Percent Composition & Empirical & Molecular Formulas

Molecular Formulas The molecular formula Shows the actual number of each type of atom in a compound Not always the simplest formula Ex: C6H12O6 = glucose Contains 6 atoms of carbon, 12 atoms of hydrogen and 6 atoms of oxygen in 1 molecule

Empirical Formulas The empirical formula Is the simplest whole number ratio of the atoms. Is calculated by dividing the subscripts in the actual (molecular) formula by a whole number to give the lowest ratio. Glucose C6H12O6  6 = C1H2O1 = CH2O molecular formula empirical formula

Some Molecular and Empirical Formulas The molecular formula is the same or a multiple of the empirical.

You Try… 1. What is the empirical formula for C4H8? 2. What is the empirical formula for C8H14? 3. Which is a possible molecular formula for CH2O? a) C4H4O4 b) C2H4O2 c) C3H6O3 4. A compound has an empirical formula SN. If there are 4 atoms of N in one molecule, what is the molecular formula? Explain.

3 Steps for determining Chemical Formulas Determine the percent composition of all elements. Convert this information into an empirical formula Find the true number of atoms/ elements in the compound (Molecular Formula)

Percent Composition Review Is the percent by mass of each element in a formula. Example: Calculate the percent composition of CO2. CO2 = 1(12.01g/mol) + 2(16.00 g/mol) = 44.01 g/mol) 12.01 g C x 100 = 27.29 % C 44.01 g CO2 32.00 g O x 100 = 72.71 % O 44.01 g CO2 100.00 %

You Try… What is the percent composition of lactic acid, C3H6O3, a compound that appears in the blood after vigorous activity? The chemical isoamyl acetate C7H14O2 gives the odor of pears. What is the percent carbon in isoamyl acetate?

Solution 5 STEP 1: Calculate Molar Mass (Mr) 3C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol STEP 2: Calculate % Composition for each element %C = 36.03 g C x 100 = 40.00% C 90.08 g %H = 6.048 g H x 100 = 6.714% H %O = 48.00 g O x 100 = 53.29% O

Solution 6 Molar mass C7H14O2 = 7C(12.01) + 14H(1.008) + 2O(16.00) = 130.18 g/mol Total C = 7C(12.01) = 84.07g % C = total g C x 100 total g % C = 84.07 g C x 100 = 64.58 % C 130.18 g

BRAIN BREAK: The Waiter Stand up. Lay a piece of paper or a spiral notebook on the top of your right hand without grabbing it. While balancing the spiral on your hand, tuck it between your right arm and waist so that the spiral will now be behind you. You will Flair your arm out away from you for this step. Again, keep the spiral balanced on top of your hand. Keep turning your arm so that the spiral will get back to the original spot. If you have mastered this, try your other hand. https://www.youtube.com/watch?v=y4HLjT1xw3M

Calculating Empirical Formula If given % composition, assume 100 g of sample Convert mass of each element to moles (gmol) Divide each of these numbers by the smallest number If necessary, multiply by the smallest number possible to make each a whole number These whole numbers are the subscripts in the empirical formula called mole ratio

Step 1: Assume 100 g of Sample 32.4% Na  32.4 g Na 22.5% S  22.5 g S The percentage composition of a compound is found to be 32.4% sodium, 22.5% sulfur, 45.1 % oxygen. Determine the empirical formula. Step 1: Assume 100 g of Sample 32.4% Na  32.4 g Na 22.5% S  22.5 g S 45.1% O  45.1 g O Step 2: Convert grams to moles      

The percentage composition of a compound is found to be 32 The percentage composition of a compound is found to be 32.4% sodium, 22.5% sulfur, 45.1 % oxygen. Determine the empirical formula. Step 3: Divide by smallest mol 1.41 mol Na 0.702 mol S 2.82 mol O       Step 4: If necessary, multiply by the smallest number possible to make each a whole number Step 5: Write Empirical Formula Na2SO4

Step 1. Assume 100 g of Sample: The percentage composition of acetic acid is found to be 39.9% C, 6.7% H, and 53.4% O. Determine the empirical formula of acetic acid. Step 1. Assume 100 g of Sample: 39.9% C  39.9 g C 6.7% H  6.7 g H 53.4% O  53.4 g O Step 2. Convert grams to moles      

The percentage composition of acetic acid is found to be 39. 9% C, 6 The percentage composition of acetic acid is found to be 39.9% C, 6.7% H, and 53.4% O. Determine the empirical formula of acetic acid. Step 3: Divide by smallest mol 3.32 mol C 6.65 mol H 3.34 mol O       Step 4: If necessary, multiply by the smallest number possible to make each a whole number Step 5: Write Empirical Formula CH2O

BRAIN BREAK: The Crab Directions: 1. Stand Up 2. Put your arms out in front of you and touch your fingers and thumbs together. 3. Now put lower your middle fingers so that the knuckles touch. Keep them flat against each other. 4. Now un-touch and retouch your thumbs. 5. Now un-touch and retouch your index fingers. 6. Now un-touch and retouch your ring fingers. 7. Lastly, un-touch and retouch your pinkies. Good luck. This one was difficult. https://www.youtube.com/watch?v=dp0sa7Z72R8

Converting Decimals to Whole Numbers When the number of moles for an element is a decimal, all the moles are multiplied by a small integer to obtain whole number.

Step 1. Assume 100 g of Sample: Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical formula. Step 1. Assume 100 g of Sample: 60.0% C  60.0 g C 4.5% H  4.5 g H 35.5% O  35.5 g O Step 2. Convert grams to moles      

Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical formula. Step 3: Divide by smallest mol 5.00 mol C 4.46 mol H 2.22 mol O       Step 4: If necessary, multiply by the smallest number possible to make each a whole number       Step 5: Write Empirical Formula C9H8O4