2. The MOLE!!!
People use words to represent specific
quantities all the time
Dozen eggs Pair of gloves Six-Pack
3. The MOLE!!!
1 mole = 6.02214199 x 1023
particles
1 mole = 6.02 x 1023
(the short form for mole is “mol”)
4. The MOLE!!!
1 mol = 6.02 x 1023
Called the Avogadro constant or
Avogadro’s number
(devised through experiments that
determined how many carbon atoms were
present in exactly 12 grams of carbon)
5. The MOLE!!!
1 mole = 602214199000000000000000 molecules
A very big
number!
6. The MOLE!!!
Converting moles to number of particles:
N = n X NA
Number of
particles
Number of
moles
Avogadro’s
number
7. The MOLE!!!
N = n X NA
A sample contains 1.25 mol of NO2.
a) How many molecules are in the
sample?
b) How many atoms are in the
sample?
8. The MOLE!!!
N = n X NA
a) A sample contains 1.25 mol of NO2.
How many molecules are in the
sample?
N = n X NA
N = (1.25mol) X (6.02 x 1023
molecules/mol)
N = 7.52 x 1023
molecules
.: there are 7.52 x 1023
molecules in 1.25 mol of NO2
9. The MOLE!!!
N = n X NA
A sample contains 1.25 mol of NO2.
b) How many atoms are in the sample?
(7.52 x 1023
molecules) x (3 atoms/molecule)
= 2.26 x 1024
atoms
.: there are 2.26 x 1024
atoms in 1.25 mol of NO2
10. The MOLE!!!
n = N
NA
Rearranging the formula…
How many moles are present in a sample of CO2
made up of 5.83 x 1024
molecules?
11. The MOLE!!!
n = N
NA
= (5.83 x 1024
molecules CO2)
(6.02 x 1023
molecules/mol)
How many moles are present in a sample of CO2
made up of 5.83 x 1024
molecules?
= 9.68 mol CO2
.: there are 9.68 mol of CO2 in the sample
12. MOLAR MASS
Molar mass of H =
1.0079 grams per mole
M = molar mass
Molar mass of Li =
6.941 grams per mole
MNa = 22.990g/mol
13. MOLAR MASS
Molar mass of compounds
MBeO = 9.01g/mol + 16.00g/mol
= 25.01g/mol
MCO = 12.01g/mol + 2x16.00g/mol
= 44.01g/mol
2
14. MOLAR MASS
m = n x M
m
n M
mass
Number of
moles
Molar mass
15. MOLAR MASS
A flask contains 0.750 mol of CO2. What mass of
CO2 is in this sample?
m = n x M
= (0.750mol) x (44.01g/mol)
GIVEN: n = 0.750mol
M = 12.01g/mol + 2 x 16.00g/mol
= 44.01g/mol
m = ?
= 33.0g
.: the mass of CO2 is 33.0g
17. MOLAR MASS
How many moles of CH3COOH are in a 23.6g sample?
n = m
M
= (23.6g)/(60.06g/mol CH3COOH)
GIVEN: m = 23.6g
M = (2 x 12.01g/mol C) + (4 x 1.008g/mol H) + (2 x 16.00g/mol O)
= 60.06g
n = ?
= 0.393mol CH3COOH
.: there are 0.393mol of CH3COOH in 23.6g of CH3COOH
19. PERCENTAGE COMPOSITION
Law of Definite Proportions:
The elements in a compound are always present
in the same proportions by mass
Example:
Water = 11.2% hydrogen, 88.8% oxygen
MH2O = 18.016g/mol MH = 1.008g/mol
% of H in H2O = mass of H/mass of water
= (1.008g/mol x 2)/(18.016g/mol)
= 0.112 11.2%
20. PERCENTAGE COMPOSITION
A compound with a mass of 48.72g contains
32.69g of Zn & 16.03g of S. What is the percent
composition of the compound?
%Zn = 32.69g/48.72g
= 0.6710 67.10%
%S = 16.03g/48.72g
= 0.3290 32.90%
.: the percentage
composition for Zn is
67.10% and the
percentage composition
for S is 32.90%
21. EMPIRICAL FORMULA
A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
Molecular formula = Actual formula of compound
Empirical formula = simplest formula (shows the
lowest number ratio of the elements)
Example: Benzene
Molecular formula = C6H6
Empirical formula = CH
22. A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
STEP1: Assume the sample is 100g
STEP2: Find the number of moles of each element
STEP3: Divide all answers from STEP2 by the LOWEST
answer from STEP2
EMPIRICAL FORMULA
23. A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
STEP1: Assume the sample is 100g
So...
