1. Introduction: Matter and Measurement
First Year
Introduction: Matter and Measurement
Chapter 1
By Dr. Hisham Ezzat Abdellatef Head of Pharmaceutical Analytical Chemistry Department 1 1 1 1

2. Units of Measurement SI Units There are two types of units:
fundamental (or base) units;
derived units.
There are 7 base units in the SI system.
Derived units are obtained from the 7 base SI units.

3. Units of Measurement SI base unit length meter M mass kilogram Kg time
Base quantity
Name
Symbol
length
meter M
mass
kilogram Kg
time
second S
electric current
ampere A
thermodynamic temperature
kelvin K
amount of substance
mole
Mol
luminous intensity
candela Cd

4. Units of Measurement SI Units There are two types of units:
fundamental (or base) units;
derived units.
There are 7 base units in the SI system.
Derived units are obtained from the 7 base SI units.
Example:

5. SI derived unit Derived quantity Name Symbol area square meter m2
Derived quantity
Name
Symbol
area
square meter m2
volume
cubic meter m3
speed, velocity
meter per second
m/s
acceleration
meter per second squared
m/s2
wave number
reciprocal meter
m-1
mass density
kilogram per cubic meter
kg/m3
specific volume
cubic meter per kilogram
m3/kg
current density
ampere per square meter
A/m2
magnetic field strength
ampere per meter
A/m
amount-of-substance concentration
mole per cubic meter
mol/m3
luminance
candela per square meter
cd/m2

7. Units of Measurement Mass
Mass is the measure of the amount of material in an object.
This is not the same as weight which is dependant on gravity.

8. Units of Measurement
Temperature

9. Units of Measurement Temperature Kelvin Scale Celsius Scale
Used in science.
Same temperature increment as Celsius scale.
Lowest temperature possible (absolute zero) is zero Kelvin.
Absolute zero: 0 K = oC.
Celsius Scale
Also used in science.
Water freezes at 0oC and boils at 100oC.
To convert: K = oC
Fahrenheit Scale
Not generally used in science.
Water freezes at 32oF and boils at 212oF.

10. Units of Measurement Temperature
Converting between Celsius and Fahrenheit

11. Units of Measurement Volume
The units for volume are given by (units of length)3.
i.e., SI unit for volume is 1 m3.
A more common volume unit is the liter (L)
1 L = 1 dm3 = 1000 cm3 = 1000 mL.
We usually use 1 mL = 1 cm3.

13. Units of Measurement Density
Density – mass per unit volume of an object.

14. Dimensional Analysis
In dimensional analysis always ask three questions:
What data are we given?
What quantity do we need?
What conversion factors are available to take us from what we are given to what we need?

15. Dimensional Analysis
Method of calculation using a conversion factor.

16. Dimensional Analysis
Example: we want to convert the distance 8 in. to feet.
(12in = 1 ft)

17. Dimensional Analysis
Example: we want to convert the distance 8 in. to feet.
(12in = 1 ft)

18. Dimensional Analysis Problem
Convert the quantity from 2.3 x 10-8 cm to nanometers (nm)

19. Centimeter (cm)  Meter (m) Meter (m)  Nanometer (nm)
Dimensional Analysis
Problem
Convert the quantity from 2.3 x 10-8 cm to nanometers (nm)
First we will need to determine the conversion factors
Centimeter (cm)  Meter (m)
Meter (m)  Nanometer (nm)

20. Centimeter (cm)  Meter (m) Meter (m)  Nanometer (nm)
Dimensional Analysis
Problem
Convert the quantity from 2.3 x 10-8 cm to nanometers (nm)
First we will need to determine the conversion factors
Centimeter (cm)  Meter (m)
Meter (m)  Nanometer (nm) Or
1 cm = 0.01 m
1 x 10-9 m = 1 nm

21. Dimensional Analysis Problem
Convert the quantity from 2.3 x 10-8 cm to nanometers (nm)
1 cm = 0.01 m
1 x 10-9 m = 1 nm
Now, we need to setup the equation where the cm cancels and nm is left.

