1. 6.02 X 1023 Unit 6: The Mole Mr. Blake
To play the movies and simulations included, view the presentation in Slide Show Mode.
6.02 X 1023

2. Review: Atomic Masses
Elements occur in nature as mixtures of isotopes.
Carbon = 98.89% 12C (6 p+, 6 n) % 13C (6p+, 7n)
<0.01% 14C (6p+, 8n)
Carbon’s atomic mass = amu

3. Atomic Weights are relative
C: # and mass
H: # and mass
Mass Ratio
1 C = 12.0 amu
1 H = 1.0 amu
12 to 1
2 C = 24.0 amu
2 H = 2.0 amu
12 to 1
3 C = 36.0 amu
3 H = 3.0 amu
12 to 1
4 C = 48.0 amu
4 H = 4.0 amu
12 to 1
5 C = 60.0 amu
5 H = 5.0 amu
12 to 1
10 C = amu
10 H = 10.0 amu
12 to 1

4. What is the mole?
Not this kind of mole!

5. A. What is the Mole? VERY A large amount!!!!
A counting number (like a dozen).
1 mol = 602,000,000,000,000,000,000,000 of anything.
A large amount!!!!
VERY

6. Similar Words Pair: 1 pair of shoelaces = 2 shoelaces.
Dozen: 1 dozen oranges = 12 oranges.
Gross: 1 gross of spider rings = 144 rings.
Ream: 1 ream of paper = 500 sheets of paper.
Mole: 1 mole of Na atoms = 6.02 X 1023 Na atoms.

7. The Mole So 1 mole of any element has 6.02 X 1023 particles.
This is a really big number – It’s so big because atoms are very small!
Defined as the number of atoms in 12.0 grams of C-12. This is the standard!
12.0 g of C-12 has X 1023 atoms.

8. Avogadro’s number (NA)
The Mole
This number is named in honor of Amedeo ______________ (1776 – 1856), who studied quantities of gases and discovered that no matter what the gas was, there were the same number of molecules present
6.02 x 1023 is also called _______________________.
Avogadro
Avogadro’s number (NA)

9. I didn’t discover it. Its just named after me!
Amadeo Avogadro

10. Just How Big is a Mole?
Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles.
If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles.
If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

11. Everybody Knows Avogadro’s Number! But Where Did it Come From?
It was NOT just picked! It was measured.
One of the better methods of measuring this number was the Millikan Oil Drop Experiment.
Since then we have found even better ways of measuring using x-ray technology.

12. The Mole 1 dozen cookies = 12 cookies
1 mole of cookies = 6.02 X 1023 cookies
1 dozen cars = 12 cars
1 mole of cars = 6.02 X 1023 cars
1 dozen Al atoms = 12 Al atoms
1 mole of Al atoms = 6.02 X 1023 atoms
- Note that the NUMBER is always the same, but the MASS is very different!
Mole is abbreviated mol

13. A Mole of Particles Contains 6.02 x 1023 particles
= x 1023 C atoms
= x 1023 H2O molecules
= x 1023 NaCl “molecules”
(technically, ionics are compounds not molecules so they are called formula units)
6.02 x 1023 Na+ ions and
6.02 x 1023 Cl– ions
1 mole C
1 mole H2O
1 mole NaCl

14. Learning Check
Suppose we invented a new collection unit called a widget. One widget contains 8 objects.
1. How many paper clips in 1 widget?
a) 1 b) 4 c) 8
2. How many oranges in 2.0 widgets?
a) 4 b) 8 c) 16
3. How many widgets contain 40 gummy bears?
a) 5 b) 10 c) 20

15. Avogadro’s Number as Conversion Factor
6.02 x 1023 particles
1 mole or
1 mole
- Note that a particle could be an atom OR a molecule!

