PowerPoint to accompany Chapter 2 Part 2 Stoichiometry: Calculations with Chemical Formulae and Equations

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Moles

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

Avogadro’s Number (N A ) 6.02 x mole of 12 C has a mass of 12 g Figure 2.8

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Moles  Why does a mole of natural carbon weigh g instead of g?  What does one mole of natural Oxygen weigh and what does this mean?

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 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.

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 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

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Interconverting Masses and Numbers of Particles The mole concept can be thought of as the bridge between the mass of a substance in grams and the number of formula units. Figure 2.10

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 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

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Finding Empirical Formulae

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Empirical Formulae from Analyses Procedure for calculating an empirical formula from percentage composition. Figure 2.11

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Combustion Analysis  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 Figure 2.12

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia

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. C:= mol C H:= mol H O:= mol O g g/mol g 1.01 g/mol g g/mol

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Calculating an Empirical Formula (continued) Step 2.Calculate the mole ratio by dividing through all elements by the smallest number of moles C C 10 :::::::: H H 18 :::::::: O O i.e. the empirical formula is C 10 H 18 O

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 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

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 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. = 1.01 molecular mass empirical formula mass = In this case, the multiplier is 1 i.e. the molecular formula is the empirical formula.

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Stoichiometric Calculations 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 Dalton’s law of conservation of mass is upheld. Table 2.3

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 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

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 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

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Limiting Reactants (Limiting Reagents)

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 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)

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 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).

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount. In this example, the H 2 would be the limiting reagent. The O 2 would be the excess reagent. Figure 2.15

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Limiting Reactants For combustion reactions, we want the fuel to be the limiting reagent. Why? 2 reasons…

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Combustion Reactions  Fuel is expensive, oxygen in the ari is free.  If Oxygen is limiting, fuel is incompletely combusted.  Less energy is produced.  Carbon Monoxide or Carbon Black (soot) is produced (toxin or fouling engine)

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 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. (impurities, poor conditions…)

Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Percent Yield A comparison of the amount actually obtained to the amount it was possible to make. Actual Yield Theoretical Yield Percent Yield =x 100