Empirical & Molecular Formulas Advanced Chem. Unit #2 Empirical & Molecular Formulas

Empirical Formula - Formula giving the lowest whole number ratio of the atoms of the elements in a compound. Empirical formulas may represent either atoms or moles.

Molecular Formula - Shows the actual number of and type of atoms in a compound.

Examples: Molecular formula Empirical formula H2O2 HO N2H4 NH2 C6H6 (Benzene)(polystyrene) CH C2H2 (Acetylene) CH CO2 CO2

Examples: Molecular formula Empirical formula CH2O (formaldhyde) CH2O C6H12O6 (glucose) CH2O CH2O C2H4O2 (acetic acid) CH2O C5H10O5 (ribose) CH2O N2O5 N2O5

Formula Weight - Same as molecular weight, but of an ionic compound. A.K.A. Chart mass

Calculating mass percentages (percent composition) from the formula.

Mass of “A” in the whole Mass percent of “A” = X 100 Mass of the whole

Formaldehyde (CH2O) is a toxic gas with a pungent odor. (Uses include: Example 1: Formaldehyde (CH2O) is a toxic gas with a pungent odor. (Uses include: (1) large amounts to make plastics. (2) a H2O solution of it is used to preserve biological specimens). Calculate the percent of each element in the whole.

%C = 12.011g * 100 = 40% 30.0262g %H = 1.0079g *2 *100 = 6.73% 30.0262g %O = 15.9994g *100 = 53.3% 30.0262g

Example 2: Calculate percent of each in: KMnO4 158.03395g/mol 39.0983 158.03395 %K = *100 = 24.7% 54.93805 158.03395 %Mn = *100 = 34.8% (15.9994 *4) 158.03395 = 40.5% %O = *100 100%

Example 3: Copper(II)nitrate Cu(NO3)2 = 187.5558g/mol 63.546 187.5558 %Cu = = 33.9% *100 = 14.9% (14.0067 *2) 187.5558 %N = *100 %O = (15.9994*6) 187.5558 = 51.2% *100 100%

Example 4: Halothane (an inhalation anesthetic) CF3CHBrCl = 197.3821096 g/mol %C = (12.011*2) * 100 = 12.2% 197.3821096 %F = (18.9984032*3) *100 = 28.9% 197.3821096 %H = 1.0079 * 100 = 0.5% 197.3821096 %Br = 79.904 *100 = 40.5% 197.3821096 %Cl = 35.453 *100 = 18.0% 197.3821096 100.1%

Calculating the mass of an Element in a given mass of a compound. (% of Element “A”) x Mass of the compound = Grams of “A”

Example #1: Refer to previous example involving Formaldehyde (CH2O) 40% %H = 6.73% %O = 53.3% Formula wt. = 30.0262 g/mol

How many grams of each element are in 83.5 grams of Formaldehyde? 40% x 83.5 g = 33.4 g %H = 6.73% x 83.5 g = 5.62 g %O = 53.3% x 83.5 g = 44.5 g

How many grams of Nitrogen, Oxygen, and Hydrogen are in 48 How many grams of Nitrogen, Oxygen, and Hydrogen are in 48.5 grams of Ammonium nitrate ? NH4NO3 = 80.0432 g/mol (14.0067 x 2) %N = x 100 = 35.0% 80.0432 (1.0079 x 4) %H = x 100 = 5.04% 80.0432 (15.9994 x 3) %O = x 100 = 60.0% 80.0432

How many grams of Nitrogen are in 48.5 grams of Ammonium nitrate ? = 0.35 0.35 x 48.5 g = 17.0 g of Nitrogen

Urea is a fertilizer that is commonly purchased in a 46-0-0 formula (46% Nitrogen by mass (in the form of Ammonium nitrate), the remainder are inert ingredients). How many pounds of Nitrogen are in 3 ton of this fertilizer ?

3 ton = 6000 lb that is 46% NH4NO3 6000 lb x 0.46 = 2760 lb of NH4NO3 2760 lb of NH4NO3 x 0.35 = 966 lb of Nitrogen

Procaine hydrochloride (Novocain) (C13H21ClN2O2) is a local anesthetic. How many grams of each element are in a 0.23 cc dose of this substance ? (Assume Novocain = 1.0 g/mL)

1st: Determine the mass of the sample. Mass of sample = 0.23 grams

2nd: Determine the molecular weight of the compound. Molecular weight = 272.7741 g/mol

Which contains more carbon; 4. 71 grams of glucose, or 5 Which contains more carbon; 4.71 grams of glucose, or 5.85 grams of Ethyl alcohol (CH3CH2OH) ? 1st: Find %Carbon in each compound. 2nd: Calculate grams from percent in given masses.

Glucose Ethanol 40% %C 52.1% Grams Of Carbon 1.88 g 3.05 g

Which contains more Sulfur; 40. 8 grams of Calcium sulfate, or 35 Which contains more Sulfur; 40.8 grams of Calcium sulfate, or 35.2 grams of Sodium sulfite (Na2SO3)? 1st: Find %Sulfur in each compound. 2nd: Calculate grams from percent in given masses.

