Download The Mole - Firelands Local Schools

THE MOLE
Chapter 11
I.
Introduction
A. Measuring Matter
1. Many items require exact counting units
a.
b.
c.
d.
e.
One dozen = 12 items
One baker’s dozen = 13 items
A pair = 2 items
One gross = 12 dozen = 144 items
One ream = 500 items
Chemists also have a unit that is used to count atoms or
molecules!
2.
a.
b.
c.
d.
A mole is the SI unit that describes the amount of a substance
One mole = 6.02 x 1023 objects
“Objects” can represent number of atoms (ex: Ca), molecules (ex:
CO2), formula units (ex: KCl), or number of any type of particle. We
refer to any of these as “particles”
1 mole = 6.02 x 1023 atoms (molecules) – this is called Avogadro’s
Number
B. Avogadro’s Number
1. The mole is a HUGE unit – we use such a large number because
atoms are tiny
2. The number 6.02 x 1023 specifically is a standard chosen by
scientists: carbon!
3. A mole is defined as the number of atoms contained in exactly
12 grams of pure carbon-12
4. The number of atoms contained in exactly 12 grams of carbon12 was determined to be 6.02214 x 1023 atoms… rounds to 6.02
x 1023!
5. So, this means that…
a.
b.
c.
d.
6.
1 mole of C atoms = 6.02 x 1023 atoms
1 mole of O atoms = 6.02 x 1023 atoms
1 mole of Cl atoms = 6.02 x 1023 atoms
1 mole of anything = 6.02 x 1023 anything
Remember, a mole is NOT a weight (mass)! It is a COUNT
C. Converting Between Moles and Particles
1. To find the number of particles (atoms, ions, molecules, etc.) in a
given number of moles, use the following conversion factor & Tchart setup:
# moles given
6.02 x 1023 particles
1 mole
2.
To find the number of moles in a given number of particles
(atoms, ions, molecules, etc.), use the following conversion
factor & T-chart setup:
# particles given
1 mole
6.02 x 1023 particles
3.
In these problems, the compound or element does not matter –
it’s just about the numbers!
4.
Examples:
a. Determine the number of particles in 2.33 mol of hydrogen gas (H2).
6.02 x 1023 particles
2.33 mol
1 mol
# particles H2 = 1.40 x 1024 particles
b. Find the number of moles contained in 5.30 x 1025 carbon monoxide
(CO) molecules.
5.30 x 1025 molecules
1 mole
6.02 x 1023 molecules
# moles CO = 88.0 mol
Homework
• p. 311 #1-3
• p. 312 #4-10
II. Mass & the Mole
A. We cannot count individual atoms, so we use their
masses to find the exact number of atoms in a sample
1.
Example: A candy shop keeper knows that 10 gumballs have a
mass of 21.4 g. How can we measure out 200 gumballs without
actually counting the individual gumballs?
2.
We can use this same approach to determine the number of
particles in a sample
B. Molar mass: mass in grams of one mole of a pure
substance; measured in grams/mole (g/mol)
1.
2.
For any element, atomic mass (in amu) = molar mass (in grams)
Found in upper left-hand corner of the element symbol on your
periodic table (use 3 sig figs if <100, 4 sig figs if > 100)
C. Conversions
1. Given grams, need moles:
# grams
1 mol
molar mass (in grams)
a.
Example: Convert 153 g of silver to moles.
2.
Given moles, need grams:
# moles
molar mass (in grams)
1 mol
a. Example: Convert 7.32 mol of gold to grams.
3.
Converting from grams to particles or particles to grams requires
two steps, converting to moles in between.
a. Example: Find the number of atoms contained in 34.2 g of pure nickel.
34.2 g Ni
1 mol Ni
6.02 x 1023 atoms
58.7 g Ni
1 mol Ni
b. Example: Find the mass of 1.54 x 1024 atoms of titanium.
1.54 x 1024 atoms Ti
1 mol Ti
47.9 g Ti
6.02 x 1023 atoms
1 mol Ti
Homework
• p. 316 #11-12
• p. 318 #13-14
• p. 319 #15-19
• p. 876 #1-14 (under Chapter 11)
III. Moles of Compounds
A. Chemical Formulas & the Mole
1. In a compound, the number of moles of an element in the
compound is equal to the subscript on that element
a. Example: find the number of moles of oxygen atoms in 3.25 moles of
CO2 gas.
