Stoichiometry: Calculations with Chemical Formulae and Equationscheminnerweb.ukzn.ac.za/Files/Chem...

PowerPoint to accompany

Chapter 2

Stoichiometry:

Calculations with

Chemical Formulae

and Equations

Dr V Paideya

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

Chemical Equations

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Figure 2.4


Chemical Equations

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

CH4 (g) + 2 O2 (g) ---- CO2 (g) + 2 H2O (g)

Products appear on theright side of the equation

Reactants appear on the

left side of the equation.


Chemical Equations

The states of the reactants and products

are written in parentheses to the right of each compound.

CH4 (g) + 2 O2 (g) ------- CO2 (g) + 2 H2O (g)

Coefficients are inserted to

balance the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)


Subscripts and Coefficients Give Different Information

Subscripts indicate the number of atoms of each element in a molecule

Coefficients indicate the number of molecules

Figure 2.2


Reaction Types


Combination Reactions

Two or more

substances

react to form

one product

Examples

N2 (g) + 3 H2 (g) 2 NH3 (g)

C3H6 (g) + Br2 (l) C3H6Br2 (l)

2 Mg (s) + O2 (g) 2 MgO (s)

Figure 2.5


Decomposition Reactions

One substance breaks down into two or more substances

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)

2 NaN3 (s) 2 Na (s) + 3 N2 (g)

Figure 2.6


Combustion Reactions

Rapid reactions that

produce a flame

Most often involve

hydrocarbons

reacting with oxygen

in the air

Examples

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)

Figure 2.7


Formula Mass

Sum of the atomic masses of the atoms

in a chemical formula

The formula mass of calcium chloride,

CaCl2, would beCa: 1 x 40.1 u = 40.1 u

+ Cl: 2 x 35.5 u = 71.0 u

111.1 u


Molecular Mass

Sum of the atomic masses of the atoms

in a molecule

For the molecule ethane, C2H6, the

molecular mass would beC: 2 x 12.0 u = 24.1 u

+ H: 6 x 1.0 u = 6.0 u

111.1 u


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


Percent Composition

For example, the percentage of carbon in

ethane, C2H6, is:

% C =2 x 12.0 u

30.0 u

24.0 u

30.0 u= x 100

= 80.0 %


Example One

Calc. the mass % of H and O in H2O2

Formula mass of H2O2 = 34.02g

Which contains 2 atoms of H and 2 atoms of O

Atomic mass of H = 1.008 and O = 16.00

% H = 2 X 1.008 g/mol /34.02 g/mol x 100 = 5.926%

% O = 2 X 16.00 g/mol / 34.02 x 100 = 94.06%


Practice Exercise

Calculate the percent composition of Cl in CCl2F2

Calculate the percent composition of each element

in C12H23O6Br

Calculate the percent composition of K and Mn in

KMnO4

Calculate the percent composition of H2O in

CaCl2.2H2O


Moles


Avogadro’s Number (NA)

1 mole of 12C has a mass of 12 g

= 6.02 x 1023 atoms (Avogadro’s number)

Figure 2.8


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


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-1

32.07g of Sulfur = 1 mol S = 6.022 x 1023 S atoms

12.01g of carbon = 1 mol of C = 6.022 x 1023 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, NA = Avogadro’s number, n = no. of mols


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


Converting between grams of a compound and number of molecules

Given: 22.5g of CO2 Find: number of CO2

molecules

1 mol of CO2 = 44.01g

x mol = 22.5g

No. Of molecules of CO2

= 22.5g x 1 mol CO2 / 44.01g x 6.022 x 1023 molecules

= 3.08 x 1023 CO2 molecules


Practice Exercise

Calculate the molar mass of the following

compounds:

N2

CO

MgCl2

CaCl2.6H2O

CH3CH2CH2CH2CH2COOH


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 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 Formulae

Q.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 the law of conservation of mass is upheld.

Table 2.3



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



Q.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


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 is

the reactant present in

the smallest stoichiometric

amount.

In this example, the H2

would be the limiting

reagent.

The O2 would be the

excess reagent.Figure 2.15


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.


Percent Yield

A comparison of the amount actually

obtained to the amount it was possible to

make.

Actual YieldTheoretical YieldPercent Yield = x 100

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