plan, ppt10(ps4) 1.finish ef (tro, 3.9) from % masses from combustion analysis data 2.chemical...

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Plan, Ppt10(PS4) 1. Finish EF (Tro, 3.9) From % masses From combustion analysis data 2. Chemical Reaction (Tro, 3.10) Process (takes time!); original subs new subs Fixed mass ratio of substances react and form Whole-number ratio of FUs react and form 3. Chemical Equation (Balanced) [later] Coefficients represent ratio of FUs when reaction occurs 4. Various problem types (“stoichiometry”) [later] (Tro, 4.2) 1 Ppt10(PS4)

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Page 1: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

Plan, Ppt10(PS4)

1. Finish EF (Tro, 3.9) From % masses From combustion analysis data

2. Chemical Reaction (Tro, 3.10) Process (takes time!); original subs new subs Fixed mass ratio of substances react and form Whole-number ratio of FUs react and form

3. Chemical Equation (Balanced) [later] Coefficients represent ratio of FUs when reaction occurs

4. Various problem types (“stoichiometry”) [later] (Tro, 4.2)

1Ppt10(PS4)

Page 2: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

Empirical Formula (EF) Quick Review

• Need to find the subscripts=> lowest whole-number ratio of atoms=> lowest whole-number ratio of MOLES (of

atoms)

2Ppt10(PS4)

Page 3: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

Determining the Empirical Formula of a Compound

• From mass data (of each element A, B, C, …) of a sample:masses (of A, B, C, …) moles (of A, B, C, … atoms) ratio of moles

reduced (whole-number) ratio of moles formula (subscripts)

• From % mass data (of each element A, B, C, …)

– Get mass data, and do as above!– Assume 100 g of compound for convenience

• 20.4 % of 100 g = 20.4 g! (etc.)

NOTE: If molar mass is also given, it may be easiest to use that mass (rather than 100 g). You will then get the molecular formula first (which can then be reduced, if needed)

3Ppt10(PS4)

Page 4: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

From Green Handout Sheet (PS4 & 5 Practice)

• What is the empirical formula of tetraethyllead, the “lead” in gasoline? The composition is: 29.71% C, 64.06% Pb, 6.23% H (These percentages are by mass)

4Ppt10(PS4)

Page 5: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

Ppt10(PS4) 5

Page 6: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

PS Sign-Posting

• The concepts and skills related toproblems 14, 18, and 19 on PS4 have been covered now in this PowerPoint. Give those problems a try now!

6Ppt10(PS4)

Page 7: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

From Combustion Analysis Data

• Often cannot easily get the elements out• So, react with oxygen:

– All C atoms end up in CO2 molecules!

– All H atoms end up in H2O molecules!

• Figure out C in sample from mass of CO2

• Figure out H in sample from mass of H2O

• Figure out a 3rd atom type (O, N, S, …) by mass difference (as in Exp 4—get Cl “by difference”)

7Ppt10(PS4)

Page 8: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

Combustion Analysis Figure 3.10

e.g., Mg(ClO4)2 e.g., NaOH

e.g., excess O2, sulfur oxides, nitrogen oxides, halogen-oxides , etc.

8Ppt10(PS4)

Page 9: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

From Green Handout Sheet (PS4 & 5 Practice)

• A hydrocarbon of unknown formula CxHy was submitted to combustion analysis with the following results. What is the empirical formula of the hydrocarbon?

9Ppt10(PS4)

Page 10: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

From Green Handout Sheet (PS4 & 5 Practice)

• Coniine, a toxic substance isolated form poison hemlock, contains only carbon, hydrogen, and nitrogen. Combustion analysis of a 5.024 g sample yields 13.90 g of CO2 and 6.048 g of H2O. What is the empirical formula of coniine?

10Ppt10(PS4)

Page 11: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

PS Sign-Posting

• The concepts and skills related toproblem 17 on PS4 have been covered now in this PowerPoint. Give those problems a try now!

[end of PS4 material]

11Ppt10(PS4)

Page 12: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

Discussion of Meaning of a Balanced Chemical Equation

• Reaction vs. equation (different!)– Most chemistry texts tend to discuss this initially but then

use the words interchangeably later on(!) Be careful!

Chemical Reaction◦ Initial substances (reactants)

turn into new substances (products)

◦ Takes time (a process)◦ Involves actual amounts

• Can be expressed in a variety of ways: grams, moles, L, etc.

Chemical Equation◦ Represents (a type of) chemical

reaction (but not “literally”)

◦ Symbols (not “reality”), on paper• Formulas, coefficients, an

arrow

◦ Does not involve actual amounts! Only RATIO of combination and formation

12Ppt10(PS4)

Page 13: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

Discussion of Meaning of a Balanced Chemical Equation (Example)

Chemical Reaction4 g of methane react with 16

grams of oxygen to produce 11 grams of carbon dioxide and 9 grams of water, all as gases.

Chemical Equation

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

The reaction occurs over some period of time, during which the number of molecules of methane and oxygen steadily decrease as the number of molecules of carbon dioxide and water increase.

• Coefficients represent a ratio only, and they represent the ratio of formula units, most fundamentally.

• Here, a 1 : 2 : 1 : 2 ratio of molecules of CH4, O2, CO2, and H2O but only as reaction occurs.

13Ppt10(PS4)

Page 14: Plan, Ppt10(PS4) 1.Finish EF (Tro, 3.9)  From % masses  From combustion analysis data 2.Chemical Reaction (Tro, 3.10)  Process (takes time!); original

Discussion of Meaning of a Balanced Chemical Equation

• Analogy (hypothetical): Working at Lawson products25 boxes packed up → $15 in pocket

• How much money is in my pocket after the “reaction” occurs?

• The equation alone can’t answer that!

14Ppt10(PS4)

– How much “reaction” occurred? (not indicated by equation!)

– How much was in your pocket before the reaction?