1. Engineering Chemistry CHM 406
Stoichiometry
Engineering Chemistry
CHM 406

2. Measurements : Units Basic S.I. (Systéme Internationale) units:
Quantity
Unit
Symbol
Length
Metre m
Mass
Kilogram kg
Time
Second s
Temperature
Kelvin K
Amount of substance
Mole
mol
Electric current
Ampere A

3. Other common units Quantity Unit Symbol Derivation Temperature
Degree Celsius °C
= K
Volume
Litre L
= 10-3 m3
Pressure
Pascal
Atmosphere
Millimetre of mercury Pa
atm
mm Hg
= kg m-1 s-2
Energy
Joule J
= kg m2 s-2

4. The mole
By definition: a quantity equal to the number of atoms of 12C in exactly 12 g ( …. g) of 12C.
This number is called Avogadro’s number, NA.
NA = X 1023 mol-1.
A mole of any (pure) substance contains NA atoms, molecules, ions, particles, etc., of that substance.

5. Molar mass
Every pure substance has a molar mass (M): the mass in grams of one mole of that substance.
The molar mass of an element is numerically equal to its atomic mass in atomic mass units.
Atomic masses have been determined and tabulated. Thus if formula is known, molar mass can be calculated for any compound.

6. Example
The molecular formula of glucose is C6H12O6. The atomic masses (molar masses of atoms) of C, H, and O are 12.01, 1.008, and g mol-1. respectively. Calculate the molar mass of glucose. Answer: 6 X X X = g mol-1.

7. Relationship between mass and number of moles
Reactions occur between molecules/atoms and therefore moles
We usually measure mass
To interconvert, use the formula
mass
molar mass
number of moles

8. Percentage composition
We can determine the percentage by mass of any element from the molecular formula. Example: Ethanol, C2H6O. Similarly, %H = 13.1%, %O = 34.8%

9. Formulae from percentage composition
The composition or percentage by mass of each element in a compound can be determined by a technique called elemental analysis.
This can be used to determine the ratio of atoms of each element in the compound (empirical formula).
If the molar mass is also known, the molecular formula can be determined.

10. Example
A white powder is analysed and found to contain 43.64% P and 56.36% O by mass. Its molar mass is found to be g mol-1. What is the molecular formula? [The molar masses of P and O are and g mol-1, respectively.] Mass ratio P : O = : Mole ratio P : O = 43.64/30.97 : 56.36/16.00 = : = 2 : 5 Hence, empirical formula is P2O5

11. (contd)
Empirical formula mass = 2 X X = Actual molar mass : empirical formula mass = : = 2 : 1 Therefore the empirical formula goes into the molecular formula twice. Hence the molecular formula must be P4O10

12. Balancing chemical equations
Multiply each reactant or product by a whole number coefficient, such that
The number of atoms of each element is the same on both sides of the arrow.
The net charge is the same on both sides of the arrow
In simple cases, this can be done by inspection.

13. Examples
C2H6 + O2 → CO2 + H2O 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O NH3 + O2 → NO + H2O 4 NH3 + 5 O2 → 4 NO + 6 H2O Cr2O72- + H3O+ + Fe2+ → Cr3+ + Fe3+ + H2O Cr2O H3O+ + 6 Fe2+ → 2 Cr Fe H2O

14. Stoichiometric calculations
The balanced chemical equation tells us the molar ratios of reactants and products.
By converting to mass, we can calculate the masses of reactants required and products obtained.
Example: "Slaked lime," Ca(OH)2, is formed from "quicklime," CaO, by adding water, H2O. What mass of water is needed to convert 10.0 kg of quicklime to slaked lime? What mass of slaked lime is produced? [Atomic mass of Ca = g mol-1.]

15. (contd)
CaO + H2O → Ca(OH)2 (calcium hydroxide) Molar masses: CaO : g mol-1 H2O : g mol-1 Ca(OH)2 : g mol-1 nCaO = 10,000 g / g mol-1 = 178 mol = nH2O mH2O = 178 mol X g mol-1 = 3207 g = 3.21 kg mCa(OH)2 = 178 mol X g mol-1 = g = 13.2 kg

16. Limiting reagent
In most reactions carried out in practice, all but one of the reactants will be in stoichiometric excess.
The reactant which gets used up first is called the limiting reagent.
Usually the most scarce or valuable reactant.

17. Example
Aluminium reacts with sulphuric acid according to the equation 2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2 If 20.0 g of Al is added to a solution containing 105 g of H2SO4, what is the limiting reagent? Molar masses: Al : g mol-1 H2SO4 : g mol-1 (S : 32.07) nAl = mol, nH2SO4 = 1.07 mol nH2SO4 (required) = X (3/2) = 1.11 mol Insufficient H2SO4 – hence it is limiting.

18. Reactions in solution
For a reaction to occur efficiently, the reacting molecules or ions must rapidly come into contact with each other.
This happens best when they are gases, miscible liquids, or dissolved in a suitable solvent such as water.
In the latter case, the quantity measured most easily is not mass, but volume.

19. Relationship between volume and number of moles
Use the formula
Concentrations must be either known or measurable
number of moles
concentration
volume

20. Ionic solutions
When an ionic substance dissolves in water, it dissociates into its component ions. The solution is actually one of the component ions.
An aqueous solution containing 1 mol of CaCl2 will actually have 1 mol of Ca2+ and 2 mol of Cl-.

21. Example 5 Fe2+ + MnO4- + 8 H3O+  5 Fe3+ + Mn2+ + 12 H2O
A g sample of a mixture of FeSO4 and Fe2(SO4)3 was dissolved in dilute H2SO4, and reacted with a mol L-1 KMnO4 solution. The reaction required 15.8 mL of the KMnO4 solution. What percent by mass of the mixture was FeSO4?
5 Fe MnO H3O+ 
5 Fe Mn H2O
No. of moles of MnO4-  No. of moles of Fe2+  Mass of FeSO4  Percentage of FeSO4
Molar mass of FeSO4 = g mol-1

22. No. of moles of MnO4- = 0.0040 mol L-1 X 0.0158 L
= X mol
No. of moles of Fe2+ = 5 X No. of moles of MnO4-
= X 10-4 mol
Mass of FeSO4 = 3.16  10-4 mol  g/mol
= g
Percentage = ( / ) X 100 = % = 48%