1. Percent Composition, Empirical and Molecular Formulas & 22.4 L @ STP Courtesy www.lab-initio.com

2. STP and the ‘Magic’ 22.4 L Another useful conversion factor for chemists is the 22.4 L @ STP What this means is that if you have one mole of a gas it will occupy 22.4L in volume. We can now use this conversion factor to figure out: Volume of a gas we based on moles Volume of a gas we based on moles Moles of a gas based on its volume. Moles of a gas based on its volume. NOTE: This holds true for a gas at STP. if not at STP, we must make additional adjustments. NOTE: This holds true for a gas at STP. if not at STP, we must make additional adjustments. STP is STP is 0 °C (32 °F or 273 Kelvin) 1 atm of Pressure

3. STP and the ‘Magic’ 22.4 L SAMPLE PROBLEMS 1) How many moles of anti-nerve gas do you need if you have a 30 000 L room poisoned and there is a 1:1 ration needed? 2) How many liters of CO have leaked out of your experiment if you have lost 135g of mass?

4. Percentage Composition The ability to calculate percent composition is part of one of the most powerful tools of modern science. Using it you can figure out what unknown substances are! For example: 1.ID poison someone has taken. 2.Determine physical ailment you have based on the compounds in your blood or urine. 3.ID drugs smuggled from an empty bag 4.ID chemical residue left at a crime scene or fire. 5.Determine valuable minerals in a rock

5. How to Calculate Percentage Composition EX Calculate the percentage composition of magnesium carbonate, MgCO 3. 1. Determine Formula Mass: magnesium carbonate: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00 2. Place the elements mass over the total formula mass, divide then multiply by 100% (same thing you do for tests!)

6. Types of Formulas  molecular formula = (empirical formula) n  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

7. Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

8. Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 OC 12 H 22 O 11

9. Empirical Formula Determination 1.Base calculation on 100 grams of compound. Determine moles of each element in 100 grams of compound. 2.Divide each value of moles by the smallest of the values. 3.Multiply each number by an integer to obtain all whole numbers.

10. Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? 1.Treat % as mass, and convert grams to moles

11. Empirical Formula Determination 2. Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

12. Empirical Formula Determination 3. Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 352 Empirical formula: C3H5O2C3H5O2

13. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

14. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 2. Divide the molecular mass by the mass given by the emipirical formula.

15. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4