16. Learning Check 1. Number of atoms in 0.500 mole of Al a) 500 Al atoms
b) x 1023 Al atoms
c) x 1023 Al atoms
2.Number of moles of S in 1.8 x 1024 S atoms
a) mole S atoms
b) mole S atoms
c) x 1048 mole S atoms

17. B. Molar Mass Mass of 1 mole of an element or compound.
Equal to the numerical value of the average atomic mass (get from periodic table)
Atomic mass tells the...
atomic mass units per atom (amu)
grams per mole (g/mol)
Round to 2 decimal places (hundredths)

18. Molar Masses of Elements (Found in Periodic Table)
1 mole of C atoms = g
1 mole of Mg atoms = g
1 mole of Cu atoms = g
Carried to the hundredths place!!

19. B. Molar Mass Examples carbon aluminum zinc 12.01 g/mol 26.98 g/mol

20. Calculating Molecular Mass
Calculate the molecular mass of magnesium carbonate, MgCO3.
24.31 g g + 3(16.00 g) =
84.32 g/mol

21. Molar Mass of Molecules and Compounds
A substance’s molecular mass (molecular weight) is the mass in grams of one mole of the compound.
1 mole Ca x g/mol = g/mol
+2 moles Cl x g/mol = g/mol
Therefore,the mass of the entire compound is…
1 mole of CaCl2 = g/mol

22. B. Molar Mass Examples water sodium chloride H2O
2(1.01) = g/mol
NaCl
= g/mol

23. B. Molar Mass Examples sodium bicarbonate sucrose NaHCO3
(16.00) = g/mol
C12H22O11
12(12.01) + 22(1.01) + 11(16.00) = g/mol

24. C. Molar Conversions molar mass (g/mol) MOLES 6.02  1023IN
GRAMS
MOLES
NUMBER OF
PARTICLES
6.02  1023
(particles/mol)

25. Calculations with Molar Mass
Grams Moles

26. Converting moles to grams
Ex. #1 How many grams of lithium are in 3.50 moles of lithium?
3.50 mol Li
1 mol Li
6.94 g Li
= g Li
24.3

27. Converting Moles and Grams
Ex. #2 Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al?
Goal: 3.00 moles Al ? g Al

28. 1. Molar mass of Al 1 mole Al = 26.98 g Al
2. Conversion factors for Al
26.98 g Al or mol Al
1 mol Al g Al
3. Setup moles Al x g Al
1 mole Al
Answer = g Al

29. Molar Conversion Examples
3. How many moles of carbon are in 26 g of carbon?
26 g C
1 mol C
12.01 g C
= 2.2 mol C

30. moles of C12H22O11? 4) How many molecules are in 2.50 6.02  1023
= 1.51  1024
molecules
C12H22O11

31. Everything must go through Moles!!!
Calculations
molar mass Avogadro’s number Grams Moles particles
Everything must go through Moles!!!

32. Ex. #1 Find the mass of 2.1  1024
molecules of NaHCO3.
2.1  1024
molecules
1 mol
6.02  1023
molecules
84.01 g
1 mol
= 290 g NaHCO3

33. Ex. #2 How many atoms of Cu are present in 35.4 g of Cu?
35.4 g Cu mol Cu X 1023 atoms Cu g Cu mol Cu
= 3.4 X 1023 atoms Cu

34. V. Percent Composition: The percent by ______ of each _________ in a compound.
mass
element
What is the % composition of Ba(CN)2 ?
1( g Ba) + 2(12.01 g C) + 2(14.01 g N) = g Ba(CN)2 =
100 x
Ba(CN) g
189.37 Ba
137.33 % 2
72.52 % Ba
12.68 % C
14.80 % N

35. What is the % composition of Iron III sulfate, ___________?
Fe2(SO4)3
2(55.85 g Fe) + 3(32.07 g S) + 12 (16.00 g O) = g Fe2(SO4)3
27.93 % Fe
24.06 % S
48.01 % O

36. Percent Composition (Hydrates):
1. Ba(CN)2 ∙ 2 H2O is what % water by mass?
1(Ba) + 2(C) + 2(N) + 2(H2O) = g Ba(CN)2 • 2 H2O
15.99 % H2O

