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PRACTICE
Empirical Formulas
NAME: _____________________________________
Period: ______________________________________
1.
What is the empirical formula for a compound that is 10.12 g of aluminum and 17.93 g of sulfur?
10.12gAl →
10.12gAl 1molAl
= 0.3751molAl ÷ 0.3751 = 1.000 →1.000 × 2 = 2.00 → 2Al
26.98gAl
17.93gS 1molS
= 0.5591molS ÷ 0.3751 = 1.491 →1.491× 2 = 2.981 → 3S
32.07gS
Empirical formula = Al2S3
17.93gS →
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2.
A compound is found to contain 4.20 g of nitrogen and 12.0 g of oxygen. What is the empirical
formula for this compound?
4.20gN →
4.20gN 1molN
= 0.300molN ÷ 0.300 = 1.00 →1.00 × 2 = 2.00 → 2N
14.01gN
12.0gO 1molO
= 0.750molO ÷ 0.300 = 2.50 → 2.50 × 2 = 5.00 → 5O
16.00gO
Empirical formula = N2O5
12.0gO →
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3.
A compound is composed of 4.80 g of carbon and 0.40 g of hydrogen. Determine the empirical
formula for this compound.
4.80gC →
4.80gC 1molC
= 0.400molC ÷ 0.40 = 1.0 →1C
12.01gC
0.40gH 1molH
= 0.40molO ÷ 0.40 = 1.0 →1H
1.01gH
Empirical formula = CH
0.40gH →
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4.
Find the empirical formula for a compound which is 63.6% iron and 36.4% sulfur. Assuming the
empirical formula is the same as the molecular formula for this compound, name this compound.
63.6gFe 1molFe
63.6%Fe → 63.6gFe →
= 1.14molFe ÷1.14 = 1.00 →1Fe
55.85gFe
36.4gS 1molS
= 1.14molS ÷1.14 = 1.00 →1S
32.07gS
Empirical formula = FeS
36.4%S → 36.4gS →
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5.
A compound contains Li, N, and O. If the Li in this compound has a mass of 4.83 g, the N has a
mass of 9.80 g, and the O has a mass of 33.6g, what is the empirical formula for this compound?
4.83gL 1molLi
4.83gLi →
= 0.696molLi ÷ 0.696 = 1.00 →1Li
6.94gLi
9.80gN →
9.80gN 1molN
= 0.700molS ÷ 0.696 = 1.01 →1N
14.01gN
33.6gO 1molO
= 2.1molO ÷ 0.696 = 3.02 → 3O
16.00gO
Empirical formula = LiNO3
33.6gO →
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6.
A bright yellow precipitate is formed during a reaction. When analyzed, this compound is found
to contain 82.88 g of lead and 101.52 g of iodine. What is the empirical formula? Assume the
empirical formula is the same as the molecular formula for this compound, and name the
compound.
82.88gPb 1molPb
82.88gPb →
= 0.4000molPb ÷ 0.4000 = 1.000 →1Pb
207.2gPb
1molI
=,80000molI ÷ 0.4000 = 2.000 → 2I
126.90gI
Empirical formula = PbI2
Name = Lead(II) iodide
101.52gI →
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101.52gI
7.
A compound is composed of 19.01 g of carbon, 18.48 g of nitrogen, 25.34 g of oxygen, and 1.58 g
of hydrogen. What is the empirical formula for this compound?
19.01gC 1molC
19.01gC →
= 1.583molC ÷1.319 = 1.200 →1.200 × 5 = 6.000 → 6C
12.01gC
18.48gN →
18.48gN 1molN
= 1.319molN ÷1.319 = 1.000 →1.000 × 5 → 5N
14.01gN
25.34gO →
25.34gO 1molO
= 1.584molO ÷1.319 = 1.201 →1.201× 5 = 6.005 → 6O
16.00gO
1.58gH 1molH
= 1.56molH ÷1.319 = 1.18 →1.18 × 5 = 5.90 → 6H
1.01gH
Empirical formula = C6N5O6H6
1.58gH →
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