Chemistry Notes - The Mole Page 4
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4D. examples
E16) Calculate the empirical formula of a compound composed of 67.6%
mercury, 10.8% sulfur, and 21.6% oxygen. (check the PROBLEM PROCEDURE steps)
STEP 1… mercury: 67.6% Hg = 67.6 g Hg out of 100 g cmpd.
sulfur: 10.8% S = 10.8 g S out of 100 g cmpd.
oxygen: 21.6% O = 21.6 g C out of 100 g cmpd.
STEP 2... Hg: 67.6 g C x 1 mol Hg = 0.337 mol Hg
S: 10.8 g S x 1 mol S = 0.336 mol S
200.6 g Hg
32.1 g S
O: 21.6 g O x 1 mol O = 1.35 mol O
16.0 g
STEP 3… Hg = 0.337 = 1
S = 0.336 = 1
O = 1.35 mol = 4
0.336
0.336
0.336 mol
STEP 4… Hg
S
O
= HgSO
1
1
4
4
E17) What is the empirical formula of a compound with 25.9% nitrogen and
74.1% oxygen?
25.9% N = 25.9 g N out of 100 g cmpd.
74.1% O = 74.1 g O out of 100 g cmpd.
N: 25.9 g N x 1 mol N = 1.85 mol N
O: 74.1 g O x 1 mol O = 4.63 mol O
14.0 g N
16.0 g O
N: 1.85 = 1
O: 4.63 = 2.5
N
O
– can’t have .5 subscripts x 2 = N
O
1
2.5
2
5
1.85
1.85
IX.
Molecular Formula
A. a multiple of the empirical formula
B. still whole number ratios
C. examples
E18) A compound with an empirical formula of CH has a molecular weight of 78.0
g/mol. What is the molecular formula?
molar mass of CH = 12.0 + 1.0 = 13.0 g
78 / 13 = 6
molecular formula = C
H
6
6
E19) A compound is 75.46% carbon, 4.43% hydrogen, and 20.10% oxygen by mass.
It has a molecular mass of 318.31 g/mol. What is the molecular formula for this compound?
67.6% C = 75.46 g C out of 100 g cmpd.
4.43% H = 4.43 g H out of 100 g cmpd.
20.10% O = 20.10 g O out of 100 g cmpd.
(75.46 g C) (1 mol/ 12.0 g C) = 6.29 mol C
(4.43 g H) (1 mol/ 1.0 g H) = 4.4 mol H
(20.10 g O) (1 mol/ 16.0 g O) = 1.26 mol O
(6.29 mol C)/ (1.26) = 4.99 = 5 mol C
(4.4 mol H)/ (1.26) = 3.49 = 3.5 mol H
(1.26 mol O)/ (1.26) = 1 mol O
.5 value means multiply subscripts by 2:
empirical fmla. = C
H
O
10
7
2
Now that you have the emp.fmla., you can find the molecular fmla like in problem E18.
emp. fmla. mass = 10(12.0) + 7(1.0) + 2(16.0) = 159.1 g/mol
The problem says the molecular mass is 318.31 g per mole.
318.31 g/mol = 2.001 = 2 ratio
159.1 g/mol
Since there are two empirical units in a molecular unit, the molecular formula = C
H
O
20
14
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