STOICHIOMETRY (pg. 234)
The study of the relationship between quantities of substances (reactants and products) involved in chemical reaction
• We do this by comparing ratios between what’s given and what we need to find in balanced equations.
• The substance that you are given info about (GIVEN) and the substance about which you are finding info about (FIND).
We deal with proportions in our everyday life, not just with science problems:
Recipe for Salad:
Recipe for double cheeseburger
2 Patties + 2 slices of cheese + 1 bun à 1 double cheese burger (DCB)
Examine the proportions below:
-For every 1 DCB there are 2 patties used. Ratio = 2 patties : 1 DCB
-For every 1 DCB there is 1 bun used. Ratio = 1 bun : 1 DCB
-For every 1 DCB there are 2 slices of cheese used. Ratio = 2 Cheese : 1 DCB
-For every 2 slices of cheese used, there is 1 bun used. Ratio = 2 cheese : 1 bun
Questions:
a. How many buns were used to make 5 double cheeseburgers?
b. 1 dozen buns, 2 dozen patties, 2 dozen cheese slices would make how many burgers?
Chemical reactions behave in much the same way when we examine their balanced equations. We can observe molar proportions. 2Mg + O2 à 2MgO The coefficients tell you the proportions in which the reaction will always occur.
Mole information!!!: This equation tells you that for the production of 1 mole of magnesium oxide, 2 moles of Magnesium reacts with 1 mole of oxygen to produce 2 moles of magnesium oxide.
Particle information: This equation tells you that for the production of 1 molecule of magnesium oxide, 2 molecule of Magnesium reacts with 1 molecule of oxygen to produce 2 molecules of magnesium oxide.
e.x. N2(g) + 3H2(g) → 2NH3(g)
You can use a ratio to express the numbers of atoms in the equation:
1 molecule N2: 3 molecules H2: 2 molecules NH3
What happens if you multiply the ratio by 2?
2 molecules N2: 6 molecules H2: 4 molecules NH3
· Using a balanced chemical equation and information about one substance, you can calculate the amount of another substance by using a stoichiometric ratio (Given/Find). Use the coefficients of the balanced chemical equation as the starting ratio in order to solve for the other.
The coefficients in chemical equations show mole-mole or particle-particle relationships that exist between the reactants and products.
· Coefficients do not show mass-mass relationships, however, knowing the molar relationships between the substances involved allow us to calculate the mass relationships
• A balanced chemical equation indicates the number of moles of each REACTANT that will react and the number of moles of each PRODUCT that will be produced in the reaction
2H2(g) + O2(g) ----> 2H2O(g).
2 molecules of hydrogen react with 1 molecule of oxygen to form 2 molecules of water.
2 moles of hydrogen will react with 1 mole of oxygen to form 2 moles of water.
2 : 1 : 2
• Even if one reactant is present in excess only the amount need to react, dictated by the molar ratio, will
actually react.
• These coefficients do not tell us the ratio of mass-mass, meaning 2 grams of hydrogen do not react with one gram of oxygen to form 2 grams of water! That would not make sense!
Recall the reaction: 2Mg + O2 à 2MgO
2 : 1 : 2
For every 2 MgO produced, 1 O2 was used. For every 2 MgO produced, 2 Mg were used.
a. How many MgO would be produced if 2 O2 were used?
b. How many MgO would be produced if only 1 Mg was used?
c. If you have 4 MgO, how many Mg do you have?
d. If you have 4 MgO, how many O2 do you have?
Finding Molar ratio sample problems:
Use the following balanced equations to find the simplest molar ratios:
1) 2Mg + O2 à 2MgO
__ Mg:__MgO
__Mg:__O2
__ O2: __MgO
2) 2Na + Cl2 à 2NaCl
__ Na:__NaCl
__Na:__Cl2
__ Cl2: __NaCl
3) CH4 + 2O2 à CO2 + 2H2O
__CH4:__CO2
__CH4:__O2
__O2:__CO2
__O2:__H2O
Mole-Mole Calculations:
A balanced chemical equation gives you the ratios by moles of the reactants and products. Therefore, if you know the amount of one substance in moles, you can determine the amounts of all other substances from the balanced equation.
Example 1: Determine how many moles of oxygen are produced when 3.50 moles of water decompose.
Step 1: Write the balanced equation for the reaction 2H2O(l) à 2H2(g) + O2(g)
(decomposition of water is done by a process called electrolysis)
Step 2: Analyze the ratios
Given Required
2H2O(l) à 2H2(g) + O2(g)
2 mole : 1 mole
3.5 mole : x mole
Step 3: Calculate the amount of the unknown required in order to maintain the proportion determined in step 2.
