Transcript Slide 1

Formulas
• Indicate the chemical composition of a
compound.
– Subscripts – small numbers located after the
element symbol in a formula that indicate how
many of that element are in the formula.
– H2O - the 2 is a subscript
– Coefficients – the large number in front of a
formula that indicates the number of
molecules or formula units.
The 3 is a
– 3 H2O
coefficient!
Formula Mass
• The sum of all the atomic masses in a
compound or formula.
–Example: H2O
2 atoms of H (mass of 1) = 2 x 1 = 2
1 atom of O (mass of 16) = 1 x 16 = 16
18 amu
How many atoms of Mg, O and H are in
3Mg(OH)2?
Molar Mass
• The molar mass (in g/mol) of any
substance is always numerically equal to
its formula weight in amu.
• Molar mass is the mass of one mole of
that substance.
Mass Ratios
• The ratio between the masses that
are present in a compound.
• Example: HO No matter how many atoms of
HO there are, as long as the
• Mass of H = 1 number of H = the number of
the ratio will always stay
• Mass of O = 16 O,
the same.
3 HO = 3 H and 3 O, the ratio is 3:48 or 1:16
Percent Composition
• What is the percent of an element in a
compound? (by mass.)
• It is the mass of part of the compound divided
by the mass of the whole compound.
Example:
 What percent of ammonia is nitrogen (by mass)?
 NH3
N = 1 x 14 = 14
H= 3x1= 3
17
Nitrogen has a mass of 14 out of a total mass of 17
amu
14 X 100 = 82.3%
N =
17
Types of Chemical Formulas
• I. Molecular Formulas
– They indicate the actual number and type of
atoms in a molecular compound.
These are both:
Qualitative
What element
and
Quantitative
How many
II. Empirical Formulas
• The simplest whole number ratio of atoms in a
formula
• Empirical formulas are always used in ionic
compounds.
Molecular formula
Empirical formula
C2H6
CH3
C8H18
C4H9
C5H10
CH2
C6H12O6
CH2O
III. Structural Formulas
• Formulas that indicate the type of element
present, how many atoms are present and
how they are arranged. (Molecular)
H
H
H
C
C
H
H
H
Determining Chemical Formula
• If you know the compound’s percent
composition, you can find the formula.
– 1. Divide each percent value by that
element’s atomic mass.
– 2. Select the lowest value of all, and divide
each number by that value.
Example
A substance is 40% sulfur and 60% oxygen.
What is the chemical formula?
S 40% ÷ 32 g = 1.25
1.25 ÷ 1.25 = 1
O 60% ÷ 16 g = 3.75
3.75 ÷ 1.25 = 3
SO3
Try this!
• Ascorbic acid (Vitamin C) contains 40.92% C,
4.58% H and 54.50 % O by mass. What is the
empirical formula?
•
•
•
•
C = 40.92g/12.01g/mol = 3.407 mol C / 3.407 = 1
H = 4.58 g/1.008g/mol = 4.54 mol H / 3.407 = 1.33
O = 54.50g/16.00g/mol = 3.406 mol O / 3.407 = 1
You need a whole number ratio! Multiply 1.33 x 3 = 4.
Multiply the others by 3 as well.
• The formula is C3H4O3
Determining Molecular
Formulas
• If given the empirical formula and a
molecular mass of a compound, you can
determine the molecular formula.
Molecular weight
Whole number multiple =
Empirical formula weight
Example
The empirical formula of a compound is
CH2 and its molecular mass is 70g. What is
the molecular formula?
Step 1:
C = 12 g
H2 = 2 g
---------14 g
Step 2:
70 g ÷ 14 g =
CH2
5
C5H10
Try this!
• Mesitylene, a hydrocarbon that occurs in small
amounts in crude oil, has an empirical formula
of C3H4. The experimentally determined
molecular weight is 121 amu. What is the
molecular formula?
121 amu
40 amu
= 3.02
so, the formula is C9H12.
Let’s Talk Moles!
• 1 mole is 6.02 x 1023
OR
602,000,000,000,000,000,000,000
This unit of measurement, 6.02 x 1023 , is
called Avogadro’s number. It’s named
after the Italian scientist Amedeo
Avogadro who published it in 1811.
So . . . How much does a mole
weigh?
It depends on
what it is!
• Molar mass is the mass of 1 mole of
something (anything!)
It is the same as the gram formula mass or
the formula mass.
Molar Mass
•
•
•
•
•
1 mole of sodium (Na) = 6.02 x 1023 atoms = 23 g
1 mole of gold (Au) = 6.02 x 1023 atoms = 197 g
1 mole of lead (Pb) = 6.02 x 1023 atoms = 207 g
1 mole of carbon (C) = 6.02 x 1023 atoms = 12 g
2 moles of carbon(C) = 12.04 x 1023 atoms = 24 g
Examples
•
A sample of aluminum (Al) has a mass
of 35 grams.
•
a. How many moles of Al is this?
27 g Al
--------1 mol Al
27x = 35
-------27
27
=
35 g Al
---------x mol Al
x = 1.3 mol Al
Example 1, continued
b. How many atoms would be in this sample?
6.02 x 1023 atoms
----------------------- =
1 mol Al
x atoms
-----------------1.3 mol Al
(1.3)(6.02 x 1023) = 7.8 x 1023 atoms
Example 2
A sample of zinc (Zn) contains 0.65 moles.
a. How many grams of zinc is this?
65 g Zn
x g Zn
---------- = ---------1 mol
0.65 mol
(65)(0.65) = 42 g Zn
Example 2, continued
b. How many atoms of zinc are in this sample?
6.02 x 1023 atoms
----------------------- =
1 mol Zn
x atoms
-----------------0.65 mol Zn
(6.02 x 1023)(0.65) = 3.9 x 1023 atoms
Helpful Hints
Only use 2 variables at a time! (like moles
and mass, or moles and atoms)
Remember, moles stay underground!