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Calculations in chemistry: stoichiometry

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1 Calculations in chemistry: stoichiometry
Chapter 15

2 Return of the mole The mole is a very useful unit because it represents a ‘measurable’ number of particles – atoms, ions or molecules. The mole concept effectively acts as a scale up from individual particles to measurable amounts of particles.

3 Stoichiometry In a chemical reaction, atoms are neither created nor destroyed. Consequently, given the amount of a reactant consumed or a product formed we can calculated the amounts of other reactants and products involved. We are now applying the mole to chemical reactions and not just chemicals.

4 Balanced Chemical Equations
All particles that exist at the start of a chemical reaction must be accounted for after the reaction is complete Consider 2H2(g) + O2(g) → 2H2O(g) The equation gives the following information about the reaction it represents 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 the substances react

5 Reacting Quantities 2H2(g) + O2(g) → 2H2O(g) Consider
This reaction involves 2 moles of H2(g) reacting with 1 mole of O2(g) to form 2 moles of H2O(g). In more general terms the amount of oxygen used will always be equal to half the amount of hydrogen used and half the amount of water formed

6 What a chemical equation does not tell us
An equation does not tell us about the rate of a reaction It does not tell us whether heat is required or given off It does not tell us what temperature or pressure is required It gives no details as to how the individual atoms or molecules are transformed from reacts to products

7 Mass-Mass Stoichiometry
We generally measure substances by mass but chemical reactions depend on relative numbers of atoms, ions of molecules as reflected in the chemical amount or number of moles. So the relationships between mass and number of mole is critical. Conversion of metric units of mass: m = n x M n = Chemical amount mol mass g molar mass g mol-1 OR m M microgram µg milligram mg gram g kilogram kg tonne t ÷103 x103

8 Mass-Mass Stoichiometry
Given the amount of one substance involved in a chemical reaction, the amounts of all other substances involved can be calculated, provided a balance equation for the reaction is known. Mass-mass stoichiometry problems can be solved in four steps: Write a balanced equation for the reaction. Calculate the amount (in mol) of the substance with the know mass. Use the mole ratio from the equation to calculate the amount (in mol) of the required substance. Calculate the mass required.

9 Mass-Mass Stoichiometry
Worked Example Determine the mass of oxygen used when 2.00 kg of water is produced according to the equation 2H2(g) + O2(g) → 2H2O(l). Suggested Solution Tag the equation with the data supplied and the quantity you have to determine 2H2(g) + O2(g) → 2H2O(l) Identify the starting point (the data you’ve got) and the finishing point (the quantity you want) of the calculations. ? m 2.00kg m(H2O) m(O2)

10 Mass-Mass Stoichiometry
Link the data you’ve got and the quantity you want to their number of mole. The link between n(O2) -what you want to find- and n(H2O) –what you’ve got- is obtained from the equation. The equation tells us that n(O2)/n(H2O) = ½, so n(O2) = ½ x n(H2O) So the calculation flow chart can now be completed n(H2O) n(O2) ÷ M(H2O) x M(O2) m(H2O) m(O2) m(H2O) m(O2) n(H2O) n(O2) ÷ M(H2O) x M(O2) X 1/2

11 Mass-Mass Stoichiometry
It is then relatively quick to follow through the calculation sequence. n(H2O) produced = m(H2O) / M(H2O) = 2.00x103 g / 18.0 g mol-1 = 1.11x102 mol n(O2) reacting = ½ x n(H2O) produced = ½ x 111 mol = 55.6 mol m(O2) required = n(O2) x M(O2) = 55.6 mol x (2 x 16.0) g mol-1 = 1.78x103 g = 1.78 kg

12 Mass-Mass Stoichiometry
Question A phosphorous manufacturer is to extract 1.00 tonne of phosphorous per day by the process given by the equation: 2Ca3(PO4)2(s) + 6SiO2(s) + 10C(s) → P4(s) + 10CO(g) + 6CaSio3(s) Calculate the mass required daily of: Calcium phosphate Silicon dioxide

13 Your Turn Page 263 Question 1 - 3

14 Concentrating on Solutions
Many chemical reactions take place between substances that have been dissolved in a liquid, most commonly water, to form a solution. The liquid in which the substances are dissolved is called the solvent. The dissolved substance is called the solute. When working with solutions, the most easily measured quantity is their volume. However, for the volume of a solution to provide useful information, the solution’s concentration must be known.

15 Calculating concentrations of solutions
Amounts of solutions are usually measured by volume. Common units for volume are litres (L) and millilitres (mL), where 1L = 103 mL Just as chemical amount (mol) and mass (g) are linked by molar mass (g mol-1), then chemical amount (mol) and volume (L) are linked by molar concentration (mol L-1) Molar concentration, symbol c, which has units mol per litre (mol L-1), is often expressed as molarity (M). So 1.0 M HCl(aq), ie a 1 molar solution of hydrochloric acid is the same as saying that c(HCl) = 1.0 mol L-1 n = cV Chemical amount mol Concentration mol L-1 Volume L c = n V

16 Preparation of a solution of known concentration
A solution of known concentration is known as a standard solution. In order to prepare a particular volume of solution of known concentration, the following five steps should be followed Calculate the number of moles of solute that are needed to obtain correct concentration of solution for the volume of solvent to be used, according to the formula n = cV Calculate the mass of the solute needed, using the formula m = nM Partially fill a volumetric flask with water, and add the correct mass of solute Dissolve the solute Add water to the required volume

