Unit 6: The Mathematics of Chemical Formulas
# of H2O
molecules
# of H atoms
# of O atoms
1
2
1
2
4
2
3
6
3
100
200
100
6.02 x 1023
2 (6.02 x 1023)
6.02 x 1023
mass:
18 g
mass:
2g
mass:
16 g
molar mass: the mass of one mole of a substance
PbO2
Pb: 1 (207.2 g) = 207.2 g
O:
2 (16.0 g) =
32.0 g
H:
1 (1.0 g) =
1.0 g
N:
1 (14.0 g) =
14.0 g
O:
3 (16.0 g) =
48.0 g
ammonium
N:
3 (14.0 g) =
42.0 g
phosphate
H:
12 (1.0 g) =
12.0 g
HNO3
NH41+
PO43–
(NH4)3PO4
P:
1 (31.0 g) =
31.0 g
O:
4 (16.0 g) =
64.0 g
239.2 g
63.0 g
149.0 g
percentage composition: the mass % of each element
in a compound
% of element 
g element
x 100
molar mass of compound
Find % composition:
PbO2
Pb: 1(207.2 g) = 207.2 g Pb  239.2 g  86.6% Pb
O: 2(16.0 g) = 32 g O  239.2 g
 13.4% O
(NH4)3PO4
N: 3(14.0 g) = 42.0 g P  149.0 g  28.2% N
H: 12(1.0 g) = 12.0 g H  149.0 g  8.1% H
P: 1(31.0 g) = 31.0 g P  149.0 g  20.8% P
O: 4(16.0 g) = 64.0 g O  149.0 g  43.0% O
zinc acetate
Zn2+ CH3COO1–

Zn(CH3COO)2
Zn: 1(65.4 g) = 65.4 g Zn  183.4 g  35.7% Zn
C: 4(12.0 g) = 48.0 g C  183.4 g  26.2% C
H: 6(1.0 g)
= 6.0 g H  183.4 g 
3.3% H
O: 4(16.0 g) = 64.0 g O  183.4 g  34.9% O
183.4 g
Finding an Empirical Formula from Experimental Data
1. Find # of g of each element.
2. Convert each g to mol.
3. Divide each “# of mol” by the smallest “# of mol.”
4. Use ratio to find formula.
A compound is 45.5% yttrium and 54.5% chlorine.
Find its empirical formula.
 1 mol Y 
45.5 g Y 
  0.512 mol Y  0.512  1
 88.9 g Y 
 1mol Cl 
54.5 g Cl 
  1.535 mol Cl  0.512  3
 35.5 g Cl 
YCl3
A ruthenium/sulfur compound is 67.7% Ru. Find its
empirical formula.
 1mol Ru 
67.7 g Ru 
  0.670 mol Ru  0.670  1
101.1
g
Ru


 1 mol S 
32.3 g S 
  1.006 mol S  0.670  1.5
 32.1 g S 
RuS1.5

Ru2S3
A 17.40 g sample of a technetium/oxygen compound
contains 11.07 g of Tc. Find the empirical formula.
 1mol Tc 
11.07 g Tc 
  0.113 mol Tc  0.113  1
 97.9 g Tc 
 1mol O 
6.33 g O 
  0.396 mol O  0.113  3.5
16.0
g
O


TcO3.5

Tc2O7
A compound contains 4.63 g lead, 1.25 g nitrogen,
and 2.87 g oxygen. Name the compound.
 1mol Pb 
4.63 g Pb 
  0.0223 mol Pb  0.0223  1
 207.2 g Pb 
 1mol N 
1.25 g N 
  0.0893 mol N  0.0223  4
14.0
g
N


 1 mol O 
2.87 g O 
  0.1794 mol O  0.0223  8
16.0
g
O


PbN4O8
Pb(NO2)4 
lead (IV) nitrite
plumbic nitrite
Pb? 4 NO21–
To find molecular formula…
1). Find empirical formula.
2). Find molar mass of empirical formula.
3). Find n = mm molecular
mm empirical
4). Multiply all parts of empirical formula by n.
A carbon/hydrogen compound is 7.7% H and has a
molar mass of 78 g. Find its molecular formula.
 1 mol C 
92.3 g C 
  7.69 mol C  7.69  1
12.0
g
C


 1 mol H 
7.7 g H 
  7.7 mol H  7.69  1
1.0
g
H


mmemp = 13 g  n = 78 g  13 g = 6 
CH
C6H6
A compound has 26.33 g nitrogen, 60.20 g oxygen,
and molar mass 92 g. Find molecular formula.
 1mol N 
26.33 g N 
  1.881 mol N  1.881  1
 14.0 g N 
 1mol O 
60.20 g O 
  3.763 mol O  1.881  2
16.0
g
O


mmemp = 46 g  n = 92 g  46 g = 2 
Mole Calculations
N2O4
NO2
New Points about Island Diagram:
a. Diagram now has four islands.
b. “Mass Island” now for elements or compounds
c. “Particle Island” now for atoms or molecules
d. “Volume Island”: for gases only
1 mol @ STP = 22.4 L = 22.4 dm3
What mass is 1.29 mol ferrous nitrate?
Fe2+
 179.8 g 
X g Fe(NO3 ) 2  1.29 mol 

