Percent Composition(Section 11.4)

• Helps determine identity of unknown compound– Think CSI—they use a mass spectrometer

• Percent by mass of each element in a compound

100xcompoundofmass

elementofmassmassbypercent

To calculate percent composition

• H2O– Use molar masses to calculate mass of compound

• 2 x 1.0 + 1 x 16.0 = 18.0 g/mol

– Use molar masses to calculate mass of each individual element

• H = 2 x 1.0 = 2.0 g/mol• O = 1 x 16 = 16.0 g/mol

– Calculate using the formula

• H = (2 / 18) x 100 = 11.1 % Hydrogen• O = (16/18) x 100 = 88.9 % Oxygen• (notice to check your work: 11.1 + 88.9 = 100%)

100xcompoundofmass

elementofmassmassbypercent

Percent Composition of Water• H = (2 / 18) x 100 = 11.1 %• O = (16/18) x 100 = 88.9 %• Even though there are 2 hydrogen atoms in water, most of

the mass of water comes from the 1 oxygen atom

hydrogen

oxygen

Percent Composition Practice

• KNO2Percent composition of N– Molar mass of N = 1 x 14 = 14 g/mol– Molar mass of compound = 1 x 39.1 + 1 x 14 + 2 x 16

= 85.1 g/mol– % by mass of N = (14 / 85.1) x 100 = 16.45%

100xcompoundofmass

elementofmassmassbypercent

• Calculate the percent composition– Phosphorus in Al2(PO4)3

• P = 3 x 31.0 = 93 g/mol

• Al2(PO4)3 = 2 x 27 + 3 x 31 + 12 x 16 = 339 g/mol

• % P = (93 / 339) x 100 = 27.43%27.43%

– Chlorine in LiCl• Cl = 1 x 35.5 = 35.5 g/mol• LiCl= 1 x 6.9 + 1 x 35.5 = 42.4 g/mol• % Cl = (35.5 / 42.4) x 100 = 83.73%83.73%

– Calcium in CaSO4

• Ca = 1 x 40.1 = 40.1 g/mol

• CaSO4 = 1 x 40.1 + 1 x 32.1 + 4 x 16 = 136.2 g/mol

• % Ca = (40.1 / 136.2) x 100 = 29.44%29.44%

• Calculate the percent composition– Sodium in NaCl

• Na = 1 x 23 g/mol = 23 g/mol• NaCl = 1 x 23 + 1 x 35.5 = 58.5 g/mol• % Na = (23 / 58.5) x 100 = 39.32%39.32%

– Sulfur in H2SO4

• S = 1 x 32.1 = 32.1 g/mol

• H2SO4 = 2 x 1 + 1 x 32.1 + 4 x 16 = 98.1 g/mol

• % S = (32.1 / 98.1) x 100 = 32.72%32.72%

– Potassium in KNO3

• K = 1 x 39.1 = 39.1 g/mol

• KNO3 = 1 x 39.1 + 1 x 14 + 3 x 16 = 101.1 g/mol

• % K = (39.1 / 101.1) x 100 = 38.67%38.67%

Empirical vs Molecular Formulas

• Empirical formula = formula with smallest whole number ratio of elements

• Molecular formula = formula with actual number of atoms of each element

Empirical Molecular

C4H9 C8H18

H2O H6O3

We use the percent composition to calculate the empirical and molecular formulas.

How to calculate empirical formula

1. Use mass of elements to calculate moles• If given percentages, assume sample is 100

grams. Therefore, percent by mass will equal the actual mass of element.

2. Pick which element has the smallest number of moles

3. Divide each element by the smallest number of moles to get ratio (these numbers become subscripts)

4. Write empirical formula

• A blue solid is found to contain 36.43% nitrogen and 63.16% oxygen. What is the empirical formula?

If we assume sample is 100 grams, we have 36.43 g N and 63.16 g O.

1. Calculate moles of each elementN = 36.43 / 14 = 2.60 mol

O = 63.16 / 16 = 3.95 mol

2. Smallest number of moles = nitrogen with 2.60 mol

3. Divide by smallest number of molesN = 2.60 / 2.60 = 1 x 2 = 2

O = 3.95 / 2.60 = 1.5 x 2 = 3

we must get whole numbers as our answer—that’s why we multiplied by two

4. Empirical formula = N2O3

• Determine the empirical formula for a compound that contains 35.98% aluminum and 64.02% sulfur

If we assume sample is 100 grams, we have 35.98 g Al and 64.02 g S.

1. Calculate moles of each elementAl = 35.98 / 27 = 1.33 mol

S = 64.02 / 32.1 = 2.00 mol

2. Smallest number of moles = aluminum with 1.33 mol

3. Divide by smallest number of molesAl = 1.33 / 1.33 = 1 x 2 = 2

S = 2.00 / 1.33 = 1.5 x 2 = 3

we must get whole numbers as our answer—that’s why we multiplied by two

4. Empirical formula = Al2S3

• Find the empirical formula for a substance that contains 48.64% carbon, 8.16% hydrogen, and 43.20% oxygen

If we assume sample is 100 grams, we have 48.64 g C, 8.16 g H, and 43.20 g O.

1. Calculate moles of each elementC = 48.64 / 12 = 4.05 molH = 8.16 / 1 = 8.16 molO = 43.20 / 16 = 2.70 mol

2. Smallest number of moles = oxygen with 2.70 moles3. Divide by smallest number of moles

C = 4.05 / 2.70 = 1.5 x 2 = 3H = 8.16 / 2.70 = 3.0 x 2 = 6O = 2.70 / 2.70 = 1.0 x 2 = 2we must get whole numbers as our answer—that’s why we multiplied by two

4. Empirical formula = C3H6O2