Unit 6 Chemical AnalysisChapter 8

Objectives39 Perform calculations using the mole to

calculate the molar mass 40 Perform calculations using the mole to

convert between grams, number of particles, volume, and moles

41 Perform advanced calculations using empirical formula, molecular formula, and percent composition

The MoleSociety prefers to work with simple

numbers.When a very large or very small

numbers are involved, society converts them to easier numbers.

For example:◦2 shoes = 1 pair◦12 eggs = 1 dozen◦10 years = 1 decade◦100 years = 1 century

The MoleScience operates under the same idea. As that atoms are so small, their

average atomic mass is very hard to work with in lab.

A scientist by the name of Amedeo Avogadro proposed an idea that a given volume of gas is proportional to the number of atoms or molecules regardless of the chemical identity of the gas assuming a constant temperature and pressure were held.

The MoleAvogadro’s idea inspired scientists to

look for the exact number of atoms or molecules in a given volume.

In the mid-1800s, scientists first determined the number.

Over time, the number was modified slightly to be more accurate and is now:

6.022141 x 1023

The MoleThe number was called Avogadro’s

constant. Because the number was so large,

the number was set equal to 1 mole and thus we can say:

1 mole = 6.02 x 1023 particles

Often Avogadro’s constant is written as 6.02 x 1023 instead of writing out all of the known numbers.

The MoleOne of the more useful aspects of

the mole is what it can do for masses.

Since Avogrado’s number is a proportionality factor molar mass to actual mass, it allows for conversions from amu’s to grams.

For instance,◦It is known that a proton has a mass of

1 atomic mass unit. ◦If there were 6.02 x 1023 protons, it

would have a mass of 1 gram.

Calculating molar massThe molar mass of a chemical looks at

how much mass it would have if there was 6.02 x 1023 units of that chemical.

For instance,◦Assume, we have 1 atom of sodium.◦Sodium has a mass of 22.99 amu’s◦If there were 6.02 x 1023 sodium atoms (or

1 mole) it’s mass would be 22.99 grams.◦We would say that sodium has a molar

mass of 22.99 grams/mole.

Calculating molar massThe molar mass of a compound or

molecule works the same way. Assume, we had carbon monoxide.

◦ If we had one molecule, there would be 1 carbon atom and 1 oxygen atom.

◦To detemine the mass of the molecule, we would add the mass of each atom (12.01 amu + 16.00 amu = 28.01 amu’s)

◦Now if we had 6.02 x 1023 molecules of carbon monoxide, we would have a mass of 28.01 grams.

◦Thus the molar mass of carbon monoxide is 28.01 grams/mole.

Practice

Molar Masses

Calculate the molar masses of the following:1. Mg

2. H2O

3. CaCl2

4. Mg(NO3)2

Molar Masses

Calculate the molar masses of the following:Mg 24.31 grams/mole

(1 Mg x 24.31g)

H2O 18.02 grams/mole

(2 H x 1.01 g) + (1 O x 16.00 g)

CaCl2 110.98 grams/ mole

(1 Ca x 40.08 g) + (2 Cl x 35.45 g)

Mg(NO3)2 148.33 grams/mole

(1 Mg x 24.31g) + (2 N x 14.01g) + (6 O x 16g)

Return

Using the moleIn science, it is rare to find exactly

one mole of a substance or to have the exact molar mass.

Therefore, it is necessary to be able to convert from molar mass to moles or to particles or to volume depending on what you are looking for.

In order to do this, we use dimensional analysis.

Dimensional AnalysisIn Unit 1, we learned a technique for

converting known as dimensional analysis.

It had the basic setup shown below:

The conversion factor to remember is:1 mole = 6.02 x 1023 particles = Molar Mass (g) = 22.4 L*Particles can stand for atoms, molecules, or formula units.

Conversion Factor

Using the moleTo see how dimensional analysis works, assume

we have 15 grams of water, and we would want to know how many moles this is.◦ We know the given is 15 grams of water◦ We can calculate the water has a molar mass of 18.02

grams/mole (our conversion factor)◦ Since we want moles, that unit will go on top.◦ We have grams so that unit will go on the bottom.◦ At this point, we can calculate.

❑ 𝑥❑=¿

15 g water

18.02 g water

1 mole water0.83 moles water

Using the moleAs long as you remember the

conversion factor and the setup for dimensional analysis, you should be able to convert.

Click the link to the left for additional practice.

Practice

Mole ConversionsConvert 2.5 moles to atoms:

Convert 24 grams CO2 to moles:

Convert 6.75 x 1022 molecules to moles:

Convert 48 liters to grams:

❑ 𝑥❑=¿

❑ 𝑥❑=¿

❑ 𝑥❑=¿

❑ 𝑥❑=¿

Mole ConversionsConvert 2.5 moles to atoms:

Convert 24 grams CO2 to moles:

Convert 6.75 x 1022 molecules to moles:

Convert 48 liters to grams of hydrogen gas:

2.5𝑚𝑜𝑙𝑒𝑠❑ 𝑥

6.02 𝑥1023𝑎𝑡𝑜𝑚𝑠1𝑚𝑜𝑙𝑒

=1.5 𝑥1024𝑎𝑡𝑜𝑚𝑠

24𝑔𝐶𝑂2❑ 𝑥

1𝑚𝑜𝑙𝑒𝐶𝑂 244.01𝑔𝐶𝑂 2

=0.55𝑚𝑜𝑙𝑒𝑠𝐶𝑂 2

6.75𝑥 1022𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠❑ 𝑥

1𝑚𝑜𝑙𝑒6.02𝑥 1023𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠

=0.112𝑚𝑜𝑙𝑒𝑠

4.3 g H2Return

Chemical AnalysisA large part of chemistry is

analyzing unknown materials. There are several analytical

instruments that help with this process that will be talked about later in the year.

