1. CHAPTER 3: STOICHIOMETRY
AP CHEMISTRY

2. Chapter 3 Objective
In this chapter, we will look at the quantities of materials consumed and produced in chemical reaction. The study of these quantities is called chemical stoichiometry.

3. 3.1 Counting By Weighing
If we were to take the individual masses of five nails, each nail would probably have a small deviation in mass, but would be relatively the same, let’s say an average mass of 2.5g per nail. If someone needed 1,000 nails for a building project, it is easier to take the mass of 1,000 nails, or 2,500g, than to count out all those nails. Objects do not have to have identical masses to be counted by weighing; they behave as if they are all identical. This is very lucky for chemists.

4. Average Atomic Mass % Isotope A (mass of A) + % Isotope B (mass of B) + …= avg. atomic mass
Just like our nails deviated slightly, atoms of the same element may have different masses. We call these isotopes.
Average Atomic Mass- an average of each isotope of an element, based on their percent abundance
Ex. Find the average atomic mass of element “X” if 1.40% of the sample is and isotope with a mass of amu, % is an isotope with a mass of amu, 22.10% is an isotope with a mass of amu and 52.40% is an isotope with a mass of amu.
0.0140( ) ( ) ( ) ( ) =
amu (probably lead)

5. 3.2 Atomic Masses
So if scientists can’t see atoms with their eyes, how do they know all these isotopes exist? They use a mass spectrometer.
The mass spectrometer heats a sample of atoms into a gas, passes them through a beam of high- speed electrons, which knock off some of the atom’s electrons giving them positive charges, and then these positively charged atoms pass through an electric field. The atoms are deflected based on their size, and then detected on a computerized plate.

7. Figure 3.2 Neon Gas
Note that these spikes on the readout reflect how much of each isotope is present in a sample.

8. Figure 3.3
When a sample of natural copper is vaporized and injected into a mass spectrometer, the results are shown. Use the data to calculate average atomic mass of natural copper. The mass values are Cu-63 (62.93) and Cu-65 (64.94).

9. Figure 3.3 Mass Spectrum of Natural Copper

10. Calculating the Mass of an Element
Example. Use the mass spectrum shown on the left to calculate the average atomic mass of this element, then identify the element.
Mass Percent Abundance 50 11 18 19
0.50(90) (91) (92) (94) (96) = or 90
Its zirconium!

11. 3.3 The Mole
The mole is the chemist’s way of counting atoms. Remember that the mole (or Avogadro’s number) is equal to x atoms of any element, which is then equal to the average atomic mass on the periodic table.
The mole is standardized as the number of carbon atoms in 12.0g of pure carbon-12.
6.022 x 1023 amu = 1 g
The mass of one mole of an element is equal to its atomic mass in grams.

12. Taken by Paul Jung (AP ’09) at the Golden Gate Bridge in San Francisco

13. Ex. Cody found a gold nugget that had a mass of 1. 250 oz
Ex. Cody found a gold nugget that had a mass of oz. How many moles was this? How many atoms?
1.250 oz Au 1 lb g 1 mol Au 16 oz 1 lb g Au = moles moles Au x 1023 atoms 1 mol Au = x 1023 atoms

14. 3.4 Molar Mass mass in grams of one mole of a substance
may also be called molecular weight

15. Ex. Calculate the molecular mass of cisplantin, Pt(NH3)2Cl2.
Pt 1 x = N 2 x = H 6 x 1.01 = 6.06 Cl 2 x = g or amu
The compound cis-PtCl2(NH3)2 cisplatinum is a platinum-based chemotherapy drug used to treat various types of cancers, including sarcomas, some carcinomas (e.g. small cell lung cancer, and ovarian cancer), lymphomas and germ cell tumors.

16. Ex. How many grams are 3.25 moles of cisplantin?
3.25 mol cisplantin g
1 mol cisplantin
= 975g

17. Percent Composition
“mass percent” or “percent by mass”

18. Ex. Find the percent composition of all elements in cisplantin, Pt(NH3)2Cl2.
Pt x 100 = 65.01% N x 100 = 9.34% H 6.06 x 100 = 2.02% Cl x 100 = 23.63%

19. 3.7 Determining the Formula of a Compound
Empirical Formula Poem
Percent to mass,
Mass to moles,
Divide by smallest,
And round (or multiply) ‘till whole
Empirical formula- simplest whole number ratio of the various types of atoms in a compound
Molecular formula- the exact formula or (empirical formula)x

20. Ex. A sample of a compound contains 11. 66g of iron and 5
Ex. A sample of a compound contains 11.66g of iron and 5.01g of oxygen. What is the empirical formula of this compound?
11.66g Fe 1 mol Fe = mol Fe 55.85g Fe 5.01g O 1 mol O = mol O 16.00g O Fe:O : :1.5 or 2: Fe2O3

21. Ex. What is the empirical formula of hydrazine, which contains 87
Ex. What is the empirical formula of hydrazine, which contains 87.5% N & 12.5%H? What is the molecular formula if the gram formula mass is 32.07g?
87.5gN 1 mol N = mol N 14.01g N 12.5g H 1 mol H = mol H 1.01 g H N:H = : : 1.98 = 1 : NH2

22. Combustion Analysis

23. Suppose you isolate an acid from clover leaves and know that it contains only the elements C, H, and O. Heating 0.513g of the acid in oxygen produces 0.501g of CO2 and 0.103g of H2O. What is the empirical formula of the acid? Given that another experiment has shown that the molar mass of the acid is 90.04g/mol, what is its molecular formula?
C + O2  CO g CO2 1 mol CO2 1 mol C 12.01gC 44.01g CO2 1 mol CO2 1 mol C = g C in compound

