1. STOICHIOMETRY Chemical Calculations
Faculty of Biotechnology
General Chemistry
Lecture 4
STOICHIOMETRY Chemical Calculations
Dr. M. Abd-Elhakeem

2. 1- Molecular weight 2- Mole 3- Avogadro’s number 4- Equivalent weight 5- Chemical equation stoichiometry 6- Gas’s laws 7-Element percent 8- Oxidation number 9- Solution stoichiometry ( Concentration – Titration)

3. Atomic weight:
The atomic weight of an atom is equal to the number of protons in the nucleus of the atom plus the number of neutrons in the nucleus of the atom (one amu each). Therefore, a carbon atom with six protons and six neutrons has an atomic weight of 12.
Molecular weight:
The weight of a molecule is the sum of the weights of the atoms of which it is made.

4. Mole: Unit of amount of chemical substance
Calculate the molecular weight of each of the following Na2CO3 , HCl , NaOH
Na2CO3 = (23 X 2) (16 X 3) = 106 g
Mole: Unit of amount of chemical substance
One mole is the molecular weight of a substance in gram. It equals the summation of atomic weight of the forming atoms.

5. Avogadro's number
Avogadro's number, also known as Avogadro's constant, is defined as the quantity of atoms in precisely 12 grams of 12C. The designation is a recognition of Amedeo Avogadro, who was the first to state that a gas' volume is proportional to how many atoms it has. Avogadro's number is given as x 1023 mol-1.

6. Number of moles: weight in grams/ molecular weight
How many moles in 56 g Na2CO3
Number of moles: weight in grams/ molecular weight
Then = 56/106 = mole
F2 Mole contain …….fluorine atoms
= 2 X Avogadro’s number

7. Mole -------Wight --------- number of atoms
Keep in your mind
Mole Wight number of atoms

8. Chemical equation
is the symbolic representation of a chemical reaction where the reactants are given on the left hand side and the products on the right hand side.

9. Balancing chemical equations
NaOH + HCl
NaCl + H2O
Na2CO3 + HCl 2 2
NaCl + H2O + CO2

10. coefficients give MOLAR RATIOS
Stoichiometry (more working with ratios)
Ratios are found within a chemical equation.
2HCl + Ba(OH)2  2H2O BaCl2 1 1
coefficients give MOLAR RATIOS
2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

11. How many moles of water are produced for each mole of P4O10 (s) when this equation is balanced? _
PH3 (g) + O2 (g) --> P4O10 (s) + H2O (l)

12. Mole – Mole Conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol
2 mol N2O5
4 mol NO2
= moles NO2
8.6
4.3 mol N2O5
?? mol NO2
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
= mole O2
2.2
4.3 mol
? mol

13. gram ↔ mole and gram ↔ gram conversions
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of N2O5 were used if 210g of NO2 were produced?
2N2O5(g)  4NO2(g) + O2(g)
? moles
210g
2 moles N2O5
(46 x4)g NO2
= moles N2O5
2.28
?? moles N2O5
210 g NO2
b. How many grams of N2O5 are needed to produce 75.0 grams of O2?
2N2O5(g)  4NO2(g) + O2(g)
= grams N2O5
506
? grams
75.0 g

14. Concentration
Extensive quantities such as mass and volume are often used when preparing solutions. We can talk about adding a teaspoon of sugar when we make ice tea, for example, or filling a 2-L pitcher with ice tea.
But when we want to compare solutions, we need intensive quantities that tell us how much sugar has been added to a given volume of lemonade or ice tea. We need to know the concentration of the solution, which is the ratio of the amount of solute to the amount of either solvent or solution.

15. Percent Concentration
Will be Studied in practical part
It is
w/w concentration (g / g)
w/v concentration (g / ml)
v/v concentration (ml / ml)

16. What is the number of moles in 50 ml of 6 M HCl
Molarity
Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M)
M = number of moles / Volume (L)
What is the number of moles in 50 ml of 6 M HCl

17. N = Number of equivalent / Volume (L)
Normality
Normality is a term used to express concentration. The units of normality are number of equivalents per liter (It is abbreviated as a capital N)
N = Number of equivalent / Volume (L)

18. Equivalent weight equal the molecular weight / e No of H in acids
No. of OH in base
No. of acid molecules that react with basic salt, and No. of base molecules that react with acidic salt.

