Transcript EMPIRICAL FORMULA

EMPIRICAL FORMULA
ANSWERS
• 1. Determine the empirical formula of
a compound that contains
2.94 g oxygen and 1.96 g sulfur.
1 mol
• 2.94 g oxygen x --------------- = 0.184 mol
16 g
1 mol
• 1.96 g sulfur x ----------------- = 0.0613 mol
32 g
THIS COLOR - USE THE PERIODIC TABLE
• Divide each by the smallest amount of mol.
0.0613
S = --------- = 1
0.0613
0.184
O = -------- = 3
0.0613
SO3
1 mol
• 40.3 g K x --------------- =
39 g
1.03 mol
1 mol
• 26.8 g Cr x ----------------- = 0.525 mol
52 g
1 mol
• 32.9 g O x ---------------- =
16 g
2.06 mol
THIS COLOR - USE THE PERIODIC TABLE
• 2. What is the empirical formula of a
compound that contains 40.3% potassium,
26.8% chromium, and 32.9% oxygen?
• Divide each by the smallest amount of mol.
1.03
K = --------- = 2
2.06
0.525
O = -------- = 4
0.525
0.525
Cr = -------- = 1
0.525
2
4
K CrO
1 mol
• 65.5 g C x --------------- =
12 g
5.46 mol
1 mol
• 5.5 g H x ----------------- =
1g
5.5 mol
1 mol
• 29.0 g O x ---------------- = 1.8 mol
16 g
THIS COLOR - USE THE PERIODIC TABLE
• 3. A compound is 65.5% carbon,
5.5% hydrogen, and 29.0% oxygen.
Find the empirical formula.
• Divide each by the smallest amount of mol.
5.45
C = --------- = 3
1.8
1.8
O = -------- = 1
1.8
5.5
H = -------- = 3
1.8
3 3
CHO
1 mol
• 18.7 g Li x --------------- =
7g
1 mol
• 16.3 g C x ----------------- =
12 g
2.67 mol
1.36 mol
1 mol
• 65 g O x ---------------- = 4.06 mol
16 g
THIS COLOR - USE THE PERIODIC TABLE
• 4. What’s the empirical formula of a molecule
containing 18.7% lithium, 16.3% carbon,
and 65% oxygen?
• Divide each by the smallest amount of mol.
2.67
Li = --------- = 2
4.06
1.36
O = -------- = 3
1.36
1.36
C = -------- = 1
1.36
2
3
Li CO
MOLECULAR FORMULA
ANSWERS
• 1) A compound with an empirical formula
of C2OH4 and a molar mass of 88 grams
per mole.
• C = 12 g x 2 = 24 g
Molar mass
• O = 16 g
= 16 g
--------------• H=1g x4= 4g
Empirical
Formula
•
+____
mass
•
44 g
C4H2O8
88 g
------ = 2
44 g
• 2) A compound with an empirical formula
of C4H4O and a molar mass of 136 grams
per mole.
• C = 12 g x 4 = 48 g
Molar mass
• H=1g x4= 4g
--------------• O = 16 g
= 16 g
Empirical
Formula
•
+____
mass
•
68 g
C8H8O2
136 g
------ = 2
68 g
• 3) A compound with an empirical formula
of CFBrO and a molar mass of
254.7 grams per mole.
• C = 12 g
= 12 g
Molar mass
• F = 19 g
= 19 g
--------------• Br = 80 g
= 80 g
Empirical
Formula
• O = 16 g
= 16 g
mass
•
+____
•
127 g
C2F2Br2O2
254.7 g
-------- = 2
127 g
• 4) A compound with an empirical formula
of C2H8N and a molar mass of 46 grams
per mole.
• C = 12 g x 2 = 24 g
Molar mass
• H=1g x8= 8g
--------------• N = 14 g
= 14 g
Empirical
Formula
•
+____
mass
•
46 g
C2H8N
46 g
------ = 1
46 g