You can measure mass, or volume, or you can count pieces.
We measure mass in grams.
We measure volume in liters.
We count pieces in

For modern systems; 12 C is taken as a standard. So; 12 C is assumed to be 12 a.m.u.
1 12 C = 12 a.m.u.
1 a.m.u.=1/12 of the mass of a single 12 C atom
=1/12 (12/6.02 X 10 23 )
=1/6.02 X 10 23 grams
=mass of 1 1 H atom
Then by comparing the masses of each atom with the mass of 12 C atomic masses of other atoms can be easily calculated “ RELATIVE ATOMIC MASSES”.
Relative atomic masses easily calculated by MASS
SPECTROMETER.
1 16 O = 16 a.m.u
1 40 Ca = 40a.m.u.
1 1 H = 1 a.m.u.

Defined as the number of carbon atoms in exactly 12 grams of carbon-12.
1 mole contains 6.02 x 10
23 particles.
Treat it like a very large dozen
6.02 x 10
23 is called
Avogadro’s number.

The smallest pieces of a substance.
For a molecular compound it is a molecule.
For an ionic compound it is a formula unit.
For an element it is an atom.

Units of measurement used with equations.
We use a mole concept to bring together the concepts of counting numbers and atomic weights of elements.
The mole is derived from the following information.
Atomic weights are an average of the relative masses of all of the isotopes of the given element.
The number of C-12 atoms in exactly 12.00 g of C-12 is 6.02 X 10 23 .
This called Avogadro’s number.

An amount of a substance that contains
Avogadro’s number of atoms, ions, molecules, or any other chemical unit is called A MOLE .
A mole of C-12 atoms is defined as having a mass of exactly 12.00 g, a mass that is equal to its atomic weight.

1 mole (mol):
Amount of matter that contains as many objects
(atoms, molecules) as the number of atoms in isotopically pure 12 C.
6.022 x 10 23
Also known as Avogadro’s number.
1 mol 12 C atoms = 6.02 x 10 23 12 C atoms
1 mol H
2
O molecules = 6.02 x 10 23 molecules
1 mole NO
3
ions = 6.02 x 10 23 NO
3
ions.



The mole concept for
(A) elements, (B) compounds, and (C) molecular substances.
A mole contains
6.02 X 10 23 particles.
Since every mole contains the same number of particles, the ratio of the mass of any two moles is the same as the ratio of the masses of individual particles making up the two moles.

Molar Mass
The atomic mass of any substance expressed in grams corresponds to 1 mol of the substance.
Atomic mass of a substance expressed in grams is the molar mass.
The molar mass of a diatomic substance is equal to twice its atomic mass.

The generic term for the mass of one mole.
The same as gram molecular mass, gram formula mass, and gram atomic mass.

The mass of 1 mole of an element in grams.
12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and
55.85 grams of iron.
We can right this as
12.01 g C = 1 mole
We can count things by weighing them.

Calculate the number of sodium atoms in 0.120 mol
Na?
0.120 mol Na x 6.02 x 10 23 atoms Na = 7.22 x 10 22
1 mol Na

Mole Calculations
Calculate the number of moles of potassium in
1.25 x 10 21 atoms of K.
1.25 x 10 21 atoms K x 1 mol K = 2.08 x 10 -3 mol K
6.02 x 10 23 atoms K

What is the mass in grams of 2.01 x 10 22 atoms of sulfur?
2.01 x 10 22 atoms S x 1 mol S x 32.07 g S
6.02 x 10 23 atoms S 1 mol S
= 1.07g S

Mole Calculations
How many O
2 oxygen gas?
molecules are present in 0.470g of
0.470 g O
2 x 1mol O
2 x 6.02 x 10 23 molecules O
2
32.00 g O
2
1 mol O
2
= 8.84 x 10 21 molecules O
2

In 1 mole of H
2
O molecules there are two moles of H atoms and 1 mole of O atoms
To find the mass of one mole of a compound determine the moles of the elements they have
Find out how much they would weigh add them up.

