1. How you measure how much?
You can measure mass,
or volume,
or you can count pieces.
We measure mass in grams.
We measure volume in liters.
We count pieces in MOLES.

2. For modern systems; 12C is taken as a standard
For modern systems; 12C is taken as a standard. So; 12C is assumed to be 12 a.m.u. 1 12C = 12 a.m.u.
1 a.m.u.=1/12 of the mass of a single 12C atom
=1/12 (12/6.02 X 1023)
=1/6.02 X 1023 grams
=mass of 1 1H atom
Then by comparing the masses of each atom with the mass of 12C atomic masses of other atoms can be easily calculated “RELATIVE ATOMIC MASSES”. Relative atomic masses easily calculated by MASS SPECTROMETER.
1 16O = 16 a.m.u
1 40Ca = 40a.m.u.
1 1H = 1 a.m.u.

3. Moles
Defined as the number of carbon atoms in exactly 12 grams of carbon-12.
1 mole contains 6.02 x particles.
Treat it like a very large dozen
6.02 x is called Avogadro’s number.

4. Representative particles
The smallest pieces of a substance.
For a molecular compound it is a molecule.
For an ionic compound it is a formula unit.
For an element it is an atom.

5. Units of measurement used with equations
Units of measurement used with equations. We use a mole concept to bring together the concepts of counting numbers and atomic weights of elements. The mole is derived from the following information. Atomic weights are an average of the relative masses of all of the isotopes of the given element. The number of C-12 atoms in exactly g of C-12 is 6.02 X This called Avogadro’s number.

6. An amount of a substance that contains Avogadro’s number of atoms, ions, molecules, or any other chemical unit is called A MOLE.
A mole of C-12 atoms is defined as having a mass of exactly g, a mass that is equal to its atomic weight.

7. Also known as Avogadro’s number.
Mole Calculations
1 mole (mol):
Amount of matter that contains as many objects (atoms, molecules) as the number of atoms in isotopically pure 12C.
6.022 x 1023
Also known as Avogadro’s number.
1 mol 12C atoms = 6.02 x C atoms
1 mol H2O molecules = 6.02 x molecules
1 mole NO3- ions = 6.02 x 1023 NO3- ions.

8. The mole concept for (A) elements, (B) compounds, and (C) molecular substances.
A mole contains
6.02 X 1023 particles. Since every mole contains the same number of particles, the ratio of the mass of any two moles is the same as the ratio of the masses of individual particles making up the two moles.

14. Molar Mass The atomic mass of any substance expressed in
grams corresponds to 1 mol of the substance.
Atomic mass of a substance expressed in grams is the molar mass.
The molar mass of a diatomic substance is equal to twice its atomic mass.

15. The generic term for the mass of one mole.
Molar Mass
The generic term for the mass of one mole.
The same as gram molecular mass, gram formula mass, and gram atomic mass.

16. Gram Atomic Mass The mass of 1 mole of an element in grams.
12.01 grams of carbon has the same number of pieces as grams of hydrogen and grams of iron.
We can right this as g C = 1 mole
We can count things by weighing them.

17. 0.120 mol Na x 6.02 x 1023 atoms Na = 7.22 x 1022 1 mol Na
Mole Calculations
Calculate the number of sodium atoms in mol Na?
0.120 mol Na x 6.02 x 1023 atoms Na = 7.22 x mol Na

18. Mole Calculations
Calculate the number of moles of potassium in 1.25 x 1021 atoms of K.
1.25 x 1021 atoms K x 1 mol K = 2.08 x 10-3 mol K
6.02 x 1023 atoms K

19. Mole Calculations
What is the mass in grams of 2.01 x 1022 atoms of sulfur?
2.01 x 1022 atoms S x 1 mol S x g S
6.02 x 1023 atoms S mol S
= 1.07g S

20. Mole Calculations
How many O2 molecules are present in 0.470g of oxygen gas?
0.470 g O2 x 1mol O2 x x 1023 molecules O2
32.00 g O mol O2
= 8.84 x 1021 molecules O2

21. What about compounds?
In 1 mole of H2O molecules there are two moles of H atoms and 1 mole of O atoms
To find the mass of one mole of a compound determine the moles of the elements they have
Find out how much they would weigh add them up.

