Percent Composition and Empirical Formulas
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Percent Composition and Empirical Formulas
Percent Composition
Percent Composition
The percent composition shows the relative percent (by mass) of each element in a compound.
Percent Composition
The percent composition shows the relative percent (by mass) of each element in a compound.
The percent composition is determined by dividing the mass of the individual elements in a compound by the entire formula mass of the compound.
Percent Composition =
Mass of individual element (g)
× 100 %
= % of element
Formula Mass of Compound (g)
For example, when the correct percent composition for HF is determined, the process is as follows:
For example, when the correct percent composition for HF is determined, the process is as follows:
Find the total formula molar mass: 1 mol H = 1.0079 g/mol1 mol F = + 18.9884 g/mol total = 19.9963 g/mol
Take the individual molar mass of each element and divide by the total formula mass, and turn it into a percent:
Take the individual molar mass of each element and divide by the total formula mass, and turn it into a percent:
for H1.0079 g/mol H
× 100 % = 5.04% H19.9963 g/mol HF
for F18.9884 g/mol F
× 100 % = 94.96% F19.9963 g/mol HF
A good way to quickly check the answers is to sum the percentages, which should equal 100% (or 1). There will be cases where the percentages might not equal exactly 100% because of rounding, but the total should always be VERY close to 100%.
As another example, consider sulfuric acid:H2SO4 :
As another example, consider sulfuric acid:H2SO4 :
2 mol H × 1.008 g/mol = 2.016 g/mol
1 mol S × 32.066 g/mol = 32.066 g/mol
4 mol O × 15.999 g/mol = 63.996 g/mol
total = 98.078 g/mol
As another example, consider sulfuric acid:H2SO4 :
for H2.016 g/mol H
× 100% = 2.06% H98.078 g/mol H2SO4
for S 32.066 g/mol S× 100% = 32.69% S
98.078 g/mol H2SO4
for O63.996 g/mol O
× 100% = 65.25% O98.078 g/mol H2SO4
Empirical Formula
Empirical FormulaOnce the percent composition of a compound is known, the empirical formula of the compound can be determined. An empirical formula shows the lowest whole-number ratio of the elements in a compound
1. Turn the percent composition information into mass. This is made simple by assuming a theoretical amount of 100 grams.
Thus 50% composition is turned into 50 grams, and 36.8% composition is turned into 36.8 grams, etc.
2. Calculate the number of moles for each element that would contain the amount of mass from step 1.
This involves dividing the mass from step 1 by the molar mass shown for the element on the periodic table.
3. The simplest whole-number ratio of each element needs to be found.
One of the ways to get a good start on this is to divide each number of moles from step 2 by the smallest amount of moles.
This will guarantee at least one whole number to start with (a “1” amount).
3. a. If the other molar amounts are within 0.15 of a whole number, it is usually safe to round up or down to that whole number.
3. b. If the other molar amounts cannot be rounded, it will be necessary to multiply ALL the molar amounts by a whole number to obtain a whole number (or a number close to a whole number.) Thus, if a molar amount had the decimal value of 0.20, it would be necessary to multiply by 5. If the decimal value is 0.25,it would be necessary to multiply by 4, and it would if the decimal value is 0.33, it would be necessary to multiply by 3, etc.
Example:White gold is 75.0% gold, 10.0% palladium, 10.0% nickel,
and 5.00% zinc. What would be the empirical formula of white gold?
75.0% Au → 75.0 g Au 1 mole Au
= 0.3807 moles Au
197.0 g
10.0% Pd → 10.0 g Pd 1 mole Pd= 0.09398 moles Pd
106.4 g
10.0% Ni → 10.0 g Ni 1 mole Ni= 0.1704 moles Ni
58.69 g
5.00% Zn → 5.00 g Zn 1 mole Zn= 0.07646 moles Zn
65.39 g
Dividing by the lowest amount of moles from above (0.07646 mol): 0.3807 moles Au
= 4.979 moles Au ≈ 5 moles Au0.07646
0.09398 moles Pd = 1.229 moles Pd
0.07646
0.1704 moles Ni = 2.229 moles Ni
0.07646
0.07646 moles Zn = 1 moles Zn
0.07646
The gold and zinc are already expressed in a whole number, but to express the palladium and nickel as a whole number, it will be necessary to multiply everything by 4.
This would make the palladium and nickel 4.916 moles and 8.916 moles (respectively), which are now close enough to round. Do not forget to multiply everything, even the ones that are already whole numbers!
The gold and zinc are already expressed in a whole number, but to express the palladium and nickel as a whole number, it will be necessary to multiply everything by 4.
This would make the palladium and nickel 4.916 moles and 8.916 moles (respectively), which are now close enough to round. Do not forget to multiply everything, even the ones that are already whole numbers!
Thus the final relative amount of moles is 20 Au, 5 Pd, 9 Ni, 4 Zn. The empirical formula is Au20Pd5Ni9Zn4.
Practice Exercise:Find the empirical formula for purple gold, Purple Gold =
80% Au, 20% Al
mass 1 mol
=
answer
= answer × factor = whole #P.T. mass lowest #
mass 1 mol
=
answer
= answer × factor = whole #P.T. mass lowest #
80 g 1 mol=
0.4062 mol
= answer × factor = whole #
196.97 g lowest #
20 g 1 mol=
0.7413 mol
= answer × factor = whole #
26.98 g lowest #
80 g 1 mol=
0.4062 mol
= 1 × factor = whole #
196.97 g 0.4062 mol
20 g 1 mol=
0.7413 mol
= 1.825 × factor = whole #
26.98 g 0.4062 mol
80 g 1 mol=
0.4062 mol
= 1 × 5 = 5196.97 g 0.4062
mol
20 g 1 mol=
0.7413 mol
= 1.825 × 5 = 926.98 g 0.4062
mol
80 g 1 mol=
0.4062 mol
= 1 × 5 = 5196.97 g 0.4062
mol
20 g 1 mol=
0.7413 mol
= 1.825 × 5 = 926.98 g 0.4062
mol
Empirical Formula = Au5Al9
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