Average Atomic Mass

Since atoms are so small and the mass of individual atoms is also very small, it is not useful to use the units of grams or kilogram.

A new unit called the atomic mass unit (amu) was developed to deal with the very small units of mass for particles like the atom.

1 amu = 1.66 x 10-24 g

It has been found to be useful that instead of using absolute masses, it is best compare the relative masses of atoms using a reference isotope as a standard.

The carbon – 12 isotope has been set as this standard.

The carbon – 12 isotope has been assigned a mass of exactly 12 amu.

For example, the oxygen atom has a relative atomic mass of 15.999 amu.

The Mole

Another way to measure the amount of a

substance is to count the number of particles in that substance.

When talking about the number atoms, the number of individual atoms in the sample is ridiculously large.

Counting the individual atoms is NOT PRACTICAL.

Just as a dozen represents 12, 1 MOLE represents 6.02 x 1023 particles of a substance.

23

23

1002.6

1

1

1002.6

x

moleor

mole

x

This number (Avogadro’s Number) is named in honor of the Italian scientist Amedo Avogadro di Quarenga from whose work the concept was based.

How can we use Avogadro’s Number in a calculation?

Example #1 How many moles of magnesium make up 1.25 x 1023 atoms of magnesium? Answer: In order to solve this problem, we must realize that the conversion needed is

23

23

1002.6

1

1

1002.6

x

moleor

mole

x

Which form do we use???

Remember the techniques that we used when we did conversions factors

“What you want goes on top.”

1.25 x 1023 atoms Mg Mgatomsx

Mgmole231002.6

1

= .208 moles Mg

Example #2 How many atoms of sulfur are present in 0.75 moles of sulfur? Answer: In order to solve this problem, we must again realize that the conversion needed is

23

23

1002.6

1

1

1002.6

x

moleor

mole

x

Which form do we use???

0.75 moles S Smole

Satomsx

1

1002.6 23

= 4.5 x 1023 atoms S

Molar Mass

One atom of aluminum has a mass of 4.48 x 10-23 g.

This mass is determined by adding up the masses of the protons, neutrons and electrons within the aluminum atom.

What would be the mass of 1 mole of aluminum atoms (or 6.02 x 1023 atoms of aluminum)?

6.02 x 1023 atoms 4.48 x 10-23 g

1 atoms=

= 26.98 g 26.98 g is the mass of 6.02 x 1023 atoms

of aluminum OR

26.98 g is the mass of 1 mole of aluminum

THEREFORE,

26.98 g is the mass of 1 mole of Aluminum

More commonly stated, 26.98 g/mole is the MOLAR MASS of aluminum.

The Atomic Weights of the elements on

the Periodic Table also represent the Molar Masses of every element. Let’s find the molar masses of some atoms: Molar Mass O = ? Molar Mass W = ? Molar Mass Mg = ? How do we use Molar Mass?

Example #3 How many grams of iron are present in 1.78 moles of iron? Answer: In order to solve this problem, we must realize that the conversion needed is the molar mass of iron:

g

moleor

mole

g

85.55

1

1

85.55

Which form do we use???

Femole

FegFemoles

1

85.5578.1

= 99.4 g Fe Example #4 How many moles of calcium are present in 5.78 g of calcium?

Answer: In order to solve this problem, we must realize that the conversation needed is the molar mass of calcium:

g

moleor

mole

g

08.40

1

1

08.40

Which form do we use???

Cag

CamoleCag

08.40

178.5

= 0.0144 moles Ca

How can we determine the molar mass of a compound? 1) You must know the chemical formula of the compound. 2) Add the molar masses of the individual atoms in the molecule. Example: SO3 The molar mass of sulfur is

32.06 g/mole. The molar mass of oxygen is

16.00 g/mole. Therefore, molar mass of SO3 =

+ 3(16.00 g/mole)1(32.06 g/mole)

80.06 g/mole) Try this: What is the molar mass of BaBr2?

Mole – Mass Conversions The molar mass of an element or

compound can be used to the convert mass of a substance into moles. Example #1: How many moles are in 6.59 g of Nickel? 1. Determine the molar mass of Ni.

mole

g69.58

2. Calculate the moles of Ni.

Nig

NimolesNig

69.5859.6

= .112 moles Ni

Example #2: How many moles are in 92.2 g of FeO? 1. Determine the molar mass of FeO.

+ 16.00 g/mole)55.85 g/mole

71.85 g/mole

2. Calculate the moles of FeO.

FeOg

FeOmolesFeg

85.712.92

= 1.28 moles FeO

The molar mass of an element or compound is used to convert moles of a substance into mass.

Example #3: How many grams are in 3.84 moles of NO2? 1. Determine the molar mass of NO2.

+ 2(16.00) g/mole)14.01 g/mole

46.01g/mole 2. Calculate the moles of NO2.

2

22

01.4684.3

NOmoles

NOgNOmoles

= 177 g NO2

Percent Composition

All substances are composed of elements.

Those substances that contain several elements, do so with specific amounts of each component element.

