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Percent Composition & Empirical & Molecular Formulas

Percent Composition & Empirical & Molecular Formulas

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Percent Composition &Empirical & Molecular

Formulas

Molecular FormulasThe molecular formula

•Shows the actual number of each type of atom in a compound

•Not always the simplest formula

Ex: C6H12O6 = glucoseContains 6 atoms of carbon, 12 atoms of hydrogen and 6 atoms of oxygen in 1 molecule

Empirical FormulasThe 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.

4

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

1. Determine the percent composition of all elements.

2. Convert this information into an empirical formula

3. Find the true number of atoms/ elements in the compound (Molecular Formula)

Percent Composition Review Percent composition • 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…5. What is the percent

composition of lactic acid, C3H6O3, a compound that appears in the blood after vigorous activity?

6. The chemical isoamyl acetate C7H14O2 gives the odor of pears. What is the percent carbon in isoamyl acetate?

Solution 5STEP 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 90.08 g

%O = 48.00 g O x 100 = 53.29% O 90.08 g

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 Formula1. If given % composition, assume 100 g of

sample

2. Convert mass of each element to moles (gmol)

3. Divide each of these numbers by the smallest number

4. If necessary, multiply by the smallest number possible to make each a whole number

5. These whole numbers are the subscripts in the empirical formula called mole ratio

The percentage composition of a compound is found to be 32.4% sodium, 22.5% sulfur, 45.1 % oxygen. Determine the empirical formula.

32.4% Na 32.4 g Na22.5% S 22.5 g S45.1% O 45.1 g O

Step 1: Assume 100 g of Sample

Step 2: Convert grams to moles

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

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.

39.9% C 39.9 g C6.7% H 6.7 g H53.4% O 53.4 g O

Step 1. Assume 100 g of Sample:

Step 2. Convert grams to moles

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 Up2. 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 adecimal, all the moles are multiplied by a smallinteger to obtain whole number.

Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical formula.

60.0% C 60.0 g C4.5% H 4.5 g H35.5% O 35.5 g O

Step 1. Assume 100 g of Sample:

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