determination of an empirical formula

DETERMINATION OF AN EMPIRICAL FORMULA

1

Empirical Formulas• Empirical formula (“formula unit”) -the lowest

whole number ratio of atoms in an ionic compound.

• There are two ways to determine the empirical formula for an ionic compound.1. Using charges (use your ions and this is

easy). C CRISS O S S

2. Mathematically (yippee!!!) 2

Method #1 – ChargesEX: Write the empirical formula

for the compound formed byNa & P c. K &

Ne

Sr & Cl d. Cu & Cl

No compoundNa3P

SrCl2 CuCl or CuCl23

Method #2 – Mathematically

• Step 1:Use the information given in the problem and dimensional analysis with the atomic mass of the element (from periodic table- round to 3 SDs) to find the number of moles you have

• Step 2:Take all the mole values and divide them by the SMALLEST one to figure out a ratio

• Step 3: Use the answers as subscripts in the empirical formula

4

Reference-Empirical Formulas 7.06 g of silver combine with an excess of fluorine to produce 8.30 g of a compound

Silver + Fluorine Ag?F?

7.06g 1.24g 8.30g7.06 g Ag

1.24 g F

X

X

= .0654 moles Ag

= .0653 moles F

= 1.00 or 1

= 1.00 or 1

Found by subtracting!

Ag1F1

ANSWER = AgF 5

Reference-Empirical Formulas A compound contains 24.58% K, 35.81% Mn, and 40.50% O. Find the empirical formula (assume working with 100 grams of the compound and change percentages to grams)

24.58 g K

35.81 g Mn

40.50 g O

X

X

X

= .629 mole K

= .652 moles Mn

= 2.53 moles O

= 1.00 or 1

= 1.04 or 1

= 4.02 or 4

ANSWER = KMnO46

Uneven Empirical Formulas

• When figuring empirical formulas mathematically, sometimes the resulting numbers don’t come out so clean

• You can’t just assume and round how you choose

7

Reference- .05 Rule• Values used in these problems are obtained by

experimentation. The 0.05 rule allows for experimental error:– If the value is within .05 of a whole number

(+0.05 or - 0.05), then the value may be rounded to that whole number

– Examples: 1.96 can be rounded to 2 • 1.07 cannot be rounded to 1• 3.02 could be rounded to 3 • 1.93 cannot be rounded to 2

• IF one of the values is not within .05 of a whole number, all the values must be multiplied by an integer so that all values fall within .05 of whole numbers 8

Reference-Uneven Empirical Formulas

4.35 g sample of zinc is combined with an excess of the element phosphorus. 5.72 g of

compound are formed. Calculate the empirical formula.

Zinc + Phosphorous Zn?P?

4.35g 1.37g 5.72gFound by subtracting!

4.35 g Zn

1.37 g P

X

X

= .0665 moles Zn

= .0442 moles P

= 1.50

= 1.00

Not within .05 of a whole number

X 2 = 3.00 or 3

X 2 = 2.00 or 2

ANSWER = Zn3P2

9

Let’s Do It!!!

A compound is found to contain 72.3% Fe and 27.7% O by weight. Calculate the empirical formula.

Assume in 100 g of compound there would be 72.3 g Fe and 27.7 g O

10

72.3g Fe 1 mole FeX —————— 55.8 g Fe

= 1.30 mole Fe

27.7g O 1 mole O X —————— 16.0 g O

= 1.73 mole O

1.30 mole 1.30 mole

1.73 mole 1.30 mole

= 1.00

=1.33

X 3 = 3.00 = 3

X 3 = 3.99 = 4

Fe3O4

11

Review • Number your paper from 1-5 and answer

the following questions. Two will be cumulative review!

• 1. Which of these is the correct symbol for Chlorine-35?

