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Empirical & Molecular Formulas

Empirical & Molecular Formulas. Chapter 8. Empirical & Molecular Formulas. 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

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Empirical & Molecular Formulas

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  1. Empirical & Molecular Formulas • Chapter 8

  2. Empirical & Molecular Formulas • 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 • Formaldehyde Acetic Acid Glucose C6H12O6 CH2O CH3CO2H (CH3COOH) (CH2O)2 CH2O (CH2O)6 (CH2O)n (CH2O)n (CH2O)n

  3. Empirical Formula Molecular Formula • the formula that gives the composition of the molecules present • the formula of a compound that expresses the smallest whole-number ratio of the atoms present • the simplest 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 moles • 60.3g 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 value • 2.48mol Mg/2.48 = 1mol Mg • 2.48mol O/2.48 = 1mol O • MgO

  6. 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

  7. 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

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

  9. 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

  10. 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

  14. 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

  15. 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? • 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

  16. 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 • Percents to grams to moles • 48g(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

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