6.02 Formulas and Composition

6.02
Formulas and
Composition
Using the Mole in Analysis Calculation
Dr. Fred Omega Garces
Chemistry 152
Miramar College
1
Formulas and Composition
05.2015
Counting by Mass: A Review
One mole substance equivalent to:
Atomic Weight (in grams) - atoms
Molecular Weight (in grams) - molecules
Formula Weight (in grams) - ionic Compounds
Atoms
Compound
C6H12O6
mole
C6H12O6
mass
180 g
molecule
1 mole
(1 mol)
C
6-C atoms
6 mol C
6(12.0) = 72g C
H
12-H atoms
12 mol C
12(1.0) = 12g H
O
6-O atoms
6 mol O
6(16.0) = 96g O
180 g total
From Chemical Formula we obtain the conversion factors:
1 molecule C6H12O6 ! 6C atoms ≅ 12H atoms ≅ 6 O atoms
or"
#1 mole C6H12O6 ! 6 C moles ≅ 12 H moles ≅ 6 O atoms
2
Formulas and Composition
05.2015
Percent Composition by Mass
What is meant by % ?
Consider the % of pennies in 20¢ that contains 3 nickels and 5 pennies
a) % pennies in terms of coins.
b) % pennies in terms of $
c) % pennies in terms of mass (1¢ = 2.5 g, 5¢ = 5.0 g)
% Composition ! [part/Whole] •100
a) % pennies in terms of coins
3"nickels"+""5"pennies"="8"coins
%"pennies"by"coins"="
5"pennies
3100"="62.5%
8"coins
b) % pennies in terms of $:
5"pennies"="5"¢,""Total"$"amount"="20"¢
5"¢
%"pennies"by"$""="
"8"100"="25%
20"¢"
3
c) % pennies in terms mass:
1"pennies"="2.5"g",""1"nickel"="5.0"g
2.5"g"
"="12.5g
1"penny"
5.0"g"
""""""mass"of"nickel""="3"nickel⋅
"="15.0g
1"nickel"
"mass"of"pennies""="5"pennies⋅
"%"mass"of"pennies""="
""""""""""=""5"pennies⋅
Formulas and Composition
12.5"g"pennies"
⋅100"="45.5"%
(12.5"g"+"15.0"g)"total"
05.2015
Percent Composition by Mass
% composition of elements by mass.
Similarly for a compound the % composition (by mass) of each element can be
determine by part/whole analysis.
C6H12O6 - glucose Consider 100 g Sample What is % C, %H and %O ?
i) Atomic basis:
% C atom = [6/24]•100 = 25%; % H atom = [12/24]•100 = 50%; % O atom = [6/24]•100 = 25%
ii) Mass basis:
Consider 1 mol of C6H12O6 : 6 C, 12 H, 6 O
C: 6 (12.0g/mol) = 72 g C / 1 mol
H: 12 (1.0g/mol) = 12 g H /mol
%C =
%H =
O: 6 (16.0g/mol) = 96 g O /mol
Total mass
4
= 180 g Total /mol
%O =
Formulas and Composition
€
72g/mol
180 g/mol
12g/mol
180 g/mol
96g/mol
180 g/mol
• 100 = 40%
• 100 = 6.7%
• 100 = 53%
05.2015
Formula Composition from % Composition
Suppose given % composition , What is the chemical formula?
Actually can only calculate Empirical (simplest) formula.
What is the Chemical formula for with 40% C; 6.7%H and 53% O
-Assume 100 g mass.
100 g CxHyOz g 40 g C; 6.7g H; 53g O
mol C = 40 g C
mol
= 3.3 mol C
12.0 g
mol
mol H = 6.7 g C
= 6.7 mol H
1.0 g
mol
mol O = 53 g C
= 3.3 mol O
16.0 g
€
Note: 1.0 g of H not H2 !!
Ratio: C3.3 H6.7 O
3.33
Reduce to simplest whole # ratio
divide by 3.33
$ C1 H2 O1
Note the best formula that can be obtain is the simplest (Empirical) formula without
other information such as the Molar mass or the total number of atoms in the
compound.
5
Formulas and Composition
05.2015
Chemical Formula from % Composition and Molar Mass
Suppose given % composition and molar mass, what is the chemical formula?
What is the chemical formula of the compound with an empirical formula of CH2O if
the molar mass for this compound was determined to be 180.0 g/mol ?
MolarMass'='n' ⋅ 'Empirical'Mass
Empirical'Mass'='Weight'of'atoms'in'the'empirical'formula
n'='multiplier'of'subscript'to'give'correct'chemical'formula.
Empirical'Mass'(C1H2O1 )'='12.01'g'(C)'+'2.0'g'(H)'+'16.0'g'(O)'='30.01'g
n"="
MolarMass
180.0"g/mol
"="
"="6
Empirical"Mass" 30.01"g/mol"
Correct"formula:
C1(x6) "H2(x6) "O1(x6) "="C6H12O6
6
Formulas and Composition
05.2015
Empirical Formula: Determination
Calculate the Empirical formula for the following sample:
12.1 % C;
16.2% O;
mol C = 12.1 g C •
sample
mol
= 1.0125 mol H
16 g
mol Cl = 71.7 g Cl •
COCl 2
Assume 100 g
mol
= 1.008 mol C
12.0 g
mol O = 16.2 g O •
€
71,7% Cl
mol
= 2.02 mol Cl
35.5 g
Cl
C
O
Cl
MWt = 98.91 g/mol
7
Formulas and Composition
05.2015
Empirical Formula: Combustion Analysis
In determining the empirical formula for an unknown sample combustion
analysis is used.