1
mol


NO31–
 Fe(NO3)2
232 g Fe(NO3 ) 2
How many molecules is 415 L sulfur dioxide at STP?
23
 1 mol   6.02 x 10 m' cules 
25
X m' cules SO 2  415 L 

1.12
x
10


1 mol
 22.4 L  

m’cules SO2
What mass is 6.29 x 1024 m’cules aluminum sulfate?
Al3+ SO42–  Al2(SO4)3
1 mol

  342.3 g 
X g  6.29 x 1024 m' c 

  3580 g Al 2 (SO 4 ) 3
23
 6.02 x 10 m' c   1 mol 
At STP, how many g is 87.3 dm3 of nitrogen gas?
 1 mol   28.0 g 
X g N2  87.3 dm3 
  109 g N2
3 
 22.4 dm   1 mol 
How many m’cules is 315 g of iron (III) hydroxide?
Fe3+
OH1–  Fe(OH)3
 1 mol   6.02 x 1023 m' cules 
  1.78 x 1024
X m' cules  315 g 
 
1 mol
m’cules
 106.8 g  

Fe(OH)3
How many atoms are in 145 L of CH3CH2OH at STP?
23
 1 mol   6.02 x 10 m' cules 
24
X m' cules  145 L 

3.90
x
10
m' c


1 mol
 22.4 L  

 9 atoms 
25
X atoms  3.90 x 1024 m' c 
  3.51 x 10 atoms
 1 molecule 
Hydrates and Anhydrous Salts
anhydrous salt: an ionic compound (i.e., a salt) that
attracts water molecules and forms loose chemical
bonds with them; symbolized by MN
“anhydrous” = “without water”
“dessicants” in leather goods,
Uses:
electronics, vitamins
hydrate: an anhydrous salt with the water attached
-- symbolized by MN . ? H2O
Examples:
H2O H2O
H2O MN H2O
H2O H2O
CuSO4 . 5 H2O
BaCl2 . 2 H2O
Na2CO3 . 10 H2O
FeCl3 . 6 H2O
HEAT
hydrate
MN
+
H2O
H2O
H2O
H2O H O
2
H2O
anhydrous salt
water
Finding the Formula of a Hydrate
1. Find the # of g of MN and # of g of H2O.
2. Convert g to mol.
3. Divide each “# of mol” by the smallest “# of mol.”
4. Use the ratio to find the hydrate’s formula.
Find formula of hydrate for each problem.
sample’s mass before heating = 4.38 g
(MN . ? H2O)
sample’s mass after heating = 1.93 g
(MN)
molar mass of anhydrous salt = 85 g
 1 mol MN 
1.93 g MN 
  0.0227 mol MN  0.0227  1
85
g
MN


 1mol H2O 
  0.1361 mol H2O  0.0227  6
2.45 g H2O 
 18 g H2O 
MN . 6 H2O
A. beaker = 46.82 g
A
B. beaker + sample before heating = 54.35 g
B
C. beaker + sample after heating = 50.39 g
C
molar mass of anhydrous salt = 129.9 g
 1 mol MN 
3.57 g MN 
  0.0275 mol MN  0.0275  1
 129.9 g MN 
 1mol H2O 
  0.22 mol H2O  0.0275  8
3.96 g H2O 
18
g
H
O


2
MN . 8 H2O
A. beaker = 47.28 g
A
B. beaker + sample before heating = 53.84 g
B
C. beaker + sample after heating = 51.48 g
C
molar mass of anhydrous salt = 128 g
 1 mol MN 
4.20 g MN 
  0.0328 mol MN  0.0328  1
 128 g MN 
 1mol H2O 
  0.1311 mol H2O  0.0328  4
2.36 g H2O 
 18 g H2O 
MN . 4 H2O
Find % water and % anhydrous salt (by mass).
% H2O 
4 (18 g H2O)
 36.0% H2O (and 64.0% MN)
4 (18 g )  128 g
OR…
% H2O 
2.36 g H2O
 36.0% H2O (and 64.0% MN)
2.36 g  4.20 g
Find % comp. of iron (III) chloride.
Fe3+
Cl1–

FeCl3
Fe: 1(55.8 g) = 55.8 g Fe  162.3 g  34.4% Fe
Cl: 3(35.5 g) = 106.5 g Cl  162.3 g  65.6% Cl
162.3 g
A compound contains 70.35 g C and 14.65 g H. Its
molar mass is 58 g. Find its molecular formula.
 1 mol C 
70.35 g C 
  5.863 mol C  5.863  1
12.0
g
C


 1 mol H 
14.65 g H 
  14.65 mol H  5.863  2.5
1.0
g
H


empirical
CH2.5

C2H5
mmemp = 29 g  n = 58 g  29 g = 2 
At STP, how many g is 548 L of chlorine gas?
C4H10
 1 mol   71.0 g 
X g Cl 2  548 L 

  1740 g Cl 2
22.4
L
1
mol



Strontium chloride is an anhydrous salt on which the
following data were collected. Find formula of hydrate.
beaker = 65.2 g
beaker + sample before heating = 187.9 g
beaker + sample after heating = 138.2 g
formula of salt  Sr2+ Cl1–  SrCl2
molar mass = 158.6 g
 1mol 
73.0 g SrCl2 
  0.460 mol SrCl 2  0.460  1
 158.6 g 
 1mol H2O 
  2.761 mol H2O  0.460  6
49.7 g H2O 
18
g
H
O


2
SrCl2 . 6 H2O