• a. 35 Cl 17 • b. 17 Cl 37 • c. 17 Cl 35

12

Review• A• 2. Which is the correct answer with the

right number of SD’s if we add 75g and 25.0g?– a. 105 g– b. 10Ō g – c. 100.0 g– d. 105.0 g

13

Review• B• 3. Which is correct if we consider our

rounding rule?– a. 4.02 couldn’t be rounded to 4 but

1.93 can be rounded to 2– b. 4.02 couldn’t be rounded to 4 and

1.93 can’t be rounded to 2– c. 4.02 can be rounded to 4 and 1.93

can be rounded to 2– d. 4.02 can be rounded to 4 and 1.93

can’t be rounded to 2 14

Review• D• 4. Zinc + Phosphorous Zn?P?

4.35g ?g 5.72g– a. 4.35 g– b. 2.37 g– c. 1.37 g– d. 1.47 g

15

Review• C• 5. If I have 32.0g of O, then how many

moles is this?– a. 2.0 moles– b. 2.5 moles– c. 2 moles– d. 20 moles

16

Review• A

17

Molar Mass• KC4H5O6 is the empirical formula

for cream of tartar• How many atoms of oxygen are in

1 formula unit of cream of tartar?6 atoms O

18

Molar Mass• KC4H5O6• How many atoms are in 1 formula unit of cream of tartar?

16 atoms 1 potassium4 carbon5 hydrogen6 oxygen 19

Molar Mass• KC4H5O6 • How many formula units are in 1

mole of cream of tartar? 6.02 x 1023 formula units

20

Molar Mass• KC4H5O6 • How many atoms of oxygen are in

(PER) 1 mole of cream of tartar?• ? atoms O/mole KC4H5O6 =

6 atoms O x 6.02 x1023 formula unit KC4H5O6 1 formula unit 1 mole KC4H5O6 KC4H5O63.61 x 1024 atoms O/mole KC4H5O6 21

Molar Mass• KC4H5O6• How many moles of oxygen atoms

are in 1 mole of cream of tartar?• Let’s put this in perspective……….

22

How many wheels are in a dozen bicycles?

1 bicycle = 2 wheels1 dozen bicycles = 2 dozen wheels

1 mole bicycles = 2 mole wheels23

How many dozen hydrogen atoms are in 1 dozen water molecules?

1 water = 2 hydrogen1 dozen water = 2 dozen hydrogen1 mole water = 2 mole hydrogen

H2O24

Molar Mass• So………• KC4H5O6 • How many moles of oxygen atoms

are in 1 mole of cream of tartar?

6 moles O

25

Molar Mass• KC4H5O6 • What is the mass of the oxygen

atoms in (PER) 1 mole of cream of tartar?

? g O/ mole KC4H5O6

6 moles O x 16.0 g O = 96.0 g O1 mole KC4H5O6 1 mole O mole

KC4H5O6 26

Let’s Do It!!!• How many atoms in one formula

unit?Magnesium Acetate: Mg(C2H3O2)2

1-Mg4-C6-H4-O

15 total atoms 27

Let’s Do It!!!• How many formula units in one

mole ofMagnesium Acetate:

Mg(C2H3O2)2

6.02 x 1023

28

Let’s Do It!!!!• How many oxygen atoms are in

one formula unit?Magnesium Acetate: Mg(C2H3O2)2

4 oxygen atoms

29

Let’s Do It!!!• How many oxygen atoms are in

one mole of this substance?Magnesium Acetate: Mg(C2H3O2)2

(4) (6.02 x 1023) = 2.41 x 1024 atoms O

30

Let’s Do It!!!• How many moles of oxygen atoms are in one mole of this

substance?Magnesium Acetate: Mg(C2H3O2)2

4 moles of Oxygen31

Let’s Do It!!!• What is the mass of oxygen atoms

in one mole of this substance? Magnesium Acetate: Mg(C2H3O2)2

• ? g O=1 mole Mg(C2H3O2)2 x 4 moles O x 16.0 g O 1 mole Mg(C2H3O2)2 1 mole O

= 64.0 g O = 64.0g/ 1 mole32

Molar Mass • We can use this information about what

makes up a compound to figure out a compound’s total molar mass

• Molar mass- the mass in grams of 1 mole of a compound

• It’s the mass of one mole: grams KC4H5O6/mole KC4H5O6 • Also called formula weight, gram

formula weight, molecular weight 33

Reference- Molar Mass• Calculate the molar mass of

magnesium iodide, MgI2 from its parts. How many grams are in 1 mole?

