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 H2O2H2O2 HO N2H4N2H4 NH 2 C 6 H 6 (Benzene)(polystyrene) CH C 2 H 2 (Acetylene) CH CO 2

Examples: Molecular formula Empirical formula CH 2 O (formaldhyde) C 6 H 12 O 6 (glucose) C 2 H 4 O 2 (acetic acid) C 5 H 10 O 5 (ribose) CH 2 O N2O5N2O5 N2O5N2O5

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

Calculating mass percentages (percent composition) from the formula.

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

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

%C = g * 100 = 40% g %C = g * 100 = 40% g %H = g *2 *100 = 6.73% g %O = g *100= 53.3% g

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

Example 3: Copper(II)nitrate = g/molCu(NO 3 ) 2 %Cu = *100 = 33.9% %N = ( *2) *100 = 14.9% %O = ( *6) *100 = 51.2% 100%

Example 4: Halothane (an inhalation anesthetic) %Cl = *100 = 18.0% % %Cl = *100 = 18.0% % CF 3 CHBrCl= g/mol %C = (12.011*2) * 100 = 12.2% %F = ( *3) *100 = 28.9% %H = *100 = 0.5% %Br = *100 = 40.5%

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 (CH 2 O) % C = % H = % O = 40 % 6.73 % 53.3 % Formula wt. = g/mol

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

How many grams of Nitrogen, Oxygen, and Hydrogen are in 48.5 grams of Ammonium nitrate ? % N = % H = % O = x 100 NH 4 NO 3 = g/mol ( x 2) = 35.0 % x 100 ( x 4) = 5.04 % x 100 ( x 3) = 60.0 %

How many grams of Nitrogen are in 48.5 grams of Ammonium nitrate ? % N = 35.0 % = x 48.5 g = 17.0 g of Nitrogen

Urea is a fertilizer that is commonly purchased in a 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 6000 lb x 0.46 =2760 lb of NH 4 NO 3 that is 46 % NH 4 NO lb of NH 4 NO 3 x 0.35 = 966 lb of Nitrogen

Procaine hydrochloride (Novocain) (C 13 H 21 ClN 2 O 2 ) 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 = g/mol

Which contains more carbon; 4.71 grams of glucose, or 5.85 grams of Ethyl alcohol (CH 3 CH 2 OH) ? 1st: Find % Carbon in each compound. 2nd: Calculate grams from percent in given masses.

GlucoseEthanol %C%C Grams Of Carbon 40 % 52.1 % 1.88 g3.05 g

Which contains more Sulfur; 40.8 grams of Calcium sulfate, or 35.2 grams of Sodium sulfite (Na 2 SO 3 )? 1st: Find % Sulfur in each compound. 2nd: Calculate grams from percent in given masses.

CaSO 4 Na 2 SO 3 %S%S Grams Of Sulfur 23.6 % 25.4 % 9.63 g8.94 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: CO 2 and H 2 O

Every 1 mole of Carbon combusted will yield 1 mole of CO 2. Every 1 mole of Hydrogen combusted will yield 0.5 mole of H 2 O.

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 CO 2, and 2.54mg of H 2 O. What is mass percentage of each element in acetic acid?

Step 1:Convert mass of CO 2 to moles of CO mg = g of CO 2 Use DIMO to convert g to moles g = mol CO 2

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

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

Step 1:Convert mass of H 2 O to moles of H 2 O mg = g of H 2 O Use DIMO to convert g to moles g = mol H 2 O

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

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

% C = g of C g of Acetic acid x 100 = % Carbon in Acetic acid

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

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

A 3.87mg sample of Ascorbic acid (Vitamin C) gives 5.80mg CO 2 and 1.58mg of H 2 O when combusted. What is the percent composition of this substance ? (contains only C, H, and O)

Step 1:Convert mass of CO 2 to moles of CO mg = g of CO 2 Use DIMO to convert g to moles g = mol CO 2

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

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

Step 1:Convert mass of H 2 O to moles of H 2 O mg = g of H 2 O Use DIMO to convert g to moles g = mol H 2 O

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

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

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

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

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

Combustion of 354mg of a particular toxic compound composed of only Carbon, Hydrogen, and Oxygen; yielded 500mg of CO 2, and 308mg of H 2 O. How many grams of Oxygen were required to combust this compound ?