22. Molar Mass

23. The Mole
Calculate the number of Magnesium and Chlorine ions present in mol of MgCl2.
# Mg2+ ions:
0.450 mol MgCl2 x 6.02 x 1023 formula units x 1 Mg mole 1 formula unit
# Mg2+ ions = 2.71 x 1023
# Cl- ions:
0.450 mol MgCl2 x 6.02 x 1023 formula units x 2 Cl mole 1 formula unit
# Cl- ions = 5.42 x 1023

24. The mole concept applied to compounds
The formula weight of a species is the sum of
atomic masses (amu) of the atoms in a species.
Molecular weight of NH3 =
For an ionic cmpd formula weight MgF2 =

25. Mass of one mole of MgF2 is Mass of one formula unit of MgF2 is
Mass of x 1023 formula units of MgF2 is

26. Gram Formula Mass The mass of one mole of an ionic compound.
Calculated the same way.
What is the GFM of Fe2O3?
2 moles of Fe x g = g
3 moles of O x g = g
The GFM = g g = g

27. Molar Mass

28. What is the mass of one mole of CH4?
What about compounds?
What is the mass of one mole of CH4?
1 mole of C = g
4 mole of H x 1.01 g = 4.04g
1 mole CH4 = = 16.05g
The Gram Molecular mass of CH4 is 16.05g
The mass of one mole of a molecular compound.

29. The gram atomic weight of an element is the mass in grams of one mole of an element that is numerically equal to its atomic weight.
The gram formula weight of a compound is the mass in grams of one mole of the compound that is numerically equal to its formula weight.
The gram formula weight of a compound is the sum total of all the individual atomic weight in the formula. The gram molecular weight is the gram formula weight of a molecular compound.

30. Molar Mass
Calculate the molar mass of Ag and of magnesium nitrate, Mg(NO3)2.
Ag is an element so its molar mass equals its atomic mass = g/mol.
Mg(NO3)2 = ( x 16.00)
= g/mol

31. How many moles is 5.69 g of NaOH?
For example
How many moles is 5.69 g of NaOH?
need to change grams to moles
for NaOH
1mole Na = 22.99g 1 mol O = g 1 mole of H = 1.01 g
1 mole NaOH = g

32. 44.01 g CO2 x 1mol CO2 = 7.31 x 10-23 g/molecule
Mole Calculations
Calculate the mass in grams of a single molecule of carbon dioxide, CO2.
44.01 g CO2 x 1mol CO2 = 7.31 x g/molecule
1 mol CO x 1023 molecules

33. Molar Mass The number of grams of 1 mole of atoms,ions,or molecules.
We can make conversion factors from these.
To change grams of a compound to moles of a compound.

34. Examples How many moles is 4.56 g of CO2 ?
How many grams is 9.87 moles of H2O?
How many molecules in 6.8 g of CH4?
49 molecules of C6H12O6 weighs how much?

35. Calc the molar mass of Ca(NO3)2.
Calc the molar mass of a compound if mol of it has a mass of 152g.

36. How many grams of each are required to have 0.100 mol of
A. NaOH
B. H2SO4
C. C2H5OH
D. Ca3(PO4)2

37. How many moles are in 50.0 g of
A. CS2
B. Al2(CO3)3
C. Sr(OH)2
D. LiNO3

38. Calc the no. of C, H, and O atoms in 1.50 g of glucose (C6H12O6).
What is the average mass of one C3H8 molecule?
What is the mass of 5.00 x 1024 molecules of NH3?

39. Types of questions
How many molecules of CO2 are the in 4.56 moles of CO2 ?
How many moles of water is 5.87 x molecules?
How many atoms of carbon are there in 1.23 moles of C6H12O6 ?
How many moles is 7.78 x 1024 formula units of MgCl2?

