Real-World Chemical Reactions
Chap 3(c)
Mg(s) + O2(g) → MgO(s)
want this
Experimental Yields,
Compositions, and
Formulas
(unbalanced rxn’s)
+ N2(g) → Mg3N2(s)
but get some of this
(undesired side product)
4.5g of Mg is enough to make 7.5g MgO
% Yield =
(theoretical amount)
6.8g
∗ 100 = 91%
7.5g
But experimentally only made 6.8g MgO
(experimental amount)
Chem 111
Dr. Gentry
Yield of Chemical Reaction
Yields of Chemical Reactions
Mg(s) + O2(g) → MgO(s)
If the actual amount of product formed in a reaction
is less than the theoretical amount, we can
calculate a percentage yield.
(unbalanced rxn’s)
+ N2(g) → Mg3N2(s)
undesired side product
[May not all react… or may get side reactions
giving other products]
→
2 Mg + O2
6.8g MgO experimental result
Actual product yield
% yield =
× 100%
Theoretical product yield
(Defined on a mass basis)
% Yield =
Yield of Chemical Reaction
Methane (CH4) reacts with chlorine (Cl2) to form
dichloromethane (CH2Cl2) and hydrogen (H2).
However hydrogen chloride is also produced as an
unwelcome by-product.
How many grams of CH2Cl2 are produced from the reaction
of 1.85 g of CH4 if the yield is 43.1% ?
43.1%
yield
→
1.85 g
side
product
Actual product yield
6.8g
× 100% =
× 100% = 91%
Theoretical product yield
7.5g
Percentage Composition
• Percent Composition: The mass percent of each element
present in a compound
mass Element 1
% Element 1 =
mass Total Compound
x 100%
CH2O (formaldehyde)
CH2Cl2 + H2
?
HCl
2 MgO
7.5g ideally (no Mg3N3)
4.5g
CH4 + Cl2
Only got 91%
of what you
expected
Rest of the Mg went to form Mg3N2
1) Do stoichiometry to find
ideal yield in grams
9.80 g
2) Then multiply by 43.1%
to find actual
4.22 g
C:
1 x 12g/mol
%C =
12g / 30g
* 100%
H:
2 x 1g/mol
% H = (2 x 1) / 30g
* 100%
=
O:
1 x 16g/mol
%O =
* 100%
= 53%
CH2O:
16g / 30g
= 40%
7%
30g/mol
1
Problem: Different compounds can
have same % composition
Empirical Formula
• The empirical formula gives ratio of lowest possible
Glucose:
Molecular formula: C6H12O6
40% C
53% O
7% H
integer number of atoms of each element in a compound.
• This is all the information we can get from % composition
Empirical formula: CH2O
Formaldehyde:
Molecular formula: CH2O
40% C
53% O
7% H
Empirical formula: CH2O
Compound
Molecular
Formula
Glucose
C6H12O6
÷6
CH2O
H2O2
÷2
OH
Benzene
C6H6
÷6
CH
Ethylene
C2H4
÷2
CH2
Propane
C3H8
÷1
C3H8
Hydrogen peroxide
Empirical Formula
• If a compound’s empirical formula is known,
… the molecular formula can be determined if we know the
molecular molar mass
Empirical
Formula
Empirical Formula
A compound was analyzed to be 82.67% carbon and
17.33% hydrogen by mass. An osmotic pressure
experiment determined that its molecular mass is
58.11 g/mol.
Empirical Formula = HO
Empirical mass = 17g/mol
Molecular mass = 34 g/mol
Ratio of masses:
What is the empirical formula and molecular formula
(from separate experiment)
for the compound?
molecular mass 34g / mol 2
=
= ratio
empirical mass 17g / mol 1
Molecular formula: 2 x empirical formula = H2O2
Empirical Formula
If given % composition for a compound, how can one find
the empirical formula?
Empirical Formula
Compound is 82.67% carbon and 17.33% hydrogen by mass.
Molecular mass is 58.11 g/mol. Empirical & molecular formulas?
1) Find # moles of each element if have 100g of material
Strategy to find Empirical Formula
82.67g C ×
1) Find # moles of each element if have 100g of material
2) Divide each of the element moles by the smallest mole #
3) Multiply by smallest number that makes all elements integers
Can then convert from empirical formula to molecular
formula.
4) Multiply empirical formula by ratio of
molecular mass
empirical mass
1mol C
12.01g C
= 6.88mol C
17.33g H ×
1mol H
1.01g H
= 17.2mol H
2) Divide each by the element moles by the smallest mole #
C:
6.88
= 1 .0 C
6.88
H:
17.2
= 2.5 H
6.88
3) Multiply by smallest number that makes both elements integers
H : 2 .5 × 2 = 5 H
Empirical formula = C2H5
C : 1 .0 × 2 = 2 C
Empirical mass = 29.1 g/mol
4) Multiply empirical formula by ratio of empirical to molecular mass
Molecular mass 58.1g / mol
= 2.0
=
Empirical mass 29.1g / mol
Molecular formula = 2 x C2H5 = C4H10
2
Vitamin C (ascorbic acid) is 40.92% carbon, 4.58% hydrogen, and
54.50% oxygen by mass. The molecular mass is 176 g/mol.
What is the empirical formula and what is the molecular formula?
 1mol C 
40.92g C × 
 = 3.41mol C
 12.01g C 
C:
3.41mol C 1.0mol C
=
3.41mol C 1.0mol C
 1mol H 
4.58g H × 
 = 4.53mol H
 1.01g H 
H:
4.53mol H 1.3mol H
=
3.41mol C 1.0mol C
 1mol O 
54.50g O × 
 = 3.41mol O
 16.00g O 
O:
3.41mol O 1.0mol O
=
3.41mol C 1.0mol C
Empirical formula = C3H4 O3
C1H1.3O1
x3
C3H4O3
Empirical mass = 88 g/mol
4) Multiply empirical formula by ratio of empirical to molecular mass
Molecular mass
88g / mol
= 2.0
=
Empirical mass 176 g / mol
Molecular formula = 2 x C3H4O3
= C 6H 8O 6
3