Clues to solving previous problem: 1st:Determine the percent composition of the “Toxic” compound.

2nd:Using the percent composition, determine the mass in grams of the Oxygen in the “Toxic” compound.

3rd:Using percent composition of the products determine the total grams of Oxygen inthese compounds.

4th:Calculate the answer to this question usingthe answers from clues # 2 and # 3.

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

Molecular formula -The formula of the molecule. Recall:Compounds with different molecular formulas can have the same Empirical Formula. CH 2 O (formaldhyde) C 6 H 12 O 6 (glucose) C 2 H 4 O 2 (acetic acid) C 5 H 10 O 5 (ribose) CH 2 O

Example 1: How to calculate empirical formulas from percent composition. What is the empirical formula of a compound that is 94.1 % oxygen and 5.9 % Hydrogen ?

Realize that percent composition is a ratio of masses of the elements in the problem. Calculate the lowest whole number ratio of the moles of the elements.

Step 1:Assume 100 grams of the sample. Therefore you have how many grams of: Oxygen (94.1 % ) = 94.1 grams Hydrogen (5.9 % ) = 5.9 grams

Step 2:Turn grams into moles (use DIMO). Oxygen = 94.1g = 5.88 mol of Oxygen Hydrogen = 5.9g =5.85 mol of Hydrogen

Step 3:Set up mole ratio, and divide by the smallest number of moles. H 5.85 mol O 5.88 mol 5.85 mol H 1 O 1.01 = HO

Example 2: How to calculate empirical formulas from percent composition. What is the empirical formula of a compound that is 67.6 % Mercury, 10.8 % Sulfur, and 21.6 % Oxygen ?

Step 1:Assume 100 grams of the sample. Therefore you have how many grams of: Hg (67.6 % ) = 67.6 grams S (10.8 % ) = 10.8 grams O (21.6 % ) = 21.6 grams

Step 2:Turn grams into moles (use DIMO). Hg = 67.6g = mol of Hg S = 10.8g =0.337 mol of S O = 21.6g = 1.35 mol of O

Step 3:Set up mole ratio, and divide by the smallest number of moles. Hg mol S mol O 1.35 mol mol Hg 1 S 1 O 4.01 =HgSO mol

Example 3: How to calculate empirical formulas from percent composition. What is the empirical formula of a compound that is 17.6 % Sodium, 39.7 % Chromium, and 42.7 % Oxygen ?

Step 1:Assume 100 grams of the sample. Therefore you have how many grams of: Na (17.6 % ) = 17.6 grams Cr (39.7 % ) = 39.7 grams O (42.7 % ) = 42.7 grams

Step 2:Turn grams into moles (use DIMO). Na = 17.6g = mol of Na Cr = 39.7g =0.764 mol of Cr O = 42.7g = 2.67 mol of O

Step 3:Set up mole ratio, and divide by the smallest number of moles. Na mol Cr mol O 2.67 mol mol Na 1 Cr 1 O 3.49 =Na 2 Cr 2 O mol

So far in this unit, you have learned: 1.How to calculate percent composition from the molecular formula. 2.How determine grams of an element in a given mass of a compound from it’s percent composition.

So far in this unit, you have learned (con’t): 3.How to calculate grams of Carbon, Hydrogen, and Oxygen by combustion. 4.How determine the empirical formula of a compound from it’s percent composition.

The last topic of this unit: How to determine the Molecular formula of a compound from its percent composition and molecular weight.

This process requires you to calculate the “factor, n”. “n” = Molecular weight of the compound Empirical weight of the compound Multiplying the ratios of the empirical formula by “n” will result in the Molecular formula.