24. 2H2 + O2  2H2O 0. 103g H2O 1 mol H2O 2 mol H 1. 01g H 18
2H2 + O2  2H2O 0.103g H2O 1 mol H2O 2 mol H 1.01g H 18.02g H2O 1 mol H2O 1 mol H = g H 0.513g cmpd - ( )= 0.364g O 0.364g O 1 mol O = mol O 16.00g O 0.137g C 1 mol C = mol C 12.01g C g H 1 mol H = mol H 1.01 g H

25. C:H:O : : : : : 1: :1:2 CHO2 efm = = /45.02 = 2 C2H2O4

26. 3.8 Chemical Equations
During a chemical reaction, atoms are rearranged when existing bonds are broken and reformed in a different arrangement.
Since atoms are not lost or gained during a reaction, a balanced equation must have the same number of atoms of each element on each side.
Symbols representing physical states: (s) (l) (g) (aq)

27. Observe the diagram to the right
Observe the diagram to the right. Balance the equation and then draw in the correct number of expected products.

28. 3.9 Balancing Chemical Equations
We balance equations by adding coefficients, never by changing formulas.
Most equations can be balanced by inspection. Some redox reactions require a different method.

29. Let’s Practice! Try these on your own!
Ex.
Al(s) Cl2 (g)  AlCl3(s)
2Al(s) Cl2 (g)  2AlCl3(s)
N2O5(s) H2O(l)  HNO3(l)
N2O5(s) H2O(l)  2 HNO3(l)
Pb(NO3)2(aq) + NaCl(aq)  PbCl2(s) + NaNO3(aq)
Pb(NO3)2(aq) + 2 NaCl(aq)PbCl2(s) +2 NaNO3(aq)

30. PH3(g) + O2(g)  H2O(g) + P2O5(s)
You should be able to write and balance an equation when given the reactants and products in words.
Ex. Phosphine, PH3(g) is combusted in air
to form gaseous water and solid diphosphorus pentoxide. Try this one!
PH3(g) + O2(g)  H2O(g) + P2O5(s)
2PH3(g) + 4O2(g)  3H2O(g) + P2O5(s)

31. NH3(g) + Na(l)  H2(g) + NaNH2(s) 2NH3(g) + 2Na(l)  H2(g) + 2NaNH2(s)
When ammonia gas is passed over hot liquid sodium metal, hydrogen is released and sodium amide, NaNH2, is formed as a solid product.
NH3(g) + Na(l)  H2(g) + NaNH2(s) 2NH3(g) + 2Na(l)  H2(g) + 2NaNH2(s)

32. 3.10 Stoichiometric Calculations
Although coefficients in balanced chemical equations tell us how many atoms will react to form products, they do not tell use the actual masses of reactants and products we will use or expect to produce. For this, we need to relate the reactants and products in terms of their mole ratios.
The mole ratio = moles required/moles given.

33. Ex. What mass of NH3 is formed when 5
Ex. What mass of NH3 is formed when 5.38g of Li3N reacts with water according to the equation: Li3N(s) + 3H2O  3LiOH(s) + NH3(g)?
5.38g Li3N 1 mol Li3N 1 mol NH g NH3
34.83g Li3N 1 mol Li3N 1 mol NH3
= 2.63g NH3

34. 3.11 The Concept of Limiting Reagent
Limiting reagent- reagent that restricts or determines the amount of product that can be formed
If in a stoichiometry problem you are given amounts of two or more reactants and asked to determine how much product forms, you must determine which reactant is limiting and use it to work the problem.

35. Ex. How many moles of Fe(OH)3(s) can be produced by allowing 1
Ex. How many moles of Fe(OH)3(s) can be produced by allowing 1.0 mol Fe2S3, 2.0 mol H2O and 3.0 mol O2 to react? 2Fe2S3(s) + 6H2O(l) + 3O2(g)  4Fe(OH)3(s) +6S(s)
1.0 mol 2.0 mol 3.0 mol limiting
1.0 mol Fe2S3 4 mol Fe(OH)3 = 2.0 mol Fe(OH)3
2 mol Fe2S3
2.0 mol H2O 4 mol Fe(OH)3 = 1.3 mol Fe(OH)3
6 mol H2O
3.0 mol O mol Fe(OH)3 = 4.0 mol Fe(OH)3
3 mol O2
In this case, we can see that H2O gave use the smallest amount of product, so it is the limiting reagent. Thus, we use that answer as our overall answer.

36. Ex. If 17. 0g of NH3(g) were reacted with 32
Ex. If 17.0g of NH3(g) were reacted with 32.0g of oxygen in the following reaction, how many grams of NO(g) would be formed? NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)
17.0g g Xg
17.0g NH3 1 mol NH3 = 1 mol NH3
17.0g NH3
32.0 g O mol O2 = 1 mol O2
32.0g O2
We need 5 mol O2 to every 4 mol NH3, so O2 is limiting.
1 mol O2 4 mol NO g NO = 24.0g NO
5 mol O mol NO

37. Once we know how much of a product if formed in a reaction, we can find the percent yield of the reaction.
Theoretical yield- amount of product that should form according to stoichiometric calculations
Percent yield = Actual yield x 100
Theoretical yield
Actual yield- experimental yield  

38. Ex. In the reaction of 1. 00 mol of CH4 with an excess of Cl2, 83
Ex. In the reaction of 1.00 mol of CH4 with an excess of Cl2, 83.5g of CCl4 is obtained. What is the theoretical yield, actual yield and % yield?
Actual yield = 83.5g CCl4 CH4 + 2Cl2  CCl4 + 2H mol CH4 1 mol CCl g CCl4 1 mol CH4 1 mol CCl4 = 154g CCl4 (theoretical yield) 83.5 x 100 = 54.2% yield 154