Download WKS-Empirical formulas key

WKS
Empirical Formula
Name
Period
Answer Key
Date
Procedure for finding empirical formulas:
You are trying to determine the subscripts for each element in the formula, so you must determine the
number of moles of each element in the sample, then reduce them to their lowest while number. Follow the
following process:
(1) Determine mass of each element in g.
a. If given % composition, simply change units from % to g
(2) Convert g of each element to actual # of moles using molar mass.
(3) Find the whole number ratio of moles of each element relative to one another by dividing all by the
lowest mole value (this assumes at least one subscript will be 1). If the ratio is close to a whole number,
round to that whole number (e.g. 1.99→2, 3.01→3)
(4) If any of the relative moles end in .5, multiply them all by 2 to make them integers (e.g. 1 and 1.5
multiply to 2 and 3)
Flowchart: % → mass → moles → divide by smallest → subscripts
Example:
An unknown sample has 47.08 % C, 6.59 % H, and 46.33 % Cl. What is its empirical formula?
1 mol C
C: 47.08% ⇒ 47.08 g ×
= 3.920
mol C ÷ 1.307
= 2.999 ≈ 3
12.01 g C
1 mol H
H: 6.59% ⇒ 6.59 g ×
= 6.54
mol H ÷ 1.307
= 5.00
1.008 g H
1 mol Cl
Cl: 46.33% ⇒ 46.33 g ×
= 1.307
mol Cl ÷ 1.307
= 1
35.45 g Cl
Empirical Formula = C3H5Cl
Problems:
1. An unknown compound was found to have a percent composition as follows: 56.6% potassium, 8.69%
carbon, and 34.7% oxygen. What is its empirical formula?
1 mol K
K: 56.6% ⇒ 56.6 g ×
= 1.45 mol K ÷ 0.724 = 2.00
39.10 g K
1 mol C
C: 8.69% ⇒ 8.69 g ×
= 0.724 mol C ÷ 0.724 = 1
12.01 g C
1 mol O
O: 34.7% ⇒ 34.7 g ×
= 2.17 mol O ÷ 0.724 = 3.00
16.00 g O
Empirical Formula = K2CO3
2. Rubbing alcohol was found to contain 60.0 % carbon, 13.4 % hydrogen, and the remaining mass was
due to oxygen. What is the empirical formula of rubbing alcohol?
1 mol C
C: 60.0% ⇒ 60.0 g ×
= 5.00 mol C ÷ 1.66 = 3.01 ≈ 3
12.01 g C
1 mol H
H: 13.4% ⇒ 13.4 g ×
= 13.3 mol H ÷ 1.66 = 7.98 ≈ 8
1.008 mol H
1 mol O
O: 26.6% ⇒ 26.6 g ×
= 1.66 mol O ÷ 1.66 = 1
16.00 g O
Empirical Formula = C3H8O
3. Isobutylene is a raw material for making synthetic rubber. A sample with a mass of 0.6481 grams was
found to contain 0.5555 grams of carbon; the rest was hydrogen. What is the empirical formula of
isobutylene?
Mass H = 0.6481 g – 0.5555 g = 0.0926 g
1 mol C
C: 0.5555 g ×
= 0.0463 mol C ÷ 0.0463 = 1
12.01 g C
H: 0.0926 g ×
1 mol H
= 0.0921 mol H ÷ 0.0463 = 1.99 ≈ 2
1.008 g H
Empirical Formula = CH2
4. There are two oxides of iron, FeO and Fe2O3. A sample of an iron oxide was found to contain 60.68 g Fe
and 26.07 g O. Determine its empirical formula and identify which oxide it is.
1 mol Fe
Fe: 60.68 g ×
= 1.086 mol Fe ÷ 1.086 = 1× 2 = 2
55.85 g Fe
O: 26.07 g ×
1 mol O
= 1.629 mol O ÷ 1.086 = 1.5 × 2 = 3
16.00 g O
Empirical Formula = Fe2O3
5. Determine the empirical formula for a compound that contains 35.98% aluminum and 64.02% sulfur.
1 mol Al
Al: 35.98% ⇒ 35.98 g ×
= 1.334 mol Al ÷ 1.334 = 1× 2 = 2
26.98 g Al
S: 64.02% ⇒ 64.02 g ×
1 mol S
= 1.996 mol S ÷ 1.334 = 1.5 × 2 = 3
32.07 g S
Empirical Formula = Al2S3
6. Propane is a hydrocarbon, a compound composed only of carbon and hydrogen. It is 81.82% carbon and
18.18% hydrogen. What is the empirical formula (numbers ending in .33 or .66 must be multiplied by
3)?
1 mol C
C: 81.82% ⇒ 81.82 g ×
= 6.813 mol C ÷ 6.813 = 1× 3 = 3
12.01 g C
1 mol H
H: 18.18% ⇒ 18.18 g ×
= 18.04 mol H ÷ 6.813 = 2.647 × 3 = 7.942 ≈ 8
1.008 H
Empirical Formula = C3H8