3.25 mol CO2
2 mol O
1 mol CO2
b. This also applies to ions of a compound in aqueous solution
c. Example: find the number of moles of hydrogen ions in 1.72 mol of
phosphoric acid (H3PO4)
1.72 mol H3PO4
3 mol H+
1 mol H3PO4
2.
Molar mass of compounds
a. Also referred to as molecular
mass or formula mass
b. The molar mass of the compound
is the sum of the molar masses of
each element times the number of
atoms of that element in the
compound
i.
Example: calculate the molar mass
of CaCl2
1 Ca: 1 x 40.1 g = 40.1 g
2 Cl: 2 x 35.5 g = 71.0 g
1 mol = 111.1 g
ii.
Example 2: find the molar mass of C6H12O6
6 C: 6 x 12.0 g =
72.0 g
12 H: 12 x 1.01 g = 12.1 g
6 O: 6 x 16.0 g =
96.0 g
180.1 g = 1 mol
iii. Example 3: find the molar mass of Na2SO4
2 Na: 2 x 23.0 g = 46.0 g
1 S: 1 x 32.1 g =
32.1 g
4 O: 4 x 16.0 g =
64.0 g
142.1 g = 1 mol Na2SO4
iv. Example 4: find the molar mass of Al(NO3)3
1 Al: 1 x 27.0 g = 27.0 g
3 N: 3 x 14.0 g = 42.0 g
9 O: 9 x 16.0 g = 144 g
213 g = 1 mol Al(NO3)3
Converting moles of a compound to mass
3.
a. To find the mass of a certain number of moles of a compound:
i. Calculate the molar mass of the compound
ii. Multiply the number of moles by the molar mass of the compound in a Tchart
b.
Example: convert 12.5 moles of water to grams
i.
ii.
Molar mass of water = (2 x 1.01 g) + 16.0 g = 18.02 g = 18.0 g
12.5 mol H2O
18.0 g
1 mol H2O
c.
Example 2: convert 9.48 moles of Sn(CO3)2 to grams
i.
ii.
Molar mass = 118.7 g + (2 x 12.0 g) + (6 x 16.0 g) = 238.7 g
9.48 mol Sn(CO3)2
238.7 g
1 mol Sn(CO3)2
Homework
• p. 321 #20-24
• p. 322 #25-26
• p. 323 #27-29
Converting mass of a compound to moles
4.
a. To find the number of moles of a certain mass of a compound:
i. Calculate the molar mass of the compound
ii. Multiply the mass by the reciprocal of the molar mass in a T-chart
b.
Example: find the number of moles in 132 g of Ca(OH)2 powder.
i.
ii.
Molar mass of Ca(OH)2 = 40.1 g + (2 x 1.01 g) + (2 x 16.0 g) = 74.12 g = 74. 1 g
132 g Ca(OH)2
1 mol Ca(OH)2
74.1 g
Converting mass of a compound to particles
5.
a. To find the number of particles of a certain mass of a compound:
i. Calculate the molar mass of the compound
ii. In a T-chart, multiply the mass given by the reciprocal of the molar mass, then
multiply by Avogadro’s number
Example: find the number of molecules in a 4.00 g packet of sucrose
(C12H22O11)
b.
i.
Molar mass of C12H22O11 = (12 x 12.01 g) + (22 x 1.01 g) + (11 x 16.0 g)
= 342.2 g
4 g C12H22O11
= 7.04 x 1021 molecules
1 mol C12H22O11
6.02 x 1023 molecules
342 g C12H22O11
1 mol
Example 2: using the answer to example 1, find the number of atoms of
carbon in one packet of sucrose
c.
i.
ii.
We know that 1 packet of sucrose is 7.04 x 1021 molecules, and that there are 12
C atoms in each molecule of sucrose
So we can determine # of C atoms by:
7.04 x 1021 molecules C12H22O11
12 atoms C
1 molecule C12H22O11
= 8.45 x 1022 atoms C
d.
Example 3: find the number of calcium & chlorine ions in solution when
15.2 g of CaCl2 is dissolved in water.
i.
ii.
iii.
