1. Chemistry XXI Unit 1 How do we distinguish substances? M1. Searching for Differences Identifying differences that allow us to separate components. M2. Modeling Matter Using 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.

2. Chemistry XXI 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.

3. Chemistry XXI The Challenge Analysis What is this? According to our models, differences between substances are the result of differences in the composition and structure of their particles. N2N2 O2O2 H2OH2O CO 2 C 4 H 10 O3O3 How can we determine what atoms and how many of each type comprise the molecules of any given substance ? C H O N Color Code

4. Chemistry XXI Finding 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).

5. Chemistry XXI Electrical 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.

6. Chemistry XXI 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(n 0 ) ~ 1800 m(e - )

7. Chemistry XXI Mass 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!

8. Chemistry XXI Atomic 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 + + # n 0

9. Chemistry XXI Isotopes 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 20 Ne 90.48% 21 Ne 0.27% 22 Ne 9.25% Average relative atomic mass m r (Ne) =19.99 x 0.9048 + 20.99 x 0.0027 + 21.99 x 0.0925 m r (Ne) = 20.18 amu

10. Chemistry XXI From Atoms to Molecules Mass 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. Methane CH 4 CH 4 + CH 3 + CH 2 + CH + C+C+ ?

11. Chemistry XXI From Atoms to Molecules This is the mass spectrum for methanol (CH 4 O): How do you explain the different peaks? Let′s think!

12. Chemistry XXI This is the mass spectrum of an elemental gaseous substance (X 2 ) 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?

13. Chemistry XXI 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

14. Chemistry XXI 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?

15. Chemistry XXI 1.Take 100.0 g of your sample and calculate how much of each element you have: 92.26 g C7.74 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 C 7.74 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): C 7.68 H 7.68  CH Empirical Formula Finding the Formula 92.26% C and 7.74% H M(X) = 78.1 g/mol

16. Chemistry XXI 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 C6H6C6H6 92.26% C and 7.74% H. M(X) = 78.1 g/mol Benzene

17. Chemistry XXI 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.

18. Chemistry XXI Assess what you know Let′s apply!

19. Chemistry XXI Let′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

20. Chemistry XXI Gas Chromatography Let′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?

21. Chemistry XXI After the components are separated, their masses are analyzed by mass spectrometry. Mass Spectrometry Let′s apply!

22. Chemistry XXI 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 Air Let′s apply! 30.4% N 69.6% O 50.0% O 50.0% S Find the empirical and molecular formulas of these two substances.

23. Chemistry XXI If you had to describe what this module was about, what would you say?

24. Chemistry XXI 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.

25. Chemistry XXI 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 C 5 H 3 Cl 4 Molecular Formula C 10 H 6 Cl 8 Chlordane M(X) = 410. g/mol 29.31% C, 1.476% H, 69.22% Cl.

26. Chemistry XXI Are You Ready?

27. Chemistry XXI 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 1 How do we distinguish substances? Can you use measurements and models to separate and identify the different substances present in a system?

28. Chemistry XXI 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 (O 2 ) and generates enough heat to bring the mixture to the boiling point and vaporize about a fifth of it.

29. Chemistry XXI Analysis of the reaction mixture (an homogeneous solution) has revealed the presence of two major substances. One of them is hydrogen peroxide, H 2 O 2, which is a liquid (T b = 150.2 o C) fully miscible in water. H 2 O 2 easily decomposes into H 2 O and O 2 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?

30. Chemistry XXI 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 o C)? What are the normal melting and boiling points of this substance?

31. Chemistry XXI 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

32. Chemistry XXI Let′s think! The separation process yields 0.028 g of C 6 H 6 O 2 and 0.25 g of H 2 O 2 per mL of aqueous solution. If the density of water (H 2 O) is close to 1.0 g/mL, how many moles of water are there per mole of C 6 H 6 O 2 and H 2 O 2 in the solution?

33. Chemistry XXI Let′s think! Use all of the information provided to build a microscopic representation of the reacting solution inside the bombardier beetle.

34. Chemistry XXI Let′s think! Use the particulate model of matter to explain how the defense mechanism of the bombardier beetle works. The reaction between C 6 H 6 O 2 and H 2 O 2 in a closed chamber inside the beetle releases a gas (O 2 ) and generates enough heat to bring the mixture to the boiling point and vaporize about a fifth of it.