1. Unit 3: Stoichiometry
The Mole, Molar Mass, % Composition, Balancing Chemical Equations, Limiting Reagent, Percent Yield

2. Unit 3 Topics
3.1 Counting by Weighing 3.2 Atomic Masses 3.3 The Mole 3.4 Molar Mass 3.5 Learning to Solve Problems (Reading) 3.6 Percent Composition of Compounds 3.7 Determining the Formula of a Compound 3.8 Chemical Equations 3.9 Balancing Chemical Equations 3.10 Stoichiometric Calculations: Amounts of Reactants and Products 3.11 The Concept of Limiting Reagent

3. Chemical Stoichiometry
Stoichiometry – The study of quantities of materials consumed and produced in chemical reactions.
Section 3.1 – Counting by Weighing

4. Elements occur in nature as mixtures of isotopes. Carbon = 98
Elements occur in nature as mixtures of isotopes. Carbon = 98.89% 12C 1.11% 13C < 0.01% 14C
Section 3.2 – Atomic Mass

5. Schematic Diagram of a Mass Spectrometer
A mass spectrometer is used to compare the masses of atoms. Atoms are passed into a beam of high speed electrons, changing them into ions. An applied electric field accelerates these ions into a magnetic field. The ions are deflected based on their mass. The more massive ions are deflected the smallest amount.

6. Carbon 12 is defined as having a mass of exactly 12 atomic mass units (amu)
All other atoms were compared to this standard
Example
Mass 13C =
Mass 12C
Therefore, Mass 13C = ( ) (12 amu)
= amu

7. Average Atomic Mass (for Carbon)
Average atomic mass: The weighted average of the masses of the isotopes for that element considering their relative abundance.
= 98.89% of 12 amu % of amu
exact number
= (0.9889)(12 amu) + (0.0111)( amu)
=12.01 amu

8. Average Atomic Mass for Carbon
Even though natural carbon does not contain a single atom with mass 12.01, for stoichiometric purposes, we can consider carbon to be composed of only one type of atom with a mass of
This enables us to count atoms of natural carbon by weighing a sample of carbon.

9. Example
An element consists of 62.60% of an isotope with mass amu and 37.40% of an isotope with mass amu. Calculate the average atomic mass and identify the element.
( amu)(.6260) + ( amu)(.3740)
= amu, Rhenium

10. Example:
Gallium has two naturally occurring isotopes, 69Ga and 71Ga, with masses of amu and amu, respectively. Calculate the percent abundances of these isotopes of gallium.
69.72amu = ( amu)(1-x) + ( amu)(x)
x =
Ga-71 = 39.73%
Ga-69 = 60.27%

11. The Mole Section 3.3 – The Mole
The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.
1 mole of anything = x 1023 units of that thing (Avogadro’s number).
1 mole C = x 1023 C atoms = g C
Section 3.3 – The Mole

12. Example
Calculate the number of iron atoms in a 4.48 mole sample of iron mole x 6.02 x 1023 atoms = 1 mole 2.70×1024 Fe atoms

13. Example
A silicon chip has a mass of 5.68 mg. How many silicon atoms are present in the chip?
1 g .
1000mg
1 mole .
28.09 gSi
6.02x1023 atoms
1 mole
5.68 mg x x x =
1.22x1020 atoms Si

14. Molar Mass Section 3.4 – Molar Mass
Mass in grams of one mole of the substance: Examples: Molar Mass of N = g/mol Molar Mass of H2O = g/mol (2 × 1.01 g) g Molar Mass of Ba(NO3)2 = g/mol g + (2 × g) + (6 × g)
Section 3.4 – Molar Mass

15. Percent Composition (%)
Mass percent of an element:
mass % =
mass of element in compound
mass of compound
x100
Section 3.6 – Percent Composition of Compounds

16. Example
Calculate the mass percent for iron in iron(III) oxide, (Fe2O3):
2 (55.85g)
2 (55.85g) + 3 (16.00g)
Mass % Fe =
= 69.94% Fe

17. Section 3.7 – Determining Formula of a Compound
Formulas
Empirical formula = CH
Simplest whole-number ratio
Molecular formula = (empirical formula)n , [n = integer]
Molecular formula = C6H6 = (CH)6
Actual formula of the compound
Section 3.7 – Determining Formula of a Compound

18. Determination of Empirical and Molecular Formulas
If given grams of each element, go to step 2. If given percent of each element, assume 100.0g of the compound (change % into grams).
Convert g moles for each element using molar mass.
Divide each mole answer by the smallest mole answer.
If you do not have a whole number, multiply by the smallest whole # to get a whole # ratio.
These whole #s are the subscripts for each element.
To find the molecular formula, divide the molecular molar mass by the empirical molar mass. Then multiply each of the subscripts by that answer.