Molar Mass

Calculate the number of Magnesium and Chlorine ions present in 0.450 mol of MgCl
2
.
# Mg 2+ ions:
0.450 mol MgCl
2 x 6.02 x 10 23 formula units x 1 Mg 2+
1 mole 1 formula unit
0.450 mol MgCl
2
# Mg 2+ ions = 2.71 x 10 23
# Cl ions: x 6.02 x 10 23 formula units x 2 Cl -
1 mole 1 formula unit
# Cl ions = 5.42 x 10 23

The formula weight of a species is the sum of atomic masses (amu) of the atoms in a species.
Molecular weight of NH
3
=
For an ionic cmpd formula weight MgF
2
=

Mass of one mole of MgF
2 is
Mass of one formula unit of MgF
2 is
Mass of 6.022 x 10 23 formula units of MgF
2 is

The mass of one mole of an ionic compound.
Calculated the same way.
What is the GFM of Fe
2
O
3
?
2 moles of Fe x 55.85 g = 111.70 g
3 moles of O x 16.00 g = 48.00 g
The GFM = 111.70 g + 48.00 g = 159.70g

What is the mass of one mole of CH
4
?
1 mole of C = 12.01 g
4 mole of H x 1.01 g = 4.04g
1 mole CH
4
= 12.01 + 4.04 = 16.05g
The Gram Molecular mass of CH
4 is 16.05g
The mass of one mole of a molecular compound.

The gram atomic weight of an element is the mass in grams of one mole of an element that is numerically equal to its atomic weight.
The gram formula weight of a compound is the mass in grams of one mole of the compound that is numerically equal to its formula weight.
The gram formula weight of a compound is the sum total of all the individual atomic weight in the formula.
The gram molecular weight is the gram formula weight of a molecular compound.

Molar Mass
Calculate the molar mass of Ag and of magnesium nitrate, Mg(NO
3
)
2
.
Ag is an element so its molar mass equals its atomic mass = 107.87g/mol.
Mg(NO
3
)
2
= 24.31 + 2(14.01 + 3 x 16.00)
= 148.33 g/mol

How many moles is 5.69 g of NaOH?
5 69 g


1 mole
40.00 g


= 0.142 mol NaOH
 need to change grams to moles
 for NaOH
 1mole Na = 22.99g 1 mol O = 16.00 g
1 mole of H = 1.01 g
 1 mole NaOH = 40.00 g

Mole Calculations
Calculate the mass in grams of a single molecule of carbon dioxide, CO
2
.
44.01 g CO
2 x 1mol CO
2
= 7.31 x 10 -23 g/molecule
1 mol CO
2
6.02 x 10 23 molecules

The number of grams of 1 mole of atoms,ions,or molecules.
We can make conversion factors from these.
To change grams of a compound to moles of a compound.

How many moles is 4.56 g of CO
2
?
How many grams is 9.87 moles of H
2
O?
How many molecules in 6.8 g of CH
4
?
49 molecules of C
6
H
12
O
6 weighs how much?

Calc the molar mass of Ca(NO
3
)
2
.
Calc the molar mass of a compound if 0.372 mol of it has a mass of 152g.

How many grams of each are required to have
0.100 mol of
A. NaOH
B. H
2
SO
4
C. C
2
H
5
OH
D. Ca
3
(PO4)
2

How many moles are in 50.0 g of
A. CS
2
B. Al
2
(CO
3
)
3
C. Sr(OH)
2
D. LiNO
3

Calc the no. of C, H, and O atoms in 1.50 g of glucose (C
6
H
12
O
6
).
What is the average mass of one C
3 molecule?
H
8
What is the mass of 5.00 x 10 24 molecules of
NH
3
?

How many molecules of CO
2 are the in 4.56 moles of CO
2
?
How many moles of water is 5.87 x 10
22 molecules?
How many atoms of carbon are there in 1.23 moles of C
6
H
12
O
6
?
How many moles is 7.78 x 10
24 formula units of MgCl
2
?