17 Preparation of a solution of known concentration
Example Calculate the number of moles of sodium chloride needed to prepare 500 mL of a mol L-1 salt solution. What mass of sodium chloride would be weighed out. Solution List the known information volume (V) = 500 mL = 0.5 L concentration (c) = mol L-1 x V(NaCl) x M(NaCl) c(NaCl) n(NaCl) m(NaCl)

18 Preparation of a solution of known concentration
Calculate number of moles (n) of NaCl needed n = cV = x 0.500 = mol Calculate the mass represented by the number of moles. m = n x M = x 58.5 = 2.34 g

19 Worked Example 15.1d What volume of M sulfuric acid reacts completely with 17.8 ml of M potassium hydroxide?

20 Your Turn Page 263 Questions 6 - 8

21 Limiting Reactant Calculations
In many chemical reactions an excess of one reactant is added to ensure complete reaction of another – generally more valuable – reactant. The reaction stops when one reactant is used up (the limiting reagent), even though some of the other substance is unreacted. The other reactant is said to be in excess (the excess reagent) – some of it remains when the reaction has finished In associated problems it is often necessary to determine which reactant is in excess before amounts of product can be determined. Again there are logical calculation techniques, which will lead to efficient solutions of such problems Chemical reactions may be considered to involved three stages: an initial stage where the reactants are added, a reacting stage where the reactants combine in the mole ration suggested by the equation, and a final stage where the reactions appear to be complete

22 Limiting Reactant Calculations
Example 1 A shop advertises the sale of dining settings, consisting of a table and four chairs. Using chemical terminology, a dining setting would be written as TaCh4. The arrangement can be set out as an equation: Ta + 4Ch → TaCh4 The shop does some stocktaking and discovers it has 16 chairs and five tables, how many complete settings can it sell?

23 Example 2 A container has a mixture of eight molecules of hydrogen gas and six molecules of oxygen gas. When ignited, hydrogen gas and oxygen gas react to form water according to the equation: 2H2(g) + O2(g) → 2H2O(g) How many water molecules can be formed?

24 Limiting Reactant Calculations
A real example A gaseous mixture of 25.0g of hydrogen gas and 100.0g of oxygen gas are mixed and ignited. The water produced is collected and weighed. What is the expected mass of water produced? Suggested answer Tag the equation with the data supplied and the quantity you have to determine 2H2(g) + O2(g) → 2H2O(g) 25.0 g 100.0 g ? g

25 Limiting Reactant Calculations
Calculate the amount of each reactant. n(H2) = m(H2)/M(H2) and n(O2) = m(O2)/M(O2) = 25.0 g / 2 x 1.01 g mol = g / 2 x 16.0 g mol-1 = 12.4 mol = 3.13 mol i) Work out which reactant is fully used (the other must be in excess). This requires comparison of initial amounts of reactants with the mole ratios suggested in the equation. Assume all the H2 reacts, then 2H2(g) O2(g) → 2H2O(g) Initially Reacting This is not possible because n(O2) = 2 x n(H2) – according to the equation – and there is simply not enough O2 present. So H2 is in excess and O2 will be fully used. 12.4 mol 3.13 mol 12.4 mol 6.2 mol

26 Limiting Reactant Calculations
ii) Show that all the O2 does react and find amounts present at end of reaction 2H2(g) O2(g) → 2H2O(g) Initially mol mol Reacting mol mol → mol Finally mol mol Complete the calculations m(H2O) = n(H2O) x M(H2O) = 6.26 mol x g mol-1 = 113 g Therefore, 113 g of water is produced from this hydrogen – oxygen mixture

27 Worked Example 15.2b 2.50 g of aluminium is mixed with 5.00 g of iodine and allowed to react according to the equation: 2Al(s) + 3I2(s) ―› 2AlI3(s) What mass of aluminium iodide would be produced. What is the mass of the reactant in excess?

28 Your Turn Page 267 Question 9, 10 and 11

29 Volumetric analysis This is what your first outcome is all about.
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.

30 Volumetric Analysis. This involves reacting the solution of an unknown concentration with a solution of accurately known concentration. A solution of accurately known concentration is a standard solution.

31 Volumetric Analysis If we want to find the concentration of a solution of hydrochloric acid, the hydrochloric acid can be reacted with a basic standard solution. This is more of an exact procedure than anything you have done to date in the lab so we need to use precisely calibrated equipment.

32 Precise Equipment A volumetric flask –
is used to prepare the standard solution. An accurately weighed sample is placed in the flask and dissolved in de-ionised water to form a specific volume of solution.

33 A pipette Is used to measure accurately a specific volume of the solution. This known volume, or aliquot, is then poured into a conical flask ready for analysis.

34 A Burette A burette delivers accurately known, but variable volumes of solution (titres). A volume of acid or base is slowly titrated to the solution of base or acid in the conical flask until the reactants are present in stoichiometrically equivalent amounts. This is called the equivalent point.

35 The equivalent point For the reaction:
Na2CO3(aq) + 2HCl(aq) ―› 2NaCl (aq) + H2O(l) + CO2(g) The equivalence point is reached when exactly 2 mol HCl has been added for each 1 mol Na2CO3 This whole process is called a titration. When the equivalence point is reached the reaction is complete.

36 Your Turn Page 268 Question 12

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