When showing the amounts of each element present
in a compound using percentage by mass
% mass =
grams of element
x 100%
total grams of compound
Part divided by total
Multiply by 100
Round till whole.
Please Make Rice!
Purple Monkeys Rule!
Px100
T
Percent Composition of CaCl2.
Ca = 40.1 g
Cl = 35.5 x 2 = 71.0 g
Total = 111.1 g
Ca = 40.1g/111.1 x 100 = 36%
Cl = 71.0g/111.1 x 100 = 64%
Exactly 9.63g of Mg combines with 3.48
g of Nitrogen.
9.63 g x 100 = 73% Magnesium
13. 11g
100% – 73% = 27% Nitrogen
(Or to check do 3.48/13.11 x 100)


Number 2
Exactly 29.0g of Argon combine with 4.30g
of Sulfur.
Numbers 3-6 on your Percent
Composition Practice worksheet.
Catalysts:
◦ Molar Mass Practice WS
◦ Moles Practice WS #2
◦ Moles to Particles WS
If you do not fill out the table of contents or staple
your work in the correct order, 5 points will be taken
off your score.
 1/21/14


Empirical Formulas
EQ: What is an empirical formula?
Calculate the percent composition of each
element in Na2O.
 Na2O


= 2(23.0) + 16.0 =
46.0 + 16.0 = 62.0 g/mol
The formulas for compounds can be
expressed as an empirical formula and as a
molecular(true) formula.
Empirical
CH
CH
CO2
CH2O
Molecular (true) Name
C2H2
acetylene
C6H6
benzene
CO2
carbon dioxide
C5H10O5
ribose
9
Write your own one-sentence definition for
each of the following:
Empirical formula
Molecular formula
10

An empirical formula represents the simplest

The molecular formula is the true or actual
whole number ratio of the atoms in a
compound.
ratio of the atoms in a compound.
11
A. What is the empirical formula for C4H8?
1) C2H4
2) CH2
3) CH
B. What is the empirical formula for C8H14?
1) C4H7
2) C6H12
3) C8H14
C. What is a molecular formula for CH2O?
1) CH2O
2) C2H4O2
3) C3H6O3
12
Pretend that you have a 100 gram
sample of the compound.
Step 1: That is, change the % to grams.


Ex: Suppose a compound whose
molecular mass is 695g is analyzed to
contain 26.7% phosphorus, 12.1%
nitrogen, and 61.2% chlorine.

◦
◦
◦
Phosphorus = 26.7g
Nitrogen = 12.1g
Chlorine = 61.2g




Convert the grams to mols for each element.
26.7 g P 1 mol P
31.0 g P
12.1 g N 1 mol N
14.0 g N
61.2 g Cl 1 mol Cl
35.5 g Cl
= 0.862 mols P
= 0.864 mols N
= 1.726 mols Cl

Because!! Divide each number by the
least number of moles.

0.862 mols P/0.862mols = 1

0.864 mols N/0.862 mols = 1

1.726 mols Cl/0.862/mols = 2


Write the number of mols as a
subscript in a chemical formula.
Multiply the result to get rid of any
fractions.
P

= 1, N = 1, Cl = 2
Therefore the Empirical formula is
PNCl2

Step
Step
Step
Step

Empirical Formula:



1:
2:
3:
4:
Change the % to grams.
Divide by molar mass
Divide by the least number answer.
Write answer as a subscript.
◦ Change, Divide, Divide, Write


Mole Test: Tuesday, Jan. 28
Mini Quiz Friday (one question for each of the
following)
◦ Percent composition, empirical formula, molar mass

A compound is 43.64 % P and 56.36 % O.
What is the empirical formula?
1mol P
43.64 g P 
 1.4 mole P
31.0 g P
1 mol O
56.36 g O 
 3.5 mole O
16.0 g O
1.4 mol P
 1 mol P
1.4 mol



3.5 mol O
 2.5 mol O
1.4 mol
The ratio is P1O2.5
Get rid of decimals by multiplying to get a
whole number. Multiply by 2.
The empirical formula is P2O5.

