What do I
What do I
know about want to
know about
moles?
moles?
What have I
learned
about moles
today?
 This
1.
2.
3.
4.
5.
6.
unit test contains 6 types of problems:
Molar mass and % composition—must be
able to write a chemical formula
Grams to moles (using molar mass)
Moles to particles/atoms/molecules (using
Avogadro’s #)
Grams to moles to particles (using molar
mass and Avogadro’s #)
Empirical formulas
Molecular formulas
 The
SI (metric) unit used to measure the
amount of a substance
 1 mole is always equal to:
--Molar mass (g/mole)
--Avogadro’s number of particles (6.02 x 1023)
--22. 4 Liters of a GAS (AKA molar volume)
These may be used as conversion factors when
working mole problems.
Define
molar mass AND
Avogadro’s number.
1 mole of
this
substance
table sugar
(sucrose)-C12H22O11
# of
atoms/
molecules/
particles
Mass of 1
mole
(molar
mass)
g/mole
342.0
Sodium
chloride-NaCl
58.5
Cupric sulfate-CuSO4
159.6
Sulfur--S
32.1
Iron--Fe
55.8
Water—H2O
18.0
Volume of
1 mole
 Ex:
Titanium 47.867 = 47.9 g/mole
(this sample contains Avogadro’s number of
atoms)
 Ex:
oxygen 15.999 = 16.0 g/mole
(this sample contains Avogadro’s number of
atoms)

Multiply the # of atoms for each element by the
atomic mass from periodic table
Ex:Magnesium hydroxide
Mg (OH)2
Mg 1(24.3) = 24.3
O
2(16.0) = 32.0
H
2(1.0) = 2.0
58.3 g/mole (this mass also
contains Avogadro’s number of molecules)
 Find
the molar mass of aluminum sulfate
Al 2 (SO4)3
Al 2(27.0) = 54.0
S 3 (32.1)= 96.3
O 12(16.0) = 192.0
342.3 g/mole (this mass also
contains Avogadro’s number of molecules)
 Calculate
trioxide.
the molar mass of diarsenic
 Stannic
carbonate
 Diarsenic
pentasulfide
 Hydrofluoric
acid
 Sucrose
Shows the % of each element that makes up a
compound
 Must be calculate molar mass first.
Ex: magnesium hydroxide Mg (OH)2
Mg 1 x 24.3 =24.3
24.3/58.3 x 100 = 41.7%
O
2 x 16= 32.0
32.0/58.3 x 100 = 54.9%
H
2x1.0 = 2.0
58. 3 g/mole

2.0/58.3
x 100 = 3.4%
 Calculate
the % composition of sulfurous acid
 SAVE
YOUR WRAPPER FOR ENTIRE LAB!!
 DO
NOT START CHEWING UNTIL YOU SIT
DOWN BACK AT YOUR DESK.
 CHECK
BALANCE TO MAKE SURE IT’S OK
BEFORE YOU START!!
 READ
PROBLEM:
 MAKE
A HYOTHESIS:
 PROCEDURE
 DATA
1—4
TABLE 1—5
 After
chewing: (KEEP YOUR SAME BALANCE)
 Procedure
 Data
5—8
table 6—8
 Conclusion
 Questions
1—2
 Calculate
the % composition for a sugar
substitute called SUCRALOSE
C
12( 12.0) = 144.0
 H 19 (1.0) = 19.0
 Cl 3 (35.5) = 106.5
O
8 (16.0) = 128.0
397. 5 g/mole
% C= 144.0 / 397.5 x 100 = 36.2%
% H= 19.0 / 397.5 x 100 = 4.8%
% Cl= 106.5 / 397.5 x 100 = 26.8%
% O= 128.0/ 397.5 x 100 = 32.2%
 1.
Calculate the % composition of carbonic
acid.
 2.
Calculate the % composition of diantimony
trioxide.
 1.
H2CO3
H-2 (1.0) = 2.0
3.2%
C- 1 (12.0) = 12.0
19.4%
O – 3(16.0) = 48.0
77.4%
62.0 g/mole
Sb2O3
Sb- 2(121.8) =243.6
O – 3(16.0) = 48.0
291.6 g/mole
2.
83.5%
16.5%
 Calculate
the molar mass AND % composition
of:
1.
C12H22O11
2.
Cupric sulfate
 C:
12 (12.0) = 144.0
 H: 22 (1.0) = 22.0
 O: 11(16.0) = 176.0
CuSO4
Cu: 1(63.5)= 63.5
S: 1 (32.1) = 32.1
O: 4(16.0) = 64.0
Will need to use unit conversion(cancellation)
and molar mass will be used for the
conversion factor.
Ex: 2.50 grams of hydrochloric acid =
____moles
H Cl
2.50 grams x 1 mole =
0.0685 moles(3sigfigs)
36.5 grams
Ex: 2.50 moles of HCl = __________grams
2.5 moles x 36.5 grams =
1 mole
91 grams
(2 sig figs)
 Particles,
atoms, molecules (synonyms)
 Will
have to use Avogadro’s number as a
conversion factor
 Ex:
5.25 x 1025 atoms of Mg = _____moles
5. 25 x 1025 atoms x 1 mole = 87.2 moles
6.02 x 1023
(3 sig figs)
2.50 moles MgO = _________molecules
2.50 moles x 6.02 x 1023 molecules
1 mole
= 1.50 x 1024 molecules (3 sig figs)

