Chapter 7: Part 2
Percent composition
&
Empirical formula
Using Chemical
Formulas
Chemical formulas used to calculate: molar mass and percent
composition.
A compound’s molar mass is calculated by summing all atomic
masses in a compound.
EXAMPLE: NaCl (ionic – a metal and a nonmetal) Na atomic
mass = 23 g/mol; Cl atomic mass = 35.5 g/mol; so the molar mass
= 23 + 35.5 = 58.5 g/mol
PRACTICE: what is the molar mass of H2SO4 (sulfuric acid)?
Using Chemical
Formulas
H2SO4
(SULFURIC ACID)
• ANSWER: H = 1 g/mol but there are two atoms of hydrogens
per one molecule of sulfuric acid. S = 32 g/mol and only one
atom of sulfur in a molecule of sulfuric acid. O = 16 g/mol but
there are 4 atoms of oxygen per one molecule of sulfuric acid.
CALCULATION: 1(2) = 2 for hydrogen; 32 for sulfur; 16(4) = 64
g/mol for oxygen. 2 + 32 + 64 = 98 g/mol. The molar mass of a
molecule of sulfuric acid = 98 g/mol.
Using Chemical
Formulas
The molar mass of a compound is a CONVERSION FACTOR. We can
determine the number of moles in a given quantity of a compound
by knowing the molar mass. Further, we can calculate the number
of molecules in a given sample by using the conversion factor
___________________.
PERCENT COMPOSITION is the percent by atomic mass present in a
molecule of sample. Percent is calculated by _______ over
___________ times 100.
EXAMPLE: in a molecule of KClO3 , calculate the percent
composition of chlorine in the molecule. Of oxygen? Of potassium?
BUT FIRST, Refer to table 2 on page 226. Use your nomenclature
skills – what is the NAME of this compound?
Using Chemical
Formulas
FIRST STEP: using your periodic table, determine the atomic
masses of all atoms in the compound. DON’T FORGET to
consider “how many” of each atom is present in the compound
(indicated by the subscripts).
SECOND STEP: use percentage formula part/whole (part-overwhole) times 100%.
CALCULATION: K = 39 g/mol; Cl = 35.5 g/mol; O = 16 g/mol (but
there are 4 atoms of oxygen in a molecule of potassium
chlorate) 16(3) = 48 g/mol.
SUMMATION: 39 + 35.5 + 48 = 122.5 g/mol – this is the molar
mass of a molecule of potassium chlorate.
Using Chemical
Formulas
Calculated KClO3 molar mass = 122.5 g/mol which is also the
whole in the percent composition formula.
THIRD STEP: plug-n-chug. Part for chlorine is simply the atomic
mass of the atom and since there is only one atom per
molecule, the Cl part = 35.5 g/mol.
CALCULATION: part/whole X 100% = % composition of an
element in a compound. 35.5/122.5 X 100 = 28.9% Cl in a
molecule of KClO3
PRACTICE: now calculate the % composition of potassium and
oxygen in the compound potassium chlorate.
Using Chemical
Formulas
ANSWER: for potassium, the part is simply the atomic mass of
one atom of K = 39 g/mol. The whole is the sum of the entire
molar mass of the compound KClO3 = 122.5 g/mol.
CALCULATION: % composition = part/whole X 100%. 39/122.5 X
100 = 31.8% K.
ANSWER: for oxygen, the part is 3 times the mass of one
oxygen atom since there are three atoms of oxygen per
molecule of KClO3.
CALCULATION: 48/122.5 X 100% = 39.2% O.
Using Chemical
Formulas
All percentage calculations should add up to ______.
CHECK YOURSELF: confirm that your calculation adds up to
what it should. Does it?
PERCENTAGE COMPOSITION is the percentage by mass of
each element in a compound. It is essential to be able to
calculate empirical formula of an unknown compound.
PERCENTAGE COMPOSITION answers the question: “what
percentage does an element comprise a compound?”
Determining Chemical
Formulas
An empirical formula consists of the symbols for the elements
combined in a compound, with subscripts showing the
______________________________________________ of the
different atoms in the compound.
When given the percent composition of a substance, diborane
for example, we assume that there are 100g of the substance.
FIRST STEP: convert %comp to grams. HOW? Simply change % to
grams. If there is 78% boron (B) and 22% hydrogen (H), since we
assume there is 100g of substance (diborane), we change 78% to
78g and 22% to 22g.
QUESTIONS??
Determining Chemical
Formulas
SECOND STEP: convert grams to moles. How do we convert
grams to moles?
The atomic mass of boron is 10.8 g/mol while the atomic mass
of hydrogen is 1 g/mol. Set up your dimensional analysis…
THIRD STEP: determine which element has the smallest
number of MOLES present in the compound. In this case,
which element has the smallest number of moles in the
compound? Then simply take the mole value of each element
and divide by the smallest mole value.
FOURTH STEP: now you have a ratio.
Determining Chemical
Formulas
EXAMPLE: compound has 78% boron and 22% hydrogen. So, 78g
of boron and 22 grams of hydrogen. Convert 78g of boron to
moles of boron. 78g B X 1 mol B = 7.22 mol B
10.8 g B
What happens to the units g of boron, B?
NEXT: 22g H X 1 mol H = 22 mol H
1g H
FINALLY: Once you’ve converted your %  grams  moles, you will
simply divide both mole values by the SMALLEST calculated value .
