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Determining Chemical Formulas

Determining Chemical Formulas. empirical formula- consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole number mole ratio of the different atoms in the compound CH 3 = empirical formula (does not exist)

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Determining Chemical Formulas

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  1. Determining Chemical Formulas • empirical formula- consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole number mole ratio of the different atoms in the compound CH3 = empirical formula (does not exist) C2H6 = molecular formula (ethene)

  2. Empirical Formulas • The formulas of ionic compounds are empirical formulas by the definition of ionic formulas. • The formulas of molecular compounds may or may not be the same as its empirical formula.

  3. Calculating an Empirical Formula • If the elements are in % composition by mass form, covert the percentages to grams. • Convert the masses of each element to moles by dividing the mass of the element by its molar mass. • Select the element with the smallest number of moles and divide the number of moles of each element by that number which will give you a 1:---:--- ratio. • IF the ratio is very close to a whole number ratio, apply the numbers to each element. If one of the number is not close to a whole number, use a multiplier to convert the ratio to a whole number ratio.

  4. 1- If the elements are in % composition by mass form, covert the percentages to grams. e.g. C = 40.0%  40.0 g H = 6.67%  6.67 g O = 53.3%  53.3 g

  5. 2- Convert the masses of each element to moles by dividing the mass of the element by its molar mass. e.g. C = 40.0/12 = 3.33 mol H = 6.67/1 = 6.67 mol O = 53.3/16 = 3.33 mol

  6. 3- Select the element with the smallest number of moles and divide the number of moles of each element by that number which will give you a 1:---:--- ratio. e.g. C = 3.33/3.33 = 1 H = 6.67/3.33 = 2 O = 3.33/3.33 = 1

  7. 4- IF the ratio is very close to a whole number ratio, apply the numbers to each element. If one of the number is not close to a whole number, use a multiplier to convert the ratio to a whole number ratio. e.g. 1:2:1 ratio  CH2O

  8. Calculating an Empirical Formula • Sample Problem • 32.38% Na, 22.65% S, & 44.99% O. 1- convert to 32.38 g Na, 22.65 g S, & 44.99 g O 2- 32.38 ÷ 22.99 = 1.408 mol Na 22.65 ÷ 32.07 = 0.7063 mol S 44.99 ÷ 16.00 = 2.812 mol O 3- 1.408 ÷ 0.7063 = 1.993 mol Na  2 0.7063 ÷ 0.7063 = 1 mol S 2.812 ÷ 0.7063 = 3.981 mol O  4 4- Rounding  2:1:4  Na2SO4

  9. Calculating an Empirical Formula • Review sample problem M on page 247. • Do practice problems #1, 2, & 3 on page 247.

  10. Practice problem #1 page 247 63.52% iron (Fe) 36.48% sulfur (S) Convert % to grams: Fe = 63.52g S = 36.48g Divide each element by its molar mass: Fe = 63.52/55.8 = 1.14 mol S = 36.48/32.1 = 1.14 mol Divide each number of moles by the smallest number: Fe = 1.14/1.14 = 1 S = 1.14/1.14 = 1 Ratio = 1:1 so FeS is the empirical formula

  11. Practice problem #2 page 247 K = 26.56% Cr = 35.41% O = 38.03% K = 26.56/39.1 = 0.679 mol Cr = 35.41/52.0 = 0.681 mol O = 38.03/16.0 = 2.38 mol K = 0.679/0.679 = 1 Cr = 0.681/0.679 = 1.003 O = 2.38/0.679 = 3.51 1:1:3.5 ratio Double the ratio to get whole numbers  2:2:7 Empirical formula is K2Cr2O7

  12. Practice problem #3 page 247. 20.0 g calcium & bromine 4.00 g Ca so 16.00 g Br Already in grams so divide by molar mass: 4.00/ 40.1 = .0997 mol Ca 16.00/79.9 = .2003 mol Br Ca = .0997/.0997 = 1 Br = .2003/.0997 = 2.009  2 Empirical formula is CaBr2

  13. Ch 7 quiz #4Empirical Formulas 1- A compound is 27.3% carbon and 72.7% oxygen by mass. What is the empirical formula of the compound? 2- A compound is 11.1% hydrogen and 88.9% oxygen. What is its empirical formula?

  14. Calculating a Molecular Formula • molecular formula- the actual formula of a molecular compound (it may or may not be the same as the empirical formula of the compound) • The molar mass of a compound is determined by analytical means & is given. • Calculate the formula mass of the empirical formula. Divide the molar mass of the compound by its empirical mass. • “Multiply” the empirical formula by this factor.

