1. Chapter 3 Stoichiometric
Atomic Masses, Mole concept, and Molar Mass (Average atomic mass).
Number of atoms per amount of element.
Percent composition and Empirical formula of molecules.
Chemical equations, Balancing equations, and Stoichiometric calculations including limiting reagents.

2. Chemical Stoichiometry
Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.
Counting by weight – when masses of substance to be counted are same.
Lets assume 1 Jelly Bean is exactly 5 g, so 1000 beans will be how many g?

3. Measuring Atomic Mass
Figure 3.1: (left) A scientist injecting a sample into a mass spectrometer. (right) Schematic diagram of a mass spectrometer.

4. Atomic mass is the mass of an atom in atomic mass units (amu)
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
Mass 13C/Mass 12C =
Based on the mass spect ratios
Mass of 13C = x 12 amu
= amu
On this scale
1H = amu
16O = amu

5. Atomic Masses Elements occur in nature as mixtures of isotopes
Carbon = 98.89% 12C
1.11% 13C
<0.01% 14C
Carbon, average atomic mass = amu
None of the C atoms is but it is an average of all three.

6. Spectrum
Most Abundant Isotope

7. Average atomic mass of lithium will be:
Natural lithium has two isotopes:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium will be:
7.42 x x 7.016
100
6.941 amu

8. The element europium exists in two natural isotopes:
151Eu has a mass of amu, and 153Eu has a mass of amu. The average atomic mass of the europium is amu. Calculate the relative abundance of the two europium isotopes.

9. The Mole
The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.
1 mole of anything =  1023 units of
that thing

10. Avogadro’s number equals 6.022  1023 units

11. CO2 = (12.01+16+16) = 44.01 grams per mole
Molar Mass
A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.
C= O=16
CO2 = ( ) = grams per mole

12. atomic mass (amu) = molar mass (grams)
eggs
cars
Molar mass is the mass of 1 mole of in grams
marbles
atoms
1 mole 12C atoms = x 1023 atoms = g
1 12C atom = amu
1 mole 12C atoms = g 12C
1 mole lithium atoms = g of Li
For any element
atomic mass (amu) = molar mass (grams)

13. 1 amu = 1.66 x g or 1 g = x 1023 amu

14. Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = x 1023 atoms H
1 mol C3H8O
60 g C3H8O x
8 mol H atoms
1 mol C3H8O x
6.022 x 1023 H atoms
1 mol H atoms x =
72.5 g C3H8O
5.82 x 1024 H atoms

15. Percent composition of an element in a compound =
n x molar mass of element
molar mass of compound
x 100%
n is the number of moles of the element in 1 mole of the compound. For example in C2H6O
%C =
2 x (12.01 g)
46.07 g
x 100% = 52.14%
C2H6O
%H =
6 x (1.008 g)
46.07 g
x 100% = 13.13%
%O =
1 x (16.00 g)
46.07 g
x 100% = 34.73%
52.14% % % = 100.0%

16. Determining Elemental Composition
(Formula)
Figure 3.5: A schematic diagram of the combustion device used to analyze substances for carbon and hydrogen.

17. The masses obtained (mostly CO2 and H2O and sometimes N2) will be used to determine:
% composition in compound
Empirical formula
Chemical or molecular formula if the molar mass of the compound is known or given.

18. Example of Combustion Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
What is the Empirical Formula of Ethanol??

19. = 6g C 16 x 2 12 1mol CO2 44g CO2 1mol C 12g C 22gCO2 mc % C in CO2
Collect 22.0 g CO2 and 13.5 g H2O
16 x 2 12
1mol CO2
44g CO2
1mol C
12g C
22gCO2 mc
% C in CO2
collected
22gCO2 x 12gC
= 6g C
mc =
44gCO2

20. Convert g to mole:
6g x 1molC
nc =
12 gC
1mol C 12g C
nmol C 6g
= 0.5 mol
Repeat the same for H from H2O
1mol H2O
18g H2O
2 mol H
2g H
13.5g H2O
nmol H mH
2x13.5
nmol H = = 1.5 mol H
18 mol H
Faster H but
still need O

21. 2x13.5
mH = = 1.5 g H 18
mO = 11.5g – mC – mH
= 11.5 – 6 – 1.5 = 4g
m 4
nO = = = 0.25 mol O
MM 16
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O

22. OR for C mC 12.01 1. Fraction of C in CO2 = = = MMCO2 44.01
2. Mass of C in compound = mass of CO2 x Fraction of C mC
3. % C in compound = x 100
msample
4. If N then % N = % C - % H

23. Empirical Formula
Caffeine contains 49.48% C, 5.15% H, 28.87% N and O by mass. Determine the empirical formula.
Solution:
Always assume Total mass of the sample is 100 g.
Convert mass to moles.
Divide all mole values by the lowest value of mole
Determine empirical formula.
C8H10N4O2

24. Molecular Formula
Caffeine contains 49.48% C, 5.15% H, 28.87% N and 16.49% O by mass and has a molar mass of g/mol. Determine the molecular formula.
Solution:
Convert mass to moles.
Determine empirical formula.
Determine actual formula.
C8H10N4O2

25. Then Empirical Formula
Using the previously calculated % in compound:
% in gram
a. Number of mole of C =
Atomic mass of C
b. Number of mole of H =
Atomic mass of H
a b c
Then divide by the smallest number:
smallest smallest smallest : :

26. NOTE When mole ratios are : 0.99 : 2.01 : 1.00
Then round-off ratios to the nearest whole numbers:
1 : 2 : 1
CH2N
When mole ratios are : 1.49 : :
Then you have to multiply all mole ratios by 2:
3 : 6 : 2
C3H6N2
Hence, empirical formula is the simplest formula of a compound
Mole ratios 2.25 must NOT be rounded to whole number
rather must be multiplied by 4 and ratios 2.33 & 2.72 by 3.
[See page 96 first left side note of the text book]

