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

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Chemical

Quantities

The Mole, % Composition,

Empirical and Molecular Formulas

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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.

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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.

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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.

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Types of questions How many oxygen atoms in the

following?–CaCO3

–Al2(SO4)3 How many ions in the following?

–CaCl2–NaOH–Al2(SO4)3

12

3

3

2

5

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Types of questions using the equality;1 mole = 6.02 x 1023

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

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Types of questions using the equality;1 mole = 6.02 x 1023

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

4.44x1024 atoms

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Measuring Moles 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.

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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 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.

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Examples 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

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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

8.60x1022 atoms

13.6 g

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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

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What about compounds? 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.

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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 56 g = 112 g 3 moles of O x 16 g = 48 g The GFM = 112 g + 48 g = 160g

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Molar Mass The generic term for the mass of one

mole. The same as gram molecular mass,

gram formula mass, and gram atomic mass.

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Examples 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

Gram Formula Mass

28 + 64 = 92gGram Molecular Mass12g

1mole Na2S = 78g

1 mole N2O4 = 92g

1 mole C = 12 g Gram Atomic Mass

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Molar Mass Cont. 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

Gram Formula Mass

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Using Molar Mass

Finding moles of compounds

Counting pieces by weighing

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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.

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For example

How many moles is 5.69 g of NaOH?

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For example

How many moles is 5.69 g of NaOH?

5 69. g

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For example

How many moles is 5.69 g of NaOH?

5 69. g mole

g

need to change grams to moles

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For example

How many moles is 5.69 g of NaOH?

5 69. g mole

g

need to change grams to moles for NaOH

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For example

How many moles is 5.69 g of NaOH?

5 69. g mole

g

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

1 mole of H = 1 g

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For example

How many moles is 5.69 g of NaOH?

5 69. g mole

g

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

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For example

How many moles is 5.69 g of NaOH?

5 69. g 1 mole

40.00 g

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

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For example

How many moles is 5.69 g of NaOH?

5 69. g 1 mole

40.00 = 0.142 mol NaOH

g

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

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Examples 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

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Examples 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

= 2.56x1023 mc

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Gases and the Mole

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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 Scientists compare gases at

Standard Temperature and Pressure

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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.

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Examples 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 CH4 x 1mole x 22.4L

1 16g CH4 1mole

= 102.816 = 103L

= .523moles

= 12.32 = 12.3L

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We have learned how to 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

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Moles

MassPeriodic Table

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Moles

MassVolume Periodic

Table

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Moles

MassVolume 22.4 L Periodic

Table

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Moles

MassVolume

Representative Particles

22.4 L Periodic Table

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6.02 x 1023

Moles

MassVolume

Representative Particles

22.4 L Periodic Table

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Moles

MassVolume

Representative Particles

6.02 x 1023

Atoms

22.4 L Periodic Table

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Moles

MassVolume

Representative Particles

6.02 x 1023

Atoms Ions

22.4 L Periodic Table

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Percent Composition Like all percents Part x 100 %

whole Find the mass of each component, divide by the total mass.

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Example 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%

/33.3 = .1291 x 100 = 12.91%

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Getting % from the formula If we know the formula, assume you

have 1 mole. Then you know the pieces and the

whole.

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Examples Calculate the percent composition of

C2H4?

C 2(12g)=24

H 4(1g) = +4 28g

/28

/28

= .8571 x 100 = 85.71%

= .1429 x 100 = 14.29%

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Example Calculate the percent composition of

Aluminum carbonate. Al2(CO3)3

Al 2(27g)= 54

C 3(12g)= 36

O 9(16)= 144

234g

/234

/234

/234

= .2308 x 100 = 23.08%

= .15.38 x 100 = 15.38%

= .6154 x 100 = 61.54%

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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

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Calculate the mass of nitrogen in 25g of (NH4

)2

CO3

.

Calculate the mass of nitrogen in 25g of (NH4

)2

CO3

.

N 2(14g) = 28

H 8(1g) = 8

C 1(12g) = 12

O 8(16g) = +128

176g (NH4

)2

CO3

%N = 28/176 x 100 = 15.91%

15.91% x 25g = 4.0g N

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Calculate the mass of magnesium in 97.4g of Mg(OH)2.

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

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Empirical Formula

From percentage to formula

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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

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Calculating Empirical 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 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.

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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 moles.

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Example 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

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Example 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

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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

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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

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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

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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.

Caffeine has a molar mass of 194 g. what is its molecular formula?

C4H5N2O1 = 97g 194/97 = 2 Molecular Formula = C8H10N4O2

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Example 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 Cl 35 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

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The ratio is 2.047 mol Cl = 1 mol Cl 2.023 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