Many of the chemicals we deal with are gases.
They are difficult to weigh.
Need to know how many moles of gas we have.
Two things effect the volume of a gas
Temperature and pressure
Compare at the same temp. and pressure.

0ºC and 1 atm pressure abbreviated STP
At STP 1 mole of gas occupies 22.4 L
Called the molar volume
Avogadro’s Hypothesis
- at the same temperature and pressure equal volumes of gas have the same number of particles.

Molar Volume
Avogadro's Theory: Two gases containing equal numbers of molecules occupy equal volumes under similar conditions.
Standard Temperature and Pressure: 0 O C and 1 atm.
Molar Volume at STP = 22.4 L

What is the volume of 4.59 mole of CO
2 at STP?
gas
How many moles is 5.67 L of O
2 at STP?
What is the volume of 8.8g of CH
4
STP? gas at

Molar Volume
Density of gas = molar mass in grams molar volume in liters
Calculate the density of ammonia gas, NH
3
, at
STP.
Density of NH
3
= molar mass NH
3 =
17.04g
molar volume NH
3
22.4 L
= 0.761 g/L

D = m /V for a gas the units will be g / L
• We can determine the density of any gas at STP if we know its formula.
•
To find the density we need the mass and the volume.
If you assume you have 1 mole then the mass is the molar mass (PT)
At STP the volume is 22.4 L.

Find the density of CO
2 at STP.
Find the density of CH
4 at STP.

22.4 L
PT
6.02 x 10
23

Mole Calculations
What is the mass of 3.36 L of ozone gas, O
3
, at STP?
3.36 L O
3 x 1 mol O
3 x 48.00g O
3
= 7.20 g O
3
22.4 L O
3
1 mol O
3

Mole Calculations
How many molecules of hydrogen gas, H
2
, occupy
0.500 L at STP?
0.500 L H
2 x 1 mol x 6.02 x 10 23 molecules H
2
22.4 L 1 mol
= 1.34 x 10 22 molecules H
2

Percent Composition
Percentage Composition: Percentage by mass contributed by each element in a substance.
% Element (E) = (# atoms of Element)(AW of E) x 100%
Formula Weight of Compound
According to the law of definite composition the elements in a compound are always present in the same proportion by mass.

Percentage composition by mass of a compound

Like all percents
Part x 100 % whole
Find the mass of each component, divide by the total mass.

Percent Composition
Calculate the percentage of Nitrogen, by mass in
Ca(NO
3
)
2
:
Formula Weight = 1 x Ca + 2 x N + 6 x O
= 1 x 40.1 + 2 x 14.0 + 6 x 16.0
= 164.1 amu
% N = (2)(14.0) = 17.1%
164.1

Percent Composition
Calculate the percent composition of trinitrotoluene (TNT), C
7
H
5
(NO
2
)
3.
Assume that there is 1 mol of TNT.
1 mol of TNT contains:
7 mol carbon atoms = 84.07 g
5 mol hydrogen atoms = 5.05 g
3 mol of nitrogen atoms = 42.03 g
6 mol oxygen atoms = 96.00 g

Percent Composition
Calculate the molar mass of TNT:
Molar mass = 227.15 g/mol.
Compare the mass of each element to the molar mass of the compound:
84.07 g C x 100 = 37.01% C
227.15 g C
7
H
5
(NO
2
)
3

Percent Composition
5.05g H x 100 = 2.22% H
227.15 g C
7
H
5
(NO
2
)
3
42.03 g N x 100 = 18.50% N
227.15 g C
7
H
5
(NO
2
)
3
96.00 g 0 x 100 = 42.26% O
227.15 g C
7
H
5
(NO
2
)
3

Empirical Formula
Empirical formula gives: the ratio of atoms in a molecule: H
2
O.
the ratio of moles of atoms on the molar level.
The molar ratio of the elements in a compound allows the calculation of the subscripts in the empirical formula of a compound.