Step 1: Molar mass of CaCl2
Step 2: Determine # of formula units in 15.2 g of CaCl2 (see example 1)
Step 3: Determine # of Ca2+ & Cl- ions by multiplying the # of formula units by the
subscripts of each ion
Molar mass CaCl2: 40.1 g + (2 x 35.5 g) = 111.1 g = 111 g
15.2 g CaCl2
1 mol CaCl2
6.02 x 1023 molecules
111 g CaCl2
1 mol
= 8.25 x 1022 formula units CaCl2
# of Ca2+ ions:
8.25 x 1022 formula units CaCl2
1 atom Ca
1 formula unit CaCl2
= 8.25 x 1022 ions Ca2+
# of Cl- ions:
8.25 x 1022 formula units CaCl2
2 atoms Cl
1 formula unit CaCl2
= 1.65 x 1023 ions Cl-
Homework
• p. 324 #30
• p. 326 #31-35
• p. 327 #36-41
**Quiz on block day over everything we’ve covered so far**
IV. Empirical & Molecular Formulas
A. Percent Composition
1. Definition: percent by mass of each element in a compound
2. Often used to identify the formula of compounds
3. Always adds up to 100%! (may be .01 - .1 off depending on sig figs)
4. Calculation to determine an element’s percent by mass:
mass of element
mass of compound
5.
x 100% = percent by mass
Can also be used to determine a polyatomic ion’s percent by mass:
mass of polyatomic ion
mass of compound
x 100% = percent by mass
6.
Examples:
a. A compound has a total mass of 200 g. 110 g are of element X, and 90
are of element Y. Calculate each element’s percent by mass.
110 g X
200 g
x 100% = 55% X
90 g Y
200 g
x 100% = 45% Y
b. Calculate the percent by mass of each element in water.
i. To get total mass, calculate molar mass of compound to 4 sig figs:
H: 2 x 1.008 = 2.016
O: 16.00
2.016 + 16.00 = 18.016 = 18.02 g
ii. Use molar mass values to plug into the equation:
2.016 g H
18.02 g H2O
16.0 g O
x 100% = 11.18% H
18.02 g H2O
x 100% = 88.79% O
c. Calculate the percent by mass of each element in Ce2(CO3)3
B. Empirical Formula
1. Percent composition can be used to find the empirical formula of
a compound
a. Definition: formula with the smallest whole number ratio of the number
of moles of the element
b. The mole ratio provides the subscripts for the empirical formula
c. Sometimes the empirical formula is the same as the actual formula
(ex: H2O), but sometimes they differ (ex: H2O2’s empirical formula is
HO)
2.
Steps to determine a compound’s empirical formula:
a. If given percent composition, assume the compound is 100g. If given
grams, skip this step.
b. Convert grams to moles for each element using molar mass to 4 sig
figs
c. Using the #s of moles calculated above, find the mole ratio of one
element to the other elements.
i.
ii.
iii.
To do so, divide all #s of moles by the lowest number of moles & round to
whole numbers.
If these do not round nicely to whole #s (see example 2 below), multiply all
the values by the smallest # that will make all values whole #s.
These numbers will become the subscripts on each element
d. Write the empirical formula -- you did it!
3.
Examples:
a. Find the empirical formula for a compound containing 40.05% sulfur
and 59.95% oxygen.
40.05 g S
1 mol
32.07 g S
59.95 g O
1 mol
16.00 g O
empirical formula: SO3
= 1.25 mol S
1.25 mol
= 3.75 mol O
1.25 mol
=1
=3
b. Find the empirical formula for a compound containing 48.64% carbon,
8.16% hydrogen, and 43.20% oxygen.
48.64 g C
1 mol
12.01 g C
8.16 g H
= 4.05 mol C
2.70 mol
= 1.5
x2=3
=3
x2=6
=1
x2=2
1 mol
1.008 g H
= 8.10 mol H
2.70 mol
43.20 g O
1 mol
16.00 g O
empirical formula: C3H6O2
= 2.70 mol O
2.70 mol
b. Find the empirical formula for a compound containing 40.0% carbon,
6.70% hydrogen, and 53.3% oxygen.
40.0 g C
1 mol
12.01 g C
6.70 g H
= 3.33 mol C
3.33 mol
=1
1 mol
1.008 g H
= 6.65 mol H
=2
3.33 mol
53.3 g O
1 mol
16.00 g O
empirical formula: CH2O
= 3.33 mol O
3.33 mol
=1
Homework
• p. 331 #42-45
• p. 333 #46-50
• Do Pre-Lab on a separate sheet of paper!