We can do Step 3 a different way: Equating the ratios!
*Step 3: Calculate the amount of the unknown by equating the ratios of coefficients to actual amounts used.
Given Required Another strategy: Equate mole ratios
2H2O(l) à 2H2(g) + O2(g)
2 mole : 1 mole
3.5 mole : x mole
Mole to Mole Stoichiometry
1. Given the following reaction: Na + Cl2 à 2NaCl
a) How many moles of Cl2 will react with 6 moles of Na? (answer: 6 mol)
b) How many moles of NaCl will be produced if 12 moles of Cl2 react with an excess amount of Na?
(answer: 24 mol)
2. Given the following reaction: 2H2O2 à 2H2O + O2
a) How many moles of H2O2 must decompose in order to produce 8 moles of O2? (answer: 16 mol)
b) If 14 moles of H2O were produced, how many moles of O2 were produced? (answer: 7 mol)
3. What amount (in moles) of potassium phosphate, K3PO4 (aq), is required to react with 0.069 mol of silver nitrate, AgNO3 (aq), according to the equation: 3AgNO3(aq) + K3PO4(aq)àAg3PO4(s) + KNO3(aq)
(answer: 0.023 mol)
4. What amount (in moles) of copper is formed when 0.25 mol of iron reacts with copper (II) chloride, CuCl2 (aq), in this reaction? 2Fe(s) + 3CuCl2(aq)à2FeCl3(aq) + 3 Cu(s) (answer: 26.9 mol)
5. How many moles of ammonium sulfate (NH4)2SO4(aq), can be made when 35.0 mol of ammonia, NH3(g) reacts with sulfuric acid, H2SO4(aq), according to the unbalanced equation? (Hint: write the balanced chemical equation first.) NH3(g) + H2SO4 (aq) à (NH4)2SO4(aq) ( answer: 14.9 mol)
6. What amount (in moles) of potassium hydroxide is required to react with 0.0225 mol of sulfuric acid? (hint: write the balanced chemical equation first) (answer: 12.6 mol)
Mass to Mass Calculations
· You are given the mass of one substance and must determine the mass of another.
· You must convert the mass into moles/molecules in order to use mole/molecular ratios to determine the mole of the compound of unknown mass. You must then convert the newly found moles/molecules into mass using the formula: m = n x M. Mass ratios will not work!
Example 2: Calculate the mass of hydrogen needed to produce 15.5 g of ammonia in the presence of an abundant amount of nitrogen.
Step 1: Write a balanced equation: N2(g) + 3H2 (g) à 2NH3(g)
Step 2: Convert the mass given into number of moles using the formula: n = m/M
15.5 g of NH3(g) x 1 mol = 0.909 moles
17.031 g
Step 3: Analyze the molar ratios of the balanced equation:
Required Given
N2(g) + 3H2 (g) à 2NH3(g)
3 : 2
x : 0.909
For every 2 moles of NH3 produced, 3 moles of H2 were used. This is a 2:3 ratio. This means that 1.5x more H2 is used than NH3 is produced.
Set-up solution with either method:
Method 1
Method 2
number of moles of given = number of moles of unknown
coefficient of given coefficient of unknown
Step 4: Using this ratio, determine the number of moles of the unknown substance:
Method 1:
Step 5: Convert the newly found mole amount into mass (g) using the formula: m = n x M.
m = 1.36 mole x 2.02 g/mol = 2.76 g of H2 were used
Step 6: statement: This reaction requires 2.76 g of H2.
Example 3: Determine the number of molecules of ammonia that are produced when 43.5 g of nitrogen reacts with an excess of hydrogen. The balanced equation for this reaction is:
N2(g) + 3H2(g) à 2NH3(g)
Step 1: Write a balanced equation: Already given in this question N2(g) + 3H2(g) à 2NH3(g)
Step 2: Convert the mass given into number of moles using the formula: n = m/M
43.5 g of N2(g) x 1 mol = 1.55 moles N2
17.031g
Step 3: Analyze the molar ratios of the balanced equation:
Given Required
N2(g) + 3H2(g) à 2NH3(g)
1 : 2
1.55 : x
For every 1 mole of N2 that react, 2 moles of NH3 are produced. This is a 1:2 ratio. This means that there are 2 x more NH3 than there are N2.
Step 4: Using this ratio, determine the number of moles of the unknown substance:
number of moles of given = number of moles of unknown
coefficient of given coefficient of unknown
1 mol N2 = 2 mol NH3
1.55 mol N2 x mol NH3
1.55 mole N2 x 2 mole NH3 = 3.10 mole of NH3 is made.
1 mole N2
Step 5: Convert the newly found mole amount into molecules using the formula: N = n x NA.
