NOTES: 10.3 Empirical and Molecular Formulas
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1 NOTES: 10.3 Empirical and Molecular Formulas
2 What Could It Be?
3 Empirical Formulas Indicate the lowest whole number ratio of the atoms in a compound: 1) Determine moles of each element present in the compound 2) Divide molar amounts by the smallest number of moles present 3) Obtain whole numbers by multiplying by integers if necessary
4 Calculating the Empirical Formula Example: A compound is found to contain the following g Copper g Oxygen Calculate it s empirical formula.
5 Calculating the Empirical Formula Step 1: Convert the masses to moles. Copper: 1moleCu 2.199g Cu g Cu moles Cu Oxygen: 1mole O 0.277g O g O moles O
6 Calculating the Empirical Formula Step 2: Divide all the moles by the smallest value. This gives the mole ratio mol Cu mol Cu molo mol 1.000O
7 Calculating the Empirical Formula Step 3: Round off these numbers, they become the subscripts for the elements. Cu 2 O
8 Empirical Formula Example Problems: 1) A compound is analyzed and found to contain 79.8 g C and 20.2 g H. Determine the empirical formula. 2) What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen?
9 1) A compound is analyzed and found to contain 79.8 g C and 20.2 g H. Determine the empirical formula. 1moleC 79.8g C g C moles C 1mole H 20.2g H g H moles H
10 1) A compound is analyzed and found to contain 79.8 g C and 20.2 g H. Determine the empirical formula molc 6.65 mol 1.0 C 20.2 mol H 6.65 mol 3.0H
11 1) A compound is analyzed and found to contain 79.8 g C and 20.2 g H. Determine the empirical formula. EMPIRICAL FORMULA = CH 3
12 2) What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen? 1mole N 25.9g N g N moles N 1moleO 74.1g O g O moles O
13 2) What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen? 1.85 mol N 1.85 mol 1.0 N 4.63 molo 1.85 mol 2.5O
14 2) What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen? EMPIRICAL FORMULA = NO 2.5 Is this formula acceptable?
15 2) What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen? EMPIRICAL FORMULA = NO 2.5 Is this formula acceptable? NO! (must be WHOLE NUMBERS) So, multiply by 2.
16 2) What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen? EMPIRICAL FORMULA: 2 x (NO 2.5 ) = N 2 O 5
17 Molecular Formulas Indicates the actual formula for a compound; the true formula Will be a multiple of the empirical formula EXAMPLE: glucose -empirical: CH 2 O -molecular: C 6 H 12 O 6
18 Molecular Formula Example: Calculate the molecular formula of the compound whose molar mass is 60.0 g and the empirical formula is CH 4 N.
19 Molecular Formula Example: Calculate the molecular formula of the compound whose molar mass is 60.0 g and the empirical formula is CH 4 N. CH 4 N has a MW of: (4) + 14 = 30.0 g Actual MW is: 60.0 g
20 Molecular Formula Example: Calculate the molecular formula of the compound whose molar mass is 60.0 g and the empirical formula is CH 4 N / 30.0 = 2 So, multiply the empirical form. by 2 2 x (CH 4 N) = C 2 H 8 N 2
21 Example: A material is found to be composed of 38.7% Carbon, 51.6% Oxygen, and 9.7% Hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol Calculate the empirical and molecular formula for the compound. If you assume a sample weight of 100 grams, then the percents are really grams.
22 Example: A material is found to be composed of 38.7% Carbon, 51.6% Oxygen, and 9.7% Hydrogen. By other means, it is known that the molecular weight is 62.0 Calculate the empirical and molecular formula for the compound. Carbon: 1mole 38.7grams 3. 23mol 12.0grams Oxygen: 1mole 51.6grams 3. 23mol 16.0grams Hydrogen: 1mole 9.7grams 9. 7mol 1.0grams
23 Carbon: Oxygen: 1mole 38.7grams 3. 23mol 12.0grams 3.23 Now, divide all the moles by the smallest one, mole 51.6grams 3. 23mol 16.0grams Hydrogen: 1mole 9.7grams 9. 7mol 1.0grams
24 Example: A material is found to be composed of 38.7% Carbon, 51.6% Oxygen, and 9.7% Hydrogen. By other means, it is known that the molecular weight is 62.0 Calculate the empirical and molecular formula for the compound. So, the empirical formula must be: CH 3 O The molecular weight of the empirical formula is. C 12 x 1 = 12 g/mol H 1 x 3 = 3 g/mol O 16 x 1 = 16 g/mol 31 g/mol
25 Remember, the empirical formula is not necessarily the molecular formula! MW of the empirical formula = 31 MW of the molecular formula = 62 Multiplyin g Factor Molecular Empirical
26 2 x ( CH 3 O ) = C 2 H 6 O 2 Empirical Formula Molecular Formula
27 Remember, the molecular formula represents the actual formula.
28 What if the mole ratios don t come out even?
29 Example: A compound is analyzed and found to contain 2.42g aluminum and 2.15g oxygen. Calculate its empirical formula. Aluminum: 1mol 2.42g. 0896mol 27g X 2 = 2 Oxygen: 1mol 2.15g. 134mol 16g X 2 = 3 Al 2 O 3
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