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Chemical Quantities The Mole, % Composition, Empirical and Molecular Formulas

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ChemicalQuantities

The Mole,

% Composition, Empirical and Molecular Formulas

- 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.

- Defined as the number of carbon atoms in exactly
12 grams of carbon-12.

- 1 mole is 6.02 x 1023 particles.
- Treat it like a very large dozen
- 6.02 x 1023 is called Avagadro’s number.

- 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.

- How many oxygen atoms in the following?
- CaCO3
- Al2(SO4)3

- How many ions in the following?
- CaCl2
- NaOH
- Al2(SO4)3

3

12

3

2

5

- How many molecules of CO2 are the in 4.56 moles of CO2?
- 4.56 mole x 6.02x1023 mc =
1 1 mole

- How many moles of water is 5.87 x 1022 molecules?
- 5.87 x 1022 mc x 1 mole =
1 6.02x1023 mc

2.75x1024mc

0.0975 mole

4.44x1024 atoms

- How many atoms of carbon are there in 1.23 moles of C6H12O6?
- 1.23 moles x 6.02x1023 mc x 6 atoms =
1 1 mole 1 mc

- How many moles is 7.78 x 1024 formula units of MgCl2?
7.78x1024 FU x 1 mole = 12.9 mole

1 6.02x1024 FU

- Remember relative atomic mass?
- The amu was one twelfth the mass of a carbon 12 atom.
- Since the mole is the number of atoms in 12 grams of carbon-12,
- the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.

- The mass of 1 mole of an element in grams.
- 12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron.
- We can right this as 12.01 g C = 1 mole
- We can count things by weighing them.

- How much would 2.34 moles of carbon weigh?
- 2.34 moles C x 12 g C
1 1mole

- How many moles of magnesium in 24.31 g of Mg?
24.31 g Mg x 1 mole =

1 24g Mg

= 28.08 g

1.013 mole

8.60x1022 atoms

13.6 g

How many atoms of lithium in 1.00 g of Li?

1.00 g Li x 1 mole x 6.02x1023 atoms

1 7 g Li 1 mole

How much would 3.45 x 1022 atoms of U weigh?

3.45x1022 atoms U x 1 mole x 238 g U

1 6.02x1023atoms 1 mole

- 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

- What is the mass of one mole of CH4?
- 1 mole of C = 12 g
- 4 mole of H x 1 g = 4g
- 1 mole CH4 = 12 + 4 = 16g
- The Gram Molecular mass of CH4 is 16.05g
- The mass of one mole of a molecular compound.

- The mass of one mole of an ionic compound.
- Calculated the same way.
- What is the GFM of Fe2O3?
- 2 moles of Fe x 56 g = 112 g
- 3 moles of O x 16 g = 48 g
- The GFM = 112 g + 48 g = 160g

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

- Calculate the molar mass of the following and indicate what type it is.
- Na2S
- 2 (23) + 32 =
- N2O4
- 2(14) + 4(16) =
- C

46 + 32 = 78g

1mole Na2S = 78g

Gram Formula Mass

28 + 64 = 92g

1 mole N2O4 = 92g

Gram Molecular Mass

12g

1 mole C = 12 g

Gram Atomic Mass

Gram Formula Mass

- Ca(NO3)2
- 40 + 2(14) + 6(16) = 40 + 28 + 96 = 164g
- 1 mole Ca(NO3)2 = 164g
- C6H12O6
- 6(12) + 12(1) + 6(16) = 70 + 12 + 96 = 180g
- 1 mole C6H12)6 = 180g Gram Molecular Mass
- (NH4)3PO4
- 3(14) + 12(1) + 31 + 4(16) = 42+12+31+64 = 149g
- 1 mole (NH4)3PO4 = 149g Gram Formula Mass

Using Molar Mass

Finding moles of compounds

Counting pieces by weighing

- 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.

- How many moles is 5.69 g of NaOH?

- How many moles is 5.69 g of NaOH?

- How many moles is 5.69 g of NaOH?

- need to change grams to moles

- How many moles is 5.69 g of NaOH?

- need to change grams to moles
- for NaOH

- How many moles is 5.69 g of NaOH?

- need to change grams to moles
- for NaOH
- 1mole Na = 23g 1 mol O = 16.00 g 1 mole of H = 1 g

- How many moles is 5.69 g of NaOH?

- need to change grams to moles
- for NaOH
- 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g
- 1 mole NaOH = 40 g

- How many moles is 5.69 g of NaOH?

- need to change grams to moles
- for NaOH
- 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g
- 1 mole NaOH = 40 g

- How many moles is 5.69 g of NaOH?

- need to change grams to moles
- for NaOH
- 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g
- 1 mole NaOH = 40 g

- How many moles is 4.56 g of CO2?
- 4.56g CO2 x 1 mole
1 44gCO2

- How many grams is 9.87 moles of H2O?
- 9.87 moles x 18g H2O
1 1 mole

= 0.104 moles

= 178g

= 2.56x1023 mc

- How many molecules in 6.8 g of CH4?
- 6.8g CH4 x 1 mole x 6.02x1023 mc
1 16g CH4 1mole

- 49 molecules of C6H12O6 weighs how much?
- 49 mc x 1 mole x 180g C6H12O6 =
1 6.02x1023 mc 1 mole

- 8820 g____= 1.5x10-20 g
6.02x1023

Gases and the Mole

- 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
- Scientists compare gases at 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.