27. Formulas molecular formula = (empirical formula)n
[n = any integer = 2, 3,4…]
empirical formula = CH, if n = 6, then
molecular formula = (CH)6 = C6H6
Then

28. Examples of substances whose empirical and molecular formulas differ.
Figure 3.6:
Examples of substances whose empirical and molecular formulas differ.
molecular formula = (empirical formula)n, where n is a integer.
Integer
Molecular
Empirical

29. Chemical Equations
Chemical change involves a reorganization of the atoms in one or more substances.

30. Chemical Equation A representation of a chemical reaction:
C2H5OH + O2  CO H2O
reactants products
Unbalanced !

31. Chemical Equation C2H5OH + 3O2  2CO2 + 3H2O The equation is balanced.
1 mole of ethanol reacts with 3 moles of oxygen
to produce
2 moles of carbon dioxide and 3 moles of water

33. Writing & Balancing an Chemical Equation
Identify Reactants & Products and their Physical states
Write the Unbalanced Equation
Start Balancing Equation with most Complicated/
Difficult molecules
Determine what Coefficients are needed to keep same
number of atoms on both sides i.e. reactants & products
Do not change the formulas (subscript) of any reactants/products
(NH4)2Cr2O7(s) Cr2O3(s) + N2(g) + H2O(g)
Fe2S3(s) + HCl(g) FeCl3(s) + H2S(g)

34. Calculating Masses of Reactants and Products
1. Balance the equation.
2. Convert mass to moles.
3. Set up mole ratios.
4. Use mole ratios to calculate moles of desired reactant or product.
5. Convert moles to grams, if necessary.

35. Methanol burns in air according to the following equation
2CH3OH + 3O CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
moles H2O
grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
1 mol CH3OH
32.0 g CH3OH x
4 mol H2O
2 mol CH3OH x
18.0 g H2O
1 mol H2O x =
209 g CH3OH
235 g H2O

36. In the balanced Equations:
2CH3OH + 3O CO2 + 4H2O
2 mol CH3OH = 3 mol O2 = 2 mol CO2 = 4 mol H2O
2 mol
4 mol
2x( ) g
4x(2+16) g
209 g m
209 x 4(2+16)
m =
2( )

37. Limiting Reactant
The limiting reactant is the reactant that is completely consumed or finished first, controlling the amounts of products formed.
Example-1
Example-2

38. Limiting Reagents
6 red left over
6 green used up

39. Limiting Reactant 5 cars + 200 drivers Limiting cars or drivers?
50 chairs + 15 students Limiting chairs or students?

40. Solving a Stoichiometry Problem
1. Balance the equation.
2. Convert masses to moles.
3. Determine which reactant is limiting.
4. Use moles of limiting reactant and mole ratios to find moles of desired product.
5. Convert from moles to grams.

41. Limiting Reactant Calculations
What weight of molten iron is produced by 1 kg each of the reactants?
Fe2O3(s) Al(s)  Al2O3(s) + 2 Fe(l)
1 mol 2 mol
6.26mol 18.52
Ratio:
> 0.108
Limiting Excess
The 6.26 mol Fe2O3 will
Disappear first

42. Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted. Its amount is
calculated using the balanced equation.
Actual Yield is the amount of product actually obtained
from a reaction. It is always given.

43. Percent Yield
Actual yield = quantity of product actually obtained Theoretical yield = quantity of product predicted by stoichiometry using limiting reactant

44. Percent Yield Example
14.4 g
excess
Actual yield = 6.26 g

45. 3. For the reaction: (Molar Masses: Cr = 52.00; Cl = 35.45 g/mol)
2 Cr (s) Cl2 (g) ========> 2 CrCl3 (s)
(a) How many grams of CrCl3 are produced by 1.00 g of Cr ?
(b) How many grams of CrCl3 are produced by 2.00 g of Cl2?
(c) If the Actual yield of CrCl3 is 2.65 g, then find limiting reactant and Percent Yield of the CrCl3?

46. Sample Exercise
Titanium tetrachloride, TiCl4, can be made by combining titanium-containing ore (which is often impure TiO2) with carbon and chlorine - TiO2 (s) Cl2 (g) + C (s) TiCl4 (l) + CO2 (g) If one begins with 125 g each of Cl2 and C, but plenty of titanium-containing ore, which is the limiting reagent in the reaction? What quantity of TiCl4 can be produced?

47. Virtual Laboratory Project

48. Practice Example
A compound contains C, H, N. Combustion of 35.0 mg of the compound produces 33.5 mg CO2 and 41.1 mg H2O. What is the empirical formula of the compound?
Solution:
1. Determine C and H, the rest from 33.5 mg is N.
2. Determine moles from masses.
3. Divide by smallest number of moles.

49. Practice Example
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant? How many grams of N2 will be formed?

50. Practice Example
Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. Suppose 68.5Kg CO(g) is reacted with 8.60Kg H 2(g). Calculate the theoretical yield of methanol. If 3.57x104g CH3OH is actually produced, what is the percent yield of methanol?

51. SnO2(s) + 2 H2(g) ® Sn(s) + 2 H2O(l)
Practice Example
SnO2(s) + 2 H2(g) ® Sn(s) + 2 H2O(l)
a) the mass of tin produced from moles of hydrogen gas.
b) the number of moles of H2O produced from 339 grams of SnO2.
c) the mass of SnO2 required to produce 39.4 grams of tin.
d) the number of atoms of tin produced in the reaction of 3.00 grams of H2.
e) the mass of SnO2 required to produce 1.20 x 1021 molecules of water.