1. Unit 6 : Stoichiometry Stoichi - element & metry – to measure

2. Empirical Formulas When a compound’s percent composition is known, its formula can be calculated. First, determine the smallest whole-number ratio of the moles of elements in the compound. – This ratio gives the subscripts in the empirical formula. The empirical formula for a compound is the formula with the smallest whole-number mole ratio of the elements. – The empirical formula might or might not be the same as the actual molecular formula! – The empirical formula = molecular formula for IONIC COMPOUNDS - ALWAYS!

3. Molecular Formulas If the empirical formula is different from the molecular formula, the molecular formula will always be a simple multiple of the empirical formula. – EX: The empirical formula for hydrogen peroxide is HO; the molecular formula is H 2 O 2. – In both formulas, the ratio of oxygen to hydrogen is 1:1

4. Calculating Empirical Formulas Use the following poem to remember the steps: Percent to mass Mass to moles Divide by small Multiply ‘til whole

5. EXAMPLE PROBLEM Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis: 48.64% carbon, 8.16% hydrogen, and 43.20% oxygen.

6. Step One: Percent to Mass Let’s assume we have a 100. g sample of methyl acetate. This means that each element’s percent is also the number of grams of that element. 48.64% C = 48.64 g C 8.16% H = 8.16 g H 43.20% O = 43.20 g O

7. Step Two: Mass to Moles Convert each mass into moles using the molar mass of each element. 48.64 g C x 1 mol C = 4.053 mol C 12.0 g C 8.16 g H x 1 mol H = 8.08 mol H 1.01 g H 43.20 g O x 1 mol O = 2.700 mol O 16.0 g O

8. Step Three: Divide by Small Oxygen accounts for the smallest number of moles in the formula, so divide each element by oxygen’s number of moles: 2.700 mol Carbon: 4.053 mol / 2.700 mol = 1.501 = 1.5 Hydrogen: 8.08 mol / 2.700 mol = 2.99 = 3 Oxygen: 2.700 mol / 2.700 mol = 1.000 = 1 Remember, we will want whole-number ratios

9. Step Four: Multiply ‘Til Whole In the previous slide, the ratio of C:H:O is 1.5:3:1 We need a whole-number ratio, so we can multiply everything by 2 to get rid of the 1.5 C  3, H  6, O  2 So, the final empirical formula is C 3 H 6 O 2

10. Calculating Molecular Formulas Sorry, no silly poem this time! Step 1 – Calculate the empirical formula (if needed) Step 2 – GIVEN molecular mass (experimental)/empirical formula molar mass = multiplier Step 3 – Multiply the empirical formula subscripts by the multiplier found in Step 2

11. EXAMPLE PROBLEM Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% carbon, 5.08% hydrogen, and 52.24% oxygen and has a molar mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid.

12. Step One: Find Empirical Formula 40.68 % C = 40.68 g C x 1mol C = 3.390 mol C 12.0 g C 5.08 % H = 5.08 g H x 1 mol H = 5.03 mol H 1.01 g H 54.24% O = 54.24 g O x 1 mol O = 3.390 mol O 16.0 g O C: H:O 3.390 mol: 5.03 mol: 3.390 mol 3.390 mol 3.390 mol 3.390 mol 1 : 1.48: 1  1 : 1.5 : 1, multiply by 2  C 2 H 3 O 2

13. Step Two: Divide Molar Masses Molar mass empirical formula = (2 x 12.0 g/mol) + (3 x 1.01 g/mol) + (2 x 16.0 g/mol) = 59.0 g/mol Given molar mass = 118.1 g/mol Multiplier = 118.1 g/mol = 2 59.0 g/mol

14. Step Three: Use Multiplier Empirical Formula = C 2 H 3 O 2 x 2 from step two Molecular formula = C 4 H 6 O 4

15. HOMEWORK QUESTIONS 1) What information must a chemist obtain in order to determine the empirical formula of an unknown compound? 2) What information must a chemist have to determine the molecular formula of a compound? 3) A compound containing barium, carbon, and oxygen has the following percent composition: 69.58% Ba, 6.09% C, 24.32% O. What is the empirical formula for this compound?

16. Balanced Equations Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction. The calculation of quantities in chemical reactions is a subject of chemistry called stoichiometry.

17. Interpreting Chemical Equations A balanced chemical equation can be interpreted in terms of different quantities, including numbers of atoms, molecules, or moles; mass; and volume.

18. Mole Ratios Mass and atoms are conserved in every chemical reaction. In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product, between moles of reactants, or between moles of products. A mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles.

19. What the Symbols Mean x = given or calculated number G = given unit a = number of mols of G b = number of mols of W W = mols of wanted

20. Mole-to-Mole Calculations x = amount of given in problem b = coefficient of wanted a = coefficient of given xb = multiplication of given amount times wanted coefficient W = answer to problem

21. Mass-to-Mass Calculations

22. Steps to Stoichiometric Success 1.Convert to moles, convert to moles, convert to moles. 2. Use mole ratio from the balanced equation to calculate the number of moles of what you are given to what you want. 3.Convert moles to any other unit as the problem requires. 4.Record your answer and don’t forget the units.

23. Conversion Map

24. Sample Problem Calculating Moles of a Product How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? List the known and the unknown. KnownUnknown – moles of nitrogen 0.60 mol N 2 moles of ammonia = ? mol NH 3 The conversion is mol N 2 →mol NH 3. According to the balanced equation, 1 mol N 2 combines with 3 mol H 2 to produce 2 mol NH 3. To determine the number of moles of NH 3, the given quantity of N 2 is multiplied by the form of the mole ratio from the balanced equation that allows the given unit to cancel. This mole ratio is 2 mol NH 3 /1 mol N 2. Calculate and solve for the unknown. 0.60 mol N 2 x 2 mol NH 3 /1 mol N 2 = 1.2 mol NH 3

25. Limiting Reagent The limiting reagent is the reagent that determines the amount of product that can be formed by a reaction. In the reaction of nitrogen and hydrogen, hydrogen is the limiting reagent. Nitrogen is the reagent that is not completely used up in the reaction. The reagent that is not used up is called the excess reagent.

26. Limiting Reagent & Percent Yields

27. Yields The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants. In contrast, the amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield. The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent.