Author: Bryan G
11/10/14
Chapter 11 Chemistry Notes: Stoichiometry
1) Definition of Stoichiometry
a) The study of quantitative relationships between the amounts of reactants
used and the amounts of products formed by a chemical reaction
b) Based on the law of conservation of mass
i) States that mass is neither created nor destroyed, but rather changed
or converted
ii) The amount of matter present at the beginning of a reaction is the
same at the end of a reaction
2) Mole Ratios
a) Coefficients in a chemical reaction indicate the relationships between
moles of reactants and products
b) These relationships can help derive conversion factors called mole ratios
c) A mole ratio is the ratio between any two substances in a balanced
chemical equation
d) Example:
i) 2K(s) + Br2(l)  2KBr(s)
(1) 2 mol K/1 mol Br2
(2) 2 mol K/2 mol KBr
(3) 1 mol Br2/2 mol K
(4) 1 mol Br2/2 mol KBr
(5) 2 mol KBR/2 mol K
(6) 2 mol KBR/1 mol Br2
e) The number of mole ratios between n amount of substances can be
found by using the following formula
i) (n)(n-1)
3) Stoichiometric Mole to Mole Conversions
a) Finding the amount of moles of a substance when given moles
b) Example:
i) 2K(s) + 2H2O(l)  2KOH(aq) + H2(g)
(1) .0400 mol of potassium is used; how much hydrogen is produced?
(2) Use the following formula:
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
(a) 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑘𝑛𝑜𝑤𝑛 × 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑘𝑛𝑜𝑤𝑛 = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
(b) . 0400 𝑚𝑜𝑙 𝐾 ×
1 𝑚𝑜𝑙 𝐻2
2 𝑚𝑜𝑙 𝐾
= .0200 𝑚𝑜𝑙 𝐻2
4) Stoichiometric Mole to Mass Conversions
a) Finding the mass of a substance when given moles
b) Example:
i) 2Na(s) + Cl2(g)  2NaCl(s)
(1) There are 1.25 moles of Cl2; what is the mass of the NaCl produced?
Author: Bryan G
11/10/14
(2) Mole ratio:
(a) 2 mol NaCl/1 mol Cl2
(3) Use the following formula:
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
(a) 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑘𝑛𝑜𝑤𝑛 × 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑘𝑛𝑜𝑤𝑛 = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
(b) . 1.25 𝑚𝑜𝑙 𝐶𝑙2 ×
2 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙2
1 𝑚𝑜𝑙 𝐶𝑙2
= 2.50 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
(4) Convert to grams:
58.44 𝑔 𝑁𝑎𝐶𝑙
(a) 2.50 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 × 1 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 = 146 𝑔 𝑁𝑎𝐶𝑙
5) Stoichiometric Mass to Mass Conversions
a) Finding the mass of one substance when given the mass of another
substance
b) Example:
i) NH4NO3(s)  N2O(g) + 2H2O
(1) The mass of ammonium nitrate is 25.0 grams; what is the mass of
water when the chemical breaks down?
(2) Convert 25.0g NH4NO3 to moles
1 𝑚𝑜𝑙 𝑁𝐻4𝑁𝑂3
(a) 25.0 𝑔 𝑁𝐻4𝑁𝑂3 × 80.04 𝑔 𝑁𝐻4𝑁𝑂3 = 0.312 𝑚𝑜𝑙𝑒𝑠 𝑁𝐻4𝑁𝑂3
(3) Find mole ratio:
(a) 2 moll H2O/1 mol NH4NO3
(4) Use the following formula:
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
(a) 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑘𝑛𝑜𝑤𝑛 × 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑘𝑛𝑜𝑤𝑛 = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
(b) . 312 𝑚𝑜𝑙 𝑁𝐻4𝑁𝑂3 ×
2 𝑚𝑜𝑙 𝐻2𝑂
1 𝑚𝑜𝑙 𝑁𝐻4𝑁𝑂3
= .624 𝑚𝑜𝑙 𝐻2𝑂
(5) Convert to grams:
18.02 𝑔 𝐻2𝑂
(a) . 624 𝑚𝑜𝑙 𝐻2𝑂 × 1 𝑚𝑜𝑙 𝐻2𝑂 = 11.2 𝑔𝑟𝑎𝑚𝑠 𝐻2𝑂
6) Why Do Reactions Stop?
a) In nature and in the lab, reactants are seldom found in exact ratios.
Because of this, when a chemical reaction occurs, there is usually some
reactant left over (is in excess)
b) Limiting Reactants
i) Limits the extent of a reaction and determines how much product is
formed during the reaction
c) Excess Reactants
i) Reactants that aren’t used up (left behind) after the limiting reactant is
fully used up
d) Calculating the Amount of Product When There is a Limiting Factor
i) Example:
(1) S8(l) + 4Cl2(g)  4S2Cl2(l)
(a) 200 g of sulfur react with 100 g of chlorine; how much disulfur
dichloride is created?
Author: Bryan G
11/10/14
(b) Find the limiting factor
(i) Calculate the number of moles of reactants
1. 1.41 mol Cl2
2. .7797 mol S8
(ii) Use mole ratios
1. Find the real mole ratio of the balanced equation
a. 4 chlorines to 1 sulfur
2. Determine the actual mole ratio by using the number of
moles of reactants that you just calculated
a. 1.41 mol Cl2/.7797 mol S8 = 1.808 mol Cl2/1 mol S8
3. Compare the theoretical mole ratio with the actual
a. There is only 1.8 moles of Chlorine available for every 1
mole of Sulfur, but there need to be a 4:1 ratio. Since
there isn’t enough chlorine to satisfy the reaction, it
“limits” the reaction and is the limiting factor.
(c) Calculate the amount of product created by using the limiting
factor just found
(i) Use the following formula:
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
× 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑙𝑖𝑚𝑖𝑡𝑖𝑛𝑔 𝑓𝑎𝑐𝑡𝑜𝑟
4 𝑚𝑜𝑙 𝑆2𝐶𝑙2
135.0 𝑔 𝑆2𝐶𝑙2
×
= 190.4 𝑔 𝑆2𝐶𝑙2
4 𝑚𝑜𝑙 𝐶𝑙2
1 𝑚𝑜𝑙 𝑆2𝐶𝑙2
1. 𝑙𝑖𝑚𝑖𝑡𝑖𝑛𝑔 𝑓𝑎𝑐𝑡𝑜𝑟 ×
2. 1.410 𝑚𝑜𝑙 𝐶𝑙2 ×
(d) Calculate the amount of your excessive factor remaining
(i) Figure out the amount of moles that would be needed to
completely react with the limiting factor
1 𝑚𝑜𝑙 𝑆8
1. 1.41 𝑚𝑜𝑙 𝐶𝑙2 × 4 𝑚𝑜𝑙 𝐶𝑙2 = .3535 𝑚𝑜𝑙 𝑆8
(ii) Figure out the mass of the amount of moles needed to
completely react with the limiting factor
265.5 𝑔 𝑆8
1. . 3525 𝑚𝑜𝑙 𝑆8 × 1 𝑚𝑜𝑙 𝑆8 = 90.42 𝑔 𝑆8
(iii) Solve for the amount of excess remaining
1. 200.0𝑔 𝑆8 − 90.42𝑔 = 109.6 𝑔 𝑆8
7) Percent Yield
a) In a lot of Stoichiometric reactions, the calculated amount of reactants
and products rarely matches the theoretical yield—the maximum amount
of product that can be produced. The actual yield—the amount of
product that is produced in reality—hardly ever matches the theoretical
yield.
b) Percent yield can be calculated by using the following formula:
i) 𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑦𝑖𝑒𝑙𝑑 =
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
× 100
Author: Bryan G
11/10/14