Empirical & Molecular Formulas

Chapter 8

Empirical & Molecular Formulas

Formaldehyde Acetic Acid Glucose• colorless gas with a

characteristic pungent odor

• known Human Carcinogen

• colorless liquid that when undiluted is also called glacial acetic acid

• main component of vinegar

• simple sugar• product of

photosynthesis• the cell’s

primary energy source

CH2O CH3CO2H (CH3COOH) C6H12O6

CH2O

(CH2O)n

(CH2O)6(CH2O)2

(CH2O)n (CH2O)n

Empirical Formula

• the formula of a compound that expresses the smallest whole-number ratio of the atoms present

• the simplest formula

• the formula that gives the composition of the molecules present

Molecular Formula

CH2O C6H12O6

4

Magnesium oxide: Find the empirical formula

• Data: 0.519g Mg, 0.341g O

• 0.519g(1mol/24.31g) = 0.0213mol Mg

• 0.341g(1mol/16.00g) = 0.0213mol O

• Divide by the smallest mole value (in this case they are the same)

0.0213mol Mg/0.0213 = 1mol Mg

0.0213mol O/0.0213 = 1 mol O

• MgO

5

Using Percent Composition (by mass) to find empirical formula

• % Mg and O in MgO?

• 60.3% Mg and 39.7% O

• Treat the % values as mass values and change to moles60.3g Mg and 39.7g O

60.3g(1mol/24.31g Mg) = 2.48mol Mg

39.7g(1mol/16.00g O) = 2.48mol O

• Divide by the smallest mole value2.48mol Mg/2.48 = 1mol Mg

2.48mol O/2.48 = 1mol O

• MgO

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An iron, oxygen, hydrogen compound was analyzed to be 52.2 % iron, 44.9 % oxygen and the rest hydrogen. Determine the empirical formula for this compound. What is its name?• Determine the % of hydrogen.

✓ 100% - 52.2 - 44.9 = 2.9%

• Treat the % values as mass values.✓ Change to moles

‣ Fe: 52.2 g*1mole/55.8g = 0.935 mole‣ O: 44.9 g*1mole/16g = 2.81 mole‣ H: 2.9 g*1mole/1.01g = 2.9 mole

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An iron, oxygen, hydrogen compound was analyzed to be 52.2 % iron, 44.9 % oxygen and the rest hydrogen. Determine the empirical formula for this compound. What is its name?

• Divide by the mole value by the smallest mole value✓ Fe: 0.94 / 0.94 = 1✓ O: 2.81 / 0.94 = 3.0✓ H: 2.9 / 0.94 = 3.1

• Round to the nearest whole number and Voilà: FeO3H3

• Consider the polyatomic ions can you find something in the chart??✓ Fe(OH)3

• Determine the name.✓ iron (III) hydroxide

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A compound is determined to be 38.76 % calcium, 19.97 % phosphorus and the rest oxygen. Determine the empirical formula.Then name this compound.

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A compound is determined to be 38.76 % calcium, 19.97 % phosphorus and the rest oxygen. Determine the empirical formula. Then name this compound

• Determine the % of oxygen.✓ 100% - 38.76% - 19.97% = 41.27%

• Treat the % values as mass values and change to moles✓ Ca: 38.76*1mole/40g = 0.969 mole ✓ P: 19.97*1mole/31g = 0.64✓ O: 41.27*1mole/16g = 2.58 mole

• Divide by the smallest mole value✓ Ca: 0.969 / 0.64 = 1.5✓ P: 0.64 / 0.64 = 1✓ O: 2.58 / 0.64 = 4

• Multiply them all by a factor of 2 to get whole numbers and Voilà: Ca3P2O8

• Can we clean this up to look more like a familiar formula?• Ca3(PO4)2 Calcium Phosphate

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A carbon, hydrogen, oxygen compound was analyzed to be 40.0 % carbon, 53.0 % oxygen and the rest hydrogen. Determine the empirical formula for this compound. The compound was further analyzed and has molar mass of 180 g/mole, determine the molecular formula.