- What is the volume of 4.59 mole of CO2 gas at STP?
- 4.59mole x 22.4L
1 1mole

- How many moles is 5.67 L of O2 at STP?
- 5.67L x 1 mole
1 22.4L

- What is the volume of 8.80g of CH4 gas at STP?
- 8.80g CH4x 1mole x 22.4L
1 16g CH4 1mole

= 102.816 = 103L

= .523moles

= 12.32 = 12.3L

- change moles to grams
- moles to atoms
- moles to formula units
- moles to molecules
- moles to liters
- molecules to atoms
- formula units to atoms
- formula units to ions

Mass

Periodic Table

Moles

Mass

PeriodicTable

Volume

Moles

Mass

Periodic Table

Volume

22.4 L

Moles

Mass

Periodic Table

Volume

22.4 L

Moles

Representative

Particles

Mass

Periodic Table

Volume

22.4 L

Moles

6.02 x 1023

Representative

Particles

Mass

Periodic Table

Volume

22.4 L

Moles

6.02 x 1023

Representative

Particles

Atoms

Mass

Periodic Table

Volume

22.4 L

Moles

6.02 x 1023

Representative

Particles

Ions

Atoms

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

- Calculate the percent composition of each element in a compound that is 29.0 g of Ag with 4.30 g of S.
- Ag 29.0g
- S + 4.30g
- 33.3g

/33.3

= .8709 x 100 = 87.09%

= .1291 x 100 = 12.91%

/33.3

- If we know the formula, assume you have 1 mole.
- Then you know the pieces and the whole.

- Calculate the percent composition of C2H4?
- C 2(12g)=24
- H 4(1g) = +4
- 28g

/28

= .8571 x 100 = 85.71%

/28

= .1429 x 100 = 14.29%

/234

= .2308 x 100 = 23.08%

/234

= .15.38 x 100 = 15.38%

/234

= .6154 x 100 = 61.54%

Calculate the percent composition of

Aluminum carbonate.

Al2(CO3)3

Al 2(27g)= 54

C 3(12g)= 36

O 9(16)= 144

234g

You can also calculate the mass of an element in a given amount of a compound using % composition.

•Step 1: calculate the % comp. only of the element you want to find the mass of.

•Step 2: Multiply the elements %, by the mass of the compound given.

Example: Calculate the mass of sulfur in 3.54g of H2S.

MM of H2S = H 2 (1) = 2

S 1(32) = +32

34g H2S

% S = 32/34 x 100 = 94.1% S

94.1% x 3.54g = 3.33g S

Calculate the mass of nitrogen in 25g of (NH4)2CO3.

N 2(14g) = 28

H 8(1g) = 8

C 1(12g) = 12

O 8(16g) = +128

176g (NH4)2CO3

%N = 28/176 x 100 = 15.91%

15.91% x 25g = 4.0g N

Mg 1 (24g) = 24

O 2(16g) = 32

H 2 ( 1g) = +2

58g Mg(OH)2

%Mg = 24/58 x 100 = 41.38%

41.38% x 97.4g = 40.3g Mg

Empirical Formula

From percentage to 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

- Just find the lowest whole number ratio
- C6H12O6
- CH4N
- It is not just the ratio of atoms, it is also the ratio of moles of atoms.
- In 1 mole of CO2there 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.

- 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 moles.

- Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.
- Assume 100 g so
- 38.67 g C x 1mol C = 3.223 mole C 12 g C
- 16.22 g H x 1mol H = 16.22 mole H 1 g H
- 45.11 g N x 1mol N = 3.222 mole N 14 g N

- The ratio is 3.223 mol C = 1 mol C 3.222 mol N 1 mol N
- The ratio is 16.00 mol H = 5 mol H 3.222 mol N 1 mol N
- C1H5N1

A compound is 43.64 % P and 56.36 % O. What is the empirical formula?

43.64 g P x 1mol P= 1.408 mole P 31 g P

56.36 g O x 1mol O= 3.523 mole O 16 g O

The ratio is 3.523 mol O= 2.5 mol O 1.408 mol P 1 mol P

Can not have 2.5 atoms! Double

P2O5

Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

49.48 g C x 1mol C = 4.123 mole C 12 g C

5.15 g H x 1mol H = 5.15 mole H 1 g H

28.87 g N x 1mol N = 2.062 mole N 14 g N

16.49 g O x 1mol O = 1.031 mole O 16 g O

The ratio is 4.123 mol C = 4 mol C 1.031 mol O 1 mol O

The ratio is 5.15 mol H = 5 mol H 1.031 mol O 1 mol O

The ratio is 2.062 mol N = 2 mol N 1.031 mol O 1 mol O

C4H5N2O1

- 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.
- Caffeine has a molar mass of 194 g. what is its molecular formula?
- C4H5N2O1 = 97g
- 194/97 = 2
- Molecular Formula = C8H10N4O2

- A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be 98.96 g. What is its molecular formula?
- 71.65 g Cl x 1mol Cl = 2.047 mole Cl35 g Cl
- 24.27 g C x 1mol C = 2.023 mole C 12 g C
- 4.07 g H x 1mol H = 4.07 mole H 1 g H

The ratio is 2.047 mol Cl = 1 mol Cl2.023 mol C 1 mol C

The ratio is 4.07 mol H = 4 mol H 1.031 mol C 1 mol C

Empirical Formula = CClH4

12+35+4 = 51g

98.96/51 = 2

Molecular Formula = C2Cl2H8