22. Dimensional Analysis Problem
Convert the quantity from 2.3 x 10-8 cm to nanometers (nm)
1 cm = 0.01 m
1 x 10-9 m = 1 nm
Now, fill-in the value that corresponds with the unit and solve the equation.

23. Dimensional Analysis Problem
Convert the quantity from 2.3 x 10-8 cm to nanometers (nm)

24. Dimensional Analysis Problem
Convert the quantity from 31,820 mi2 to square meters (m2)

25. Mile (mi)  kilometer (km) kilometer (km)  meter (m)
Dimensional Analysis
Problem
Convert the quantity from 31,820 mi2 to square meters (m2)
First we will need to determine the conversion factors
Mile (mi)  kilometer (km)
kilometer (km)  meter (m)

26. Mile (mi)  kilometer (km) kilometer (km)  meter (km)
Dimensional Analysis
Problem
Convert the quantity from 31,820 mi2 to square meters (m2)
First we will need to determine the conversion factors
Mile (mi)  kilometer (km)
kilometer (km)  meter (km) Or
1 mile = km
1000m = 1 km

27. Dimensional Analysis Problem
Convert the quantity from 31,820 mi2 to square meters (m2)
Now, we need to setup the equation where the mi cancels and m is left.
1 mile = km 1000m = 1 km

28. Dimensional Analysis Problem
Convert the quantity from 31,820 mi2 to square meters (m2)
Now, we need to setup the equation where the mi cancels and m is left.
1 mile = km 1000m = 1 km
Notice, that the units do not cancel, each conversion factor must be “squared”.

29. Dimensional Analysis Problem
Convert the quantity from 31,820 mi2 to square meters (m2)

30. Dimensional Analysis Problem
Convert the quantity from 31,820 mi2 to square meters (m2)

31. Dimensional Analysis Problem
Convert the quantity from 31,820 mi2 to square meters (m2)

32. Dimensional Analysis Problem
Convert the quantity from 14 m/s to miles per hour (mi/hr).

33. Dimensional Analysis Problem
Convert the quantity from 14 m/s to miles per hour (mi/hr).
Determine the conversion factors
Meter (m)  Kilometer (km) Kilometer(km)  Mile(mi)
Seconds (s)  Minutes (min) Minutes(min)  Hours (hr)

34. Dimensional Analysis Problem
Convert the quantity from 14 m/s to miles per hour (mi/hr).
Determine the conversion factors
Meter (m)  Kilometer (km) Kilometer(km)  Mile(mi)
Seconds (s)  Minutes (min) Minutes(min)  Hours (hr) Or
1 mile = km 1000m = 1 km
60 sec = 1 min 60 min = 1 hr

35. Dimensional Analysis Problem
Convert the quantity from 14 m/s to miles per hour (mi/hr).
1 mile = km 1000m = 1 km
60 sec = 1 min 60 min = 1 hr

36. Dimensional Analysis Problem
Convert the quantity from 14 m/s to miles per hour (mi/hr).
1 mile = km 1000m = 1 km
60 sec = 1 min 60 min = 1 hr

37. Dimensional Analysis Problem
Convert the quantity from 14 m/s to miles per hour (mi/hr).
1 mile = km 1000m = 1 km
60 sec = 1 min 60 min = 1 hr

38. Calculation with Chemical Formulas and Equations:

39. The Mole: mol
The quantity of a given substance that contains as many molecules or formula units the number of atoms in exactly 12g of carbon -12.
e.g. number of ethanol molecules in
1 mole ethanol = number of carbon atoms in 12g carbon -12
(which is called Avogadro's number NA).
NA = 6.02 X1023

40. A mole of O atoms contains 6.02 X 1023 O atoms.
A mole of O2 molecules contains 6.02 X 1023 molecules, that is 2 X 6.02 X 1023 O atoms,
1 mole Na2CO3= 6.02 X1023 Na2CO3 molecules
= 2 X6.02 X 1023 Na+ ions and 6.02 X 1023 CO32- ions.

41. Molar mass gram formula weight or the gram molecular weight
Ethanol, C2H6O has a molecular weight of 46.1 amu and a molar mass of 46.1 g / mol.

