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)

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.

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.

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

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

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

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

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

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

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

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

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

Ch 7 quiz #4 Empirical 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?

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.

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

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

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

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

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

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

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

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

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.

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

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.

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.

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 ?

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

3- What is the mass of 2. 25 moles of H2O. H = 2 x 1. 0 = 2 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%

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