Mg 1 mole (24.3g Mg/mole )= 24.3 g Mg

I 2 mole (127 g I/mole) = 254 g I

278 g/mole

MgI2 (for SD use place

value!)

34

Reference- Molar MassCalculate the molar mass (grams in

one mole) of ammonium sulfite, (NH4)2SO3

In 1 mole of the compound there are: 2 moles of N X 14.0 g N/mole= 28.0g N 8 moles of H X 1.01 g H/mole = 8.08gH 1 mole of S X 32.1 g S/mole =32.1 g S 3 moles of O X 16.0 g

O/mole=48.0g O SD by place value 116.2gunits always g/mole 1mole

(NH4)2SO3

35

• Molar mass can be used to convert between moles and grams

• For an element: 1 mole = 6.02x1023 atoms = atomic mass • For a compound: • 1 mole = 6.02x1023 formula units =

molar mass

Molar Mass

36

• What is the mass of 1.35 moles of (NH4)2SO3?

• Earlier we found the molar mass: 116.2g (NH4)2SO3 = 1mole (NH4)2SO3

? g (NH4)2SO3 = 1.35 moles (NH4)2SO3 x 116.2 g (NH4)2SO3

1 mole (NH4)2SO3

= 157 g (NH4)2SO3

Reference- Molar Mass as a Conversion Factor

37

• 75.2 g (NH4)2SO3 is how many moles of (NH4)2SO3?

? moles (NH4)2SO3 =

1 mole (NH4)2SO3

75.2 g (NH4)2SO3 X ---------------------- 116.2 g (NH4)2SO3

= .647 moles (NH4)2SO3

Reference- Molar Mass as a Conversion Factor

38

Let’s Do It!!!!• How many moles are in 87.4 g of

aluminum monohydrogen phosphate Al2(HPO4)3 ?

• This is a two part problem: remember you first have to find the molar mass!

39

Let’s Do It!!!• Find the molar mass of aluminum

monohydrogen phosphate, Al2(HPO4)32 moles Al

3 moles H3 moles P12 moles O

(27.0g Al/mole) =(1.01g H/mole) =(31.0 g P/mole) =(16.0 g O/mole)=

54.0 g Al3.03 g H93.0 g P192 g O_

342 gmole Al2(HPO4)3MASS OF ONE MOLE!!!

So 342 g Al2(HPO4)3 = 1 mole Al2(HPO4)3 40

Let’s Do It!!!!• Then find the number of moles in

87.4 g of aluminum monohydrogen phosphate• Molar mass = 342 g/mole• ? moles Al2(HPO4)3 =87.4 g Al2(HPO4)3 x 1 mole Al2(HPO4)3

342 g Al2(HPO4)3

= .256 moles Al2(HPO4)3

Obviously, this is less than one mole!

41

Review• Number your paper from 1-5 and

answer the following questions. Two will be cumulative review!

• 1. What does Avogadro’s number, 6.02 x 1023, mean?– a. There are that many grams in a

mole– b. There are that many moles in a

gram– c. There are that many moles in an

atom– d. There are that many atoms in a

mole

42

Review• D• 2. Which of these is false?