40. Gases Many of the chemicals we deal with are gases.
They are difficult to weigh.
Need to know how many moles of gas we have.
Two things effect the volume of a gas
Temperature and pressure
Compare at the same temp. and pressure.

41. Standard Temperature and Pressure
0ºC and 1 atm pressure
abbreviated STP
At STP 1 mole of gas occupies 22.4 L
Called the molar volume
Avogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

42. Molar Volume
Avogadro's Theory: Two gases containing equal numbers of molecules occupy equal volumes under similar conditions.
Standard Temperature and Pressure: 0OC and 1 atm.
Molar Volume at STP = 22.4 L

43. Examples What is the volume of 4.59 mole of CO2 gas at STP?
How many moles is L of O2 at STP?
What is the volume of 8.8g of CH4 gas at STP?

44. Molar Volume Density of gas = molar mass in grams
molar volume in liters
Calculate the density of ammonia gas, NH3, at STP.
Density of NH3 = molar mass NH3 = g
molar volume NH L
= g/L

45. Density of a gas D = m /V for a gas the units will be g / L
We can determine the density of any gas at STP if we know its formula.
To find the density we need the mass and the volume.
If you assume you have 1 mole then the mass is the molar mass (PT)
At STP the volume is 22.4 L.

46. Find the density of CO2 at STP. Find the density of CH4 at STP.
Examples
Find the density of CO2 at STP.
Find the density of CH4 at STP.

47. Mass Volume Moles 6.02 x 1023 Representative Particles Ions AtomsPT
Moles
6.02 x 1023
Representative
Particles
Ions
Atoms

48. What is the mass of 3.36 L of ozone gas, O3, at STP?
Mole Calculations
What is the mass of 3.36 L of ozone gas, O3, at STP?
3.36 L O3 x 1 mol O3 x g O3 = 7.20 g O3
22.4 L O mol O3

49. Mole Calculations
How many molecules of hydrogen gas, H2, occupy L at STP?
0.500 L H2 x 1 mol x x 1023 molecules H2
22.4 L mol
= 1.34 x 1022 molecules H2

50. Mole Calculations

51. Percent Composition
Percentage Composition: Percentage by mass contributed by each element in a substance.
% Element (E) = (# atoms of Element)(AW of E) x 100%
Formula Weight of Compound
According to the law of definite composition the elements in a compound are always present in the same proportion by mass.

52. Percentage composition by mass of a compound
It shows us how many grams of each element exist in 100 gram of a compound.

53. Percent Composition Like all percents Part x 100 % whole
Find the mass of each component,
divide by the total mass.

54. Calculate the percentage of Nitrogen, by mass in Ca(NO3)2:
Percent Composition
Calculate the percentage of Nitrogen, by mass in Ca(NO3)2:
Formula Weight = 1 x Ca + 2 x N + 6 x O
= 1 x x x 16.0
= amu
% N = (2)(14.0) = 17.1%

55. Percent Composition
Calculate the percent composition of trinitrotoluene (TNT), C7H5(NO2)3.
Assume that there is 1 mol of TNT.
1 mol of TNT contains:
7 mol carbon atoms = g
5 mol hydrogen atoms = 5.05 g
3 mol of nitrogen atoms = g
6 mol oxygen atoms = g

56. Calculate the molar mass of TNT:
Percent Composition
Calculate the molar mass of TNT:
Molar mass = g/mol.
Compare the mass of each element to the molar mass of the compound:
84.07 g C x 100 = 37.01% C
g C7H5(NO2)3

57. Percent Composition 5.05g H x 100 = 2.22% H 227.15 g C7H5(NO2)3
42.03 g N x 100 = 18.50% N
96.00 g 0 x 100 = 42.26% O

58. Empirical Formula Empirical formula gives:
the ratio of atoms in a molecule: H2O.
the ratio of moles of atoms on the molar level.
The molar ratio of the elements in a compound allows the calculation of the subscripts in the empirical formula of a compound.

