Download Report

Atomic mass units and relative atomic mass
Atomic mass and molar mass
The atomic mass of an element is the mass of a single atom of that element
expressed in atomic mass units (amu)
By definition:
The mass of one carbon-12 atom is exactly 12 atomic mass units (amu)
• 1 amu = 1/12 the mass of one carbon-12 atom
Remember that Avogadro’s number is defined as the number of atoms
present in exactly 12 grams of carbon-12
12.000 g carbon-12
6.022 x 1023 atoms carbon-12
1 amu =
Example: Carbon
relative atomic mass = 12.01 amu
atomic
number
6
C
= 1.9927 x 10-23 g per atom of carbon-12
element
symbol
1
mass of one
12 carbon-12 atom
=
1 amu =
1
12
12.01
This is the mass of one carbon atom
(average mass weighted for isotopic abundance)
1.9927 x 10-23 g
1.6606 x 10-24 g
Atomic mass and molar mass
The molar mass of an element is the mass of 1 mole of that element
(6.022 x 1023 atoms) expressed in grams
• the molar mass has the same value in grams as the atomic mass in amu
Sample problems
What is the mass of 3.50 moles of sodium (Na)?
Change moles of Na to grams of Na
Molar mass of Na = 22.99 g / mol
12.01 amu
6.022 x 1023 atoms C
1.6606 x 10-24 g
1 atom C
1 mol C
1 amu
Example: Carbon
atomic
number
element
symbol
= 12.01 g / mol C
molar mass = 12.01 g
6
C
12.01
This is the mass of one mole of carbon
(average mass weighted for isotopic abundance)
3.50 moles Na
x
22.99 g Na
1 mol Na
=
80.5 g Na
Sample problems
Sample problems
How many magnesium atoms are contained in 5.00 g of Mg?
How many moles of iron (Fe) does 25.0 g of Fe represent?
Change grams of Fe to moles of Fe
Change grams of Mg to moles Mg
Then change moles Mg to atoms of Mg
Molar mass of Fe = 55.85 g / mol
Molar mass of Mg = 24.31 g / mol
1 mol Mg
5.00 g Mg x
1 mol Fe
25.0 g Fe x
=
=
0.206 mol Mg
24.31 g Mg
0.448 mol Fe
55.85 g Fe
6.022 x 1023 atoms Mg
= 1.24 x 1023 atoms Mg
0.206 mol Mg x
1 mol Mg
Calculating the molar mass of a compound
H H
H H
H H
H H
H H
H
O
H
H
O
H
O
O
O
O
H
O
H
H H
H
H H
H H
H H
H H
H H
H
H
1 mol Na
11.5 g Na
O
H
=
O
O
H
1 mol O
6.022 x 1023 O atoms
= 0.500 mol Na
22.99 g Na
O
+
6.022 x 1023 atoms Na
0.500 mol Na x
O
H
O
H
H
3.01 x 1023 atoms Na x
O
mass = 1.008 g
O
H
H
1 mol Na
H2O molecules
=
O
H
O
6.022 x
1023
O
H
H
1 mol H2O
O
H
O
1 mol H
6.022 x 1023 H atoms
Molar mass of Na = 22.99 g / mol
O
O
O
H
Change atoms of Na to moles of Na
Then change moles of Na to grams of Na
H H
What is the mass of 3.01 x 1023 atoms of sodium (Na)?
H
Sample problems
mass = 16.00 g
2 moles H
2 x (1.008 g)
=
1 mole O
=
16.00 g
1 mole H2O
=
18.016 g
2.016 g
1 mol H
6.022 x 1023 H atoms
mass = 1.008 g
18.02 g
Molar mass of H2O
18.02 g / mol
Molar mass of compounds
Molar mass of compounds
What is the mass of 1 mol of sulfuric acid?
What is the mass of 1 mol of diatomic oxygen (O2)?