19. Example Calculation CH4
Determine the empirical formula of methane given that 6.00g of methane can be decomposed into 4.49g of carbon and 1.51g of hydrogen.
1 mol C
12.01gC
.374
4.49gC
1.51gH x = = 1
.374
1 mol H
1.01 gH
1.50 x = =
4.01 ≈ 4
.374
CH4

20. Example P2O5 Empirical P4O10 Molecular 1 mol P 30.97gP 1.409 43.64gP
A white powder is analyzed and found to contain 43.64% phosphorous and 56.36% oxygen by mass. The compound has a molar mass of g/mol. Determine the compound’s empirical and molecular formula.
1 mol P
30.97gP
1.409
43.64gP
56.36gO x = = 1
(2) = 2
1.409
P2O5
Empirical
1 mol O
16.00gO
3.523 x = =
2.500
(2) = 5
1.409
2P + 5O = 2(30.97) + 5(16.00) = g
283.88g
141.94g
P4O10
Molecular = 2

21. Section 3.8 – Chemical Equations
A representation of a chemical reaction:
C2H5OH + 3O2  2CO H2O
reactants products
Reactants are only placed on the left side of the arrow, products are only placed on the right side of the arrow.
Section 3.8 – Chemical Equations

22. Section 3.9 – Balancing Chemical Equations
C2H5OH + 3O2  2CO H2O
The equation is balanced.
All atoms in the reactants are accounted for in the products.
1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.
Section 3.9 – Balancing Chemical Equations

23. Animation Link: Balancing an Equation
The balanced equation represents an overall ratio of reactants and products, not what actually “happens” during a reaction.
Use the coefficients in the balanced equation to decide the amount of each reactant that is used, and the amount of each product that is formed.

24. Notice
The number of atoms of each type of element must be the same on both sides of a balanced equation.
Subscripts must not be changed to balance an equation.
A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction.
Coefficients can be fractions, although they are usually given as lowest integer multiples.

25. Concept Check
Which of the following are true concerning balanced chemical equations? There may be more than one true statement. I. The number of molecules is conserved. II. The coefficients tell you how much of each substance you have. III. Atoms are neither created nor destroyed. IV. The coefficients indicate the mass ratios of the substances used. V. The sum of the coefficients on the reactant side equals the sum of the coefficients on the product side.

26. Stoichiometric Calculations
Chemical equations can be used to relate the masses of reacting chemicals.
Section 3.10 – Stoichiometric Calculations

27. Calculating Masses of Reactants and Products in Reactions
1. Balance the equation for the reaction. 2. Convert the known mass of the reactant or product to moles of that substance. 3. Use the balanced equation to set up the appropriate mole ratios. 4. Use the appropriate mole ratios to calculate the number of moles of desired reactant or product. 5. Convert from moles back to grams if required by the problem.

28. Example
Consider the following reaction: P4 (s) + 5 O2 (g)  2 P2O5 (s) If 6.25 g of phosphorus is burned, what mass of oxygen does it combine with?
6.25 gP4 1
1 mol P4
123.88gP4
5 mol O2
1 mol P4
32.00 gO2
1 mol O2 x x x
= 8.07 gO2

29. Section 3.11 – Limiting Reagents
Limiting Reactants
Limiting reactant – the reactant that is consumed first and therefore limits the amounts of products that can be formed.
Excess reactant – the reactant that is left over at the end of the reaction.
Determine which reactant is limiting to calculate correctly the amounts of products that will be formed.
Section 3.11 – Limiting Reagents

30. Example
Consider the reaction of 80.0g of copper with 20.0 g of oxygen according to the following equation. What is the limiting reactant? 4 Cu + O2 → 2 Cu2O LR
80.0g Cu
x 1 mole Cu
63.55g Cu
= 1.26 mole Cu
= 0.315
4 mol Cu
20.0g O2
x 1 mole O2
32.00g O2
= mole O2
= 0.625
1 mol O2

31. Example
If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant and what mass of copper can be produced? The unbalanced equation is: NH3 + CuO → N2 + Cu + H2O 2 3 3 3
18.1g NH3
x 1 mole NH3
17.04g NH3
= 1.06 mole NH3
= 0.531
2 mol NH3
90.4g CuO
x 1 mole CuO
79.55g CuO
= 1.14 mole CuO
= 0.379 LR
3 mol CuO
Now use the LR to find the mass of Cu:
90.4g CuO
x 1 mole CuO
79.55g CuO
x 3 mole Cu
3 mole CuO
x 63.55g Cu
1 mole Cu
= 72.4 g Cu

32. To Calculate how much ER is left over:
How much excess reactant remains after the reaction is complete? 2 NH3 + 3 CuO → N2 + 3 Cu + 3 H2O
Use the LR to determine how much of the ER reacts with it. Subtract this from the original amount of the ER.
90.4g CuO
x 1 mole CuO
79.55g CuO
x 2 mole NH3
3 mole CuO
x 17.04g NH3
1 mole NH3
= 12.9 g NH3 reacts
18.1g -12.9g = 5.2 g unreacted

33. Percent Yield Actual yield x 100 % yield = Theoretical yield
An important indicator of the efficiency of a particular laboratory or industrial reaction.
Theoretical yield is the amount of product formed when the LR is completely consumed according to stoichiometric calculations.
Actual yield is the amount that is actually produced in a laboratory setting.
% yield =
Actual yield
Theoretical yield
x 100

34. Example
Consider the following reaction: P4(s) + 6F2(g)  4PF3(g) What mass of P4 is needed to produce 85.0 g of PF3 if the reaction has a 64.9% yield?
85.0 gPF3 1
1 mol PF3
87.97gPF3
1 mol P4
4 mol PF3
gP4
1 mol P4 x x x =
29.9 gP4
64.9% = (85.0 g PF3 / x)(100); x = g PF3
( g PF3)(1 mol PF3 / g PF3)(1 mol P4 / 4 mol PF3)( g P4 / 1 mol P4) = 46.1 g of P4 needed
x = 46.1 gP4
29.9 gP4 = 0.649 x