19. Compound e
HCl
H2SO4
H3PO4
NaOH
Ca(OH)2 1 2 3
CaCl2
Na2HPO4
Na2SO4
KMnO4
H2O2
Na2CO3
1, 2
1,3,5

20. Calculate the equivalent weight of each of the following
HCl, H2SO4, CaOH, Li2CO3
HCl = molecular wt/ 1 = (1+35.5)/1 = 36.5

21. You must learn how to prepare
1- percent solution
2- Molar solution
3- Normal solution
4- Transform between them

22. 1- w/w % to Molarity or Normality
M = 10pd/M. wt. (Eq. wt.)
2- w/v % to Molarity or Normality
M = 10 p/M. wt. (Eq. wt.)
3- Molarity to Normality
M = N/e

23. To dilute any concentrated solution use the relation
Dilution
To dilute any concentrated solution use the relation
C X V (before dilution) = C X V (after dilution)
What is the needed volume of concentrated HCl (11 M) to prepare 500 ml 1 M solution

24. Solution Stoichiometry
50.0 ml of 6.0 M H2SO4 were spilled and solid NaHCO3 is to be used to neutralize the acid. How many grams of NaHCO3 must be used?
H2SO4(aq) NaHCO3  2H2O(l) + Na2SO4(aq) + 2CO2(g)
50.0 ml
? g
Our Goal
6.0 M =
Look!
A conversion factor!

25. Solution Stoichiometry
H2SO4(aq) NaHCO3  2H2O(l) + Na2SO4(aq) + 2CO2(g)
50.0 ml
? g
Our Goal
6.0 M =
NaHCO3
2 mol
NaHCO3
84.0 g
= g NaHCO3
50.4
H2SO4
50.0 ml
1 mol
H2SO4
mol
NaHCO3

26. Solution Stoichiometry:
Determine how many mL of M NaOH solution are needed to neutralize 35.0 mL of M H2SO4 solution.
____NaOH + ____H2SO4  ____H2O ____Na2SO4
First write a balanced
Equation.

27. Solution Stoichiometry:
Determine how many mL of M NaOH solution is needed to neutralize 35.0 mL of M H2SO4 solution.
N X V (acid) = N X V (base)
N of NaOH = M X e = X 1
N of H2SO4 = M X e = X 2
0.102 X V = X 35
V = ml

28. Solution Stoichiometry
What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
1st write out
a balanced chemical
equation

29. Solution Stoichiometry
2HCl(aq) Ba(OH)2(aq)  2H2O(l) + BaCl2
47.1 mL
0.75 M
0.40 M
? mL
N X V (acid) = N X V (base)

30. We must first write a balanced equation.
Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of M HNO3. Determine the molarity of the Ca(OH)2 solution.
We must first write a balanced equation.

31. Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of M HNO3. Determine the molarity of the Ca(OH)2 solution.
Ca(OH)2(aq) HNO3(aq)  2 2
H2O(l)
+ Ca(NO3)2(aq)
48.0 mL
19.2 mL
? M
0.385 M

32. Equivalent weight NaOH + HCl NaCl + H2O Na2CO3 + HCl 2 2
NaCl + H2O + CO2

33. Theoretical yield vs. Actual yield
Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield.
Theoretical yield = 19.5 g based on limiting reactant
Actual yield = 12.3 g experimentally recovered

34. Try this problem (then check your answer):
Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO3)3 in 455 ml of solution.

35. Element’s percent
If the bacterial medium additives listed below are priced according to their nitrogen content, which will be the least expensive per 50 kg. bag?
NH4Cl, (NH4)2SO4, NH2CONH2
Percent of element in its compounds: Simply to calculate the percent of any element in a compound
Atomic weight of the element X number of element X 100
Molecular weight of the compound

36. Then to calculate the Nitrogen percent in NH4Cl
1- M. wt of salt = = 53.5 g 2-
= 14 X 1X = 26.1%
53.5

37. Gas’s Law
Boyle's Law: for a fixed amount of gas at a constant temperature, the volume of the gas varies inversely with its pressure
PV = constant
P1 V1 = P2 V2

38. Charles's Law: the volume of a fixed amount of a gas at constant pressure is directly proportional to its temperature (Kelvin)
V/T = constant
V1/ T1= V2/ T2

39. One mole of an ideal gas at STP occupies 22.4 liters.
Avogadro's Law: At fixed temperature and pressure, the volume of a gas is directly proportional to the amount of the gas (either No. of moles or No. of molecules)
V/n = constant
V1/ n1= V2/ n2
One mole of an ideal gas at STP occupies 22.4 liters.

40. The combined gas law( Ideal gas)
PV/nT = constant = R
P1V1/N1T1 = P2V2/N2T2
R = L atm /mol K
= L Torr/ mol K
= j/mol K

41. Na2CO3 + H2SO4 Na2SO4 + CO2 + H2O
How many grams of Na2CO3 are needed to neutralize 0.5 mole H2SO4
How many moles of H2O are produced when we start with 200 g Na2CO3
How many liters of CO2 are produced when we start with 250 ml of 2N H2SO4 (at STP).
How many moles of Na2CO3 are consumed when we need to produce 200 L of CO2 at ( 20oC and 1 atm).
What is the volume of 6 M H2SO4 was needed to neutralize 106 g Na2CO3