Formula of sulfuric acid: H2SO4
1 mole of H2SO4
=
2 moles H atoms
1 mole S atoms
4 moles O atoms
element
# of atoms
H
2
molar mass
1.008 g
S
1
32.07 g
32.07 g
O
4
16.00 g
64.00 g
Formula of diatomic oxygen: O2
Remember that 1 molecule of O2 contains 2 oxygen atoms
So 1 mole of O2 molecules contains 2 moles of oxygen atoms
total mass
2.016 g
element
# of atoms
O
2
molar mass
16.00 g
Molar mass of diatomic oxygen =
Molar mass of sulfuric acid =
total mass
32.00 g
32.00 g
98.09 g
Stoichiometry
STOICHIOMETRY
Quantitative
relationships
between
reactants and
products in
chemical
reactions
stoichiometry -- the quantitative relationships between
reactants and products in a chemical reaction
You can think of stoichiometry as a kind of accounting
system for chemistry
• empirical formulas and mass percent compositions
of compounds
• moles of substances consumed and produced in
a chemical reaction
• masses of substances consumed and produced
in a chemical reaction
• limiting reactants
• theoretical yields
Percent composition of compounds
Percent composition of compounds
The percent composition of a compound refers to the mass percent
of each of the elements in the compound
The percent composition of a compound can be determined from
experimental data without knowing the formula of the compound
H H H H H H H H H
H H H H H H H H H H
O
1 mol H
O
O
O
O
O
O
O
O
O
O
O
O
H H H H H H H H H
H
H
H
H
H
H
O
H
O
H
H
O
H
H H H H H H H H H H
18.02 g
O
6.022 x 1023 H2O molecules
+
O
H
=
1 mol H2O
O
mass = 1.008 g
O
O
6.022 x 1023 H atoms
1 mol O
6.022 x 1023 O atoms
mass = 16.00 g
1 mol H
6.022 x
1023
Zn
1.63 g
H atoms
mass = 1.008 g
H 2O
18.02 g (100%)
H
1.008 g (5.6%)
H
2.016 g (11.2%)
H
1.008 g (5.6%)
Example: When heated in air, 1.63 g of zinc combines with 0.40 g of
oxygen to form zinc oxide. Calculate the percent composition of the zinc
oxide product.
O
16.00 g (88.8%)
O
16.00 g (88.8%)
Empirical formula versus molecular formula
There are two types of formulas that can be used to describe a
compound: the empirical formula and the molecular formula
+
zinc oxide
(formula: ZnO? ZnO2? other?)
O2
0.40 g
mass of the reactants
2.03 g
=
mass of the products
Calculate the mass percent of each element in the compound
mass % of zinc = 100 x ( 1.63 g / 2.03 g ) = 80.3% Zn
mass % of oxygen = 100 x ( 0.40 g / 2.03 g ) = 20.% O
Example of empirical formula determination
When heated in air, 1.63 g of zinc combines with 0.40 g of oxygen to form
zinc oxide. What is the empirical formula of the zinc oxide product?
empirical formula (simplest formula) -- gives the smallest whole-number
ratio of atoms of each element in the compound
-- i.e., the relative number of atoms of each element in the compound
The word empirical has a similar meaning to the word experimental
-- empirical formulas are based on results from laboratory techniques
(known as quantitative analysis) that determine the amount of a given
element in a substance
Note: The empirical formula does not give any information about;
• the exact number of atoms present in a molecule of the compound
• the structural arrangement of the atoms in a molecule of the compound
Zn
1.63 g
+
O2
0.40 g
zinc oxide
(formula: ZnxOy)
Convert grams to moles
1.63 g Zn x ( 1 mol Zn / 65.39 g Zn ) = 0.0249 moles of Zn atoms
0.40 g O x ( 1 mol O / 16.00 g O ) = 0.025 moles of O atoms
Divide by smallest number of moles
Zinc: ( 0.0249 mol / 0.0249 mol ) = 1.0
Oxygen: ( 0.025 mol / 0.0249 mol ) = 1.0
Assign subscripts
Zn1O1
ZnO
H
Calculating empirical formulas
To calculate the empirical formula of a compound, you need to know
the following:
-- the elements that combine to form the compound
-- the molar masses for each of the elements
Calculating empirical formulas
Example: A compound is formed by combining 2.233 g of iron and 1.926
g of sulfur. What is the empirical formula of the compound?