• These are the reactions that will occur in lab:
• KCl (aq) + AgNO3 (aq) → KNO3 (aq) + AgCl (s)
• NaCl (aq) + AgNO3 (aq) → NaNO3 (aq) + AgCl (s)
• BaCl2 (aq) + AgNO3 (aq) → Ba(NO3) 2 (aq) + AgCl (s)
• AlCl3 (aq) + AgNO3 (aq) → Al(NO3)3 (aq) + AgCl (s)
• In every reaction, AgCl will precipitate
• Dichlorofluorescein turns pink when all of the chloride has
reacted – but we will use sodium chromate, which turns a gross
pink/red/orange color (looks like nasty tomato soup)
• The # of drops that it takes for all of the chloride to react will help
determine the cation to anion ratio
• Theoretically, the ratio for KCl will be 1:1
• The same volume of BaCl2 should take twice as many drops for all of the
chloride to react since there is twice as many chloride ions – that would give
it a cation:anion ratio of 1:2. (this will help you with Hypothesis #2)
• Hypothesis #1 is asking you about how many drops of AgNO3
you think it will take for the chloride to completely react. Which
solution do you think will need the most AgNO3? The least?
Say that in your hypothesis.
• Ms. Sparks’ numbers, using 1 mL of each solution & 3-4
drops indicator:
Sample
T1 Drops
T2 Drops
Ave. Drops
Ratio
KCl
29
28
28.5
1:1
NaCl
38
33
35.5
1:1.25 (4:6)
BaCl2
55
47
51
1:1.8 (5:9)
AlCl3
57
64
60.5
1:2.1 (10:21)
• What numbers should have been:
Sample
T1 Drops T2 Drops
Ave. Drops
Ratio
KCl
21
19
20
1:1
NaCl
19
21
20
1:1
BaCl2
38
40
39
1:2
AlCl3
59
61
60
1:3
• Possible causes for error:
• Every group should list the indicator as a cause for error – it
worked, but inconsistently, possibly because the salt that I made it
from was very old
• Type of pipette used – could affect drop sized & number of drops
•
•
•
•
needle-tip pipette
transfer pipette
Did you measure exactly 1 mL of each chloride solution?
Did the same person do the dropping every time?
Any possible contamination?
Was the pipette held vertically above the test tube? This could
affect size of drops
C. Molecular Formula
1. The empirical formula is not always the same as the molecular
formula! Compounds with the same empirical formula but
different molecular formulas can have very different properties
2. Definition: actual number of atoms of each element in one
molecule or formula unit of a substance
3. Molecular formula can be determined using a compound’s
empirical formula & molar mass
Steps to find molecular formula:
4.
a.
b.
c.
d.
Find mass of empirical formula (use values for molar mass of each
element)
Find the number of times the empirical formula mass will divide into
the molar mass & round to nearest whole number
Multiply the subscripts in the empirical formula by the number found
above
Write molecular formula!
Examples:
5.
a.
•
•
•
•
Find the molecular formula for a compound with the empirical formula
CH2O and a molar mass of 180.1 g/mol.
Mass of empirical formula:
1 C + 2 H + 1 O = 12.01 + (2 x 1.008) + 16.00 = 30.03 g
# of times empirical mass goes into molar mass:
180.1 g / 30.03 g = 5.997 = 6
Multiply subscripts in the empirical formula by number found above
1Cx6=6
2 H x 6 = 12
1Ox6=6
Molecular formula: C6H12O6
b.
Succinic acid is 40.68% carbon, 5.08% hydrogen, and 54.24%
oxygen with a molar mass of 118.1 g/mol. What are the empirical and
molecular formulas?
• Find empirical formula
40.68 g C
1 mol
12.01 g C
5.08 g H
1 mol
1.008 g H
54.24 g O
= 3.387 mol C
/ 3.387 = 1
x2=2
= 5.040 mol H
/ 3.387 = 1.5 x 2 = 3
= 3.390 mol O
/ 3.387 = 1
1 mol
16.00 g O
empirical formula: C2H3O2
x2=2
• Mass of empirical formula:
2 C + 3 H + 2 O = (2 x 12.01) + (3 x 1.008) + (2 x 16.00) = 59.04 g
• # of times empirical mass goes into molar mass:
118.1 g / 59.04 g = 2.0003 = 2
• Multiply subscripts in the empirical formula by number found above
2Cx2=4
3Hx2=6
2Ox2=4
• Molecular formula: C4H6O4
c.
