Download Composition and formulae

Composition and formulae
Of moles and men
Learning objectives
 Count atoms in formula
 Define the mole
 Determine numbers of atoms or molecules in
molar quantities
 Determine molar mass from chemical formula
 Determine moles from mass of substance
 Perform calculations of:
 Percent composition
 Empirical formula
 Molecular formula
Molecules or moles
 The numbers (coefficients) in chemical equations
can refer to molecules
 But for practical applications, we need a more
useful number: we cannot count molecules
Counting particles: The Mole
 The mole is a unit of quantity used in
chemistry to measure the number of atoms
or molecules
 DEFINITION:
 The number of atoms in exactly 12 g of 12C
 A mole of anything always has the same
number of particles: atoms, molecules or
potatoes – 6.02 x 1023 – Avogadro’s number
Mole conversions
Weighing molecules and moles
 Two scales:
 Atomic mass unit scale
 The mass of an individual atom or molecule in
atomic mass units (amu)
 Molar mass scale
 The mass of a mole of atoms or molecules in
grams
 Confusing?
The Good News
 The mass of a single atom or molecule in amu has
same numerical value as molar mass in grams
 The atomic mass of carbon is 12 amu
 The molar mass of carbon is 12 g/mol
 The same is true for molecules and compounds
 The formula mass of H2O is 18 amu (1+1+16)
 The molar mass of H2O is 18 g/mol
Calculations with molar mass
 Moles =
mass
molar mass
 How many moles are in 13.88 g of lithium if
the atomic mass of Li is 6.94 amu?
 2.00
Particle – mole conversions
No particles
Moles (mol) =
1
23
6.02 x 10 (
)
mol
1
No particles = Moles (mol) x 6.02 x 10 (
)
mol
23
Gram – mole conversions
Mass (g)
Moles(mol) =
Molar mass (g/mol)
Mass(g) = Moles(mol) x Molar mass (g/mol)
Particle – gram conversions
Mass (g)
Particles =
x 6.02 x 1023 (/mol)
Molar mass (g/mol)
Particles
Mass (g) =
x Molar mass (g/mol)
23
6.02 x 10 (/mol)
Significance of formula unit
 Ionic compounds do not contain molecules. Simplest
formula is the formula unit
 Covalent compounds, the molecular formula is the
formula unit
Percent composition and empirical
formula
 Chemical analysis
gives the mass % of
each element in the
compound
 Molar masses give the
number of moles
 Obtain mole ratios
 Determine empirical
formula
Determining percent composition
 Percent composition is obtained from the actual
masses.
Example:
Sample contained 0.4205 g of C and 0.0795 g of H.
Total mass = 0.5000 g (0.4205 + 0.0795)
Therefore: in 100 g there are:
100 g
x0.4205 = 84.10 g C (84.10 %)
.5000 g
100 g
x0.0795 = 15.90 g H (15.90 %)
.5000 g
Percent composition: 84.10 % C, 15.90 % H
Percent composition from formula
 What is percent composition of C5H10O2?
 1 mol C5H10O2 contains 5 mol C, 10 mol H and 2 mol O atoms
 Mass of each element (multiply subscript by molar
mass)
12.01 g C
= 60.05 g C
1 mol C
1.008 g H
10 mol H
= 10.08 g H
1 mol H
16.00 g O
2 mol O
= 32.00 g O
1 mol O
5 mol C

Total mass of one mole = 102.13 g
Convert mass of elements into
percents
%C=
60.05 g C
x100 = 58.80 % C
102.13 g C5 H10 O2
%H=
10.08 g H
x100 = 9.870 % H
102.13 g C5 H10 O2
%O=
32.00 g O
x100 = 31.33 % O
102.13 g C5 H10 O2
 Percent composition:
58.80 % C + 9.870 % H + 31.33 % O = 100.00%
(Check: adds to 100 %)
Empirical formula from percent
composition: 84.1 % C, 15.9 % H
1.
2.


Convert percents into moles (divide by molar mass)
Determine
84.10 g C
84.10 g of C ≡ 7.00 mol C 12.00 g/mol
15.9 g H
15.9 g of H ≡ 15.8 mol H
1.008 g/mol
mole ratio
15.8 mol H
 2.26 :1
Mole ratio H:C =
7.00 mol C
Simplest formula (decimal form): C1H2.26
Make smallest integers by multiplying
C4H9
May require rounding. Errors in real data cause problems

Do percent composition and empirical formula exercises
Empirical formula with more than two elements

Percent composition of vitamin C is:

40.9 % C, 4.58 % H, 54.5 % O
1. Convert into moles
2. Determine mole ratios
3. Find lowest whole numbers
Practice empirical formula problem
 A compound contains 62.1 % C, 5.21 % H, 12.1 % N and
20.7 % O. What is the empirical formula?
Inaccuracy can lead to ambiguous or
incorrect formulas
 What if H:C is 2.20 rather than 2.26? An
error of only 3 %
 Formula becomes C5H11 rather than C4H9
 What if H:C is 2.30 rather than 2.26? An
error of only 2 %
 Formula becomes C3H7
 Sometimes chemical intuition is required: we
know there is FeO, Fe3O4 and Fe2O3; so a
formula FeO3 would indicate an error
Empirical and molecular formula
 Percent composition gives the empirical
(simplest) formula. It says nothing about the
molecular formula.
 Molecular formula describes number of
atoms in the molecule
 May be much larger than the empirical formula
in the case of molecular covalent compounds
 For ionic compounds empirical formula =
“molecular” formula
Elements and compounds can have molecular
formula different from simplest formula
Substance Empirical
formula
Molecular
formula
Substance Empirical
formula
Molecular
formula
Sulphur S
S8
Phosphorous
P4
Benzene CH
C6H6
Acetylene CH
C2H2
Ethylene CH2
C2H4
Cyclohexane
C6H12
P
CH2
Which substances have same
empirical and molecular formula?
Determination of molecular formula
 Require:
1. Empirical formula from percent composition
analysis
2. Molar mass from some other source
 Number of empirical formula units in molecule:
Molar mass
n
Empirical formula mass


There are n (AaBbCc) in molecule:
Molecular formula is AnaBnbCnc
Molecular formula of vitamin C
 Empirical formula of vitamin C is C3H4O3
 Molar mass vitamin C is 176.12 g/mol
 Mass of empirical formula = 88.06 g/mol
 (3 x 12.01 + 4 x 1.008 + 3 x 16.00)
 Number of formula units per molecule =
Molar mass vitamin C
176.12
n

2
Empirical formula mass vitamin C 88.06
 Molecular formula = 2(C3H4O3) = C6H8O6
Practice molecular formula problem
 Ibuprofen contains 75.69 % C, 8.80 % H and 15.51 % O.
What is the molecular formula if molar mass is 206 g/mol?
Molarity
 Concentration is usually expressed in terms
of molarity:
 Moles of solute/liters of solution (M)
 Moles of solute = molarity x volume of solution
Example
 What is molarity of 50 ml solution containing
2.355 g H2SO4?
 Molar mass H2SO4 = 98.1 g/mol
 Moles H2SO4 = .0240 mol
 Volume of solution = 50/1000 = .050 L
 Concentration = moles/volume
= .0240/.050 = 0.480 M
Dilution
 More dilute solutions are prepared from
concentrated ones by addition of solvent
M1V1 = M2V2
Molarity of new solution M2 = M1V1/V2
To dilute by factor of ten, increase volume by factor of
ten