37. Formulas molecular formula = (empirical formula)n [n = integer]
Empirical formula: the lowest whole number
ratio of atoms in a compound.
Molecular formula: the true number of atoms
of each element in the formula of a compound.
molecular formula = (empirical formula)n [n = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH

38. Formulas (continued)
Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).
Examples:
K2CO3
NaCl
MgCl2
Al2(SO4)3

39. Formulas (continued)
Formulas for molecular compounds might be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11

40. Empirical Formula Determination
Base calculation on 100 grams of compound.
Determine moles of each element in 100 grams of compound.
Divide each value of moles by the smallest of the values.
Multiply each number by an integer to obtain all whole numbers.

41. Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid? ( )
49.32g C
4.107
12.01
molC gC =
1 mol C ( )
6.851
6.78
1.01
gH 1 molH
molH gH = ( )
43.84
2.74
16.00
g O 1molO
molO gO =

42. Empirical Formula Determination (part 2)
Divide each value of moles by the smallest of the values.
4.107
1.50
2.74
molC
molO =
Carbon:
6.78
2.47
2.74
molH
molO =
Hydrogen:
2.74
1.00
molO =
Oxygen:

43. Empirical Formula Determination (part 3)
Multiply each number by an integer to obtain all whole numbers.
Carbon: 1.50
Hydrogen: 2.50
Oxygen: 1.00
x 2
x 2
x 2 3 5 2
Empirical formula:
C3H5O2

44. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
1. Find the empirical mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = g

45. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
2. Divide the molecular mass by the mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = g
146 2 73 =

46. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
3. Multiply the empirical formula by this number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = g
146 2 73 =
(C3H5O2) x 2 =
C6H10O4

47. Hydrates mol H O mol anhydrous salt ionic H2O Hydrated salt
A. An _______ compound that has _______ trapped in the crystal lattice structure.
1. ________________: A compound with water (i.e. BaI2 •2 H2O).
2. __________________: A compound without water (i.e. BaI2).
B. Determination of a hydrate:
1. Determine the mass of ________.
2. Determine the mass of the _________________.
3. Find the __________ of water and anhydrous salt.
4. Find the mol ratio.
X = * Whole number
5. Write the formula.
Hydrated salt
Anhydrous salt
H2O
anhydrous salt
moles
mol H O 2
mol
anhydrous
salt

48. A barium iodide hydrate is heated in a crucible. Determine
C. Example:
A barium iodide hydrate is heated in a crucible. Determine
the formula of the hydrate given the following data:
Before heating: g BaI2 • X H2O (Hydrated Salt)
After heating: g BaI2 (Anhydrous Salt)
____________ _
Step 1: Mass of water?
0.887 g H2O
Step 2: Mass of Anhydrous salt?
9.520 g BaI2
Step 3: Calculate the mol of the water & the anhydrous salt.
0.887 g H2O
1 mol H2O
= mol H2O
18.02 g H2O
9.520 g BaI2
1 mol BaI2
= mol BaI2
g BaI2
Step 4: Calculate the number of waters.
salt
anhydrous
mol O H 2 = x 2
BaI
mol
0.0243 O H
0.0492 =
= 2
Step 5:
BaI2• 2 H2O

49. 2. Data: Before heating: 5.20 g ZnSO3 • X H2O (Hydrated salt)
After heating: g ZnSO3 (Anhydrous salt)
____________ _
2.60 g H2O
Step 1: Mass of water?
Step 2: Mass of Anhydrous salt?
2.60 g ZnSO3
Step 3: Calculate the mol of the water & the anhydrous salt.
2.60 g H2O
1 mol H2O
= mol H2O
18.02 g H2O
2.60 g ZnSO3
1 mol ZnSO3
= mol ZnSO3
g ZnSO3
Step 4: Calculate the number of waters.
= 8
Step 5:
ZnSO3 • 8 H2O

50. IQ #2
Using your knowledge of mole calculations and unit conversions, determine how many atoms there are in 1 gallon of gasoline. Assume that the molecular formula for gasoline is C6H14 and that the density of gasoline is approximately 0.85 grams/mL.