Percent composition - the mass percent of each element present in a compound.
1. Calculate the chemical formula from percent composition.
2.Find the relative numbers of moles of each element in the compound.
3.Use molar masses of the elements as conversion factors.

Empirical Formula
Find the ratio of the numbers of moles by dividing the larger number of moles by the smaller number.
Multiply the subscripts by small integers in a trial-and-error procedure until whole numbers are found.

Empirical Formula
• The flow diagram summarizes the
• relationship among mass percent, moles, mole ratios and subscripts in the formula of a compound used in the determination of empirical formula from percent composition .

Just find the lowest whole number ratio
C
6
H
12
O
6
CH
2
O
It is not just the ratio of atoms, it is also the ratio of moles of atoms.
In 1 mole of CO
2 there is 1 mole of carbon and 2 moles of oxygen.
In one molecule of CO
2 there is 1 atom of C and
2 atoms of O.

Means we can get ratio from percent composition.
Assume you have a 100 g.
The percentages become grams.
Can turn grams to moles.
Find lowest whole number ratio by dividing by the smallest.

Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11
%N.
Assume 100 g so
38.67 g C x 1mol C = 3.220 mole C
12.01 gC
16.22 g H x 1mol H = 16.09 mole H
1.01 gH
45.11 g N x 1mol N = 3.219 mole N
14.01 gN

The ratio is 3.220 mol C = 1 mol C
3.219 mol N 1 mol N
The ratio is 16.09 mol H = 5 mol H
3.219 mol N 1 mol N
C
1
H
5
N
1
• A compound is 43.64 % P and 56.36 % O.
What is the empirical formula?
• Caffeine is 49.48% C, 5.15% H, 28.87% N and
16.49% O. What is its empirical formula?

Empirical Formula
Q.
The percent composition of a solid is known to be 68.4% Ba, 10.3% P, and 21.3% O. What is the empirical formula of the compound?
A.
Assuming a 100 g sample gives us 68.4 g Ba,
10.3 g P, and 21.3 g O.
Convert these masses to numbers of moles: 0.498 moles Ba, 0.332 mol P, 1.33 mol O

Empirical Formula
Knowing the relative numbers of moles, find the ratio by dividing the two larger numbers by the smaller number. Answer = 1.5
The O:Ba:P ratio of 4:1.5:1 gives an empirical formula of Ba
1.5
PO
4
. However, we must multiply the subscripts by a small integer (in this case 2) to find whole numbers for the formula. Ba
3
P
2
O
8
, or
Ba
3
(PO
4
)
2
.

Empirical Formula
Q. What is the percent composition of sodium hydrogen carbonate?
A. The formula for sodium hydrogen carbonate is
NaHCO
3
. The Na:H:C:O mole ratio is 1:1:1:3.
Convert this mole ratio into a mass ratio by assuming there is a 1 mole sample present. Answer = 23g Na,
1.0g H, 12g C, 48 g O

Empirical Formula
To determine the percent composition, divide the mass of each element present by the total mass of the compound and multiply by 100.
Total mass of 1 mole of NaHCO
3
= 84 g
Answer = 27% Na, 1.2% H, 14% C, 57% O

Empirical Formula
0.500 g of scandium was heated and allowed to react with oxygen. The resulting product oxide had a mass of 0.767g. What is the empirical formula for scandium oxide?
Calculate the number of moles of scandium:
0.500 g Sc x 1 mol Sc = 0.0111 mol Sc
44.96 g Sc

Calculate the moles of oxygen:
Mass of oxygen = 0.767g– 0.500g O = 0.267 g O
Number of moles of oxygen:
0.267g x 1 mol 0 = 0.0167 mol O
16.00 g
Mole ratio of Sc:O is 0.0111:0.0167
Divide by smallest number: 1:1.50

Empirical Formula
Sc
1.00
O
1.50
Using the law of multiple proportions we need a whole number ratio:
Sc
2.00
O
3.00

Empirical Formula
Calculate the empirical formula of glycine given that it contains 32.0% carbon, 6.7% hydrogen,
18.7% nitrogen and 42.6% oxygen.
Assume a 100g sample. Then the percent composition equals number of grams.

