Transcript Simplest & Molecular Formula PowerPoint

Simplest formula calculations
Q- a compound is found to contain the following % by mass:
69.58% Ba, 6.090% C, 24.32% O. What is the simplest (i.e.
empirical) formula?
Step 1: imagine that you have 100 g of the substance. Thus, %
will become mass in grams. e.g. 69.58 % Ba becomes
69.58g Ba. (Some questions will give grams right off,
instead of %)
Step 2: calculate the # of moles (mol = g ¸ g/mol)
Step 3: express moles as the simplest ratio by dividing through
by the lowest number.
Step 4: write the simplest formula from mol ratios.
Simplest formula: sample problem
Q- 69.58% Ba, 6.090% C, 24.32% O.
What is the empirical (a.k.a. simplest) formula?
1: 69.58 g Ba, 6.090 g C, 24.32 g O
2: Ba: 69.58 g
¸ 137.33 g/mol = 0.50666 mol Ba
C:
6.090 g ¸ 12.01 g/mol = 0.50708 mol C
O: 24.32 g
¸ 16.00 g/mol = 1.520 mol O
3:
Ba
C
O
mol
0.50666
0.50708
1.520
mol
0.50666/
0.50708/
1.520/
(reduced) 0.50666 = 0.50666 = 0.50666 =
1
1.001
3.000
4: the simplest formula is BaCO3
Mole ratios and simplest formula
Given the following mole ratios for the hypothetical
compound AxBy, what would x and y be if the mol ratio
of A and B were:
A = 1 mol, B = 2.98 mol
AB3
A = 1.337 mol, B = 1 mol
A4B3
A = 2.34 mol, B = 1 mol
A7B3
A = 1 mol, B = 1.48 mol
A2B3
1. A compound consists of 29.1 % Na, 40.5 % S, and 30.4 %
O. Determine the simplest formula.
2. A compound is composed of 7.20 g carbon, 1.20 g
hydrogen, and 9.60 g oxygen. Find the empirical formula
for this compound
3. A compound consists of 28.9 % K, 23.7 % S, and 47.7 %
O. Determine the simplest formula.
Question 1
1: Assume 100 g: 29.1 g Na, 40.5 g S, 30.4 g O
2: Na: 29.1 g ¸ 22.99 g/mol = 1.266 mol Na
S: 40.5 g ¸ 32.06 g/mol = 1.263 mol S
O: 30.4 g ¸ 16.00 g/mol = 1.90 mol O
3:
Na
S
O
mol
1.266
1.263
1.90
mol
1.266/ 1.263 1.263/ 1.263 1.90/ 1.263
(reduced)
= 1.00
=1
= 1.50
4: the simplest formula is Na2S2O3
For instructor: prepare molecular models
Question 2
1: 7.20 g C, 1.20 g H, 9.60 g O
2: C: 7.20 g ¸ 12.01 g/mol
= 0.5995 mol C
H: 1.20 g ¸ 1.01 g/mol
= 1.188 mol H
O: 9.6 g
¸ 16.00 g/mol
= 0.60 mol O
3:
C
H
O
mol
0.5995
1.188
0.60
mol
0.5995/
1.188/
0.60/ 0.5995
(reduced)
0.5995 = 0.5995 =
= 1.0
1
1.98
4: the simplest formula is CH2O
Question 3
1: Assume 100 g: 28.9 g K, 23.7 g S, 47.7 g O
2: C: 7.20 g ¸ 12.01 g/mol = 0.5995 mol C
H: 1.20 g ¸ 1.01 g/mol = 1.188 mol H
O: 9.6 g ¸ 16.00 g/mol = 0.60 mol O
3:
C
H
O
mol
0.5995
1.188
0.60
mol
0.5995/
1.188/
0.60/ 0.5995
(reduced)
0.5995 = 0.5995 =
= 1.0
1
1.98
4: the simplest formula is CH2O
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Molecular formula calculations
There is one additional step to solving for a molecular
formula. First you need the molar mass of the compound.
e.g. in Q2, the molecular formula can be determined if we
know that the molar mass of the compound is 150 g/mol.
First, determine molar mass of the simplest formula. For
CH2O it is 30 g/mol (12+2+16).
Divide the molar mass of the compound by this to get a
factor: 150 g/mol ¸ 30 g/mol = 5
Multiply each subscript in the formula by this factor:
C5H10O5 is the molecular formula. (models)
Q- For OF, give the molecular formula if the compound is 70
g/mol
O2F2 70 ¸ 35 = 2
7.
Combustion analysis gives the following:
26.7% C, 2.2% hydrogen, 71.1% oxygen.
If the molecular mass of the compound is 90 g/mol,
determine its molecular formula.
8. What information must be known to determine a) the
empirical formula of a substance?
b) the molecular
formula of a substance?
9. A compound’s empirical formula is CH, and it weighs
104 g/mol. Give the molecular formula.
10. A substance is decomposed and found to consist of 53.2%
C, 11.2% H, and 35.6% O by mass. Calculate the
molecular formula of the unknown if its molar mass is 90
g/mol.
Question 7
1: Assume 100 g total. Thus:
26.7 g C, 2.2 g H, and 71.1 g O
2: C: 26.7 g ¸ 12.01 g/mol
= 2.223 mol C
H: 2.2 g
¸ 1.01 g/mol
= 2.18 mol H
O: 71.1 g ¸ 16.00 g/mol
= 4.444 mol O
3:
C
2.223
mol
mol
2.223/2.18 =
(reduced)
1.02
H
O
2.18
4.444
2.18/ 2.18
=1
4.444/2.18
= 2.04
4: the simplest formula is CHO2
5: factor = 90/45=2. Molecular formula: C2H2O4
Question 8, 9
• For the empirical formula we need to know the moles of
each element in the compound (which can be derived from
grams or %).
