NaCl H2O C6H12O6 Chemical Formulas NaHCO3

What information is in a chemical formula? A ratio of ATOMS. In 1 molecule of H2O there are 2 atoms of H and 1 atom of O. A ratio of MOLES. In 1 mole of H2O there are 2 moles of H and 1 mole of O. Formulas are NOT ratios of grams. In 1 mole H2O there are NOT 2 g of H for every g of O.

How do chemists determine the formula of a compound? Step 1: Elemental Analysis Identify ELEMENTS present Determine % COMPOSITION by MASS for each element.

How do chemists determine the formula of a compound? Step 2: Determination of empirical formula Empirical Formula = SIMPLEST Ratio of atoms in a formula

IONIC COMPOUND (+ metal ion/ nonmetal – ion or polyatomic ion) Empirical Formula = “Final” Formula Ions bond together in a lattice – giant “molecule” Actual formula would be to difficult write, e.g. Na 1023 Cl 1023 So formula is written as simplest ratio, NaCl

Covalent Compound (2 or more nonmetals sharing pairs of electrons exist as molecules. Empirical Formulas (simplest ratio) may or may not be true ratio True or Actual Formula is called MOLECULAR Formula Example: H2O: empirical = molecular Benzene : empirical ≠ molecular Empirical = CH Molecular C6H6

MOLECULAR FORMULA DETERMINATION Requires empirical formula + MOLAR MASS Molar Mass is Determined by Experiment (in problems, molar mass must be provided).

Empirical Formula vs Molecular Formula Empirical Formula = SIMPLEST ratio of atoms in a compound. Molecular Formula = ACTUAL ratio of atoms in a compound. Ionic compound → Empirical Formula is “actual” formula Molecular compound (covalent bonds) → Molecular formula is usually different than empirical formula. Determining empirical formula is still a key step in the process of the determining the formula of a Molecular Compound.

Examples of Empirical and Molecular Formulas for covalently bonded compounds: Carbon dioxide: Empirical = CO2 Molecular = CO2 Benzene: Empirical CH Molecular = C6H6 Glucose: Molecular = C6H12O6 Empirical = CH2O

% Composition by Mass Mass of Element x 100% Total Mass of Compound Definition of %: Part out of ; To calculate % : x 100 % Definition of Mass % = 100 Part Total Mass of Element x 100% Total Mass of Compound Mass % of each element must add up to 100

Example #1: % from mass data What is the % by mass of a compound containing 28.0 g Fe and 8.0 g O? % Fe = % O = Mass of Fe x 100 % Total Mass of Compound (Fe + O) 28.0 g Fe x 100% = 77.8% 28.0 + 8.0 = 36.0 8.0 g O x 100% = 22.2% 36.0 g Fe + O

Example #2: % from a chemical formula What is the % by mass of each element in water, H2O? Step 1 – Calculate grams of each element and molar mass: H: 2 mol x 1.01 g/mol = 2.02 g O: 1 mol x 16.00 g/mol = 16.00 g 18.02 g Step 2 – Calculate % H: 2.02 / 18.02 x 100 % = 11.2 % O: 16.00/ 18.02 x 100 % = 88.8 % 2nd method to calculate % O = 100 – 11.2 = 88.8% HW Self check 8.7

Determining Empirical Formula from Elemental Analysis Data Example #1: Elemental Analysis (% by mass) for a compound is: Al: 32.13 % F: 67.87% Recall = Simplest Ratio of Atoms: AlxFy, where x,y = integers Step 1: Find masses of elements. Since data in mass %, assume 100 g sample; then can replace % with g. Assume 100 g sample: 32.13% of 100 = 32.13 g Al 67.87% of 100 = 67.87 g F

Step #2: Convert to moles ( 1 mole Al ) Al: 32.13 g F: 67.87 g = 1.191 moles Al 26.98 g Al 1 mole F ) ( = 3.572 moles F 19.00 g F Can we leave final Answer as: Al 1.911 F3.572 ? No! Final Answer must be simplest ratio!

Step 3: Divide by smallest # of moles: Al: 1.191 moles F: 3.572 moles F 3.572 > 1.191 moles so divide both by 1.191 Al: 1.191 ÷ 1.191 = 1 F: 3.572 ÷ 1.191 = 3 Final Answer: AlF3 HW: Self check 8-8

Determining Empirical Formula from Elemental Analysis Data Example #2: An oxide of aluminum is formed by the reaction of 4.151 g of al with 3.692 g of oxygen. Calculate the empirical formula of the compound. Al: 4.151 g O: 3.692 g Recall = Simplest Ratio of Atoms: AlxOy, where x,y = integers Step 1: Find masses of elements: Given here

Step #2: Convert to moles ( 1 mole Al ) Al: 4.151 g O: 3.692 g = 0.1539 moles Al 26.98 g Al 1 mole O ) ( = 0.2308 moles O 16.00 g O Can we leave final Answer as: Al 01539 O0.2308 ? No! Final Answer must be simplest ratio!

Step 3: Divide by smallest # of moles: Al: 0.1539 moles O: 0.2308 moles O 0.2308 > 0.1539 moles so divide both by 0.1539 Al: 0.1539 ÷ 0.1539 = 1 O: 0.2308 ÷ 0.1539 = 1.5 Can’t leave answer as: AlO1.5

1.500 moles O x 2 = 3 moles O 1.000 moles Al x 2 = 2 moles Al Step 4: Multiply the numbers from step 3 by the smallest integer that convert them both to whole numbers. 1.5 = 3/2 so x 2 will get rid of 2 in denominator 1.500 moles O x 2 = 3 moles O 1.000 moles Al x 2 = 2 moles Al Final Answer: Al2O3 HW: Self check 8-9 and 8.10

Common Decimals/Fractions in our problems 1/4 1/3 1/2 2/3 3/4 0.25 0.33 0.50 0.67 0.75

Moles X Moles Y Multiply by Final Answer 1.00 2.00 NA XY2 1.25 1.00 4 X5Y4 1.33 1.00 3 X4Y3 1.50 1.00 2 X3Y2 1.67 1.00 3 X5Y3 X7Y4 1.75 1.00 4 1.99 1.00 Round to 2 X2Y 1.89 1.00 Error check your math