Determining Chemical
Formulas
So which of the two values will be in the denominator of your
next calculation, moles of boron, B, or moles of hydrogen, H?
7.22 mol B
7.22
&
22 mol H
7.22
The ratio now is 1 mol boron, B : 3 mol hydrogen, H
WE NOW HAVE AN EMPIRICAL FORMULA!!!
BH3
DIGITAL INSERT: Comparing Molecular and Empirical Formulas.
Note** we will get to molecular formula shortly.
Determining Chemical
Formulas
But what if we aren’t given % comp but only the mass of the
sample?
Convert grams  moles.
Divide by the smallest number of moles for all elements in the
compound.
Determine atomic ratios.
Write empirical formula.
QUESTIONS?
Determining Chemical
Formulas
PRACTICE: a 10.150g sample of a compound know to contain only
phosphorus, P, and oxygen, O, indicates a phosphorus content of
4.433g. What is the E.F.?
FIRST STEP: determine how much oxygen is present in the
compound by mass by method of differences.
We know the compound has only P and O. We also know that the P
content is 4.433g. How can we calculate the O content by mass?
METHOD OF DIFFERENCES. The total sample is 10.150g – P content
4.433g = 5.717g of O.
Now we know the mass content of both components of this
compound. Simply follow the steps of finding the E.F.
Determining Chemical
Formulas
Convert both masses in grams to moles using dimensional
analysis and a conversion factor. WHAT CONVERSION FACTOR?!
4.433g P
X 1 mol P
=
31 g P
0.1431 mol P
NOW FIND OXYGEN
5.717g O
X 1 mol O
16 g O
= 0.3573 mol O
NOW DIVIDE EACH MOLE VALUE BY THE SMALLEST
0.1431 mol P
0.1431
&
0.3573 mol O
0.1431
Determining Chemical
Formulas
The ratio becomes 1 mol P : 2.497 mol O. When you have a ratio
that is not close to a WHOLE NUMBER, you multiply the E.F. by
the SMALLEST factor that will give a whole number ratio. In this
example, what factor can we multiply both of the numbers in the
above ratio to get a whole number ratio?
E.F. = P2O5 What factor did I multiply both numbers in the above
ratio to arrive at this E.F.?
Empirical formula is not always the molecular formula. The
molecular formula is most important in determining the actual
amounts of atoms present in a compound or an ionic formula
unit. Lets talk about molecular formula.
Determining Chemical
Formulas
Empirical formulas contain the __________ possible whole
numbers to describe atomic ratios; molecular formulas are the
_________________ of a molecular compound.
To calculate the molecular formula of a compound, you must
know the empirical formula.
EXAMPLE: If the E.F. for a compound containing only phosphorus
and oxygen was found to be P2O5 , but the molar mass of a
compound containing only these two elements is found to be
283.89 g/mol, determine the compound’s molecular formula
(MF).
FIRST STEP: determine the molar mass of the E.F.
Determining Chemical
Formulas
P atomic mass = 31 g/mol
O atomic mass = 16 g/mol
How many atoms of phosphorus are present in the EMPIRICAL
FORMULA?
How many atoms of oxygen are present in the EMPIRICAL FORMULA?
Don’t forget to multiply the atomic masses of each element by the
subscripts that indicate how many atoms of each element are present.
P2O5 is the E.F. therefore 31(2) = 62g/mol of P in this compound and
16(5) = 80g/mol of O in this compound so the EMPIRICAL FORMULA
WEIGHT = 142g/mol
Determining Chemical
Formulas
SECOND STEP: now that we have calculated the empirical
formula weight, EFW, and we are given the molar mass or the
molecular formula weight, MFW, we simply divide the MFW over
the EFW.
Given: MFW = 283.89 g/mol
Calculated: EFW = 142 g/mol
Divide & Determine: MFW/EFW  plug-n-chug  284/142 = 2
THIRD AND FINAL STEP: 2 is your whole number. Take your
whole number and multiply it by the empirical formula P2O5. So
(P2O5)(2) = P4O10 = your molecular formula.
Determining Chemical
Formulas
Using your nomenclature skills, what is the name of
the compound with the molecular formula just
determined?
Chapter 7, Part 2 Quiz
1. What is the difference between formula mass and molar mass?
2. How is molar mass determined?
3. What is the equation for determining percent composition?
4. What IS percent composition?
5. What conversion factor is used to convert a sample in moles to a sample in
molecules?
6. An empirical formula consists of the _____________mole ratio of different
atoms in a compound.
7. What are the steps to calculating the empirical formula of a compound?
8. What are the extra steps required to determine the molecular formula from
the empirical formula?
ANSWERS
1. Units of measurement. Formula mass = amu and molar mass = g/mol
2. By adding all atomic masses in the compound while taking into
consideration the subscripts that indicate how many atoms of an element are
present in the compound.
3. % comp = Part/Whole X 100%
4. The percentage of an element present in a compound, by mass.
5. Avogadro’s number
6. smallest whole number
7. convert %  grams  moles then divide each mole value by the smallest
mole value and determine an elemental ratio. If the ratio does not have whole
numbers, multiply by the smallest factor that will give a whole number ratio.
8. Divide the molecular formula weight by the empirical formula weight to
calculate a whole number. Then take that whole number and multiply it by the
empirical formula.