• a. protons repel protons• b. protons attract neutrons• c. protons attract electrons• d. neutrons don’t repel anything

because they have no charge

43

Review• B• 3. How many elements are in this

compound? KC4H5O6 – a. 4– b. 15– c. 16– d. 10

44

ReviewA• 4. How many moles of oxygen would

be in 3 moles of KC4H5O6 ?– a. 3– b. 6– c. 18– d. 48

45

Review• C• 5. If H2O has a molar mass of 18, what

does this mean?– a. There are 18 moles of water in a

gram– b. There are 18 moles of oxygen in a

mole of water – c. There are 18 grams in a mole of

water– d. There are 10 moles of oxygen and

8 moles of hydrogen in each water molecule

46

Review• C

47

% Composition• As we learned before, when we

have multiple elements that make up a compound, each one has a certain ratio

• We call this the compound’s empirical formula

• From an empirical formula we can figure out each element’s percent composition 48

Reference-% Composition• Find percent composition of

Al(C2H3O2)3(1 moles Al)(6 moles C)(9 moles H)(6 moles O)

(27.0 g/mole) =(12.0 g/mole) =(1.01 g/mole) =(16.0 g/mole) =

27.0 g Al72.0 g C9.09 g H96.0 g O

204.1 g Al(C2H3O2)3/204.1 g Al(C2H3O2)3 x 100 =

/204.1 g Al(C2H3O2)3 x 100 =/204.1 g Al(C2H3O2)3 x 100 =/204.1 g Al(C2H3O2)3 x 100 =

13.2 %35.3 %4.45 %47.0 %~100%

27.0 g Al72.0 g C9.09 g H96.0 g O 49

% Composition• We can use this information further• How many grams of aluminum can

be obtained from 1.50 moles of aluminum acetate?

• ? grams Al =1.50 moles Al(C2H3O2)3 x X 𝟐𝟕 .𝟎𝒈 𝑨𝒍

𝟐𝟎𝟒 .𝟏𝒈 𝑨𝒍 (𝑪𝟐𝑯𝟑𝑶𝟐 )𝟑

= 40.5 g Al

1.50 moles Al(C2H3O2)3 x X 27.0 g Al 1 mole Al

= 40.4 g Al

OR

50

Let’s Do It!!!• Find the percent composition of iron

(III) dichromate, Fe2(Cr2O7)3

(2 moles Fe)(6 moles Cr)(21 moles O)

(55.8 g/mole) =(52.0 g/mole) =(16.0 g/mole) =

112 g Fe312 g Cr336 g O

76Ō g Fe2(Cr2O7)3

/76Ōg Fe2(Cr2O7)3 x 100 =/76Ōg Fe2(Cr2O7)3 x 100 =/76Ōg Fe2(Cr2O7)3 x 100 =

14.7 %41.1 %44.2 %~100%

112 g Fe312 g Cr336 g O

51

Let’s Do It!!!• How many moles of iron can be

obtained from 525 g of iron (III) dichromate, Fe2(Cr2O7)3?

• ? moles of Fe =525 g Fe2(Cr2O7)3 X

X = 1.38 moles Fe

525 g Fe2(Cr2O7)3 X 112 g Fe 76Ōg

X = 1.39 moles Fe

OR

52

Reference- Advanced Molar Mass

• How many grams of Mg are needed to combine with 2.00 g of N to make magnesium nitride?

Magnesium + Nitrogen Magnesium nitride (Mg3N2) ?g 2.00 g(3 moles Mg)(2 mole N)

(24.3 g Mg/mole) =(14.0 g N/mole) =

72.9 g Mg28.0 g N

100.9 g Mg3N2

2.00 g N x 72.9 g Mg 28.0 g N

= 5.21g Mg

72.9 g Mg + 28.0 g N = 100.9 Mg3N2

53


• OR THE EASY WAY!!! Magnesium + Nitrogen Magnesium nitride (Mg3N2) ?g 2.00 g

? g Mg = 2.00 g N x 1 mole N x 3 mole Mg x 24.3 g = 14.0 g N 2 mole N 1 mole Mg

= 5.21g Mg

72.9 g Mg + 28.0 g N = 100.9 Mg3N2

54

• How many grams of nitrogen are needed to combine with 5.00 g Mg to make magnesium nitride?

