 g H
35.453 g Cl
CH3Cl
Step 1: Mass of Compound
Determine molar mass of the
compound
Step 2: Mass of Elements
Find the individual masses of
each element in the compound
Step 3: Divide
Divide the mass of each element
by the total mass of compound.
Multiply by 100%.
12.011 g x 1 = 12.011 g
1.0079 g x 3 = 3.0237 g
35.453 g x 1 = 35.453 g
50.488 g
12.011 g C
3.0237 g H
35.453 g Cl
NaHSO4
Step 1: Mass of Compound
Determine molar mass of the
compound
Step 2: Mass of Elements
Find the individual masses of
each element in the compound
Step 3: Divide
Divide the mass of each element
by the total mass of compound.
Multiply by 100%.
22.99 g x 1 = 22.99 g
1.0079 g x 1 = 1.0079 g
32.06 g x 1 = 32.06 g
16.00 g x 4 = 64.00 g
120.06 g
22.99 g Na
1.0079 g H
32.06 g S
64.00 g O
NaHSO4
Step 1: Mass of Compound
Determine molar mass of the
compound
Step 2: Mass of Elements
Find the individual masses of
each element in the compound
Step 3: Divide
Divide the mass of each element
by the total mass of compound.
Multiply by 100%.
22.99 g x 1 = 22.99 g
1.0079 g x 1 = 1.0079 g
32.06 g x 1 = 32.06 g
16.00 g x 4 = 64.00 g
120.06 g
22.99 g Na
1.0079 g H
32.06 g S
64.00 g O



What is the percent composition of calcium acetate
(Ca(C2H3O2)2)?
25.34% Ca, 30.37%C, 3.823% H, 40.46% O
Aspartame, C14H18N2O5, is the artificial sweetener that
is 160 times sweeter than table sugar when dissolved in
water. Calculate the percent composition of
aspartame.
57.13% C, 6.16% H, 9.52% N, 27.18% O
What is the percent composition of magnesium
phosphate?
27.73% Mg, 23.57% P, 48.70% O
 Empirical
formula - lowest whole number ratio
of the elements in a compound
• It may or may not be the same as the molecular
formula.
 Ex. H2O is both the empirical & molecular formula
 H2O2 is a molecular formula while HO is the
empirical formula for H2O2.
 Which
of the following are empirical?
N2O4
P2O5
C10H22
NO3
C2H6O
C2H6
A compound is 79.3% C and 20.2 % H. Find its
empirical formula.
Step 1: Percent to Grams
If you have percentage composition,
assume 100 g and turn percent into
grams
Step 2: Grams to Moles
For each element in the compound,
convert from grams to moles by using
molar mass.
79.3 g C
20.2 g H
A compound is 79.3% C and 20.2 % H. Find its
empirical formula.
Step 3: Divide by Small
Look at the number of moles for each
element. Divide each number by the
smallest number of moles.
Step 4: Multiply to get Whole
If you don’t have whole number
ratios, multiply to get whole
numbers. Need to multiply each by
the same number!
1 mol C
3 mol H
A compound is 79.3% C and 20.2 % H. Find its
empirical formula.
Step 5: Write the formula!
Use the mole ratios as the subscripts
for each element in compound.
CH3
An oxide of Al is formed by the complete
reaction of 4.151 g of Al with 3.692 g of oxygen.
Calculate the empirical formula for this
compound.
Step 1: Percent to Grams
If you have percentage composition,
assume 100 g and turn percent into
grams
Step 2: Grams to Moles
For each element in the compound,
convert from grams to moles by using
molar mass.
4.151 g Al
3.692 g O
An oxide of Al is formed by the complete
reaction of 4.151 g of Al with 3.692 g of oxygen.
Calculate the empirical formula for this
compound.
Step 3: Divide by Small
Look at the number of moles for each
element. Divide each number by the
smallest number of moles.
Step 4: Multiply to get Whole
If you don’t have whole number
ratios, multiply to get whole
numbers. Need to multiply each by
the same number!
1 mol Al x 2 = 2 mol Al
1.5 mol O x 2 = 3 mol O
An oxide of Al is formed by the complete
reaction of 4.151 g of Al with 3.692 g of oxygen.
Calculate the empirical formula for this
compound.
Step 5: Write the formula!
Use the mole ratios as the subscripts
for each element in compound.
Al2O3