59. Empirical Formula
Percent composition - the mass percent of each element present in a compound.
1. Calculate the chemical formula from percent composition.
2.Find the relative numbers of moles of each element in the compound.
3.Use molar masses of the elements as conversion factors.

60. Empirical Formula
Find the ratio of the numbers of moles by dividing the larger number of moles by the smaller number.
Multiply the subscripts by small integers in a trial-and-error procedure until whole numbers are found. 

61. Empirical Formula The flow diagram summarizes the
relationship among mass percent, moles, mole ratios and subscripts in the formula of a compound used in the determination of empirical formula from percent composition.

62. Calculating Empirical
Just find the lowest whole number ratio
C6H12O6
CH2O
It is not just the ratio of atoms, it is also the ratio of moles of atoms.
In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.
In one molecule of CO2 there is 1 atom of C and 2 atoms of O.

63. Calculating Empirical
Means we can get ratio from percent composition.
Assume you have a 100 g.
The percentages become grams.
Can turn grams to moles.
Find lowest whole number ratio by dividing by the smallest.

64. Example
Calculate the empirical formula of a compound composed of % C, % H, and %N.
Assume 100 g so
38.67 g C x 1mol C = mole C
12.01 gC
16.22 g H x 1mol H = mole H
1.01 gH
45.11 g N x 1mol N = mole N
14.01 gN

65. Example The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N
The ratio is mol H = 5 mol H mol N mol N
C1H5N1
A compound is % P and % O.
What is the empirical formula?
Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

66. Empirical Formula
Q. The percent composition of a solid is known to be 68.4% Ba, 10.3% P, and 21.3% O. What is the empirical formula of the compound?
A. Assuming a 100 g sample gives us 68.4 g Ba, 10.3 g P, and 21.3 g O.
Convert these masses to numbers of moles: moles Ba, mol P, 1.33 mol O

67. Empirical Formula
Knowing the relative numbers of moles, find the ratio by dividing the two larger numbers by the smaller number. Answer = 1.5
The O:Ba:P ratio of 4:1.5:1 gives an empirical formula of Ba1.5PO4. However, we must multiply the subscripts by a small integer (in this case 2) to find whole numbers for the formula. Ba3P2O8, or Ba3(PO4)2.

68. Empirical Formula
Q. What is the percent composition of sodium hydrogen carbonate?
A. The formula for sodium hydrogen carbonate is NaHCO3. The Na:H:C:O mole ratio is 1:1:1:3. Convert this mole ratio into a mass ratio by assuming there is a 1 mole sample present. Answer = 23g Na, 1.0g H, 12g C, 48 g O

69. Empirical Formula
To determine the percent composition, divide the mass of each element present by the total mass of the compound and multiply by 100.
Total mass of 1 mole of NaHCO3 = 84 g
Answer = 27% Na, 1.2% H, 14% C, 57% O

70. Empirical Formula
0.500 g of scandium was heated and allowed to react with oxygen. The resulting product oxide had a mass of 0.767g. What is the empirical formula for scandium oxide?
Calculate the number of moles of scandium:
0.500 g Sc x 1 mol Sc = mol Sc
44.96 g Sc

71. Empirical Formula Calculate the moles of oxygen:
Mass of oxygen = 0.767g– 0.500g O = g O
Number of moles of oxygen:
0.267g x 1 mol 0 = mol O
16.00 g
Mole ratio of Sc:O is :0.0167
Divide by smallest number: 1:1.50

72. Using the law of multiple proportions we need a whole number ratio:
Empirical Formula
Sc1.00O1.50
Using the law of multiple proportions we need a whole number ratio:
Sc2.00O3.00

73. Empirical Formula
Calculate the empirical formula of glycine given that it contains 32.0% carbon, 6.7% hydrogen, 18.7% nitrogen and 42.6% oxygen.
Assume a 100g sample. Then the percent composition equals number of grams.

