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IUnit 1

How do we distinguish substances?

M1. Searching for Differences Identifying differences that allow us to separate components.

M2. Modeling MatterUsing the particulate model of matter to explain differences.

M3. Comparing Masses Characterizing differences in particle’s mass and number.

M4. Determining Composition Characterizing differences in particle’s composition.

The central goal of this unit is to help you understand and apply basic ideas that can be used to distinguish

the different substances present in a system.

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Unit 1

How do we distinguish substances?

Module 4: Determining Composition

Central goal: To use experimental data to determine the atomic

composition of the particles that make up a

substance.

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The Challenge AnalysisWhat is this?

According to our models, differences between substances are the result of differences in the composition and structure of their particles.

N2

O2

H2O

CO2

C4H10

O3

How can we determine what atoms and how

many of each type comprise the

molecules of any given substance ?

C H O NColor Code

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IFinding the Mass

One first step could be determining the relative mass of the particles. For this purpose, we could

use a Mass Spectrometer.

1. Particles are vaporized and broken into ions (charged particles).

2. The ions are accelerated and separated using a magnetic field.

3. The ratio mass/charge is recorded (most ions have a 1+ charge).

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IElectrical Nature

Mass spectrometry is based on our knowledge and understanding of the

electrical nature of matter.

When atoms or molecules lose or gain electrons they become electrically charged. The charged particles are called ions.

We model

atoms as made by + and –

charges.

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Each type of atoms is characterized by the number of protons in the

nucleus (atomic number, Z).

In a neutral atom, the number of protons is equal to the

number of electrons.

q(e-) = q(p+)

Atomic Number

Most of the atomic mass is concentrated in the nucleus

m(p+) ~ m(n0) ~ 1800 m(e-)

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IMass Spectrometry

So, imagine we take a sample of a gas, like Neon (Ne), and we analyze it with a

mass spectrometer:

rel. int

m/z

90.48%

0.27%9.25%

Why do we get 3 peaks instead of one?

Let′s think!

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IAtomic Differences

All atoms of the same element have the same number of

protons in the nucleus, Z, but they may differ in the number

of neutrons.

To differentiate them we can use a quantity called

Mass Number A:

A = # p+ + # n0

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IIsotopes

We call isotopes to these atoms of an element with the same number of protons,

but a different number of neutrons

rel. int

m/z

20Ne90.48%

21Ne 0.27%

22Ne 9.25%

Average relative atomic mass

mr(Ne) =19.99 x 0.9048 +

20.99 x 0.0027 +

21.99 x 0.0925

mr(Ne) = 20.18 amu

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From Atoms to MoleculesMass spectrometry can also be used to determine the

mass of molecules and analyze their structure.

However, in this case it is

important to realize that the molecules will

break apart into several

fragments.

MethaneCH4

CH4+

CH3+

CH2+

CH+

C+

?

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From Atoms to Molecules

This is the mass spectrum for methanol (CH4O):

How do you explain the

different peaks?

Let′s think!

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This is the mass spectrum of an elemental gaseous substance (X2) collected from a tanker truck spill.

Let′s think!

Which is this element?

How do you explain all of the different peaks

in the spectrum?

75.78%

24.22%

34.97 amu 36.97 amu

What is the average relative atomic mass of this element?

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Combustion Analysis

Knowing the relative mass of the particles, we need a method to determine particle composition

Particle composition: Type and number of atoms of each type per particle

Elemental Analysis

Through elemental analysis we can find the percentage of each

element present in the substance

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I This is the substance’s mass

spectrum.M(X) = 78.1 g/mol

Consider this data:

The analysis of the exhaust gases from motor traffic reveals the presence of a carcinogenic

substance. Elemental Analysis reveals that the substance contains

92.26% C and 7.74% H.

Collecting Data

What is this?

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1. Take 100.0 g of your sample and calculate how much of each element you have: 92.26 g C 7.74 g H

2. Calculate the number of moles of each type of atom in the sample:

92.26 g C x 1 mole C/12.01 g = 7.68 mol C7.74 g H x 1 mole H/1.008 g = 7.68 mol H

3. Write a preliminary formula and convert to the the smallest integer subscripts that have the same proportion (divide by the smallest subscript):

C7.68H7.68 CH Empirical Formula

Finding the Formula92.26% C and 7.74% H M(X) = 78.1 g/mol

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4. Compare the actual molar mass of the compound (mass spectrum) with the molar mass calculated using the empirical formula (13.018 g/mol).