42. n= number of moles.
N= number of molecules, ions or atoms.
NA= Avogadro’s number
M.wt = molecular weight.
m = mass in grams.

43. Example: How many grams of Cu are there in 2.55 mole of Cu? Solution:
1 mole Cu = 63.5 g Cu

44. Example: What number of moles of aluminum is present in 125 g of Al ?
Solution: 1 mole Al = 27.0g Al

45. Example:
How many carbon atoms are there in a 1.0 carat diamond? Diamond is pure carbon and one carat is exactly 0.2g.
Solution: 1 mole C = g C
1 mole C = X 1023 atoms C

46. Percentage Composition of Compounds:

47. Example: What is the percentage of Fe in Fe2 O3? Solution:
One mole of Fe2 O3 contains:
2 mole Fe = 2 X 55.8g Fe = 111.6g Fe
3 mole O = 3 X16.0 g O = 48 g O the sum of the masses of one mole = g Fe2O3 the percentage of Fe in Fe2O3 is

48. Determining Molecular Formulas:
Molecular weight = n x empirical formula weight
indicates the atomic properties of the compound.

49. Example:
A 1.26 g sample of pure caffeine contains 0.624g C, 0.065g H, 0.364g N and 0.208g 0.
What is the empirical formula of Caffeine, and if the molecular weight of caffeine is 194. What is the molecular formula?
Solution: We calculate the number of moles of each element present in the sample.

51. division of each of these values by the smallest value
(0.013) gives the ratio.
4 mole C : 5 mole H : 2 mole N : 1 mol O
The empirical formula of caffeine is C4H5N2O.
The formula weight indicated by C4H5N2O is 97.
The molecular formula of caffeine is C8H10N402

52. Stoichiometry: Mole - Mass Relations in Chemical Reactions:
The quantitative relationships between reactants and products in a balanced equation are known as Stoichiometry.
C2H5OH + 3O2  2CO2 + 3H2O
1 mole C2H5OH + 3 mole O2  2mole CO2 + 3 mole H2O
1 mole C2H5OH ~ 3 mole O2.
1 mole C2H5OH ~ 2 mole CO2.
1 mole C2H5OH ~ 3 mole H2O.
3 mole O2 ~ 2 mole CO2.
3 mole O2 ~ 3 mole H2O.
2 mole CO2 ~ 3 mole H2O.

53. Example: a) How many moles of O2 are needed to burn 1.8 mole C2H5OH?
b) How many moles of water will form when 3.66 mole CO2 are produced during the combustion of C2H5OH?
Solution:
1 mole C2H5OH ~ 3 mole O2
O2 needed for combustion = 1.8 x 3/1 = 5.4 mole O2
2 mole CO2 ~ 3 mole H2O
H2O formed = 3.66 x 3/2 = 5.49 mole H2O.

54. Limiting Reactant; Theoretical and Percentage Yields;
The limiting reactant is the reactant that is entirely consumed when a reaction goes to completion. The mole of product are always determined by the mole of limiting reactant.

55. Example:
How many moles of H2 can prepared from 4 moles of Fe and 5 moles of H2O?
3 Fe + 4H2O  Fe3 O4 + 4H2
Solution: -determine which reactant limits the reaction.
3 mole Fe = 4 mole H2O.
The amount of Fe given in the problem is 4 moles which is
The 5 moles H2O given in the problem is.
 The H2O limits the extent of the reaction since a smaller proportionate.

56. Percent Yield:
Example: How many grams of N2F4 can theoretically be prepared from 4.0g of NH3 and 14.0g of F2?
The chemical equation for the reaction is
2NH3 + 5F2  N2F4 + 6HF
and if 4.8g of N2F4 is obtained from the experiment, what is the percent yields?

57. Solution:
2 mole NH3 = 5 mole F2

58. The F2, therefore is the limiting reactant, since a smaller proportionate amount has been supplied
? g N2F4 = mol F2
from the equation 1 mole N2F4 = 5mole F2
104g N2F4 = 5 mole F2

59. The end of chapter 1