PowerPoint to accompany. Chapter 2. Stoichiometry: Calculations with Chemical Formulae and Equations. Dr V Paideya

Transcription

1 PowerPoint to accompany Chapter 2 Stoichiometry: Calculations with Chemical Formulae and Equations Dr V Paideya

2 Chemical Equations CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Figure 2.4

3 Chemical Equations CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Reactants appear on the left side of the equation. CH 4 (g) + 2 O 2 (g) ---- CO 2 (g) + 2 H 2 O (g) Products appear on the right side of the equation

4 Chemical Equations CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) The states of the reactants and products are written in parentheses to the right of each compound. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) Coefficients are inserted to balance the equation.

5 Subscripts and Coefficients Give Different Information Figure 2.2 Subscripts indicate the number of atoms of each element in a molecule Coefficients indicate the number of molecules

6 Reaction Types

7 Combination Reactions Two or more substances react to form one product Figure 2.5 Examples N 2 (g) + 3 H 2 (g) C 3 H 6 (g) + Br 2 (l) 2 Mg (s) + O 2 (g) 2 NH 3 (g) C 3 H 6 Br 2 (l) 2 MgO (s)

8 Decomposition Reactions One substance breaks down into two or more substances 2 NaN 3 (s) 2 Na (s) + 3 N 2 (g) Figure 2.6 Examples CaCO3 (s) CaO (s) + CO2 (g) 2 KClO3 (s) 2 KCl (s) + 3 O2 (g) 2 NaN3 (s) 2 Na (s) + 3 N2 (g)

9 Combustion Reactions Rapid reactions that produce a flame Figure 2.7 Examples CH 4 (g) + 2 O 2 (g) C 3 H 8 (g) + 5 O 2 (g) Most often involve hydrocarbons reacting with oxygen in the air CO 2 (g) + 2 H 2 O (g) 3 CO 2 (g) + 4 H 2 O (g)

10 Formula Mass Sum of the atomic masses of the atoms in a chemical formula The formula mass of calcium chloride, CaCl 2, would be Ca: 1 x 40.1 u = 40.1 u + Cl: 2 x 35.5 u = 71.0 u u

11 Molecular Mass Sum of the atomic masses of the atoms in a molecule For the molecule ethane, C 2 H 6, the molecular mass would be C: 2 x 12.0 u = 24.1 u + H: 6 x 1.0 u = 6.0 u u

12 Percent Composition One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation: % element = (number of atoms of that element)(atomic mass of that element) (formula mass of the compound) x 100

13 Percent Composition For example, the percentage of carbon in ethane, C 2 H 6, is: % C = 2 x 12.0 u 30.0 u 24.0 u = x u = 80.0 %

14 Example One Calc. the mass % of H and O in H 2 O 2 Formula mass of H 2 O 2 = 34.02g Which contains 2 atoms of H and 2 atoms of O Atomic mass of H = and O = % H = 2 X g/mol / g/mol x 100 = 5.926% % O = 2 X g/mol / x 100 = 94.06%

15 Practice Exercise Calculate the percent composition of Cl in CCl 2 F 2 Calculate the percent composition of each element in C 12 H 23 O 6 Br Calculate the percent composition of K and Mn in KMnO 4 Calculate the percent composition of H 2 O in CaCl 2.2H 2 O

16 Moles

17 Avogadro s Number (N A ) Figure mole of 12 C has a mass of 12 g = 6.02 x atoms (Avogadro s number)

18 Molar Mass The mass of a single atom of an element (in u) is numerically equal to the mass (in grams) of 1 mole of that element. This is true regardless of the element. Examples: i) if 1 atom of Au has an atomic mass of 97 u then 1 mole of Au has a mass of 97 g. ii) if 1 NaCl unit has a molecular mass of 58.5 u then 1 mole of NaCl has a mass of 58.5 g.

19 Interconverting Masses and Numbers of Particles Moles = mass in grams molar mass in g/mol n = m M where n =moles m =mass in grams M =molar mass in g/mol

20 Interconverting Masses and Numbers of Particles Figure 2.10 The mole concept can be thought of as the bridge between the mass of a substance in grams and the number of formula units.