• Determine the % of hydrogen.✓ 100% - 53% - 40% = 7%

• Treat the % values as mass values. and change to moles✓ C: 40g * 1mole/12g = 3.33 mole ✓ O: 53g * 1mole/16g = 3.31✓ H: 7g * 1mole/1.01g = 6.9 mole

11

A carbon, hydrogen, oxygen compound was analyzed to be 40 % carbon, 53 % oxygen and the rest hydrogen. Determine the empirical formula for this compound.The compound was further analyzed and has molar mass of 180 g/mole, determine the molecular formula.

• Divide by the smallest mole value✓ C: 3.33 / 3.31 = 1.006✓ O: 3.31 / 3.31 = 1✓ H: 7 / 3.31 = 2

• Rounding to the nearest whole number and Voilà - Empirical Formula: CH2O

12

A carbon, hydrogen, oxygen compound was analyzed to be 40 % carbon, 53 % oxygen and the rest hydrogen. Determine the empirical formula for this compound.The compound was further analyzed and has molar mass of 180 g/mole, determine the molecular formula.

• Calculate the molar mass of the empirical formula✓ CH2O 12 + 2(1) + 16 = 30

• Divide the molar mass of the molecule by the molar mass of the empirical formula.✓ 180 / 30 = 6

• Drive the factor of 6 through the empirical formula to get the molecular formula✓ C6H12O6 is the molecular formula. This is

actually the molecule. It’s sugar.

13

3.4 g phosphorus was burned in oxygen and the resulting compound weighed 7.8 g. Determine the empirical and molecular formulas, then name the compound. The molar mass is 284 g/mole.• Calculate the mass of oxygen

✓ 7.8 g P?O? - 3.4 g P = 4.4 g O

• Determine the moles✓ 4.4 g O * 1mol/16g = 0.275✓ 3.4 g P * 1mol/31g = 0.1097

• Divide by the smallest✓ Oxy 0.275 / 0.1097 = 2.5✓ P 0.1097 / 0.1097 = 1

• multiply both by a factor (2x, 3x, maybe 4x)✓ Oxy 2.5 x2 = 5✓ P 1 x2 = 2

• Thus P2O5

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3.4 g phosphorus was burned in oxygen and the resulting compound weighed 7.8 g. Determine the empirical and molecular formulas, then name the compound. The molar mass is 284 g/mole.• Empirical formula: P2O5 • Determine the molecular formula.• Add the molar mass of the empirical formula

✓ 2(31) + 5(16) = 142✓ Divide into the molar mass given✓ 284/142 = 2✓ Drive the 2 factor through the empirical formula✓ P4O10 name ????✓ tetraphosphorus decoxide

What is the empirical formula of a compound if a 50.0 g sample of it contains 9.12 g Na, 20.6 g Cr, and 22.2 g O?

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• Data: 9.11g Na, 20.6g Cr, and 22.2g O

9.11g(1mol/22.99g) = 0.396mol Na

20.6g(1mol/52.00g) = 0.396mol Cr

22.2g(1mol/16.00g) = 1.39mol O

• Divide by the smallest mole value

0.396mol Na/0.396 = 1mol Mg

0.396mol Cr/0.396 = 1 mol Cr

1.39mol O/0.396 = 3.5 mol O

• Multiply them all by a factor of 2 to get whole numbers and Voilà: Na2Cr2O7

• Name: sodium dichromate

Sulfadiazine, a drug used for the treatment of bacterial infections, analyzes to 48 % carbon, 4.0 % hydrogen, 22.4 % nitrogen, 12.8 % sulfur, and 12.8 % oxygen. The molecular mass is 250.0 g/mole. Determine the empirical and molecular formulas

16

• Percents to grams to moles48g(1mol/12.01g) = 3.99mol C

4.0g(1mol/1.01g) = 3.96mol H

22.4g(1mol/14.01g) = 1.60mol N

12.8g(1mol/32.07g) = 0.399mol S

12.8g(1mol/16.00g) = 0.8mol O

• Divide by the smallest mole value 3.99mol C/0.399 = 10mol C

3.96mol H/0.399 = 9.9mol H

1.60mol N/0.399 = 4.01 mol N

0.399mol S/0.399 = 1 mol S

0.8mol O/0.399 = 2.0 mol O

• Empirical: C10H10N4SO2

• Molecular : same