Download Chapter 3 - Stoichiometry

CHAPTER 3 STOICHIOMETRY
3.1 – ATOMIC MASSES

Carbon-12, the relative standard
C-12 is assigned a mass of exactly 12 atomic mass units
(amu)
 Masses of all elements are determined in comparison to the
12C atom which is the most common isotope of carbon
 Comparisons are made using a mass spectrometer


Atomic Mass (average atomic mass, atomic
weight)
Atomic masses are the weighted average of the naturally
occurring isotopes of an element
 Atomic mass does not represent the mass of any actual
atom
 Atomic mass can be used to “weigh out” large numbers of
atoms

3.2 – THE MOLE

Avogadro’s number


6.022 x 1023 units = 1 mole
Measuring moles

An elements atomic mass expressed in grams
contains 1 mole of that element
12.01 grams of carbon = 1 mole of carbon
 12.01 grams of carbon = 6.022 x 1023 atoms of carbon
23 atoms of carbon
 1 mole of carbon = 6.022 x 10

3.3 – MOLAR MASS

Molar mass is the mass in grams of one mole
of a compound

The sum of the masses of the component atoms in a
compound is calculated using the PTE
What is the molar mass of ethane (C2H6)?
2 moles of C = 2 x 12.01 g = 24.02 g
6 moles of H = 6 x 1.008 g = 6.048 g
30.07 g/mole ethane
3.4 – PERCENT COMPOSITION

A percentage is


“The part, divided by the whole, multiplied by 100”
Percent Composition

Calculate the percent composition of each element in the
total mass of a compound (or sample)
(# atoms of the element)(atomic mass of element) x 100
(molar mass of the compound)
What is the % of sodium in sodium chloride?
NaCl has a molar mass of 58.5 g/mol
(23 /58.5) x 100 = 39%
3.5 – DETERMINING THE FORMULA OF A
COMPOUND

Empirical formula – simplest whole number
ratio of elements in a compound
Determine the percentage of each element in your
compound
 Treat % as grams and convert grams to moles using the
mass from the PTE
 Find the smallest whole number ratio of atoms (multiply by
an integer to make them whole numbers).
A compound contains 63.5% Silver, 8.2% Nitrogen and 28.2%
Oxygen. What is the empirical formula for this compound?
Ag = 63.5 g/ 108 g = .59 mol
This is a 1:1:3 ratio
N = 8.2 g / 14 g = .59 mol
Formula for this
O = 28.2 g / 16 g = 1.76 mol
compound is

AgNO3
3.5 – DETERMINING THE FORMULA OF A
COMPOUND

Molecular formula – actual ratio of elements in a
compound
Calculate the empirical formula mass
 Divide the known molecular mass by the empirical formula
mass deriving a whole number, n
 Multiply the empirical formula by n to derive the molecular
formula

Ethane gas has an empirical formula of CH3 and a molecular
mass of 30 g/mol. What is the molecular formula for
ethane?
Empirical formula mass is = 15 g/mol
30/15 = 2
2(CH3) = C2H6
3.6 – CHEMICAL EQUATIONS

Chemical Reactions
Reactants are listed on the left had side of the arrow
 Products are listed on the right hand side of the arrow
 Atoms are neither created or destroyed

All atoms present in the reactants must be accounted for among
the products, in the same number
 No new atoms may appear in the products that were not present in
the reactants

REACTANTS
yield
Zn (s) + 2 HCl (aq) 
PRODUCTS
ZnCl2(aq) + H2 (g)
3.6 – CHEMICAL EQUATIONS

The meaning of a chemical reaction
Solid - (s)
 Liquid - (l)
 Gas - (g)
 Dissolved in water (aqueous solution) - (aq)


Relative numbers of reactants and products

Coefficients give atomic/molecular/mole ratios
Zn (s) + 2 HCl (aq) 
ZnCl2(aq) + H2 (g)
3.7 – BALANCING CHEMICAL EQUATIONS

Determine what reaction is occuring


Write the unbalanced equation


Focus on writing correct atomic and compound
formulas
Balance the equation using an atom
inventory


It is sometimes helpful to write in word form
It is often helpful to work systematically from left to
right
Include phase information
Hydrogen and oxygen gases combine to form
liquid water
3.8 – STOICHIOMETRIC CALCULATIONS
Balance the chemical equation
 Convert grams of reactant (or product) to
moles, if required
 Compare moles of the known value to moles
of desired substance (mole ratio)
 Convert from moles back to grams if
required

If 5 g of Zn reacts with excess hydrochloric
acid, how many grams of zinc chloride are
formed?
3.9 – CALCULATIONS INVOLVING A LIMITING
REACTANT

A concept of limiting reactant


The limiting reactant controls the amount of product that
can form
Solving limiting reactant problems
Convert grams of both reactants to moles
 Use mole ratios to determine which reactant forms the
least amount (moles) of desired product
 Convert the least moles to grams (if required)

If 5 g of Zn reacts with 5 g of hydrochloric acid,
how many moles of hydrogen gas are formed?
3.9 – CALCULATIONS INVOLVING A
LIMITING REACTANT

Calculating percent yield
Actual yield – what you got by actually doing the
experiment
 Theoretical yield – what stoichiometric
calculations say the reaction should have produced.

Actual yield (data table)
Theoretical yield (Stoich)
x 100 = % yield
5 g of Zn reacts with excess hydrochloric acid.
9.4 g of zinc chloride are formed. What is the
percent yield for this reaction?