Interpreting Chemical Equations
Goal: Given a balanced chemical equation or information from which a chemical equation can be written, describe its meaning at the particulate, molar and macroscopic levels.
Example 1: 2H2(g) + O2(g) 2H2O(g)
Particulate Level: 2 molecules 1 molecule 2 molecules
Molar Level: 2 moles 1 mole 2 moles
Macroscopic Level: 4 g 32 g 36 g
Example 2: 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O (l)
Particulate Level: 2 molecules 7molecules 4 molecules 6 molecules
Molar Level: 2 moles 7 moles 4 moles 6 moles
Macroscopic Level: 60 g 224 g 176 g 108 g
Quantity Relationships in Chemical Reactions, (Stoichiometry)
Goal: Given a chemical equation, or a reaction for which the equation is known, and the number of moles of one species in the reaction, calculate the number of moles of any other species.
Consider the equation:
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
- How many O2 molecules are required to react with 308 molecules of NH3?
Answer: On the particulate level,
4 molecules of NH3 react with 5 molecules of O2
That is OR
Recall that: Required number and units = Given number and units x conversion factor
Therefore,
# O2 molecules = 308 NH3molecules x = 385 O2 molecules
- If 3.20 moles of NH3 react according to the above equation, how many moles of H2O will be produced?
Answer: The unit path is mol NH3 mol H2O
Therefore, use as the conversion factor
Hence, mole H2O = 3.20 molNH3 x = 4.80 mol H2O
- How many moles of oxygen are required to burn 2.40 moles of ethane, C2H6?
Answer: First, write a balanced equation for the reaction described.
Hint: Remember that combustion of hydrocarbon always forms CO2 and H2O
2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O (l)
The conversion factor from the relationship between the given and the wanted quantity using the coefficients in the equation is
Hence, mol O2 = 2.40 molC2H6 x = 8.40 mol O2
Mass Calculations
Mass calculations may involve:
- Writing a balanced equation
- Calculating molar masses from chemical formulas
- Using molar masses to change from mass to moles or from moles to mass
- Using the equation to change from moles of one species to moles of another
The mass –to- mass path is:
Mass of given moles of given moles of wanted mass of wanted
Mass given x x x
Example: Calculate the number of grams of oxygen that are required to burn 155 g of ethane in the reaction: 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O (l)
Answer:
Number of gram O2 = 155 g C2H6 x x x = 577 g O2
Example: How many grams of oxygen are required to burn 3.50 moles of liquid heptane, C7H16(l)?
Answer: First, write a balanced equation.
C7H16(l) + 11O2(g) 7CO2(g) + 8H2O(l)
g O2 = 3.50 mol C7H16 x x = 1,230 g O2
Example: How many grams of CO2 will be produced by burning 66.0 g C7H16 according to the equation: C7H16(l) + 11O2(g) 7CO2(g) + 8H2O(l)?
Answer:
g CO2 = 66.0 g C7H16 x x x = 203 g CO2
Limiting and Excess Reactants, Theoretical Yield, Actual Yield and Percent Yield
Definitions:
Limiting reactant: the reactant that is completely used up during the reaction is called the limiting reactant.
Excess reactant: the reactant that some of it remains unreacted after the reaction is complete.
Theoretical Yield: the amount of product formed from the complete conversion of the given amount of reactant to product. Theoretical yield is always a calculated quantity based on the stoichiometry, (ratio) of the reaction equation.
Actual Yield: a measured quantity determined by experiment. Factors such as impure reactants, incomplete reaction and side reactions cause the actual yield to be less than the theoretical yield.
Percent Yield: this is the actual yield expressed as a percentage of the theoretical yield.
Yield = x 100
Understanding Limiting reactant
- How many pairs of gloves can you assemble from 20 left gloves and 30 right gloves?
Answer: 20 pairs theoretical yield
- How many unmatched gloves will be left over?
Answer: 10 right gloves excess amount
- Which hand will these unmatched gloves fit?
Answer: Right excess reactant
- What prevented you from assembling more than 20 pairs?
Answer: Left gloves limiting reactant
How To Solve Limiting Reactant Problems
There are two methods:
- Smaller-Amount method
- Comparison-of-Moles Method
We shall use the smaller-amount method
Procedure:
- Calculate the amount of product that can be formed by the initial amount of each reactant
- The reactant that yields the smaller amount of product is the limiting reactant.
- The smaller amount of product is the amount (theoretical yield) that will be formed when all of the limiting reactant is used up.
- Calculate the amount of excess reactant that is used by the total amount of the limiting reactant.
- Subtract from the amount of excess reactant present initially, the amount that is used by the limiting reactant. The difference is the amount of excess reactant that is left.
Example: Calculate the mass of antimony (III) chloride, SbCl3, that can be produced by the reaction of 129 g antimony, Sb, and 106 g chlorine. How much of the excess reactant is left?
Answer:
2Sb +3Cl22SbCl3
Molar mass (g/mol) 121.8 71 228.3
Mass Used (g) 129 106
Moles Used, (mol)
1.059 mol 1.492 mol
a. From balanced equation,
2 molSb 2 mol SbCl3
1.059 molSb xmol SbCl3
Hence, xmol SbCl3 = 1.059 molSb
b. 3 mol Cl22mol SbCl3
1.492 mol Cl2xmol SbCl3
Hence, xmol SbCl3= = 0.995 mol SbCl3
Comparing amounts of products from each reactantshows that Cl2 produces less SbCl3 than Sb.
Therefore:
(i)Cl2 is the limiting reactant.
(ii)Sb is the excess reactant.
(iii)0.995 mol SbCl3 produced from the limiting reactant is the theoretical yield.
0.995 mol SbCl3 x = = 227 g SbCl3
c. Excess amount of Sb
To find out how much of Sb is left, first find how much Sb was used to react with all 106 g Cl2
Amount Sb Used:
106 g Cl2 x x x = 121 g Sb
Amount left (excess) = 129 g (initial) – 121 g (used) =8 g Sb
Percent Yield
A student carried out the reaction in the last example and obtained 210 g SbCl3. What is the student’s percent yield?
Answer:Actual yield = 210 g SbCl3
Theoretical yield = 227 g SbCl3
Yield = x 100
Yield = x 100 = 93%