Basic idea:
CxHyOz
+ O2
→
CO2
C in CO2 is
from Cx
+
H in H2O
is from Hy
Strategy:
Wt sample →
O - will be determine
by difference
Wt. C → mol
wt CxHyOz
Wt. H → mol H
Wt. CO2 → wt. C
Wt. O → mol O
Wt H2O → wt. H
Wt O = [wt sample] - [wt C] - [wt H]
8
H2O
Formulas and Composition
C
mole C
H
mole H
O
mole O
05.2015
Empirical Formula: Combustion Analysis Example1
B&L 3.53b: Menthol, the substance we can smell in mentholated cough drops, is composed C, H and O. A 0.1005-g
sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. What is the empirical formula for
menthol? If the compound has a molar mass of 156 g/mol, what is its molecular formula ?
2. Mass of O by difference
1. Mass CO2 and H2O to Mass C and H
12.0 g/mol C
mass C = 0.2829 g CO2 •
= 7.72 • 10−2 g C
44.0 g/mol CO 2
mass H = 0.1159 g H2 O •
2(1.0 g/mol) H
= 1.29 •10 −2 g H
18.0 g/mol H 2 O
Wt. C
7.72•10-2 g
Wt. H
1.29•10-2 g
Wt. O
1.05•10-2 g
6.43• 10−3 mol C
C:
= 9.80 mol C
6.56 •10 −4
1.29 •10 −2 mol H
H:
= 19.67 mol H
6.56 •10 −4
−4
O = 1.0 mol O
O : 6.56 • 10 mol
−4
6.56 •10
C 9.8H19.67O1 !round-off
!!
!→
C1 0H2 0O1 (Emp Formula)
Emp.Wt • n =
Mol. Wt.
156.0 • n = 156.0
n =1
to Molc' Formula
C1 0H2 0O1 !Emp
!!
! ! !→
9
0.1005 g
4. Simplify moles ratio of C, H and O
3. Mass of C, H and O to moles
C : 7.72 • 10-2 g • mol C = 6.43 •10−3 mol C
12.0 g C
mol H
H : 1.29 •10 -2 g • 1.0 g H = 1.29 •10−2 mol H
mol O
O : 1.05• 10-2 • 16 g O = 6.56 •10−4 mol O
5. Empirical Formula
to Molecular Formula
Wt. Sample
Formulas and Composition
C1 0H2 0O1
05.2015
Empirical Formula: Combustion Analysis Example2
Consider the following: Combustion of 0.157 g CxHyNwOz , yields 0.213 g CO2 and 0.0310 g H2O. A separate analysis
of 0.103 g sample yields 0.0230 g NH3. What is the molecular formula of this substance if it contains 12-O atoms in
its chemical formula.
12.0 g/mol C
mass C = 0.213 g CO2 • 44.0 g/mol CO = 5.81• 10−2 g C
2
2(1.0 g/mol) H
mass H = 0.0310 g H2 O • 18.0 g/mol H O = 3.44 •10 −3g H
2
1(14.0 g/mol) N
−2
mass N = 0.0230 g NH3 • 17.0 g/mol NH = 1.894 •10 g N
3
% N in sample =
1.894 • 10−2 N
•100 = 18.39% N
0.103 g sample
Wt. Sample
0.157 g
Wt. C
5.81•10-2 g
Wt. H
3.44•10-3 g
Wt. N
2.887•10-2 g
Wt. O
6.668•10-2 g
Amt N in 0.157 g sample = 0.157 g •18.39% N = 2.887• 10-2 g N
C : 5.81• 10-2 g • mol C = 4.84 • 10−3 mol C
12.0 g C
mol H
H : 3.44 •10 -3 g • 1.0 g H = 3.44 • 10−3 mol H
mol N
N : 2.887 •10 -2 • 14 g N = 2.06 •10−3 mol N
mol O
O : 6.668• 10-2 •
= 4.17 •10 −3 mol O
16 g O
10
−3
4.84
•10
mol C = 2.35mol C
C:
2.06 •10 −3
−3
3.44
•10
H:
= 1.66mol H
2.06 •10 −3
2.06 •10 −3
N:
= 1mol N
2.06 •10 −3
−3
4.17•
10
O:
= 2mol O
2.06 • 10−3
C 2.35H1.66N1O2
Formulas and Composition
x3
""
→
C 7H5N3O6
05.2015
Empirical Formula: In Class
Consider the following: Combustion of 0.157 g CxHyNwOz , yields 0.213 g CO2 and 0.0310 g H2O. A separate analysis
of 0.103 g sample yields 0.0230 g NH3. What is the molecular formula of this substance if it contains 12-O atoms in
its chemical formula.
11
Formulas and Composition
05.2015
Summary: Chem Formula of CxHyOz
Determine the mass of C in CO2 .
Determine the mass of H in H2O .
Determine the mass of O by difference Mass CxHyOz – MC – MH = MO.
Convert mass of C, H, and O to moles (subscript - x, y, and w).
Take mole ratio of Cx, Hy, and Oz and find simplest integer ratio.
Simplest ratio (or whole number ratio) is the empirical formula.
Divide Molc’ Wt . by the Emp Wt. to get multiplier n for Mol. Formula.
Burn
in O2
CxHyRw
g CO2
g H2 O
1 mol CO2
44.01 g
1 mol H2O
18.02 g
mol CO2
= mol C
mol H2O
= 2 mol H
12 g C
1 mol C
1gH
1 mol H
gC
gH
Mol H (y)
Formulas and Composition
Empirical
Formula
n
Mass by O2
difference
g O2
12
Mol C (x)
Mol O (z)
Molecular
Formula
05.2015
Summary
Moles and Chemical Formulas
C6H12O6
Glucose
13
Formulas and Composition
05.2015