  15. Calculating a Molecular Formula • empirical formula = P2O5 • molecular mass is 283.89 • empirical mass is 141.94 • Dividing the molecular mass by the empirical mass gives a multiplication factor of : 283.89 ÷ 141.94 = 2.0001  2 • 2 x (P2O5)  P4O10

  16. Chapter 7 Problems • Do practice problems #1 & 2 on page 249. • Do section review problems #1-4 on page 249.

  17. Practice problem #1 page 249 empirical formula = CH formula mass = 78.110 amu empirical mass = ? = 12.0 + 1.0 = 13.0 amu molecular mass / empirical mass = 78.110/13.0 = 6.008  multiplication factor of 6 molecular formula = CH x 6  C6H6

  18. Practice problem #2 page 249 formula mass = 34.00 amu 0.44 g H & 6.92 g O 1st find empirical formula: H = 0.44/1.0 = 0.44 O = 6.92/16.0 = 0.43 0.44/0.43  1 H & 0.43/0.43  1 O empirical formula = HO empirical mass = 17.0 formula mass / empirical mass = 34.00/17.0 = 2 HO x 2  H2O2

  19. Section review problem #1 page 249. 36.48% Na 25.41% S 38.11% O 36.48/23.0 = 1.58 mol Na 25.41/32.1 = 0.792 mol S 38.11/16.0 = 2.38 mol O 1.58/0.792 = 1.995  2 0.792/0.792 = 1  1 2.38/0.792 = 3.005  3 2:1:3  Na2SO3

  20. Section review problem #2 page 249. 53.70% Fe 46.30% S 53.70/55.8 = 0.962 mol Fe 46.30/32.1 = 1.44 mol S 0.962/0.962 = 1 Fe 1.44/0.962 = 1.50 S 1:1.5 doubled  2:3  Fe2S3

  21. Section review problem #3 page 249 1.04 g K 0.70 g Cr 0.86 g O 1.04/39.1 = .0266 mol K 0.70/52.0 = .0135 mol Cr 0.86/16.0 = .0538 mol O .0266/.0135 = 1.97  2 .0135/.0135 = 1 .0538/.0135 = 3.99  4 Empirical formula = K2CrO4

  22. Section Review problem #4 page 249 4.04 g N 11.46 g O f.m. = 108.0 amu 4.04/14.0 = .289 mol N 11.46/16.0 =.716 mol O .289/.289 = 1 .716/.289 = 2.45 double ratio  2:5  N2O5 e.f.m. = 108 f.m./e.f.m. = 108/108 = 1 empirical formula is same as molecular formula N2O5

  23. To find molar mass: add the masses of the elements in the formula of the compound. To find number of grams (mass): multiply # of moles times the molar mass of the compound. To find the number of moles: divide the number of grams by the molar mass of the compound.

  24. To calculate % composition by mass: 1- find the molar mass of a compound 2- divide the mass of each element by the molar mass of the compound 3- multiply by 100 to convert each ratio to a percent

  25. Calculating an Empirical Formula • If the elements are in % composition by mass form, convert the percentages to grams. • Convert the masses of each element to moles by dividing the mass of the element by its molar mass. • Select the element with the smallest number of moles and divide the number of moles of each element by that number which will give you a 1:---:--- ratio. • IF the ratio is very close to a whole number ratio, apply the numbers to each element. If one of the number is not close to a whole number, use a multiplier to convert the ratio to a whole number ratio.

  26. Calculating a Molecular Formula • molecular formula- the actual formula of a molecular compound (it may or may not be the same as the empirical formula of the compound) • The molar mass of a compound is determined by analytical means & is given. • Calculate the formula mass of the empirical formula. Divide the molar mass of the compound by its empirical mass. • “Multiply” the empirical formula by this factor.

  27. Final Practice- chapter 7 part 2 1- Determine the molar mass of the compound Na3PO4 . 2- How many moles of CO2 are in 198 g ? 3- What is the mass of 2.25 moles of H2O ? 4- What is the % composition of each element of the compound P4O10 ? 5- What is the empirical formulas of a compound that is 25.9% N and 74.1% O ? What is it molecular formula if its molecular mass is 216 ?

  28. Final Practice- chapter 7 part 2 1- Determine the molar mass of the compound Na3PO4 . Na = 3 x 23.0 = 69.0 P = 1 x 31.0 = 31.0 O = 4 x 16.0 = 64.0 164.0 g/mol 2- How many moles of CO2 are in 198 g ? C = 1 x 12.0 = 12.0 O = 2 x 16.0 = 32.0 44.0 g/mol 198/44.0 = 4.5 mol CO2

  29. 3- What is the mass of 2.25 moles of H2O ? H = 2 x 1.0 = 2.0 O = 1 x 16.0 = 16.0 18.0 g/mol 2.25 mol x 18.0 g/mol = 40.5 g H2O 4- What is the % composition of each element of the compound P4O10 ? P = 4 x 31.0 = 124.0 O = 10 x 16.0 = 160.0 284.0 g/mol P = 124/284(100) = 43.7% O = 160/284 (100) = 56.3%

  30. 5- What is the empirical formulas of a compound that is 25.9% N and 74.1% O ? What is it molecular formula if its molecular mass is 216 ? N = 25.9/14.0 = 1.85 O = 74.1/16.0 = 4.63 N = 1.85/1.85 = 1 O = 4.63/1.85 = 2.5 1:2.5 doubled  2:5 so empirical formula = N2O5 e.f.m. = (2 x 14) + (5 x 16) = 108 216 (mfm)/108 (efm) = 2 2 x 2:5  4:10  N4O10

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