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

PowerPoint to accompany

Chapter 2Part 2

Stoichiometry:Calculations with

Chemical Formulae and Equations

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

Moles



Avogadro’s Number (NA)

6.02 x 1023

1 mole of 12C has a mass of 12 g

Figure 2.8


Moles

Why does a mole of natural carbon weigh 12.011 g instead of 12.000 g?

What does one mole of natural Oxygen weigh and what does this mean?


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 uthen 1 mole of Au has a mass of 97 g.

ii) if 1 NaCl unit has a molecular mass of 58.5 uthen 1 mole of NaCl has a mass of 58.5 g.


Interconverting Massesand Numbers of Particles

Moles =

mass in grams

molar mass in g/mol

n =

m

M

wheren = moles

m = mass in gramsM= molar mass in

g/mol


Interconverting Massesand 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


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


Finding Empirical Formulae


Empirical Formulae from Analyses

Procedure for calculating an empirical formula from percentage composition.

Figure 2.11


Combustion Analysis

Compounds containing C, H and O are routinely analysed through combustion. C is determined from the mass of CO2 produced

H is determined from the mass of H2O produced O is determined by difference after the C and H have

been determined

Figure 2.12



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 100.00 g of material.

C: = 6.484 mol C

H: = 11.37 mol H

O: = 0.6481 mol O

77.87 g12.01 g/mol

11.76 g1.01 g/mol

10.37 g16.00 g/mol


Calculating an Empirical Formula (continued)

Step 2. Calculate the mole ratio by dividing through all elements by the smallest number of moles.

6.4840.6481

11.670.6481

0.68410.6841

C

10.00

C10

:

:

:

:

H

18.00

H18

:

:

:

:

O

1.000

O

i.e. the empirical formula is C10H18O


Calculating a Molecular Formulaefrom an Empirical FormulaeQ.Eucalyptol has the empirical formula of C10H18O.

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 C10H18O.

10(12.0 u) + 18(1.0 u) + 1(16.0 u) = 154 u


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.

= 1.01molecular mass

empirical formula mass

154

152 =

In this case, the multiplier is 1i.e. the molecular formula is the empirical

formula.


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


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


Stoichiometric CalculationsQ.From the following balanced equation, determine

how many grams of water are produced from 1.00 g of C6H12O6 (glucose)?

C6H12O6 + 6 O2 6 CO2 + 6 H2O

i) Convert grams of C6H12O6 to moles of C6H12O6

ii) Use the balanced equation to convert moles of C6H12O6 to moles of H2O

iii) Use the molar mass of H2O to convert moles of H2O to grams of H2O


Limiting Reactants(Limiting Reagents)


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 Bd2Ch)


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).


Limiting Reactants

The limiting reactant isthe reactant present in the smallest stoichiometricamount.

In this example, the H2

would be the limiting

reagent.

The O2 would be the

excess reagent.Figure 2.15


Limiting ReactantsFor combustion reactions, we want the fuel to be the limiting reagent. Why? 2 reasons…


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)


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…)


Percent Yield

A comparison of the amount actually obtained to the amount it was possible to make.

Actual YieldTheoretical YieldPercent Yield = x 100

Documents

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