Transcript Molecular Formula

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 69.58 g 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
mol
0.50666/ 0.50708/
(reduced) 0.50666 0.50666
=1
= 1.001
4: the simplest formula is BaCO3
1.520
1.520/
0.50666
= 3.000
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
A7B3
A = 2.34 mol, B = 1 mol
A2B3
A = 1 mol, B = 1.48 mol
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. - 6. Try questions 3 - 6 on page 189.
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/
1.90/
(reduced)
1.263
1.263
1.263
= 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
mol
(reduced)
0.5995
0.5995/
0.5995
=1
1.188
1.188/
0.5995
= 1.98
4: the simplest formula is CH2O
0.60
0.60/
0.5995
= 1.0
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/
(reduced) 0.5995
0.5995
0.5995
=1
= 1.98
= 1.0
4: the simplest formula is CH2O
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
O2F2 70  35 = 2
compound is 70 g/mol
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
H
O
mol
2.223
2.18
4.444
mol
2.223/2.18 2.18/ 2.18 4.444/2.18
(reduced)
= 1.02
=1
= 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
C 4H 9
114
b) ethanol
C 2H 6O
46
c) naphthalene
C 5H 4
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),
S= 21.10% (32.06/151.91 x 100), O= 42.13%
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: 70.1 g  107.87 g/mol = 0.6499 mol Ag
N: 9.1 g  14.01 g/mol = 0.6495 mol N
O: 20.8 g  16.00 g/mol = 1.30 mol O
AgNO2
Ag
N
O
Mol
0.6499
0.6495
1.30
Mol
.6499/.6495 .6495/.6495 1.30/.6495
reduced
= 1.0
=1
= 2.00
2 c) K: 26.6 g  39.10 g/mol = 0.6803 mol K
Cr: 35.4 g  52.00 g/mol = 0.6808 mol Cr
O: 38.0 g  16.00 g/mol = 2.375 mol O
K2Cr2O7
K
Cr
O
Mol
0.6803
0.6808
2.375
Mol
.6803/.6803 .6808/.6803 2.375/.6495
reduced
=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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