Finding the Formula

78.1 /13.018 = 6.00

Molecular Formula = 6 x Empirical Formula

C6H6

92.26% C and 7.74% H. M(X) = 78.1 g/mol

Benzene

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M(X) = 410. g/mol

The analysis of indoor air in some Tucson homes treated for termites reveals the presence of a harmful substance that affects the nervous system. Elemental analysis provides the following percent composition:

29.31% C, 1.476% H, and 69.22% Cl.

Let′s think!

Find the empirical and

molecular formulas of this

compound.

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I Assess what you know

Let′s apply!

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ILet′s apply!

Environmental analysis of air samples is often performed using a combined analytical technique

called GC/MS (Gas Chromatography-Mass Spectrometry)

Other applications include drug detection,

explosives investigation, and identification of unknown samples

GC/MS

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Gas ChromatographyLet′s apply!

GC is used to separate the

components of the mixture.

What differentiating

characteristics is used to separate

substances?

Why some substances move faster than others

inside the column?

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After the components are separated, their masses are analyzed by mass spectrometry.

Mass SpectrometryLet′s apply!

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A sample of Tucson’s air is analyzed by

GC/MS, followed by elemental analysis. Among the various substances in the sample, two major

pollutants are found.

Tucson AirLet′s apply!

30.4% N 69.6% O

50.0% O 50.0% S

Find the empirical and molecular

formulas of these two substances.

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If you had to describe what this module was about,

what would you say?

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Determining Composition

Summary

We take advantage of the electrical nature of matter to explore its properties and composition.

Mass spectrometry, an analytical technique that exploits the electrical nature of matter, can be used to compare the relative masses of different atoms and differentiate one from another.

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We can use MS and elemental analysis to determine the number of atoms of each type that make up the

particles of a given chemical compound.

Determining Composition

Empirical Formula

C5H3Cl4

Molecular Formula

C10H6Cl8

Chlordane

M(X) = 410. g/mol

29.31% C, 1.476% H, 69.22% Cl.

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Are You Ready?

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The central goal of this unit was to help you understand and apply basic ideas that can be used to distinguish the different substances present in a

system.

Unit 1How do we distinguish substances?

Can you use measurements and models to separate

and identify the different substances present in a system?

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The Challenge

Bombardier beetles seem to store aqueous solutions of two

different substances.

When they are threatened, the two solutions are squirted

through two tubes, where they are mixed and undergo a violent

chemical reaction.

The reaction releases a gas (O2) and generates enough heat to bring the mixture to the boiling

point and vaporize about a fifth of it.

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Analysis of the reaction mixture (an homogeneous solution) has revealed the presence of two major substances. One of them is hydrogen peroxide,

H2O2, which is a liquid (Tb = 150.2 oC) fully miscible in water. H2O2 easily decomposes into H2O and O2

when heated up.

Let′s think!

How would you propose to separate the other component?

What additional information would help you better answer this question?

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Let′s think!The analysis of the phase behavior of the isolated substance produces the following phase diagram:

What is the stable phase of this

substance at room temperature (20 oC)?

What are the normal melting and boiling

points of this substance?

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Let′s think!What would you propose to do to determine the

chemical composition of the unknown substance?

Elemental analysis of the

unknown reveals the following composition:

65.45% C, 5.49% H, 29.06% O.

What is the molecular formula

and the molar mass?

M = 110.1 g/mol

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Let′s think!The separation process yields 0.028 g of C6H6O2 and 0.25 g of H2O2 per mL of aqueous solution.

If the density of water (H2O) is close to 1.0 g/mL,how many moles of water are there per mole

of C6H6O2 and H2O2 in the solution?

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Let′s think!

Use all of the information provided to build a microscopic representation of the reacting

solution inside the bombardier beetle.

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Let′s think!

Use the particulate model of matter to

explain how the defense

mechanism of the bombardier beetle

works.

The reaction between C6H6O2 and H2O2 in a closed chamber inside the beetle releases a gas (O2) and generates enough heat to bring the mixture to the

boiling point and vaporize about a fifth of it.