Empirical Formula
32.0 g C, 6.7g H, 18.7g N, 42.6g O.
Convert grams into moles:
32.0 g C x 1 mol = 2.66 mol C
12.01 g C
6.7 g H x 1 mol H = 6.6 mol H
1.01 g H

Empirical Formula
18.7 g N x 1 mol N = 1.33 mol
14.01 g N
42.6 g O x 1 mol O = 2.66 mol O
16.00 g O
Divide the mole ratios by the smallest number: 1.33
C
2
H
5
N
1
O
2

The lowest whole number ratio of elements in a compound.
The molecular formula the actual ration of elements in a compound.
The two can be the same.
CH
2
C
2
H
4
C
3
H
6 empirical formula molecular formula molecular formula
H
2
O both

Molecular Formula
The formula obtained from percentage compositions is always the empirical formula.
Molecular formula - gives the actual numbers of atoms in a molecule.
1. May be the same as the empirical formula.
2. May be a multiple of the empirical formula.
a. Multiple = Molecular mass
Empirical Formula

Since the empirical formula is the lowest ratio the actual molecule would weigh more.
By a whole number multiple.
Divide the actual molar mass by the the mass of one mole of the empirical formula.

Molecular Formula
The empirical formula for fructose is CH
2
O. If the molar mass of fructose is 180 g/mol, what is the molecular formula?
Molar mass of the empirical formula = 30 g/mol.

Multiple = molar mass of fructose molar mass of empirical formula
Multiple = 180 g/mol = 6
30 g/mol
Multiply the empirical formula by 6:
C
6
H
12
O
6

Molecular Formula
Q. Determine the molecular formula of ethyl butyrate.
The molecular mass of the compound is 116 g/mol and the empirical formula is C
3
H
6
O.
A. The molecular formula of the compound may be the same as the empirical formula or it may be a multiple of the empirical formula.
For ethyl butyrate: Multiple = 116 =2
58
The subscripts in the empirical formula are multiplied by 2 giving a molecular formula of C
6
H
12
O
2
.

How many atoms are there in 5.10 moles of sulfur?
What’s the mass of 5.10 moles of S?
How many moles of calcium atoms are in
1.16 x 10 24 atoms of Ca? How many grams?

Which of the following has more atoms: 1.10g of hydrogen atoms or 14.7 g of chromium atoms?
How many moles are in 0.040 kg Na?

A compound is known to be composed of
71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

The atoms in a molecule are connected or chemically bonded in a precise way. A SF.
Shows how the atoms in a molecule are arranged.
For ex: H
2
O H-O-H
C
2
H
6
CH
3
H-C- C- H
H H

Empirical formula
• The simplest whole number ratio of atoms of elements in a compound, described with the use of subscripts.
• Ionic compounds are always shown as
empirical formulas.
Molecular Formula
The actual numbers of atoms in a molecule.
Structural Formula
Show the relative arrangements of atoms in a molecule

•
If you know the name of an ingredient, you can write a chemical formula, and the percent composition of a particular substance can be calculated from the formula.
This can be useful information for consumer decisions.

Solids which are found in combined form with water in definite proportion are called as HYDRATES.
When hydrates are heated, H
2
O evaporates, and only solid is obtained in amorphous .
(w/o a certain geometric structure, generally in powdered form. H
2
O molecules surround ionic substances with certain amounts.

WATER OF HYDRATION : Water molecules of a hydrate.


Na
2
CO
3
.10H
2
O Na
2
CO
3(s)
+ 10H



DEHYDRATION: Evaporation of water of hydration.
Na
2
CO
3
.10H
2
O,
CaSO
4
.2H
2
O,
CuSO
4
.5H
2
O
2
O
(g)