Chapter 15 – Calculations in Chemistry: Stoichiometry

1. Chapter 15 – Calculations in Chemistry: Stoichiometry

2. Stoichiometry • Stoichiometry means the study of ratios of substances. • It is based onthe law of conservation of mass: in a chemical reaction, the total mass of all products is equal to the mass of all reactants. • Meaning, atoms are neither created nor destroyed.

3. Reacting Quantities • Consider the equation: 2H2(g) + O2(g)  2H2O(l) • It gives us the following information: • It identifies the reactants and products • It identifies the states of the reactants and products • Because it is balanced, it gives the ratio in which substances react. The coefficients in a balanced equation give us the ratio of the amounts (in mol) of reactants and products. This mole ratio will be used in calculating masses of reactants and products.

4. Calculations based on Amount of a Reactant or Product • Given the amount of one substance involved in a chemical reaction, the amounts of all other substances can be calculated, provided a balanced equation for the reaction is known.

5. Mass-Mass Stoichiometry • In these problems, the mass of one of the reactants or products is known and is used to determine the mass of another aspect. • The process for solving mass-mass problems is: Step 1: Write a balanced equation for the reaction. Step 2: Calculate the amount (in mol) of the substance with the known mass. n = m/M Step 3: Use the mole ratio to calculate the amount (in mol) of the required substance. want/got Step 4: Calculate the mass required. m = n x M

6. Stoichiometry involving Solutions • Many chemical reactions occur between reactants in solution. • The amount of reactant present in a solution can be determined from the volume of the solution used, and its concentration. • In these questions we use the equation n = c x v. • The process for this type of reaction is very similar to mass-mass, except for the equation. Step 1: Write a balanced equation for the reaction Step 2: Calculate the amount of mol of the substance with the known concentration. n = c x v Step 3: Use the mole ratio to calculate the amount (in mol) of the required substance. want/got Step 4: Calculate the concentration required. c = n/v

7. Calculations based on amount of Two Reactants • In our earlier examples, it is assumed that any reactants were present in sufficient quantities to react completely with all of the given substance. • Now, fixed quantities of two different reactants are mixed. • It is likely that one of the reactants will be used up before the other. • The reaction stops when one reactant is used up. • This is called the limiting reagent. • The other reactant is said to be in excess, or the excess agent.

8. Example… A shop advertises a sale on a stereo/speaker set, consisting of a stereo and 2 speakers. If we were to write this as an equation it would look like: St + 2Sp  StSp2 The shop owner discovers he has 3 stereos and 8 speakers. So how many sets can they sell? You can see there are only enough stereos for three sets and one speaker is in excess. So even, though there are enough speakers for another set, we are limited by the number of stereos. Limiting Reagent = ? Excess Reagent = ?

9. Chemistry Example… • There is a mixture of 8 molecules of hydrogen gas and 6 molecules of oxygen gas. • When ignited they react to form water. • While we have more hydrogen to begin with, the equation shows that two molecules of hydrogen are required to react with one oxygen. • So when ignited, eight molecules of hydrogen will react with four molecules of oxygen. • This means that two molecules of oxygen remain in excess. • After the reaction: • No hydrogen gas remains, since this is the limiting agent • The amount of water vapour formed is limited by the amount of hydrogen gas available • 2 mol of oxygen gas remains unreacted.

10. Worked Example… A gaseous mixture of 25.0g of hydrogen gas and 100g of oxygen gas are mixed and ignited. The water produced is collected and weighed. What is the expected mass of water produced? Step 1: Write a balanced equation. Step 2: Calculate the amount of each reactant. Step 3: Identify the reactant in excess. Step 4: Use the amount of the limiting reagent and the equation to determine the amount (in mol) of the required product. Step 5: Find the mass of water produced.

11. Worked Example 2… 2.50 g of aluminium is mixed with 5.00g of iodine and is allowed to react. • What mass of aluminium iodide would be produced? • What is the excess mass of the excess reactant? Step 1: Write the balanced equation. Step 2: Calculate the amount of each reactant. Step 3: Identify the reactant in excess. Step 4: Use the amount of limiting reagent to determine the amount of products formed. Step 5 (a): Calculate the mass of aluminium iodide formed. Step 5 (b): Calculate the mass of aluminium in excess.

12. Volumetric Analysis • There are many situations when it is essential to know the exact amount of acid or base in a substance. • The concentration of solutions of acids and bases can be determined accurately by a technique called volumetric analysis. • This involves reacting the solution of unknown concentration with a solution of accurately known concentration (a standard solution).

13. Volumetric Analysis cont… • If we want to find the concentration of a solution of hydrochloric acid, the hydrochloric acid can be reacted with a standard solution of sodium carbonate. • Precisely calibrated glassware is used to carry out the analysis.

14. Calibrated Equipment • A volumetric flask is used to prepare the standard solution of sodium carbonate. • A pipette is used to measure accurately a specific volume of the solution of HCl. This known volume is called an aliquot. • The standard solution of sodium carbonate is placed in a burette, which delivers accurately known, but variable, volume of solution (titres).

15. Volumetric Analysis Process • An accurately weighed sample of sodium carbonate is placed in the volumetric flask and dissolved in de-ionised water to form a specific volume of a solution. • The aliquot is then poured into a conical flask ready for analysis. • The solution of sodium carbonate is slowly added (titrated), from the burette, to the acid solution in the conical flask until the reactants are present in stoichiometrically equivalent amounts. This is called the equivalence point.

16. Volumetric Analysis Example… • For the reaction between hydrochloric acid and sodium carbonate solution: • The equivalence point occurs when exactly 2 mol of HCl has been added for each 1 mol of Na2CO3. • This whole process is called a titration. • When the equivalence point is reached, the reaction is complete. • The volume of sodium carbonate delivered by the burette is recorded. • The concentration and volume of the sodium carbonate is now known and also the volume of the aliquot of hydrochloric acid used. • The concentration of the acid can now be found using the principles of stoichiometry.

17. Finding the Equivalence Point • Since both the acid and base solutions are often colourless, there needs to be some way to determine when the reaction is complete. • This is done by adding an indicator that changes colour at (or very close to) the equivalence point of the titration.