Percent Composition and Empirical Formulas

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:

Find the total formula molar mass: 1 mol H = g/mol 1 mol F = g/mol total = g/mol

Take the individual molar mass of each element and divide by the total formula mass, and turn it into a percent:

for H g/mol H × 100 %= 5.04% H g/mol HF for F g/mol H × 100 %= 94.96% F 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: H 2 SO 4 :

As another example, consider sulfuric acid: H 2 SO 4 : 2 mol H × g/mol= g/mol 1 mol S × g/mol= g/mol 4 mol O × g/mol= g/mol total = g/mol

As another example, consider sulfuric acid: H 2 SO 4 : for H g/mol H × 100%= 2.06% H g/mol H 2 SO 4 for S g/mol S × 100%= 32.69% S g/mol H 2 SO 4 for O g/mol O × 100%= 65.25% O g/mol H 2 SO 4

Empirical Formula

Once 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 2. This involves dividing the mass from step 2 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 3 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 Au1 mole Au = moles Au g 10.0% Pd →10.0 g Pd1 mole Pd = moles Pd g 10.0% Ni →10.0 g Ni1 mole Ni = moles Ni g 5.00% Zn →5.00 g Zn1 mole Zn = moles Zn g

Dividing by the lowest amount of moles from above ( mol): moles Au = moles Au ≈ 5 moles Au moles Pd = moles Pd moles Ni = moles Ni moles Zn = 1 moles Zn

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 moles and 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 moles and 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 Au 20 Pd 5 Ni 9 Zn 4.

Practice Exercise: Find the empirical formula for purple gold, Purple Gold = 80% Au, 20% Al

mass1 mol = answer = ×factor=whole # P.T. masslowest # mass1 mol = answer = ×factor=whole # P.T. masslowest #

80 g1 mol = mol = answe r × fact o r = whole # glowest # 20 g1 mol = mol = answe r × fact o r = whole # glowest #

80 g1 mol = mol =1× fact o r = whole # g mol 20 g1 mol = mol =1.825× fact o r = whole # g mol

80 g1 mol = mol =1×5= g mol 20 g1 mol = mol =1.825×5= g mol

80 g1 mol = mol =1×5= g mol 20 g1 mol = mol =1.825×5= g mol Empirical Formula = Au 5 Al 9