For example, consider K2CrO4 and K2Cr2O7:

The relative amounts are element expressed can be expressed as the Percent Composition.

The percent composition will be determined as a function of mass of each element in a compound.

If a compound, AB, is made of

elements A & B, then the

100% xmass

massA

AB

A

Example #1: A compound is composed of 45.98 g of sodium and 70.90 g of chlorine. What is the percent composition? Answer: 45.98 g + 70.90 g 116.88 g

%34.3910088.116

98.45% x

g

gNa

%66.6010088.116

90.70% x

g

gCl

OR % Cl = 100% - 39.34% = 60.66%)

Example #2: An 8.20 g piece of magnesium combines with 5.40 g of oxygen to form a compound. What is the percent composition of the compound? 8.20 g + 5.40 g 13.60 g

%3.6010060.13

20.8% x

g

gMg

%7.3910060.13

40.5% x

g

gO

OR (% O = 100% - 60.3% = 39.7%) If information is not given about

the individual amounts for each element, the percent composition can be determined from the chemical formula.

Example #1: What is the percent composition of C3H8?

Answer: Step 1: Determine the molar mass of C3H8.

+ 3(12.01) g/mole)8(1.01) g/mole

44.11g/mole

Step 2: Determine the molar masses for each type of element. For C: 3(12.01 g/mole) = 36.03 g/mole For H: 8(1.01 g/mole) = 8.08 g/mole Step 3: Determine the percent composition for each element.

%68.8110011.44

03.36% x

g

gC

% H = 100% - 81.68% = 18.32%

Example #2: What is the percent composition of HCN? Answer: Step 1: Determine the molar mass of HCN.

+ 12.01 g/mole1.01 g/mole

27.03g/mole14.01 g/mole

Step 2: Determine the percent compositions for each element.

%43.4410003.27

01.12% x

g

gC

%83.5110003.27

01.14% x

g

gN

% H = 100% - 51.83% - 44.43%=3.74%

Example: What is the percent water in the hydrate cobalt (II) chloride dehydrate (CoC12 ● 2H2O)? Answer: Step 1: Determine the molar mass of CoC12 ● 2H2O.

1(58.93 g/mole) 2(35.45 g/mole) 4(1.01 g/mole) 2(16.00 g/mole) 165.87 g/mole

+

Step 2: Determine the percent water.

100/87.165

/)00.16(2)01.1(4% 2 x

moleg

molegOH

= 21.72 % water

EMPIRICAL FORMULA

The simplest ratio of the atoms in a molecule.

Sometimes the molecular formula is the same as the empirical formula.

Example:

NaCl → NaCl

Molecular Empirical Formula Formula

Sometimes they are not:

C3H9 → CH3

Molecular Empirical Formula Formula

Example #1: A compound has a composition 90.10 g P 8.90 g H What is the empirical formula? Answer: 1) Convert the mass to moles.

PmolePg

PmolePg 909.2

97.30

110.90

HmoleHg

HmoleHg 81.8

01.1

190.8

2) Select the atoms with the least number of moles. In this case it is the 2.909 mole P

3) Divide each of the number of moles calculated in step #1 by the # of moles determined in step #2.

0.1

909.2

909.2:PFor

00.302.3

909.2

81.8:HFor

Giving the ratio of P : H = 1 : 3

Empirical Formula PH3

Example #2: A compound has a % composition 65.2% Sc 34.8% O What is the empirical formula? 1) Convert the percent to mass.

65.2% Sc → 65.2 g Sc

+ 34.8% O → + 34.8 g O

100.0% 100.0 g 2) Convert the mass to moles.

ScmoleScg

ScmoleScg 45.1

96.44

12.65

OmoleOg

OmoleOg 18.2

00.16

18.34

3) Select the atoms with the least number of moles. In this case it is the 1.45 mole Sc

4) Divide each of the number of moles calculated in step #2 by the # of moles determined in step #3.

0.145.1

45.1:ScFor

50.145.1

18.2:OFor

Giving the ratio of Sc : O = 1 : 1.5

Empirical Formula Sc1O1.5 Is this formula possible? Can you have 1.5 atoms???

NO! Solution: Multiply the subscripts in order to get whole numbers.

→Multiply by two.

Empirical Formula Sc2O3

Example #3: A compound is composed of 0.59 g H and 9.40 g O. It has been determined that the molar mass of the compound is 34.0 g/mole. 1. What is the empirical formula? 2. What is the molecular formula? Answer: 1. Empirical Formula

OmoleOg

OmoleOg 590.0

00.16

140.9

HmoleHg

HmoleHg 58.0

01.1

159.0

0.158.0

58.0:HFor

0.158.0

59.0:OFor

Giving the ratio of H : O = 1 : 1

Empirical Formula HO 2. Molecular Formula The molar mass of the empirical formula: H O (1.01 g/mole) + (16.00 g/mole) = 17.01 g/mole for HO If the empirical molar mass is doubled, then it will equal the molar mass of the molecular compound. Therefore, the molecular formula must be double the amount of atoms. Giving: H2O2