Convert grams to moles
2.233 g Fe x ( 1 mol Fe / 55.85 g Fe ) = 0.03998 moles of Fe atoms
-- the mass percentages for each of the elements
1.926 g S x ( 1 mol S / 32.07 g S ) = 0.06006 moles of S atoms
1) Convert the mass of each element (grams) into moles
2) Divide the number of moles of each element by the smallest
number of moles
Divide by smallest number of moles (multiply by 2 to get whole numbers)
iron: ( 0.03998 mol / 0.03998 mol ) = 1.000
x 2 = 2.000
3) If the values from the preceding step are not whole numbers,
multiply them by the smallest factor that will result in whole
numbers
sulfur: ( 0.06006 mol / 0.03998 mol ) = 1.502
x 2 = 3.004
Assign subscripts to elements in the empirical formula
4) Use the whole numbers as subscripts for the elements in the
empirical formula
Fe2S3
Empirical formula versus molecular formula
Calculating molecular formulas
empirical formula (simplest formula) -- gives the relative number of atoms
of each element in the compound
Remember that the molecular formula of a compound will either be the
same as its empirical formula or a whole number multiple of its empirical
formula
molecular formula (true formula) -- gives the exact number of atoms of
each element present in one molecule of the compound
Compound
formaldehyde
acetic acid
glucose
empirical formula
molecular formula
CH2O
CH2O
CH2O
CH2O
C2H4O2
= CH2O
= (CH2O)2
C6H12O6 = (CH2O)6
Note: The molecular formula is either the same
as the empirical formula or a whole number
multiple of the molecular formula
Molecular formula = (empirical formula)n
Compound
water
[ n = 1, 2, 3, etc. ]
empirical formula
H2O
molecular formula
H 2O
acetylene
CH
C2H2
= (CH)2
benzene
CH
C6H6
= (CH)6
The molecular formula can be calculated from the empirical formula if the
molar mass of the compound is known
Number of empirical formula units =
molar mass of compound
mass of empirical formula
Calculating molecular formulas
Calculating molecular formulas
Example (trivial): We saw that the empirical formula for water is H2O. The
Example: A sample of polypropylene contains 14.3 g of hydrogen and
85.7 g of carbon. Polypropylene has a molar mass of 42.08 g. What is
the molecular formula for polypropylene?
molar mass of water is 18.02 g. What is the molecular formula for water?
Step 1: Calculate the empirical formula
Molecular formula = (empirical formula)n = (H2O)n
Convert grams to moles
mass of empirical formula = 2 x (molar mass of H) + molar mass of O
14.3 g H x ( 1 mol H / 1.008 g H ) = 14.2 moles of H atoms
!
!
!
= 2 x ( 1.008 g ) + ( 16.00 g )
85.7 g C x ( 1 mol C / 12.01 g C ) = 7.14 moles of C atoms
!
!
!
= 18.02 g
Divide by smallest number of moles
n =!
molar mass of compound
!
!
!
mass of empirical formula
18.02 g
!
18.02 g
=!
= 1.000
hydrogen: ( 14.2 mol / 7.14 mol ) = 1.989 ! 2
carbon: ( 7.14 mol / 7.14 mol ) = 1.000 ! 1
Assign subscripts to elements in the empirical formula
Molecular formula = (H2O)1 =
H 2O
(same as empirical formula)
Calculating molecular formulas
Example: A sample of polypropylene contains 14.3 g of hydrogen and
85.7 g of carbon. Polypropylene has a molar mass of 42.08 g. What is
the molecular formula for polypropylene?
molecular formula = (empirical formula)n = (CH2)n
mass of empirical formula = (molar mass of C) + 2 x molar mass of H
= 12.01 g + ( 2 x 1.008 g )
= 14.03 g
molar mass of compound
mass of empirical formula
=
CH2
Homework assignment
Chapter 6 Problems:
6.38, 6.42, 6.49, 6.50, 6.51, 6.54, 6.56, 6.58, 6.60
Step 2: Calculate the molecular formula
n =
C 1H 2
42.08 g
14.03 g
molecular formula = (CH2)3 =
C 3H 6
= 2.999 ! 3