•
•
•
•
•
Determine the empirical & molecular formulas for a compound that is
26.7% P, 12.1% N, and 61.2% Cl with a molar mass of 695.3 g/mol.
Find empirical formula
Mass of empirical formula
# of times empirical mass goes into molar mass
Multiply subscripts in the empirical formula by number found above
Molecular formula
Homework
• p. 337 #58-62
• Worksheet: side 1
V. Hydrates
A. Definition: solid crystals that have a specific number of
water molecules bound to its atoms
1. Anhydrous/anhydrate: substance remaining when the water is removed
from a hydrate, usually through heating
2. Anhydrous forms are often used as drying agents because they can
absorb water easily
3. Some are used to store solar energy
B. Naming Hydrates
1. Formulas of hydrates contain the formula of the crystal followed
by a dot and the number of water molecules contained for each
formula unit
a.
2.
Examples:
Na2CO3 ᐧ 10H2O contains 10 water molecules per formula unit
NiSO3 ᐧ 6H2O contains 6
Prefix
#
Prefix
To name hydrates:
a. Name compound
b. Choose appropriate prefix
#
mono
1
hexa
6
di
2
hepta
7
3
octa
8
4
nona
9
5
deca
10
for the number of water molecules
tri
c. Combine prefix with the word “-hydrate” tetra
d. Ex: CaSO3 ᐧ 2H2O
penta
calcium sulfite dihydrate
C. Finding Formulas of Hydrates
1. In a lab environment, a hydrated compound will be massed,
heated to drive off the water, then massed again to determine the
mass of the water.
a. Usually both of these masses will be given
b. If you are given percent composition instead of the masses, assume
the sample is 100 g
2. Convert grams of each component (compound & water) to moles
3. Find the ratio of compound to water by dividing both amounts of
moles by the number of moles of the compound (1:n where n is
the # of water molecules per formula unit)
4. Write the formula of the hydrate with n as the coefficient on H2O!
5.
Examples:
a. A 10.4 g sample of hydrated barium iodide (BaI2) is heated strongly to
drive off the water. The dry sample has a mass of 9.52 g. Find the
formula of the hydrate.
• Find mass of water: 10.4 g – 9.52 g = 0.88 g H2O
• Convert masses to moles:
9.52 g BaI2
1 mol
391.1 g
0.88 g H2O
1 mol
18.02 g
•
•
•
•
= 0.02434 mol BaI2 / 0.02434 = 1
= 0.04883 mol H2O / 0.02434 = 2
Divide both #s by the # of moles of compound – 0.02434 in this case
Ratio is 1:2, so 2 becomes the coefficient on H2O
Formula: BaI2 ᐧ 2H2O
Name: barium iodide dihydrate
b. Find the formula of a hydrate with the following percentages by mass: 23%
zinc; 11% sulfur; 22% oxygen; 44% water.
• Same setup as finding empirical formula; do NOT break up water into H & O
23.0 g Zn
1 mol
/ 0.3431 = 1
= 0.3517
65.39 g Zn
11.0 g S
1 mol
32.06 g S
22.0 g O
1 mol
16.00 g O
44.0 g H2O
= 0.3431
/ 0.3431 = 1
= 1.375
/ 0.3431 = 4
= 2.442
/ 0.3431 = 7
1 mol
18.02 g H2O
formula: ZnSO4 ᐧ 7H2O
name: zinc sulfate heptahydrate
Homework
• p. 340 #63-64
• p. 341 #65-69
• Composition of Hydrates Worksheet (back of the one
about molecular formulas)
Review
• p. 345: Vocabulary (all) -- you already did this!
• p. 346: Concept Mapping #70 -- DRAW IT!
• p. 346: Mastering Concepts #71-88
• p. 346: Mastering Problems #89-156
• HONORS ONLY p. 349: Thinking Critically #170-173
• p. 351: Standardized Test Practice #1-10