51. Using a conversion factor of 3785 mL per gallon, we can determine that the mass of gasoline in one gallon is 3785 mL x g/mL = 3217 grams.
Because the molar mass of C6H14 is 86 g/mole, there are 3217 / 86 moles of gasoline molecules, or 37.4 moles of molecules present.
Multiplying 37.4 x 20 (the number of atoms per mole of gasoline), there are 748 moles of atoms.
Finally, multiplying 748 moles of atoms by 6.02 x 1023 atoms/mole, we can find that there are
4.50 x 1026 atoms present in the sample.

52. CHEMICAL REACTIONS
Reactants: Zn + I2
Product: Zn I2

53. Chemical Equations
Their Job: Depict the kind of reactants and products and their relative amounts in a reaction.
4 Al (s) O2 (g) ---> 2 Al2O3 (s)
The numbers in the front are called
stoichiometric ____________.
The letters (s), (g), and (l) are the physical states of compounds.
coefficients

54. Introduction
Chemical reactions occur when bonds between the outermost parts of atoms are formed or broken.
Chemical reactions involve changes in matter, the making of new materials with new properties, and energy changes.
Symbols represent elements, formulas describe compounds, chemical equations describe a chemical reaction.

55. Parts of a Reaction Equation
Chemical equations show the conversion of reactants (the molecules shown on the left of the arrow) into products (the molecules shown on the right of the arrow).
A “+” sign separates molecules on the same side
The arrow is read as “yields”
Example
C + O2  CO2
This reads “carbon plus oxygen react to yield carbon dioxide”

56. The charcoal used in a grill is basically carbon
The charcoal used in a grill is basically carbon. The carbon reacts with oxygen to yield carbon dioxide. The chemical equation for this reaction, C + O2  CO2, contains the same information as the English sentence but has quantitative meaning as well.

57. Chemical Equations Because of the principle of
Lavoisier, 1788
Because of the principle of
the conservation of matter,
an equation must be
balanced.
It must have the same
number of atoms of the
same kind on both sides.

58. Symbols Used in Equations
Solid ___
Liquid (l)
Gas ___
Aqueous solution (aq)
Catalyst H2SO4
Escaping gas ()
Change of temperature ()
(s)
(g)

59. Balancing Equations Changing the subscripts changes the compound.
When balancing a chemical reaction you may add coefficients in front of the compounds to balance the reaction, but you may not change the subscripts.
Changing the subscripts changes the compound.
Subscripts are determined by the valence electrons (charges for ionic or sharing for covalent)

60. Subscripts vs. Coefficients
The subscripts tell you how many atoms of a particular element are in a compound. The coefficient tells you about the quantity, or number, of molecules of the compound.

61. Chemical Equations 4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)
This equation means:
4 Al atoms + 3 O2 molecules ---produces--->
2 molecules of Al2O3
AND/OR
4 moles of Al + 3 moles of O produces--->
2 moles of Al2O3

62. Steps to Balancing Equations
There are four basic steps to balancing a chemical equation.
Write the correct formula for the reactants and the products. DO NOT TRY TO BALANCE IT YET! DO NOT CHANGE THE FORMULAS!
Find the number of atoms for each element on the left side. Compare those against the number of the atoms of the same element on the right side.
Determine where to place coefficients in front of formulas so that the left side has the same number of atoms as the right side for EACH element in order to balance the equation.
Check your answer to see if:
The numbers of atoms on both sides of the equation are now balanced.
The coefficients are in the lowest possible whole number ratios. (reduced)

63. Some Suggestions to Help You
Take one element at a time, working left to right except for H and O. Save H for next to last, and O until last.
If everything balances except for O, and there is no way to balance O with a whole number, double all the coefficients and try again. (Because O is diatomic as an element)
(Shortcut) Polyatomic ions that appear on both sides of the equation should be balanced as independent units

65. Balancing Equations 2 2 What Happened to the Other Oxygen Atom?????
___ H2(g) + ___ O2(g) ---> ___ H2O(l)
What Happened to the Other Oxygen Atom?????
This equation is not balanced!
Now, follow your rules for balancing equations and see if you can do it!