We are going to take a look at how to use the data these instruments provide.

Percent CompositionIn Unit 4, we discussed the relative size

of atoms. When a compound is created, such as

KF, the size can help us visualize what percent is potassium and percent is fluorine.

In this case, it is clear that even though we have one of each atom, potassium makes up a larger percentage of this compound. K+ F-

Percent CompositionA more exact percentage can be

calculated using the molar mass of the compound.

KF would have a molar mass of 58.10 g/mol.

If we divide the mass of each element by the whole, we will get the percentage of each:

Percent Composition-Review

1. Calculate the molar mass of the compound.

2. Calculate the total mass of each element in the compound.

3. Divide the total mass of each element by the molar mass of the compound.

4. Multiply that answer by 100

Practice

Percent CompositionCalculate the % composition of

CaCl2:

Calculate the % composition of CaSO4:

Percent CompositionCalculate the % composition of

CaCl2:

Calculate the % composition of CaSO4:

Return

Empirical FormulaWhen doing a chemical analysis,

the percent composition is often what is provided.

The empirical formula is the lowest whole number ratio of the elements that make up the compound.

Empirical FormulaAssume we are given the following

analysis for a chemical:◦The chemical is 13.20% magnesium.◦The chemical is 86.80% bromine.

We want to know the empirical formula.Once we have the %’s, assume you

have exactly 100. grams for the analysis.◦By doing this, we can easily convert the %’s

into grams.◦If I have 13.20% of 100. grams, I have

13.20 grams.

Empirical FormulasSince we assumed we had 100.

grams, ◦We can say that we have 13.20g of Mg

and 86.80g of Br.Our goal is to find a ratio of Mg to

Br. Since this is looking for a number of each element, we can use the mole to determine this ratio.

Empirical FormulasNow we know how many moles of

each we have:◦0.5430 moles Mg◦1.086 moles Br

We can now look at the ratio between the two. ◦Since we want a whole number ratio, we

need to modify our numbers.◦To do this, divide each by the smaller

number. 0.5430/0.5430 = 1Mg 1.086/0.5430 = 2 Br

Therefore, our empirical formula

would be MgBr2.

Empirical Formulas-Review

1. Convert the %’s to grams.◦ Assume 100 grams

2. Convert grams into moles for each element.

3. Divide each part by the smallest number from step #2.

◦ Round to nearest whole number if close i.e.: 3.99 should be written as 4

4. The answers from step #3 are the subscripts for each element.Practice

Determine the empirical formulas

11.2% H and 88.8% O

36.48% Na, 25.44% S, and 38.08% O

Determine the empirical formulas11.2% H and 88.8% O

1. 11.2 g H and 88.8 g O2. 3. 11.1 / 5.55 = 2 H

5.55 / 5.55 = 1 O

H2O

Determine the empirical formulas36.48% Na, 25.44% S, and

38.08% O1. 36.48 g Na, 25.44 g S and 38.08 g

O2. Na 3. 1.587 / 0.7933 = 2 Na

0.7933 / 0.7933 = 1 S2.389 / 0.7933 = 3 O

Na2SO3

Return

Molecular FormulasCalculating the empirical formula will

always work if your compound is ionic.However, there are a few more steps to

determining covalent or organic formulas from the percent composition.

Here is the problem:◦If there was 14.4% H and 85.6% C, the

empirical formula would be CH2.

◦Unfortunately, C2H4, C3H6, C4H8,…. will all give a percent composition of 14.4% H and 85.6 % C and they are all known molecules.

Molecular FormulasOne other piece of information that

is obtained while doing a chemical analysis is the typically the mass of the unknown compound.

If we know the mass of the compound and the empirical formula, we will be able to determine the correct molecular formula.

Molecular FormulasFor the example from before, we knew that the

empirical formula was CH2.Assume, our analysis also provided the mass of the

unknown compound to be 42.09 grams. If we divide the mass of the unknown by the molar

mass of the empirical formula we will determine how many CH2’s make up this molecule.

Therefore, the compound is 3(CH2). Since we do not write formulas in that way, multiple the 3 by each subscript to get C3H6.

Molecular Formula-Review

1. Determine the empirical formula.2. Divide the mass of the unknown

by the molar mass of the empirical formula.

3. Multiple the subscripts on the empirical by the answer to step #2.

Practice

Determine the molecular formula

Empirical formula CH3

◦Mass of unknown compound: 60.16 grams

Empirical Formula SiH3Cl◦Mass of unknown compound: 199.70

grams

Determine the molecular formula

Empirical formula CH3

◦Mass of unknown compound: 60.16 grams

C4H12

Empirical Formula SiH3Cl◦Mass of unknown compound: 199.70

grams

Si3H9Cl3Return

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