Magnesium + Nitrogen Magnesium nitride Mg3N2

5.00 g ?g

? g N =5.00 g Mg x 28.0 g N 72.9 g Mg

= 1.92 g N


72.9 g Mg + 28.0 g N = 100.9 Mg3N2

55

• OR THE EASY WAY!!! Magnesium + Nitrogen Magnesium nitride Mg3N2

5.00 g ?g ? g N =5.00 g Mg x 1 mole Mg x 2 mole N x 24.3 g = 24.3 g Mg 3 mole Mg 1 mole Mg

= 1.92 g N


56


• How many grams of Mg are needed to make 25 g of magnesium nitride?

Magnesium + Nitrogen Magnesium nitride – Mg3N2

?g 25g72.9 g Mg = 28.0 g N = 100.9 Mg3N2

? g Mg=

25 g Mg3N2 x 72.9 Mg = 18 g Mg 100.9 g Mg3N 2

57


OR THE EASY WAY!!!• How many grams of Mg are needed to

make 25 g of magnesium nitride?Magnesium + Nitrogen Magnesium nitride Mg3N2

?g 25g? g Mg = 25 g Mg3N2x 1 mole Mg3N2 x 3 mole Mg x 24.3 g = 100.9 g Mg3N2 1 mole Mg3N2 1 mole Mg molar mass = 18 g Mg

58

Let’s Do It!!!• How many grams of magnesium

nitride can be made from 12.5 g of magnesium?


12.5 g ?g

72.9 g Mg = 28.0 g N = 100.9 Mg3N2

59

Let’s Do It!!!• How many grams of magnesium



12.5 g ?g

72.9 g Mg = 28.0 g N = 100.9 Mg3N2

12.5 g Mg x 100.9 g Mg3N 2

72.9 g Mg

= 17.3 g Mg3N2

60

Let’s Do It!!! OR THE EASY WAY!!!• How many grams of magnesium


Magnesium + Nitrogen Magnesium nitride Mg3N2

12.5g ?g

? g Mg3N2 = molar mass 12.5 g Mg x 1 mole Mg x 1 mole Mg3N2 x 100.9g Mg3N2 24.0 g Mg 3 mole Mg 1 mole Mg3N2

= = 17.5 g Mg3N2

61

• Number your paper from 1-5 and answer the following questions. Two will be cumulative review!

• 1. Atoms with the same atomic number but different mass numbers are

• a. different elements• b. ions• c. isotopes of the same element• d. isotopes of different elements

Review

62

Review• D• 2.An alpha particle

• a. has a negative charge• b. can penetrate a sheet of

aluminum• c. is identical to a helium nucleus• d. all of the above

63

Review• C• 3. How many moles of Cr are in a mole

of Fe2(Cr2O7)3– a. 6– b. 3– c. 2– d. 1

64

Review• A• 4. What is the percent composition of

hydrogen and oxygen in water?– a. 80% O and 20% H– b. 88.9% O and 11.2% H– c. 75% O and 35% H– d. 66.6% H and 33.3% O

65

Review• B• 1 mole O x 16.0 g O = 16.0 g 16.0/18.0 x

100 1 mole O• 2 moles H x 1.01 g H = 2.02 2.02/18.0 x

100 1 mole H• 5. How many grams of water can be

made from 10 grams of hydrogen and excess oxygen?– a. 90 g– b. 10 g – c. 20 g– d. 0 g

66

Review• A• 5. How many grams of water can be

made from 10 grams of hydrogen? EASY WAY!!– a. 90 g– b. 10 g – c. 20 g– d. 0 g

? g H2O = molar mass 10 g H x 1 mole H x 1 mole H2O x 18.0 g H2O 1.01 g H 2 mole H 1 mole H2O

67

Hydrates• Hydrated compounds- compounds with

molecules of water held in their crystal structure

• These compounds contain an anhydrous (non-water) part and a hydrous (water based) part

68

Hydrates

Ex.) CaSO4 . 7H2O• Compounds with molecules of water held in

their crystal structure• Very common in nature• Water can be removed by heating, leaving

behind what is called the anhydrous compound.