74. Convert grams into moles:
Empirical Formula
32.0 g C, 6.7g H, 18.7g N, 42.6g O.
Convert grams into moles:
32.0 g C x 1 mol = 2.66 mol C
12.01 g C
6.7 g H x 1 mol H = 6.6 mol H
1.01 g H

75. Divide the mole ratios by the smallest number: 1.33
Empirical Formula
18.7 g N x 1 mol N = 1.33 mol
14.01 g N
42.6 g O x 1 mol O = 2.66 mol O
16.00 g O
Divide the mole ratios by the smallest number: 1.33
C2H5N1O2

76. The Empirical Formula
The lowest whole number ratio of elements in a compound.
The molecular formula the actual ration of elements in a compound.
The two can be the same.
CH2 empirical formula
C2H4 molecular formula
C3H6 molecular formula
H2O both

77. Molecular Formula
The formula obtained from percentage compositions is always the empirical formula.
Molecular formula - gives the actual numbers of atoms in a molecule.
1. May be the same as the empirical formula.
2. May be a multiple of the empirical formula.
a. Multiple = Molecular mass
Empirical Formula

78. Empirical to molecular
Since the empirical formula is the lowest ratio the actual molecule would weigh more.
By a whole number multiple.
Divide the actual molar mass by the the mass of one mole of the empirical formula.

79. Molecular Formula
The empirical formula for fructose is CH2O. If the molar mass of fructose is 180 g/mol, what is the molecular formula?
Molar mass of the empirical formula = 30 g/mol.

80. Molecular Formula Multiple = molar mass of fructose
molar mass of empirical formula
Multiple = 180 g/mol = 6
30 g/mol
Multiply the empirical formula by 6:
C6H12O6

81. For ethyl butyrate: Multiple = 116 =2 58
Molecular Formula
Q. Determine the molecular formula of ethyl butyrate. The molecular mass of the compound is 116 g/mol and the empirical formula is C3H6O.
A. The molecular formula of the compound may be the same as the empirical formula or it may be a multiple of the empirical formula.
For ethyl butyrate: Multiple = 116 =
The subscripts in the empirical formula are multiplied by 2 giving a molecular formula of C6H12O2.

82. Some problems
How many atoms are there in 5.10 moles of sulfur? What’s the mass of 5.10 moles of S?
How many moles of calcium atoms are in 1.16 x 1024 atoms of Ca? How many grams?

83. How many moles are in 0.040 kg Na?
Which of the following has more atoms: 1.10g of hydrogen atoms or 14.7 g of chromium atoms?
How many moles are in kg Na?

84. Example
A compound is known to be composed of % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be g. What is its molecular formula?

85. STRUCTURAL FORMULA
The atoms in a molecule are connected or chemically bonded in a precise way. A SF. Shows how the atoms in a molecule are arranged.
For ex: H2O H-O-H
C2H6 CH H-C- C- H
H H

86. The actual numbers of atoms in a molecule.
Empirical formula
The simplest whole number ratio of atoms of elements in a compound, described with the use of subscripts.
Ionic compounds are always shown as empirical formulas.
Molecular Formula
The actual numbers of atoms in a molecule.
Structural Formula
Show the relative arrangements of atoms in a molecule

88. If you know the name of an ingredient, you can write a chemical formula, and the percent composition of a particular substance can be calculated from the formula. This can be useful information for consumer decisions.

89. HYDRATES
Solids which are found in combined form with water in definite proportion are called as HYDRATES. When hydrates are heated, H2O evaporates, and only solid is obtained in amorphous .
(w/o a certain geometric structure, generally in powdered form. H2O molecules surround ionic substances with certain amounts.

90. WATER OF HYDRATION : Water molecules of a hydrate.
Na2CO3.10H2O Na2CO3(s) + 10H2O(g)
DEHYDRATION: Evaporation of water of hydration.
Na2CO3.10H2O,
CaSO4.2H2O,
CuSO4.5H2O