C = 81.9g H = 6.12g O = 12.1g
EMPIRICAL FORMULA
24. A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
STEP2: Find the number of moles of each element
C = 81.9g/12.01g/mol H = 6.12g/1.008g/mol
= 6.819mol = 6.0714mol
O = 12.1g/16.00g/mol
= 0.75625mol SMALLEST ANSWER
EMPIRICAL FORMULA
25. A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
STEP3: Divide all answers by the smallest answer
C = 6.819/0.75625 H = 6.0714/0.75625
= 9.0168 = 8.028
O = 0.75625/0.75625
= 1
EMPIRICAL FORMULA
26. A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
= C9H8O
C = 6.819/0.75625 H = 6.0714/0.75625
= 9.0168 = 8.028
O = 0.75625/0.75625
= 1
EMPIRICAL FORMULA
27. The empirical formula of ribose is CH2O. The molar
mass of this compound was determined to be
150g/mol. What is the molecular formula of ribose?
GIVEN: Empirical formula (1 C, 2 H, 1 O)
M = 150g/mol
MOLECULAR FORMULA
STEP1: Determine the molar mass of the empirical
formula
STEP2: Divide the given molar mass by your answer
from STEP1
STEP3: Multiply your empirical formula by your
answer from STEP2
28. The empirical formula of ribose is CH2O. The molar
mass of this compound was determined to be
150g/mol. What is the molecular formula of ribose?
MOLECULAR FORMULA
STEP1: Determine the molar mass of the empirical
formula
12g/mol + 2 x 1.008g/mol + 16g/mol = 30g/mol
STEP2: Divide the given molar mass by your answer
from STEP1
150g/mol / 30g/mol = 5
29. The empirical formula of ribose is CH2O. The molar
mass of this compound was determined to be
150g/mol. What is the molecular formula of ribose?
MOLECULAR FORMULA
C1x5H2x5O1x5 = C5H10O5
STEP3: Multiply your empirical formula by your answer
from STEP2
31. MASS SPECTROMETER
1) Upload sample
2) Same is vapourized
3) Sample is ionized
4) Ions accelerated by electric field
5) Detection to mass-to-charge ratio based on details of motion
6) Ions assorted according to mass-to-charge ratio
33. CARBON-HYDROGEN COMBUSTION ANALYZER
A 1.000g sample of pure compound, containing
only carbon and hydrogen, was combusted in a
carbon-hydrogen combustion analyzer. The
combustion produced 0.6919g of water and 3.338g
of carbon dioxide.
a)Calculate the masses of the hydrogen and the
carbon
b)Find the empirical formula of the compound.
34. CARBON-HYDROGEN COMBUSTION ANALYZER
A 1.000g sample…The combustion produced 0.6919g of
water and 3.338g of carbon dioxide.
a) Calculate the masses of the hydrogen and the carbon
Mass of H = 2.02g/mol H2 x 0.6919g H2O
18.02g/mol H2O
= 0.07756g H2
This gives you the percent
composition of hydrogen in
water
Mass of C = 12.01g/mol C x 3.338g CO2
44.01g/mol CO2
= 0.9109g C
.: there was 0.0775g of H and 0.911g of C
always
0.112
always
0.27289
35. CARBON-HYDROGEN COMBUSTION ANALYZER
A 1.000g sample…The combustion produced 0.6919g of
water and 3.338g of carbon dioxide.
b) Find the empirical formula of the compound
Moles of H = 0.07756g
1.008g/mol
= 0.07694mol
Moles of C = 0.9109g
12.01g/mol
= 0.07584mol
SMALLEST ANSWEREmpirical formula = C0.07694/0.07584H0.07584/0.07584
= C1.0H1.0
= CH
.: the empirical formula is CH
36. HYDRATED SALTS
A 50.0g sample of Ba(OH)2·XH2O contains
27.2g of Ba(OH)2.
a)Calculate the percent, by mass, of water in
the sample
b)Find the value of X
37. HYDRATED SALTS
A 50.0g sample of Ba(OH)2·XH2O contains 27.2g of Ba(OH)2.
a) Calculate the percent, by mass, of water in the sample
= (total mass of sample) – (mass of Ba(OH)2 in sample) x 100%
(total mass of sample)
= 50.0g – 27.2g x 100%
50.0g
= 45.6%
.: the percent by mass of water is 45.6%
38. HYDRATED SALTS
A 50.0g sample of Ba(OH)2·XH2O contains 27.2g of Ba(OH)2.
b) Find the value of X
nH2O = m/M
= (50.0g – 27.2g)
(18.02g/mol) this is the molar mass of H2O
= 1.27mol H2O
0.159/0.159 mol Ba(OH)2·1.27/0.159 mol H2O
Ba(OH)2·8H2O
nBa(OH)2 = m/M
= (27.2g)
(171.3g/mol)
= 0.159 mol Ba(OH)2SMALLEST ANSWER
.: X is 8