66. Rewrite these word equations as balanced chemical equations
IQ 2
Rewrite these word equations as balanced chemical equations
Hydrogen + sulfur  hydrogen sulfide
Iron III chloride + calcium hydroxide  iron III hydroxide + calcium chloride
Heating copper (II) sulfide in the presence of oxygen gas produces pure copper and sulfur dioxide gas.

67. Balancing Equations 2 3
___ Al(s) + ___ Br2(l) ---> ___ Al2Br6(s)

68. Balancing Equations
Since we can’t have a fraction for a coefficient, we must multiply it out by multiplying all of the coefficients by 2.
Take the number you need and put it over 2 (i.e. 11/2).
Need 11 Oxygen atoms but they only come in pairs! How can we fix this?
____C3H8(g) _____ O2(g) > _____CO2(g) _____ H2O(g) 5 3 4 11 2
____B4H10(g) _____ O2(g) > ___ B2O3(g) _____ H2O(g) 2
=11 1 ? 10 4 2 5

69. Learning Check! Na3PO4 + _____ Fe2O3 ----> ____ Na2O + _____ FePO4
Sodium phosphate + iron (III) oxide  sodium oxide + iron (III) phosphate
Na3PO4 + _____ Fe2O > ____ Na2O + _____ FePO4

70. Algebraic Method A Ca3(PO4)2 + B H2SO4 C CaSO4 + D H3PO4
Consider:
Ca3(PO4)2 + H2SO CaSO4 + H3PO4
1. Assign letters to unknown coefficients:
A Ca3(PO4)2 + B H2SO C CaSO4 + D H3PO4
Make a grid indicating the appearance of element (or ion) in each species of the equation. Use a whole number and the coefficient “letter” to indicate each appearance.
Ca 3a + 0 = c + 0
PO4 2a + 0 = 0 + d
H 0 + 2b = 0 + 3d
SO4 0 + b = c + 0

71. 4. Assume a = 1; solve for coefficients:
3. Reduce each equation:
Ca 3a = c
PO4 2a = d
H 2b = 3d
SO4 b = c
4. Assume a = 1; solve for coefficients:
A= 1
C= 3
D = 2
B = 3
5. Write equation with coefficients:
Ca3(PO4)2 + 3 H2SO CaSO4 +2 H3PO4

72. Example: A = 6 E = 6 F = 9 x 6 C = 2 D =4 B = 3
Na2CO3 + C + Sb2S Sb + Na2S + CO2
a = 1
A = 6
Na: 2a = 2e
e = 1
E = 6
C: a + b = f
f = 3/2
F = 9
O: 3a = 2f
x 6
c = 1/3
C = 2
Sb: 2c = d
d = 2/3
D =4
S: 3c = e
b = 1/2
B = 3
Na2CO C + Sb2S Sb + Na2S + CO2 6 3 2 4 6 9

73. Topics for Unit #5 The Mole & Quantifying Substances (ch. 10)
Avagadro’s Number (ch. 10)
Chemical Equations & Balancing (ch. 11)
Empircal vs. Molecular Formulas (ch. 11)

74. Introductory Questions #1
How do you measure matter if there’s a lot of it there? Name the three ways suggested in the text (pg. 287)
What is the mass of 90 average-sized apples if 1 dozen of apples has a mass of 2.0 kg?
How many representative parts do you find in one mole?
Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, tools, and garden furniture. How many moles of magnesium is 1.25 x1023 atoms of magnesium? (see practice problem 10.2 on pg 291)

75. IQ#1-answers Count the items, determine the mass, & Volume
We know: 1 dozen apples = 2.0 Kg
# of apples = 90
12 apples = 1 dozen
**Dimensional analysis:
90 apples x 1 doz. apples x kg
12 apples doz apples
= 15 kg
1 mol = 6.02 x 1023 parts