Example 1: A 15 gram sample of a substance that has a molecular weight of 60 AMU was determined to have a composition consisting of: 5.99g of Carbon 1.01g of Hydrogen 8.01g of Oxygen

1st:Find the empirical formula. 2nd:Using “n”, determine the molecular formula.

1st:The Empirical formula: 5.99g of Carbon= 1.01g of Hydrogen= 8.01g of Oxygen= mol of O 1.00 mol of H mol of C C H 1.00 O = CH 2 O

2nd:Find “n”, then determine the Molecular formula. “n” = Molecular weight of the compound Empirical weight of the compound “n” = 60.0 AMU 30.0 AMU = 2

“n” (Empirical formula) = Molecular formula 2 (CH 2 O) = C 2 H 4 O 2 CH 3 COOH Acetic acid

Example 2: 7.36 grams of compound “X” decomposed to give 6.93 grams of Oxygen and the rest is Hydrogen. If “X” has an atomic weight of 34.0 g/mol, what is its formula ?: 7.36 grams total of “X” 6.93g of Oxygen

1st:Find the empirical formula. 2nd:Using “n”, determine the molecular formula.

1st:The Empirical formula: 7.36g total of “X” 6.93g of Oxygen= mol of H mol of O H O = HO 0.43g of Hydrogen=

2nd:Find “n”, then determine the Molecular formula. “n” = Molecular weight of the compound Empirical weight of the compound “n” = 34.0 AMU 17.0 AMU = 2

“n” (Empirical formula) = Molecular formula 2 (HO) = H 2 O 2 Hydrogen peroxide

Example 3: Methyl butanoate smells like apples. It’s molecular weight is 102 AMU. It was determined to have a composition consisting of: 58.8 % Carbon 9.8 % Hydrogen 31.4 % Oxygen What is it’s molecular formula ?

1st:Find the empirical formula. 2nd:Using “n”, determine the molecular formula.

1st:The Empirical formula: 58.8g of Carbon= 9.8g of Hydrogen= 31.4g of Oxygen= 1.00 mol of O 5.00 mol of H 2.50 mol of C C 2.5 H 5.00 O = C 5 H 10 O 2

2nd:Find “n”, then determine the Molecular formula. “n” = Molecular weight of the compound Empirical weight of the compound “n” = 102 AMU = 1

“n” (Empirical formula) = Molecular formula 1 (C 5 H 10 O 2 ) = C 5 H 10 O 2 This time the empirical and molecular formulas are identical.

Example 4: Putrescine, a substance produced by decaying animal flesh, is 54.5 % C, 13.7 % H, and 31.8 % N. Several determinations of molecular weight gave values in the range of g/mol. What is it’s molecular formula ?: 54.5g of Carbon 13.7g of Hydrogen 31.8g of Oxygen

1st:Find the empirical formula. 2nd:Using “n”, determine the molecular formula.

1st:The Empirical formula: 54.5g of Carbon= 13.7g of Hydrogen= 31.8g of Nitrogen= 2.27 mol of N 13.6 mol of H 4.54 mol of C C 4.54 H 13.6 N = C2H6OC2H6O

2nd:Find “n”, then determine the Molecular formula. “n” = Molecular weight of the compound Empirical weight of the compound “n” = 87 AMU AMU = 1.97

2nd:Find “n”, then determine the Molecular formula. “n” = Molecular weight of the compound Empirical weight of the compound “n” = 88 AMU AMU = 2.00

2nd:Find “n”, then determine the Molecular formula. “n” = Molecular weight of the compound Empirical weight of the compound “n” = 89 AMU AMU = 2.02

2nd:Find “n”, then determine the Molecular formula. “n” = Molecular weight of the compound Empirical weight of the compound “n” = 90 AMU AMU = 2.04

Values of “n” (average) = “n”

“n” (Empirical formula) = Molecular formula 2 (C 2 H 6 N) = C 4 H 12 N 2 Putrescine

Example 5: Adipic acid is used to manufacture nylon. It is 49.3% C, 6.9% H, and 43.8% O by mass; with a molecular weight of 146 g/mol. What is it’s molecular formula ?