Chapter 3: Calculations with
Chemical Formulas
Renee Y. Becker
Valencia Community College
CHM 1045
Atomic and Molecular Mass
• The molecular mass is the sum of the masses
of the atoms making up the molecule (units
amu) (molecular compounds only!)
calculate the molecular mass:
H2O
C2H4O2
Formula Mass
• Formula Mass is the sum of atomic masses of all
atoms in a formula unit of any compound,
molecular or ionic (units amu)
calculate the formula mass:
NaCl
MgCO3
Avogadro
• One mole of a substance is the gram mass
value, molar mass, equal to the amu mass of
a substance
• Molar mass, (MM), is in units grams/mole
(conversion factor for gmol and molg)
• Avogadro’s Number (NA = 6.022  1023
particles) is the numerical value assigned to
the unit, 1 mole
Example 1: Molar Mass
Calc. the molar mass:
Fe2O3
C6H8O7
C16H18N2O4
Example 2: Avogadro’s Number
Li2SO4
1. How many Li2SO4 units are in 3 moles of
Li2SO4?
2. How many Li atoms are in 3 moles of Li2SO4?
3. How many S atoms are in 3 moles of Li2SO4?
4. How many O atoms are in 3 moles of Li2SO4?
Example 3: Molesg
Calculate
2.35 moles C  g C
4.67 x10-3 moles H2O  g H2O
83.2 moles CaO  g CaO
Example 4: g Moles
Calculate
45.8 g Fe  moles Fe
6.54 x104 g C2H4  moles C2H4
Stoichiometry
• Stoichiometry: Relates the moles of
products and reactants to each other
and to measurable quantities
Example 5: Stoichiometry
2 NaOH(aq) + Cl2(g)  NaOCl(aq) + NaCl(aq) + H2O(l)
1.
How many moles of Cl2 are needed to react with 2
moles of NaOH?
2.
How many moles of Cl2 are needed to react with 3
moles of NaOH?
3.
How many moles of NaCl are formed by this reaction?
4.
If you start with an excess of Cl2 and 5 moles of NaOH,
how many moles of H2O can you produce?
Example 6: Stoichiometry
2 NaOH(aq) + Cl2(g)  NaOCl(aq) + NaCl(aq) + H2O(l)
1. How many moles of NaOH are needed to react
with 25.0 g of Cl2?
2. How many grams of NaOH are needed to react
with 25.0 g of Cl2?
Example 7: Stoichiometry
• Let’s make up some more examples
2 NaOH(aq) + Cl2(g)  NaOCl(aq) + NaCl(aq) + H2O(l)
Stoichiometry
• Yields of Chemical Reactions: If the actual
amount of product formed in a reaction is less
than the theoretical amount, we can calculate
a percentage yield.
% Yield = Actual yield
X 100%
Theoretical yield
Example 8: Stoichiometry
• CH2Cl2 is prepared by reaction of CH4 with
Cl2 giving HCl. How many grams of CH2Cl2
result from the reaction of 1.85 kg of CH4 if
the yield is 43.1%?
CH4(g) + 2 Cl2(g)  CH2Cl2(l) + 2 HCl(g)
Stoichiometry
• Limiting Reagents: The extent to which a
reaction takes place depends on the reactant
that is present in limiting amounts--the limiting
reagent.
– I need to make a fruit salad that is 1/2
apples and 1/2 oranges. I have 10 apples
but only 7 oranges. What is the limiting
fruit? How many apples and oranges can I
use?
Example 9: Stoichiometry
• Limiting Reagent Calculation: Lithium oxide
is a drying agent used on the space shuttle.
If 80.0 kg of water is to be removed and 65 kg
of lithium oxide is available, which reactant is
limiting?
Li2O(s) + H2O(l)  2 LiOH(s)
Li2OMM = 29.88 g/mol
H2OMM = 18.02 g/mol
Example 10: Stoichiometry
K2PtCl4 + 2 NH3  Pt(NH3)2Cl2 + 2 KCl
• If 10.0 g of K2PtCl4 and 10.0 g of NH3 are
allowed to react:
a) which is limiting reagent?
b) how many grams of the excess reagent are
consumed?
c) how many grams of cisplatin are formed?
K2PtCl4 = 415.08 g/mol
NH3 = 17.04 g/mol
Pt(NH3)2Cl2 = 300.06 g/mol
Percent Composition
• Percent Composition: Identifies the elements
present in a compound as a mass percent of
the total compound mass.
• Empirical Formula: Determined from data
about percent composition, tells only the
smallest whole number ratio of atoms in a
compound.
• Molecular Formula: Tells actual numbers of
atoms in a molecule, can be same as
empirical formula or a multiple of it.
Example 11: Percent Composition
What is glucoses, C6H12O6, empirical formula, and
what is the percentage composition of glucose?
Example 12: Percent Composition
• Saccharin has the molecular formula C7H5NO3S, what is its
empirical formula and the percent composition?
Example 13: Empirical Formula
• A compound was analyzed to be 82.67%
carbon and 17.33% hydrogen by mass. The
molar mass is 58.11 g/mol.
– What is the empirical formula and
molecular formula?