N = 3.10 mole x (6.02 x 1023) molecules/mole = 1.866 x 1027 molecules.
Step 6: statement: This reaction produces 1.866 x 1027 molecules of NH3(g).
Example 4: The equation for photosynthesis is 6CO2(g) + 6H2O(l) à C6H12O6(s) + 6O2 (g). What mass of water (H2O) is required to react with 1.00 x 103 g of carbon dioxide (CO2)?
Step 1: Write a balanced equation: 6CO2(g) + 6H2O(l) à C6H12O6(s) + 6O2(g)
Step 2: Convert the mass given into number of moles using the formula: n = m/M
1.0 x 103 g of CO2(g) x 1 mol = 22.72 moles CO2
44.01 g
Step 3: Analyze the molar ratios of the balanced equation:
Required Given
6CO2(g) + 6H2O(l) à C6H12O6(s) + 6O2(g)
6 6
X 22.72
For every 6 moles of CO2 that reacts, 6 moles of H2O react. This is a 6:6 ratio (1:1). This means that 1 mole of CO2 will react with 1 mole of H2O.
Step 4: Using this ratio, determine the number of moles of the unknown substance:
Step 5: Convert the newly found mole amount into mass (g) using the formula: m = n x M.
M = 22.74 mole x 18.02 g/mol = 409.45 g of H2O used
Step 6: statement: This reaction requires 409.45 g of H2O.
Example 5. How many grams of water are produced when 7.00 grams of oxygen react with an excess of hydrogen according to the reaction 2H2(g) + O2(g) ----> 2H2O(g)?
• Note: "excess" reactant hydrogen can be ignored since there is enough of it to completely react and be consumed.
• Identify the given and unknown. (The substance that they give you information about is called the "given." The substance they are asking you about is the "unknown.")
Find the number of moles of given O2:
mass given = 7.00g O2 molar mass of oxygen = 32.0g/mole
moles of oxygen= 7.00 g O2 x 1 mol H2 =0.219 mole
32.0 g/mol O2
Find the # moles of the unknown by comparing the molar ratio;
Given unknown
2H2(g) + O2(g) ----> 2H2O(g)
mole ratio 1 mol : 2 mol
0.219 mol : x
number of moles of given = number of moles of unknown
coefficient of given coefficient of unknown
1 = 2
0.219 x
Number of moles of water (unknown) x= 0.438 mole
Change the number of moles of the unknown to mass with the formula:
# of moles of water = 0.438 mole molar mass of water = 18.0g/mole
mass of water =# of moles x molar mass= 0.438 mole x 18.0 g/mole= 7.89 g
\ 7.89 grams of water are produced when 7.00 grams of oxygen react with an excess of hydrogen.
Where did the additional 0.89 grams of mass come from?
Example 6. How many grams of sulfuric acid are required to react completely with 15.0 grams of zinc in a single displacement reaction?
Derive the balanced chemical Equation: Zn(s) + H2SO4(aq) ----> ZnSO4(aq) + H2(g)
Change the mass given to the number of moles with the formula
molar mass of zinc = 65.4 g/mole mass given= 15 g
# of moles =15.00g Zn = 0.229 mole
65.4g/mole Zn
Determine the number of moles of the unknown by comparing the molar ratio
Zn(s) + H2SO4(aq) ----> ZnSO4(aq) + H2(g)
Unknown Given
1 : 1 1 = 1
0.229 : x 0.229 x
Therefore, the number of moles of sulfuric acid (unknown) x= 0.229 mole
Change the number of moles of the unknown to mass with the formula
# of moles of sulfuric acid = 0.229 mole
molar mass of sulfuric acid = 98.1g/mole
mass = # of moles x molar mass
mass of sulfuric acid = 0.229 mole x 98.1 g/mole = 22.5 g
\ It would take 22.5 grams of sulfuric acid to completely react with 15.0 grams of zinc.
STEPS TO STOICHIOMETRY
Stochiometry problems generally involve the following steps:
1. Complete a balanced equation, if not already given.
2. Convert any given masses to moles (moles= mass ¸ molar mass)
3. Determine the number of moles of the unknown by comparing the molar ratio
number of moles of given = number of moles of unknown
coefficient of given coefficient of unknown
4. Change the number of moles of the unknown to mass with the formula; (mass = # of moles x molar mass)
Example 7: Air contains 21.0% oxygen and 78.0% nitrogen. When magnesium is burned in air, it reacts with both the oxygen and the nitrogen. If 35.0 L of air at standard temperature and pressure is used in this reaction, calculate the mass of magnesium oxide and the mass of magnesium nitride that are produced. (ASSUME 22.4 L =1 mol of gas)