21 Converting between Grams and moles of an element Mass of 1 mol of atoms of an element is called the molar mass For example Copper has a atomic mass of 63.55, therefore 1 mol of Cu atoms has a mass of 63.55g and the molar mass of copper = 63.55g/mol or gmol g of Sulfur = 1 mol S = x S atoms 12.01g of carbon = 1 mol of C = x C atoms Eg. 1 Given 0.58g of C find mols of C 0.58 X 1 mol C / 12.01g C = 4.8 X 10-2 N = no. of atoms, N A = Avogadro s number, n = no. of mols

22 Mole Relationships One mole of atoms, ions or molecules contains Avogadro s number of those particles One mole of molecules or formula units contains Avogadro s number times the number of atoms or ions of each element in the compound

23 Converting between grams of a compound and number of molecules Given: 22.5g of CO 2 Find: number of CO 2 molecules 1 mol of CO 2 = 44.01g x mol = 22.5g No. Of molecules of CO 2 = 22.5g x 1 mol CO 2 / 44.01g x x molecules = 3.08 x CO 2 molecules

24 Practice Exercise Calculate the molar mass of the following compounds: N 2 CO MgCl 2 CaCl 2.6H 2 O CH 3 CH 2 CH 2 CH 2 CH 2 COOH

25 Finding Empirical Formulae

26 Empirical Formulae from Analyses Procedure for calculating an empirical formula from percentage composition. Figure 2.11

27 Combustion Analysis Figure 2.12 Compounds containing C, H and O are routinely analysed through combustion. C is determined from the mass of CO 2 produced H is determined from the mass of H 2 O produced O is determined by difference after the C and H have been determined

28 Calculating an Empirical Formula Q. Eucalyptol, from eucalyptus oil, contains 77.87% C, 11.76% H and 10.37% O. What is the empirical formula of this compound? Step 1. Assume g of material g C: g/mol = mol C g H: 1.01 g/mol = mol H g O: g/mol = mol O

29 Calculating an Empirical Formula (continued) Step 2. Calculate the mole ratio by dividing by the smallest number of moles. C : H : O : : : : C 10 : H 18 : O i.e. the empirical formula is C 10 H 18 O

30 Calculating a Molecular Formulae from an Empirical Formulae Q.Eucalyptol has the empirical formula of C 10 H 18 O. The experimentally determined mass of this substance is 152 u. What is the molecular formula of eucalyptol? Step 1. Calculate the formula mass of the empirical formula C 10 H 18 O. 10(12.0 u) + 18(1.0 u) + 1(16.0 u) = 154 u

31 Calculating a Molecular Formulae from an Empirical Formulae (continued) Step 2. The following division will provide us with the multiplier of the subscripts of the empirical formulae. molecular mass empirical formula mass = 1.01 In this case, the multiplier is 1 i.e. the molecular formula is the empirical formula. =

32 Stoichiometric Calculations Table 2.3 The coefficients in a balanced chemical equation indicate both the relative numbers of molecules involved in the reaction and the relative number of moles. Note that the law of conservation of mass is upheld.

33 Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it s a product) or used (if it s a reactant). Figure 2.13

34 Stoichiometric Calculations Q.From the following balanced equation, determine how many grams of water are produced from 1.00 g of C 6 H 12 O 6 (glucose)? C 6 H 12 O O 2 6 CO H 2 O i) Convert grams of C 6 H 12 O 6 to moles of C 6 H 12 O 6 ii) Use the balanced equation to convert moles of C 6 H 12 O 6 to moles of H 2 O iii) Use the molar mass of H 2 O to convert moles of H 2 O to grams of H 2 O

35 Limiting Reactants

36 How many cheese sandwiches can I make from 10 slices of bread and 7 slices of cheese? First, we need to know the stoichiometry, i.e. the ratio of cheese slices to slices of bread per sandwich. Let s make an ordinary cheese sandwich consisting of two slices of bread and one slice of cheese. In other words: 2 Bread + 1 Cheese 1 Cheese Sandwich (2 Bd + 1 Ch 1 Bd 2 Ch)

37 How many cheese sandwiches can I make from 10 slices of bread and 7 slices of cheese? So, if we want to make 7 sandwiches with the 7 slices of cheese, we would need 14 slices of bread. However, with only 10 slices of bread, the bread would be the limiting reactant, because it will limit the amount of sandwiches we can make (to 5).

38 Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount. Figure 2.15 In this example, the H 2 would be the limiting reagent. The O 2 would be the excess reagent.

39 Theoretical Yield The theoretical yield is the maximum amount of product that can be made. In other words, it s the amount of product possible as calculated through the stoichiometry problem. Actual yield, on the other hand, is the amount one actually produces and measures.

40 Percent Yield A comparison of the amount actually obtained to the amount it was possible to make. Actual Yield Percent Yield = Theoretical Yield x 100