Transcript Chemical Formula Calculations Powerpoint

How Scientists Determine
Formulas
You have learned
• how atoms form bonds
• how to write formulas
• how to name compounds
• how to count the atom in
compounds.
Don’t you wonder how scientists figure
out the formulas in the first place?
Atomic Masses:
Counting Atoms by Weighing
• Let’s REVIEW a bit:
• Atoms have very tiny masses so scientists
made a unit to avoid using very small
numbers.
1 atomic mass unit (amu) = 1.66 10–24 g
TOO TINY TO WORK WITH IN A LAB!
The Mole
brings measurements to a scale we can work with!
• One mole of anything contains 6.022 × 1023 units
of that substance.
 Avogadro’s number is 6.022 × 1023
• A sample of an element with a mass equal to that
element’s average atomic mass (expressed in g)
contains one mole of atoms.
The Mole
• The average atomic mass for an element is the
weighted average of the masses of all the isotopes of
an element.
• It is found on the periodic table.
• Use digits to the hundredths place for all problems in
this class.
EX]
Oxygen 16.00 g/mol Hydrogen 1.01 g/mol
*Sometimes, in demo problems, I will round the values. I
do this so you see the concept more easily and don’t
get bogged down in numbers.
Objectives
1.
2.
3.
4.
Understand the definition of molar mass
Calculate molar mass
Learn to convert between moles and mass
Learn to calculate the mass percent of an
element in a compound
Molar Mass
• A compound is a collection of atoms bound together.
•
The molar mass of a compound is obtained by summing the masses of the
component atoms.
Molar Mass
• For compounds containing ions, the molar mass
is obtained by summing the masses of the
component ions.
Mass in grams of one mole of the substance:
Molar Mass of N = 14.01 g/mol
Molar Mass of N2 = 28.02 g/mol
(2 × 14.01 g/mol)
Molar Mass of H2O = 18.02 g/mol
(2 × 1.01 g/mol) + 16.00 g/mol
Molar Mass of Ba(NO3)2 = 261.35 g/mol
137.33 g/mol + (2 × 14.01 g/mol) + (6 × 16.00 g/mol)
Remember how to “count atoms”
1 Ba, 2 N, 6 O…because the subscript 2 applies to everything in
the parentheses that precedes it
Percent Composition of Compounds
• Percent composition consists of the mass percent of
each element in a compound:
Mass percent =
mass of a given element in 1 mol of compound
´100%
mass of 1 mol of compound
For iron in iron(III) oxide,(Fe2O3):
mass % Fe =
2(55.85 g)
111.70 g
=
2(55.85 g)
+ 3(16.00 g) 159.70 g
× 100% = 69.94%
Mass % O?
You can do it 2 ways:
3(16.00 g)
= 48.00 g =
30.06%
2(55.85 g) + 3(16.00 g)
g159.70 g
OR
100%- 69.94%= 30.06%...percent should add up to 100%
Note: You can check your work. Total percent should always be
very close to 100%, although it may be slightly off due to
rounding.
Calculate % composition from masses
of each element
• Scientists also do this calculation from actual masses
obtained from experiments. This is actually a very
important research tool!
• EX] A new compound is extracted from a plant root. It
breaks down as shown below. What is its % composition?
•
•
•
•
Total sample mass = 0.2370 g
C = 0.9480 g/ 0.2370 g x 100% = 40.00%
O = 0.1264 g/ 0.2370 g x 100% = 53.33%
H = 0.0158 g/0.2370 g x 100% = 6.67%
_______
100.00%
Exercise
Consider separate 100.0 gram samples of each of
the following:
H2O, N2O, C3H6O2, CO2
 Rank them from highest to lowest percent
oxygen by mass.
H2O, CO2, C3H6O2, N2O
SO WHAT! Why do we care?
• Because, from this information, we can figure out the
actual chemical formula!
• This is how scientists determine the chemical
formulas for substances!
• Discovery, research…SCIENCE!
Objectives
1. Understand the meaning of empirical
formula
2. Learn to calculate empirical formulas
3. Learn to calculate the molecular formula
of a compound
Empirical Formulas
The word “empirical” means it is based on
observations, experiments, or
experience…not theory or logic only
Scientists collect data in experiments and use
it to determine empirical formulas
•The empirical formula of a compound is the
simplest whole number ratio of the atoms
present in the compound.
•The empirical formula can be found from the percent composition of
the compound.
•The empirical formula can be found from masses of each element
also.
Calculation of Empirical Formulas
Refresher Course!
• You need to remember how to calculate moles
of an element from grams of the element
• Divide the given mass by the molar mass
• EXAMPLE:
• How many moles in
• 36.03 g C|1mol C
= 3.00 mol C
|12.01 g C
• Grams or percent…It doesn’t matter!
• If you are given grams, calculate number of moles by
dividing by the molar mass
• If you are given %composition, change units from
percent to grams and calculate moles by dividing by
the molar mass. (You assume a sample size of 100 g to
do this.)
• Why do we do it 2 ways?
– Because sometimes experiments yield results measured in
grams, and sometimes the results are measured as
percentages of the whole sample.
B. Calculation of Empirical Formulas
Subscripts must be WHOLE numbers
• Sometimes when you divide all mole values by
the smallest, you don’t get whole numbers!
• Multiply by the smallest whole number that will
convert them all to whole numbers. See hint
below
• Look for common decimal equivalents of
fractions
.25=1/4
.67=2/3
.33=1/3
.2=1/5
.5=1/2
.17=1/6
.75=3/4
• Multiply all values by the denominator to get
whole numbers
Calculation of Molecular Formulas
• The molecular formula is the exact formula of the molecules present in a
substance.
• Remember, only IONIC compounds are simplified to the lowest whole
number ratio
• This is because molecular compounds actually form molecules (!!groups of
atoms held together by bonds formed from shared electrons!!)
• Molecules can be multiples of the simplest formula (empirical)
• The molecular formula is always an integer multiple of the empirical
formula.
Molecular formula = (empirical formula)n
where n is a whole number
Molecular formulas
• How do you know when you need to calculate
a molecular formula?
– You must be given the actual molar mass of the
compound.
• What do you do to calculate the molecular
formula?
– Divide the given molar mass by the mass of the
empirical formula
– Multiply each SUBSCRIPT by this value
EXAMPLE: Glucose
Given molar mass of
substance
= 180 g/mol
Molar mass of empirical
formula =30 g/mol
Given mass = 180 = 6
Emp mass
30
Multiply each subscript
by the calculated integer
MOLECULAR FORMULA:
C6H12O6
Exercise
The composition of adipic acid is 49.3% C, 6.9%
H, and 43.8% O (by mass). The molar mass of the
compound is about 146 g/mol.
 What is the empirical formula?
C3H5O2
 What is the molecular formula?
C6H10O4
p223
p223
p225
p225
p225
p225