76. IQ#1-answers cont’d 4. Count the items, determine the mass, & Volume
**We know: # of atoms present = 1.25 x1023 Mg atoms
1 mol Mg = 6.02 x1023 atoms Mg
Conversion: # atoms → # moles
Use Dimensional analysis:
1.25 x1023 atoms Mg x mol Mg x1023 atoms Mg
= mol Mg

77. Answers to Homework Practice Problems #1-5 1) 5.0 kg (pg. 289)
2) 672 seeds (pg. 289)
3) mol Si (pg. 291)
4) mol Br2 (pg. 291)
5) 2.75 x1024 atoms (pg. 292)
6) mol NO2 (pg. 292)

78. Introductory Questions #2
Using your periodic table determine the molecular mass of ammonium nitrate and Potassium permanganate.
Which compound has higher molar mass ammonium nitrate or Potassium permanganate? Show your work.
If I had 38.5 g of potassium permanganate and 58.6g of ammonium nitrate which do I have more of? Convert each into the number ofmoles present to compare.

79. Homework Do Quest. #7-13 from “The Mole Worksheet I”
Practice problems: pg #7 & 8
pg #16 & 17
pg. 299 #18 & 19
Read/Review Percent Composition Pgs
*Do Practice problems #34 & 35 on pg. 307

80. Introductory Questions #3
Which of the following substances has more molecules/compounds present? Be sure to show all calculations and work.
242.38g Heptanitrogen pentoxide
85.56g Aluminum arsenate
154.28g hydrogen perioxide
How much would 1000 atoms of magnesium weigh in grams?
Determine the % composition of barium perchlorate.

81. Assignments to Check/Review
% Composition Worksheet answers
Chemistry WS II: % Comp & Empirical Formulas
Practice Problems from text: #36 & 37 (pg.310)
Hydrate Formula----Worksheet Today’s Handout
Lect/Disc: Chemical Reactions & equation writing
**Homework: Hydrate Formula WS
Chemical Equation Worksheet

82. Signs of a Chemical Reaction
Evolution of heat and light
Formation of a gas
Formation of a precipitate
Color change

83. B.Law of Conservation of Mass
Mass is neither created nor destroyed in a chemical reaction
Total mass stays the same
Atoms can only rearrange
4 H
2 O
4 H
2 O
36 g
4 g
32 g

84. C. Chemical Equations
A+B  C+D
REACTANTS
PRODUCTS

85. Chemical Symbols

86. Writing Equations 2H2(g) + O2(g)  2H2O(g)
Identify the substances involved.
Use symbols to show:
How many? - coefficient
Of what? - chemical formula
In what state? - physical state
Remember the diatomic elements.

87. D. Writing Equations 2 Al (s) +3 CuCl2 (aq)  3 Cu (s) + 2 AlCl3 (aq)
Two atoms of aluminum react with three units of aqueous copper(II) chloride to produce three atoms of copper and two units of aqueous aluminum chloride.
How many?
Of what?
In what state? 2 Al
(s) +3
CuCl2
(aq)
 3 Cu
(s)
+ 2
AlCl3
(aq)

88. E. Describing Equations
Describing Coefficients:
individual atom = “atom”
covalent substance = “molecule”
ionic substance = “unit”
3CO2 
2Mg 
4MgO 
3 molecules of carbon dioxide
2 atoms of magnesium
4 units of magnesium oxide

89. E. Describing Equations
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
How many?
Of what?
In what state?
One atom of solid zinc
reacts with
two molecules of aqueous hydrochloric acid
to produce
one unit
of aqueous zinc chloride
and one
molecule of hydrogen gas.

90. Unit 5 HW Packet Book Notes (15 total) 10-1 practice problems #1-4*
Page 315 #50-56, 58-59
Moles, Molecules, and Grams WS
Percent Composition WS
p. 315 # 63-68
Percent Comp., Emp. Formula WS
Writing Complete Equations WS
Balancing Chemical Equations WS
Balancing Equations: Algebraic Technique WS
Unit 5 Sample Test