1. STOICHIOMETRY

2. What is stoichiometry? Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

3. D.P.S.E. DECONSTRUCT- Identify what you 1)Have 2)Want 3) Need

4. D.P.S.E. Plan- Decide what type of pathway or equation you need to use.

5. D.P.S.E. Solve – Insert numbers with units in proper places in your Plan

6. D.P.S.E. Evaluate – Circle your answer with units, and check it to see if it makes sense. Is it too big, too small, does it make logical sense?

7. What You Should Expect Given : Amount of reactants Question: how much of products can be formed. Example 2 A + 2B 3C Given 20.0 grams of A and sufficient B, how many grams of C can be produced?

8. What do you need? You will need to use i. molar ratios, ii. molar masses, iii. balancing and interpreting equations, and iv. conversions between grams and moles. Note: This type of problem is often called "mass-mass."

9. Steps Involved in Solving Mass-Mass Stoichiometry Problems Balance the chemical equation correctly Using the molar mass of the given substance, convert the mass given to moles. Construct a molar proportion (two molar ratios set equal to each other) Using the molar mass of the unknown substance, convert the moles just calculated to mass.

10. Mole Ratios A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound.

11. Example Reaction between magnesium and oxygen to form magnesium oxide. ( fireworks) 2 Mg(s) + O2(g) 2 MgO(s) Mole Ratios: 2 : 1: 2

12. 1) N 2 + 3 H 2 ---> 2 NH 3 Write the mole ratios for N 2 to H 2 and NH 3 to H 2. 2) A can of butane lighter fluid contains 1.20 moles of butane (C 4 H 10 ). Calculate the number of moles of carbon dioxide given off when this butane is burned. Practice Problems

13. Mole-Mole Problems Using the practice question 2) above: Equation of reaction 2C 4 H 10 + 13O 2 8CO 2 + 10H 2 O Mole ratio C 4 H 10 CO 2 1: 4 [ bases] 1.2 : X [ problem] By cross-multiplication, X = 4.8 mols of CO 2 given off

14. Mole-Mass Problems Problem 1: 1.50 mol of KClO 3 decomposes. How many grams of O 2 will be produced? [k = 39, Cl = 35.5, O = 16] 2 KClO 3 2 KCl + 3 O 2

15. Three steps…Get Your Correct Answer Use mole ratio Get the answer in moles and then Convert to Mass. [Simple Arithmetic] Hello! If you are given a mass in the problem, you will need to convert this to moles first. Ok?

16. Let’s go! 2 KClO 3 2 KCl + 3 O 2 2: 3 1.50 : X X = 2.25mol Convert to mass 2.25 mol x 32.0 g/mol = 72.0 grams Cool!

17. Try This: We want to produce 2.75 mol of KCl. How many grams of KClO 3 would be required? Soln KClO 3 : KCl 2 : 2 X :2.75 X = 2.75mol In mass: 2.75mol X 122.55 g/mol = 337 grams zooo zimple!

18. Mass-Mass Problems There are four steps involved in solving these problems: Make sure you are working with a properly balanced equation. Convert grams of the substance given in the problem to moles. Construct two ratios - one from the problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem. Convert moles of the substance just solved for into grams.

19. Just follow mass- mass problem to the penultimate level Mass-Volume Problems

20. Like this: There are four steps involved in solving these problems: Make sure you are working with a properly balanced equation. Convert grams of the substance given in the problem to moles. Construct two ratios - one from the problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem. Convert moles of the substance just solved for into Volume.

21. Conversion of mole to volume No of moles = Volume Molar volume Can you remember a similar equation?

22. Molar volume The molar volume is the volume occupied by one mole of ideal gas at STP. Its value is: 22.4dm 3

23. Practice Problems Calculate the volume of carbon dioxide formed at STP in ‘dm 3 ' by the complete thermal decomposition of 3.125 g of pure calcium carbonate (Relative atomic mass of Ca=40, C=12, O=16) Solution: Convert the mass to mole: Molar mass of CaCO 3 = 40 + 12 + (16 x 3) = 100gmol -1 Mole = mass/molar mass 3.125/100 = 0.03125mol

24. Practice Problems As per the equation, Mole ratio1 : 1 problem0.03125mol X X = 0.03125mol of CO 2 Convert mole to volume [slide 17] Volume = ( 0.03125 x 22.4)dm 3 = 0.7dm 3