CaSO4 Na2SO3 23.6% %S 25.4% Grams Of Sulfur 9.63 g 8.94 g

Calculating the percent of C and H by combustion. Suppose you have discovered a new compound composed of only C, H, and O, whose formula you wish to determine. What are the products of combustion if the reactants consist of only C, H, and O ? Answer: CO2 and H2O

Every 1 mole of Carbon combusted will yield 1 mole of CO2. Every 1 mole of Hydrogen combusted will yield 0.5 mole of H2O.

Calculate the percent of Carbon, Hydrogen, and Oxygen by combustion. Acetic acid contains only C, H, & O. A 4.24mg sample is completely burned. It results in 6.21mg of CO2, and 2.54mg of H2O. What is mass percentage of each element in acetic acid?

Step 1: Convert mass of CO2 to moles of CO2. 6.21 mg = 0.00621 g of CO2 Use DIMO to convert g to moles. 0.00621 g = 0.000141 mol CO2

Step 2: Relate moles of CO2 to moles of C. Since 1 mol C produces 1 mol CO2 Then: 0.000141mol CO2 was produced by 0.000141 mol of C.

NOW DO THESE SAME THREE STEPS FOR HYDROGEN Step 3: Convert moles of C to grams of C (use DIMO) 0.000141 mol C = 0.001695 g of C NOW DO THESE SAME THREE STEPS FOR HYDROGEN

Step 1: Convert mass of H2O to moles of H2O. 2.54 mg = 0.00254 g of H2O Use DIMO to convert g to moles. 0.00254 g = 0.000141 mol H2O

Step 2: Relate moles of H2O to moles of H. Since 1 mol H produces 0.5 mol H2O Then: 0.000141mol H2O was produced by 0.000282 mol of H.

CALCULATE MASS PERCENTAGES OF C & H Step 3: Convert moles of H to grams of H (use DIMO) 0.000282 mol H = 0.000284 g of H TO FINISH: CALCULATE MASS PERCENTAGES OF C & H IN ACETIC ACID

0.001695 g of C %C = x 100 0.00424 g of Acetic acid = 39.98% Carbon in Acetic acid

SUBTRACT %C and %H FROM 100% 0.000284 g of H %H = x 100 0.00424 g of Acetic acid = 6.698% Hydrogen in Acetic acid TO FIND %OXYGEN: SUBTRACT %C and %H FROM 100%

100% - (39.98% C + 6.698% H) = 53.32% O TOTAL = 99.998%

(contains only C, H, and O) A 3.87mg sample of Ascorbic acid (Vitamin C) gives 5.80mg CO2 and 1.58mg of H2O when combusted. What is the percent composition of this substance ? (contains only C, H, and O)

Step 1: Convert mass of CO2 to moles of CO2. 5.80 mg = 0.00580 g of CO2 Use DIMO to convert g to moles. 0.00580 g = 0.000132 mol CO2

Step 2: Relate moles of CO2 to moles of C. Since 1 mol C produces 1 mol CO2 Then: 0.000132mol CO2 was produced by 0.000132 mol of C.

NOW DO THESE SAME THREE STEPS FOR HYDROGEN Step 3: Convert moles of C to grams of C (use DIMO) 0.000132 mol C = 0.00158 g of C NOW DO THESE SAME THREE STEPS FOR HYDROGEN

Step 1: Convert mass of H2O to moles of H2O. 1.58 mg = 0.00158 g of H2O Use DIMO to convert g to moles. 0.00158 g = 0.0000877 mol H2O

Step 2: Relate moles of H2O to moles of H. Since 1 mol H produces 0.5 mol H2O Then: 0.0000877mol H2O was produced by 0.000175 mol of H.

CALCULATE MASS PERCENTAGES OF C & H Step 3: Convert moles of H to grams of H (use DIMO) 0.000175 mol H = 0.000177 g of H TO FINISH: CALCULATE MASS PERCENTAGES OF C & H IN ACETIC ACID

0.00158 g of C %C = x 100 0.00387 g of Ascorbic acid = 40.8% Carbon in Ascorbic acid

SUBTRACT %C and %H FROM 100% 0.000177 g of H %H = x 100 0.00424 g of Ascorbic acid = 4.57% Hydrogen in Ascorbic acid TO FIND %OXYGEN: SUBTRACT %C and %H FROM 100%

100% - (40.8% C + 4.57% H) = 54.63% O TOTAL = 100%

Determining Empirical Formula From Mass Percent Remember: Empirical Formula - “Formula in lowest terms”

The formula of the molecule. Molecular formula - The formula of the molecule. Recall: Compounds with different molecular formulas can have the same Empirical Formula. CH2O (formaldhyde) C6H12O6 (glucose) CH2O C2H4O2 (acetic acid) C5H10O5 (ribose)