For the molecular formula we need the above information
& the molar mass of the compound
• Molar mass of CH = 13 g/mol
Factor = 104 g/mol ¸ 13 g/mol = 8
Molecular formula is C8H8
Question 10
1: Assume 100 g total. Thus:
53.2 g C, 11.2 g H, and 35.6 g O
2: C: 53.2 g ¸ 12.01 g/mol = 4.430 mol C
H: 11.2 g ¸ 1.01 g/mol
= 11.09 mol H
O: 35.6 g ¸ 16.00 g/mol = 2.225 mol O
3:
C
H
O
mol
4.430
11.09
2.225
mol
4.43/2.225 11.09/2.225 2.225/2.225
(reduced)
= 1.99
= 4.98
=1
4: the simplest formula is C2H5O
5: factor = 90/45=2. Molecular formula: C4H10O2
Assignment
1. Calculate the percentage composition of each
substance: a) SiH4, b) FeSO4
2. Calculate the simplest formulas for the compounds
whose compositions are listed:
a) carbon, 15.8%; sulfur, 84.2%
b) silver,70.1%; nitrogen,9.1%; oxygen,20.8%
c) K, 26.6%; Cr, 35.4%, O, 38.0%
3. The simplest formula for glucose is CH2O and its
molar mass is 180 g/mol. What is its molecular
formula?
4. Determine the molecular formula for each
compound below from the information listed.
substance
simplest formula
molar mass(g/mol)
a) octane
C4H9
114
b) ethanol
C2H6O
46
c) naphthalene
C5H4
128
d) melamine
CH2N2
126
5. The percentage composition and approximate
molar masses of some compounds are listed below.
Calculate the molecular formula of each
percentage composition
molar mass(g/mol)
64.9% C, 13.5% H, 21.6% O
74
39.9% C, 6.7% H, 53.4 % O
60
40.3% B, 52.2% N, 7.5% H
80
1 a) Si= 87.43% (28.09/32.13 x 100), H= 12.57%
b) Fe= 36.77% (55.85/151.91 x 100),
21.10% (32.06/151.91 x 100), O= 42.13%
S=
2 a) Assume 100 g. Thus: 15.8 g C, 84.2 g S.
C: 15.8 g ¸ 12.01 g/mol
= 1.315 mol C
S: 84.2 g ¸ 32.06 g/mol
= 2.626 mol S
Mol
Mol reduced
C
1.315
1.315/1.315
=1
the simplest formula is CS2
S
2.626
2.626/1.315
= 2.00
2 b) Ag:
N:
O:
AgNO2
Mol
Mol
reduced
2 c) K:
Cr:
O:
K2Cr2O7
Mol
Mol
reduced
70.1 g ¸ 107.87 g/mol = 0.6499 mol Ag
9.1 g ¸ 14.01 g/mol = 0.6495 mol N
20.8 g ¸ 16.00 g/mol = 1.30 mol O
Ag
N
O
0.6499
0.6495
1.30
.6499/.6495 .6495/.6495 1.30/.6495
= 1.0
=1
= 2.00
26.6 g ¸ 39.10 g/mol = 0.6803 mol K
35.4 g ¸ 52.00 g/mol = 0.6808 mol Cr
38.0 g ¸ 16.00 g/mol = 2.375 mol O
K
Cr
O
0.6803
0.6808
2.375
.6803/.6803 .6808/.6803= 2.375/.6495
=1
1.00
= 3.49
3 C6H12O6 (CH2O = 30 g/mol, 180/30 = 6)
4 a) C8H18 (C4H9 = 57 g/mol, 114/57 = 2)
b) C2H6O (C2H6O = 46 g/mol, 46/46 = 1)
c) C10H8 (C5H4 = 64 g/mol, 128/64 = 2)
d) C3H6N6 (CH2N2 = 54 g/mol, 126/42 = 3)
5 a) C: 64.9 g ¸ 12.01 g/mol = 5.404 mol C
H: 13.5 g ¸ 1.01 g/mol = 13.37 mol H
O: 21.6 g ¸ 16.00 g/mol = 1.35 mol O
C4H10O
C
H
O
Mol
5.404
13.37
1.35
Mol
5.404/1.35
13.37/1.35
1.35/1.35
reduced
= 4.00
= 9.90
=1
C4H10O (C4H10O = 74 g/mol, 74/74 = 1)
5 b) C: 39.9 g ¸ 12.01 g/mol = 3.322 mol C
H:
6.7 g ¸ 1.01 g/mol = 6.63 mol H
O: 53.4 g ¸ 16.00 g/mol = 3.338 mol O
CH2O
C
H
O
Mol
3.322
6.63
3.338
Mol
3.322/3.322 6.63/3.322 3.338/3.322
reduced
=1
= 2.0
= 1.00
C2H4O2 (CH2O = 30 g/mol, 60/30 = 2)
5 c)
B
N
H
Mol
3.728
3.726
7.43
Mol
3.728/3.726 3.726/3.726 7.43/3.726
reduced
= 1.00
=1
= 2.0
B3N3H6 (BNH2 = 26.84 g/mol, 80/26.84= 2.98)
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