Will need to use both molar mass AND Avogadro’s
number as conversion factors
Will be 2 steps instead of 1 step unit cancellation

Ex: 4.5 grams nitrous acid = __________molecules

HNO2

4.5 g x 1 mole x 6.02 x 1023 molecules =
47 g
1 mole
5.8 x 1022 molecules (2 sig figs)
 Ex:
9.35 x 1021 particles of carbon
tetrabromide = _____grams
C Br4
9.35 x 1021 p x 1 mole x 154 grams =
6.02x1023 p
1 mole
2.39 grams (3 sig figs)
1.
2.
3.
4
5.
6.
7.
8.
0.14 mole (gram to moles)
150 g (moles to grams)
1.1 x 1023 molecules (g to molecules)
5.30 x 1025 molecules (moles to molecules)
0.074 mole( gram to moles)
0.619 g (particles to grams)
0.49 mole (grams to moles)
0.00083 mole (particles to moles)
22 g
2. 1.53 x 1024 molecules
3. 0.014 g
4. 7.2 x 1021 molecules
5. 7.2 x 1023 molecules
6.
2.08 x 106 g
7. 56 g
8. 2.5 g
9. 31 g
10. 0.0029 mole
11. 167 g
BONUS: 3.37 x 1026 atoms
1.
If grams are converted to moles,
use _______________ to convert.
If moles are converted to
molecules, then use
______________to convert.
****Have calculator, periodic
table, and best friend chart****
All members of your group must show their
work on separate sheet of paper.
When you calculate the answer, flip the card
over to find a word.
All of your words will make a sentence.
First group to show all work and finish first,
wins bonus!
Grams-----moles-----particles(atoms or molecules)
1.
55.33 grams of sodium oxide = ____moles
2.
5.00 x 1022 particles of sodium=
_______moles
3.
2.49 x 1026 atoms of acetic acid =
_______________grams
Video sheet Problems
#1—3, 5, 6
Fill in the blanks with multiply/divide OR
molar mass/Avogadro’s number:
***When going from moles to grams,
_______________ by _____________.
***When going from moles to particles,
____________by _________________.
I AM CHECKING 5 HOMEWORK PROBLEMS!!!!
Briefly describe the steps
for calculating an
empirical formula AND
molecular formula.
Define empirical formula
***both labs due today***
1.
Convert 5.03 x 1024 molecules
of phosphoric acid to grams.
2.
Convert 35.75 grams of
dinitrogen monoxide to moles.
3.
Convert 5.0 moles of water to
molecules.
1.
819 grams of H3PO4
2.
0.8125 moles of N20
3.
3.0 x 1024 molecules of water
Tell how to solve for each:
1. G to moles
2. Moles to G
3. Particles to moles
4. Moles to particles
5. G to particles
6. Particles to G
 HYPOTHESIS,
DATA 1—8, CONCLUSION,
QUESTIONS 1—4
 3.
MASS OF SUGAR
(in grams—data #8)--------MOLES
(SUGAR = C12 H22 O11)
 4.
MOLES----------PARTICLES
 Data
Table: mass of empty vial AND
substances mass (make sure you’ve
subtracted empty vial each time!!)
 SHOW
WORK TO GET CREDIT
 CONVERT
GRAMS----------MOLES
 CONVERT MOLES---------PARTICLES
 ANSWER QUESTIONS 1---5, 6 (BONUS)
Convert 25.0 moles of
water to grams.
A molecular formula is a whole
number____________of the
empirical formula.
 Shows
the SIMPLEST, WHOLE NUMBER ratio of
elements in a compound
 Will
give you % composition of compound and
ask you to find the formulas
1. Change % sign to grams (some problems may
already give you grams instead of %)
2. Convert grams to moles (using molar mass)
**round to 4 decimals***
3. Simplify the mole ratio by dividing each one by
the smallest
4. Round to the nearest whole number and assign
these numbers to the appropriate element
A compound is 78.1% Boron and 21.9% H.
Calculate the empirical formula.
78.1 grams B x 1 mole = 7.2315 moles B
10.8 g
21.9 grams H x 1mole =21.9 moles H
1.0 g
7.2315 : 21.9
7.2315
7.2315
1: 3 = BH3
Is a WHOLE NUMBER MULTIPLE of the empirical
formula
You must then first know the empirical formula
1.
You must first have empirical formula (if not,
you will have to calculate it first!!)
2.
Find the molar mass of the empirical formula.
3.
Take the molar mass of the molecular formula
that is given in the problem divided by the
empirical formula’s molar mass. Round this to
a whole number.
Distribute this number to the numbers within
the empirical formula to get the new
molecular formula.
4.
Given the empirical formula of BH3 and the
molecular formula’s molar mass of 27.67
g/mole, find the molecular formula.
Molar mass of BH3 is 13.8 g/mole
27.67 divided by 13.8 = 2
So molecular formula is B2H6
A compound is 4.04 grams of nitrogen and 11.46
grams of oxygen. The molecular molar mass is
108.0 g/mole. Find the molecular formula.
4.04 g N x 1 mole= 0.2886 mole N
14.0 g
11.46 g O x 1 mole =0.7163 mole O
16.0 g
0.2886 : 0.7163
0.2886
0.2886
1: 2 = NO2 empirical molar mass = 46.0 g/mole
108.0 divided by 46.0 = 2
Molecular formula = N2O4