Anhydrous compound Wate

r

69

Reference-Naming Hydrated Compounds

• The following is tacked on the name obtained from the ions

H2O monohydrate 2 H2O dihydrate 3 H2O trihydrate 4 H2O

tetrahydrate 5 H2O pentahydrate 6 H2O

hexahydrate 7 H2O heptahydrate 8 H2O

octahydrate 9 H2O nonahydrate 10 H2O

dekahydrate CaSO4 7 H2O -- named as

calcium sulfate heptahydrate

70

Reference-Empirical Formula of Hydrates

• Find the empirical formula of a hydrate of CaSO4 hydrate that is 28.5% H2O

• To solve this problem, find the simplest mole ratio between the anhydrous part of the compound (CaSO4) and the water (H2O)

• H2O =28.5 % CaSO4 =71.5 %

(100% - 28.5%) 71

Reference-Empirical Formula of Hydrates 71.5 g CaSO4 x 1 mole CaSO4 = .525 mole

CaSO4 136.2 g CaSO4 molar mass of CaSO4

28.5 g H2O x 1 mole H2O = 1.58 mole H2O

18.0 g H2O

..525 = 1.00 = 1 .525

1.58 = 3.01 = 3 .525CaSO4 3 H2O

•When you are finding formulas of hydrates they ALWAYS come out even!

72

Reference- Molar Mass of a Hydrate

• Find the molar mass of CaSO4 7 H2O

• First we find the molar mass for H2O and treat the water like it’s an element!

2 mole H X 1.01 g/mole = 2.02 g

1 mole O X 16.0 g/mole = 16.0 g

18.0 g/mole

(MEMORIZE THIS)

73

Reference- Molar Mass of a Hydrate

CaSO4 7 H2O 1 mole Ca x 40.1 g/mole =

40.1 g1 mole S x 32.1 g/mole =

32.1 g4 mole O x 16.0 g/mole =

64.0 g7 mole H2O x 18.0 g/mole = 126

g 262 g/mole

(SD by place value)

74

Reference- % Composition of a Hydrate

• Find the % composition of CaSO4 7 H2O• The water is treated like an element! % Ca = 40.1 g Ca X 100% = 15.3 % Ca 262 g compound

% S = 32.1 g S X 100% = 12.3 % S 262 g compound

% O = 64.0 g O X 100% = 24.4 % O 262 g compound

% H2O = 126 g H2O X 100% = 48.1 % H2O 262 g compound

75

Using Molar Mass or % Composition of a Hydrate

• How much water do we get when we heat 2.00 g of CaSO4 7 H2O?

2.00 g x .481 = .962 g H2O

OR

2.00 g CaSO47 H2O x 126 g H2O = 0.962 g H2O

262 g CaSO47 H2O 76

Review • Number your paper from 1-5 and

answer the following questions. Two will be cumulative review! – 1. What is the actual mass of a

hydrogen-3 atom?• a. 6.02 x 1023 g• b. 1.67 x 1024 g• c. 5.01 x 1024 g• d. 5.01 x 10-24 g

77

Review• D• 2. Which of these is true about the

discovery of Millikan’s oil drop experiment– a. He discovered the electron– b. He discovered the mass of the

neutron– c. He discovered the mass and the

charge of the electron– d. He discovered the proton 78

Review• C• 3. What is CaSO4 5 H2O called?

– a. calcium sulfate pentahydrate– b. calcium sulfate heptahydrate– c. calcium sulfate hydrate– d. calcium sulfate trihydrate

79

Review• A• 4. What is the mass of water in CaSO4

5 H2O? – a. 5 g– b. 90.0 g – c. 5.0 g– d. 18 g

80

Review• B• 5 mole H2O x 18.0 g/mole = 90.0 g• 5. Which of these would have the most

water given off when heated?– a. calcium sulfate pentahydrate– b. calcium sulfate heptahydrate– c. calcium sulfate hydrate– d. calcium sulfate trihydrate

81

Review• B

82

determination of an empirical formula

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