The Mole
Ch.8
(8-1) Mole
• (mol): amt. of substance
– # of atoms in 12g of carbon-12
• Avogadro’s constant: 6.02 x 1023
particles / mol
– Atoms, molecules (covalent) ,
formula units (ionic)
Moles  # of Atoms
• 2.66 mol x 6.02 x 1023 atoms = 1.60 x 1024
1 mol
atoms
• # of Atoms  Moles
• 2.54 x 1024 atoms x
1 mol
= 4.22 mol
6.02 x 1023 atoms
Moles to Molecules
• 1.04 mol x 6.02 x 1023 molecules = 6.26 x 1023
1 mol
molecules
• Formula units to moles
3.49 x 1024 form.u. x
1 mol
= 5.80 mol
6.02 x 1023 form.u.
Moles to Atoms Practice
How many atoms are in 2.5 mol of Si?
1. List known
2.5 mol Si
2. Set up conv. factor (what you’re going to is
on top, what needs to cancel is on bottom)
2.5 mol Si x 6.02 x 1023 atoms
1 mol Si
Moles to Atoms Practice
3. Cancel units, multiply top #’s & divide by
bottom
2.5 mol Si x 6.02 x 1023 atoms = 1.5 x 1024 atoms Si
1 mol Si
Atoms to Moles Practice
Convert 3.01 x 1023 atoms of Si to mols Si
1. List known
3.01 x 1023 atoms Si
2. Set up conv. factor (what you’re going to is
on top, what needs to cancel is on bottom)
3.01 x 1023 atoms Si x
1 mol Si
=
6.02 x 1023 atoms
Atoms to Moles Practice
3. Cancel units, multiply top #’s & divide by
bottom
3.01 x 1023 atoms Si x
1 mol Si
= 0.500 mol Si
6.02 x 1023 atoms Si
Relative Atomic Mass
• Weighted avg. of each isotope’s mass
– Need mass & % abundance
Isotope
Mass
(amu)
%
Decimal
Copper-63
62.94
69.17
.6917
Copper-65
64.93
30.83
.3083
(62.94 amu)(.6917) + (64.93 amu)(.3083) = 63.55 amu
Molar Mass
• Sum of all the atomic masses of a
substance
• Ex: CO2
C = 12.01 g/mol, O = 16 g/mol
12.01 g/mol + (2)(16 g/mol) = 44.01 g/mol
Molar Mass Practice
Calculate the molar mass of Ba(NO3)2
1. List known (at. masses from PT)
Ba = 137.33 g/mol, N = 14.01 g/mol, O = 16 g/mol
2. Calculate # of each atom in the formula
(outer subscript multiplies w/ inner
subscripts)
1 Ba
1Nx2=2N
3Ox2=6O
Molar Mass Practice
3. Multiply each at. mass by the # of atoms &
add together
(1 Ba)(137.33 g/mol) + (2 N)(14.01 g/mol) +
(6 O)(16 g/mol) =
261.35 g/mol
Moles  Mass
• 2.49 mol KF x 58.1 g KF = 145 g KF
1 mol KF
• Grams  Moles
• 110 g KF x 1 mol KF = 1.89 mol KF
58.1 g KF
Moles to Mass Practice
What’s the mass in grams of 3.50 mol Cu?
1. List known
3.50 mol Cu
2. Find at. mass or calculate molar mass
63.55 g Cu
1 mol Cu
Moles to Mass Practice
3. Set up conv. factor (what you’re going to is
on top, what needs to cancel is on bottom)
3.50 mol Cu x 63.55 g Cu
1 mol Cu
4. Cancel units, multiply top #’s, & divide by
bottom
3.50 mol Cu x 63.55 g Cu = 222 g Cu
1 mol Cu
Mass to Moles Practice
Determine the # of moles in 237 g of Cu
1. List known
237 g Cu
2. Find at. mass or calculate molar mass
63.55 g Cu
1 mol Cu
Mass to Moles Practice
3. Set up conv. factor (what you’re going to is
on top, what needs to cancel is on bottom)
237 g Cu x
1 mol Cu =
63.55 g Cu
4. Cancel units, multiply top #’s, & divide by
bottom
237 g Cu x
1 mol Cu = 3.73 mol Cu
63.55 g Cu
Atoms to Grams
1.2 x 1024 atoms B x
1 mol B
x 10.81 g B = 21.55 g B
6.02 x 1023 atoms 1 mol B
(8-2) Percentage Composition
• % by mass of each element in a cmpd
– Ex: H2O = 88.7% O, 11.3% H
H
H
O
O
Determine % Comp. from
Chemical Formula
• % = mass of component x 100
mass of whole
• To verify answers all components %,
should = 100%
% Comp. Practice
Find the % Comp. of Cu2S
1. Find at. masses & multiply by # of atoms
Cu = (63.55 g/mol) (2) = 127.10 g/mol
S = (32.07 g/mol) (1) = 32.07 g/mol
% Comp. Practice
2. Divide each mass by total mass of cmpd &
multiply by 100
127.10 g/mol Cu x 100 = 79.85% Cu
159.17 g/mol Cu2S
32.07 g/mol S
x 100 = 20.15% S
159.17 g/mol Cu2S
3. Add % to verify it’s close to 100%
79.85% + 20.15% = 100%
Hydrates
• To determine % comp. of a hydrate, use
same format as before
• % = mass water x 100
mass of whole
Hydrate % Comp. Practice
Determine the % water in Na2CO3•10H2O
1. Calculate molar mass of water (coef. is
distributed)
(20)(1.01 g/mol) + (10)(16 g/mol) = 180.2 g/mol H2O
2. Divide water mass by the total hydrate mass
180.2 g/mol H2O
x 100 = 62.96%H2O
286.2 g/mol Na2CO3•10H2O
Empirical Formula
• Empirical Formula: simplest ratio
among the elements of a cmpd
– Ex: CH2O: CH2O, C2H4O2, C6H12O6
Determining Emp. Formulas
63.0% Mn, 37.0% O by mass
1. Convert to g assuming you have 100 g
63.0 g Mn, 37.0 g O
2. Convert g to mols using molar mass
63.0 g Mn x 1 mol Mn = 1.15 mol Mn
54.94 g
37.0 g O x 1 mol O = 2.31 mol O
16 g
Emp. Formula (cont.)
3. Divide by smallest amt (round to whole #)
1.15 mol Mn = 1 mol Mn
1.15
2.31 mol O = 2 mol O
1.15
4. Write empirical formula
MnO2
Molecular Formula
• Actual # of atoms in a cmpd
• Determined by emp.formula & molar
mass
Determining a Molecular
Formula
emp. formula = P2O5
molar mass of cmpd = 284 g/mol
1. Find emp. formula molar mass
(2)(30.97 g/mol) + (5)(16 g/mol) = 141.94 g/mol
2. Solve for n (round to nearest whole #)
n = molar mass of cmpd = 284 g/mol = 2
molar mass emp.
141.94 g/mol
Determining a Molecular
Formula
3. Use n to get molecular formula
n (emp.form.) = molecular form.
2 (P2O5) = P4O10
4. To verify answer, calculate molar mass &
compare to value given
(4)(30.97 g/mol) + (10)(16 g/mol) = 284 g/mol