Slide presentation
Reactants in Accordance with the Percent Yield
General Chemistry • Chemical Reactions
Reverse yield planning
Reactants in Accordance with the Percent Yield
When percent yield is less than \(100\%\), the reaction does not produce the full theoretical amount. To obtain a desired actual product mass, you must plan for a larger theoretical yield.
\[ \text{theoretical yield} = \frac{\text{actual yield}}{\text{percent yield}/100} \]Learning target: work backward from desired actual product to required theoretical product, then to reactant amounts.
Desired actual product
The mass you want to obtain in the real experiment.
Percent yield
The fraction of theoretical product usually recovered.
Required reactants
Stoichiometry converts the required theoretical product back to reactant mass.
Why it matters
Reaction planning must account for imperfect yield
In a real lab, product can be lost during filtration, transfer, drying, purification, or because the reaction does not go to completion.
If a process produces only \(80.0\%\) of the theoretical amount, making \(50.0\ \mathrm{g}\) actual NH3 requires planning for more than \(50.0\ \mathrm{g}\) theoretical NH3.
Percent-yield planning is used to
- prepare a target product mass,
- estimate how much reactant to buy or measure,
- avoid under-producing product,
- scale laboratory reactions safely,
- connect stoichiometry to real reaction efficiency.
Core concept
Work backward from actual yield to reactants
Forward yield problems calculate percent yield from an actual result. Reverse yield problems start with a desired actual result and ask how much reactant is needed.
When percent yield is less than \(100\%\), required theoretical yield is larger than desired actual yield.
Vocabulary
Yield planning terms and units
| Term | Meaning | How it is used | Common unit |
|---|---|---|---|
| Actual yield | Product mass obtained or desired in the real experiment | Starting point in reverse yield problems | g or mol |
| Percent yield | Actual yield as a fraction of theoretical yield | Converts actual target into theoretical requirement | % |
| Theoretical yield | Ideal product amount predicted by stoichiometry | \(\frac{\text{actual yield}}{\text{percent yield}/100}\) | g or mol |
| Mole ratio | Reactant-product relationship from balanced coefficients | Converts moles product to moles reactant | mol/mol |
| Reactant mass | Amount of starting material required | Calculated from moles and molar mass | g |
Main relationship
Percent yield changes the required theoretical target
Use the percent yield first, then use the balanced equation.
Interactive simulation
Plan reactants for a desired actual mass of NH3
Reaction setup
Required planning amounts
At 80.0% yield, plan for 62.5 g theoretical NH3 to obtain 50.0 g actual NH3.
Static fallback: to obtain \(50.0\ \mathrm{g}\) actual NH3 at \(80.0\%\) yield, plan for \(62.5\ \mathrm{g}\) theoretical NH3.
Dynamic relationship
Lower percent yield requires a larger theoretical target
The graph compares desired actual product with the theoretical product that must be planned.
Interpretation: the theoretical bar grows when percent yield decreases because the reaction must be planned to produce more product ideally.
Worked example
Calculate reactant masses from a desired actual yield
Problem: How much N2 and H2 are required to obtain \(50.0\ \mathrm{g}\) actual NH3 if the reaction has an \(80.0\%\) yield?
- 1. Convert desired actual yield to required theoretical yield. \[ \text{theoretical NH_3} = \frac{50.0\ \mathrm{g}}{80.0/100} = 62.5\ \mathrm{g\ NH_3} \]
- 2. Convert theoretical NH3 to moles. \[ n_{\mathrm{NH_3}} = \frac{62.5\ \mathrm{g}}{17.031\ \mathrm{g/mol}} = 3.67\ \mathrm{mol\ NH_3} \]
- 3. Use mole ratios to find reactant moles. \[ 3.67\ \mathrm{mol\ NH_3} \times \frac{1\ \mathrm{mol\ N_2}}{2\ \mathrm{mol\ NH_3}} = 1.84\ \mathrm{mol\ N_2} \] \[ 3.67\ \mathrm{mol\ NH_3} \times \frac{3\ \mathrm{mol\ H_2}}{2\ \mathrm{mol\ NH_3}} = 5.50\ \mathrm{mol\ H_2} \]
- 4. Convert reactant moles to grams. \[ 1.84\ \mathrm{mol\ N_2} \times 28.014\ \mathrm{g/mol} = 51.4\ \mathrm{g\ N_2} \] \[ 5.50\ \mathrm{mol\ H_2} \times 2.016\ \mathrm{g/mol} = 11.1\ \mathrm{g\ H_2} \]
Common mistake
Do not use the desired actual yield directly as the theoretical yield
Incorrect shortcut
A student starts stoichiometry with \(50.0\ \mathrm{g}\) NH3 even though the yield is only \(80.0\%\).
That would plan for too little reactant, because only \(80.0\%\) of the theoretical product is expected to be recovered.
Correct first step
Calculate the larger theoretical target first:
\[ \text{theoretical yield} = \frac{\text{actual yield}}{\text{percent yield}/100} \]
Then work backward through stoichiometry.
Key idea: percent yield must be handled before converting product mass back to reactant mass.
Practice check
Work backward from product to reactant
Calcium carbonate decomposes according to:
How many grams of CaCO3 are needed to obtain \(28.0\ \mathrm{g}\) actual CaO if the percent yield is \(70.0\%\)?
Show answer and reasoning
Find the required theoretical CaO:
\[ \text{theoretical CaO} = \frac{28.0\ \mathrm{g}}{70.0/100} = 40.0\ \mathrm{g\ CaO} \]
Convert CaO to moles:
\[ n_{\mathrm{CaO}} = \frac{40.0\ \mathrm{g}}{56.08\ \mathrm{g/mol}} = 0.713\ \mathrm{mol\ CaO} \]
The equation has a 1:1 ratio between CaCO3 and CaO, so \(0.713\ \mathrm{mol}\) CaCO3 is needed.
\[ 0.713\ \mathrm{mol\ CaCO_3} \times 100.09\ \mathrm{g/mol} = 71.4\ \mathrm{g\ CaCO_3} \]
Answer: \(71.4\ \mathrm{g}\) CaCO3 is required.
Apply the topic
A reliable strategy for reactant planning with percent yield
Start with actual target
Identify the product mass the experiment must actually produce.
Correct for percent yield
Divide by \(\text{percent yield}/100\) to get theoretical product.
Use stoichiometry backward
Convert product to moles, then use the coefficient ratio to get reactant moles.
Convert to reactant mass
Use molar mass to find the required grams of reactant.
If a reactant is intentionally in excess, calculate the required amount of the limiting reactant first, then decide how much excess reactant is appropriate for the experiment.
Summary
What to remember
Percent yield changes planning
Less than \(100\%\) yield means more theoretical product must be planned.
Actual target is not enough
First convert desired actual yield into required theoretical yield.
Balanced equations still control ratios
Use coefficients to move from product moles back to reactant moles.
Reactant mass is the practical answer
Convert reactant moles to grams using molar mass.