Do problems 2, 3, and 4 on your worksheet.
From empirical to molecular formula



The empirical is the formula with the lowest
whole number ratios between elements.
The molecular formula is a multiple of the
empirical formula. It represents the actual
molecule.
Examples:
CH3 is an empirical formula, C2H6 is a molecular
formula
(CH3 x 2 = C2H6)
CH2O is an empirical formula, C6H12O6 is a molecular
formula (CH2O x 6 = C6H12O6)


Since the empirical formula is the lowest
ratio, the actual molecule would weigh the
same or more.
Divide the actual molar mass by the molar
mass of the empirical formula to determine
the factor.
Step 1: Find the molar mass of the empirical
formula.
Step2: Divide the actual molar mass by the
molar mass of the empirical formula.
Step 3: Multiply the empirical formula
subscripts by your answer from step 2.


The empirical formula of a substance is CH5N.
The actual molar mass of the substance is 62
g/mol. What is its molecular formula?
CH5N = 12.0 + 5(1.0) + 14.0 = 31.0 g/mol
62.0 g/mol
2
31.0 g/mol


Multiply the empirical formula x 2.
The molecular formula is C2H10N2.

For Empirical Formula and molecular practice,
work on completing the front side of the
worksheet provided today.

Quiz tomorrow!!!
◦ Three Questions
 Percent Composition
 Empirical Formula
 Molecular Formula




1/23/14
EQ: What do I know about moles?
Pull out your worksheet from yesterday. Your
catalyst is to complete #6.
You may work it on the worksheet, and we
will give your stamp on the catalyst sheet
when you are done.

#2-10
◦ You will receive a classwork stamp for this
assignment. IT IS DUE TODAY.
After you have completed this, work on the
study guide for next week’s test and
tomorrow’s quiz.


1.
1/24/14
EQ: What do I know about moles?
What is the percent composition of KNO3?



You will need a calculator and something to
write with.
A periodic table will be provided.
When you finish your quiz, work on your
study guide.

Catalysts
◦
◦
◦
◦
Percent Composition
Empirical Formula
Empirical and Molecular Formula Practice
AKS Moles Test Review
If you do not put the Table of Contents or your name
on the front, 5 points will be deducted.

Mole Babies NEXT MONDAY
◦ Bring in fabric, ribbon, buttons, thread or any
accessory you would like to put on your mole.
◦ If you bring supplies for yourself or for others to
share with, you will get EXTRA CREDIT!!!
◦ If you need to finish your test, come in today during
your guided study!

1/31/14 Mole Conversions

EQ: How can I use balanced equations to convert
moles of one substance to another?

6H2O + 6CO2  C6H12O6 + 6O2


Observe the balanced chemical equation above.
There are 6 moles of CO2 in this chemical
equation. There are also 6 moles of oxygen gas
in this chemical equation. How many moles of
water are there?
 Li3PO4
How many mols are there of each element?
 Li = 3 mols
 P = 1 mol
 O = 4 mols


Just as you can compare elements to each
other within a compound, the same works for
chemical equations.
Because of this, we can predict the yield of a
certain substance within a reaction.
2H2

+ O2  2H2O
How many moles?



How many moles of O2 are needed to react with
one mole of C3H8?
5
How many moles of CO2 will be produced from
one mole of C3H8?
3
How many moles of H2O will be produced from
one mole of C3H8?
4



How many moles of O2 are needed to react with
six moles of C3H8?
30 (6 moles x 5 O2)
How many moles of CO2 will be produced from
six moles of C3H8?
18 (6 moles x 3 CO2)
How many moles of H2O will be produced from
six moles of C3H8?
24 (6 moles x 4 H2O)
1

How many moles of O2 are needed to react
with 16.3 moles of C3H8?
16.3 mol C3H8
5 mol O2
1 mol C3H8
= 81.5 mol O2

How many moles of H2O are produced from
12.9 moles of O2?
12.9 mol O2 4 mol H2O = 10.3 mol
5 mol O2 H2O

If 29 moles of H2O was produced, how many
moles of C3H8 were burned?
29 mol H2O
1 mol C3H8
4 mol H2O
= 7.25 mol C3H8
4 Al + 3 O2  2 Al2O3
4 mol Al: 3 mol O2
4 mol Al: 2 mol Al2O3
3 mol O2: 2 mol Al2O3
1 mol Al2O3
4 mol Al
2 mol Al2O3
= 2 mol Al

Complete practice problems 2-4.