Slide presentation
Relating the Mass of Reactants and Products
General Chemistry • Chemical Reactions
Mass relationships
Relating the Mass of Reactants and Products
Mass-to-mass stoichiometry answers a practical question: if a certain mass of reactant is used, what mass of product can form?
CH4 + 2O2 → CO2 + 2H2OLearning target: convert grams to moles, use coefficient ratios, and convert moles back to grams.
Mass is measured
Reactants and products are usually measured in grams in the laboratory.
Moles compare substances
The balanced equation compares substances using coefficients in moles.
Product mass is predicted
Theoretical yield is the mass predicted by the balanced equation.
Why it matters
A balanced equation becomes a laboratory recipe
Chemists do not count molecules one by one. They measure mass, convert that mass to moles, and then use the balanced equation to predict another mass.
The equation tells us that \(2\ \mathrm{mol}\) H2 form \(2\ \mathrm{mol}\) H2O, but the lab balance gives grams. Molar mass connects the two.
Mass-to-mass calculations help predict
- how much product should form,
- how much reactant is required,
- whether one reactant is limiting,
- how experimental yield compares to theory.
Core concept
Grams cannot be compared directly by coefficients
The coefficients in a balanced equation compare particles or moles, not grams. Before using a mole ratio, convert the given mass into moles.
For methane combustion, \(1\ \mathrm{mol}\) CH4 forms \(1\ \mathrm{mol}\) CO2 and \(2\ \mathrm{mol}\) H2O.
Vocabulary
Quantities used in mass-to-mass stoichiometry
| Quantity | Meaning | Formula or source | Typical unit |
|---|---|---|---|
| Mass | Amount measured on a balance | Given or calculated | g |
| Molar mass | Mass of \(1\ \mathrm{mol}\) of a substance | Periodic table and formula | g/mol |
| Moles | Counting unit for particles | \(n = \frac{m}{M}\) | mol |
| Mole ratio | Ratio between substances in a balanced equation | Coefficients | mol/mol |
| Theoretical yield | Maximum product mass predicted by calculation | Stoichiometry from limiting reactant | g |
Main relationship
The mass-to-mass setup is dimensional analysis
A full mass-to-mass calculation multiplies conversion factors so unwanted units cancel.
Convert grams to moles
Use molar mass of the given substance.
Use mole ratio
Use coefficients from the balanced equation.
Convert moles to grams
Use molar mass of the wanted substance.
Interactive simulation
Change reactant masses and predict product masses
Methane combustion
Predicted amounts
CH4 is limiting, so it controls the product masses.
Static fallback: 16.0 g CH4 with 80.0 g O2 gives about 43.9 g CO2 and 36.0 g H2O.
Dynamic relationship
The limiting reactant controls both product masses
The bars show the predicted mass of CO2 and H2O from the available reactants.
Interpretation: product mass changes with reactant mass, but the balanced equation and molar masses determine the exact relationship.
Worked example
Calculate grams of product from grams of reactant
Problem: If 12.0 g CH4 burns in excess O2, what mass of CO2 forms?
- 1. Convert grams CH4 to moles CH4. \[ 12.0\ \mathrm{g\ CH_4} \times \frac{1\ \mathrm{mol\ CH_4}}{16.04\ \mathrm{g\ CH_4}} = 0.748\ \mathrm{mol\ CH_4} \]
- 2. Use the mole ratio from the balanced equation. \[ 0.748\ \mathrm{mol\ CH_4} \times \frac{1\ \mathrm{mol\ CO_2}}{1\ \mathrm{mol\ CH_4}} = 0.748\ \mathrm{mol\ CO_2} \]
- 3. Convert moles CO2 to grams CO2. \[ 0.748\ \mathrm{mol\ CO_2} \times 44.01\ \mathrm{g/mol} = 32.9\ \mathrm{g\ CO_2} \]
- Final answer: The theoretical yield is \(32.9\ \mathrm{g}\) CO2.
Common mistake
Do not use coefficient ratios directly with grams
Incorrect shortcut
In CH4 + 2O2 → CO2 + 2H2O, a student says 10 g CH4 makes 10 g CO2 because the coefficient ratio is 1:1.
The 1:1 ratio is a mole ratio, not a gram ratio.
Correct thinking
Use the 1:1 ratio only after converting CH4 to moles. Then convert moles CO2 to grams CO2.
\[ \mathrm{g\ CH_4} \rightarrow \mathrm{mol\ CH_4} \rightarrow \mathrm{mol\ CO_2} \rightarrow \mathrm{g\ CO_2} \]
Practice check
Mass-to-mass stoichiometry
Hydrogen reacts with oxygen to form water:
If 8.00 g H2 reacts with excess O2, what mass of H2O forms?
Show answer and reasoning
First convert grams H2 to moles H2:
\[ 8.00\ \mathrm{g\ H_2} \times \frac{1\ \mathrm{mol\ H_2}}{2.016\ \mathrm{g\ H_2}} = 3.97\ \mathrm{mol\ H_2} \]
The mole ratio is \(2\ \mathrm{mol}\) H2 to \(2\ \mathrm{mol}\) H2O, so \(3.97\ \mathrm{mol}\) H2 forms \(3.97\ \mathrm{mol}\) H2O.
\[ 3.97\ \mathrm{mol\ H_2O} \times 18.015\ \mathrm{g/mol} = 71.5\ \mathrm{g\ H_2O} \]
Answer: \(71.5\ \mathrm{g}\) H2O.
Apply the topic
A reliable strategy for reactant-product mass problems
Balance the equation
The coefficients must be correct before using any mole ratio.
Convert grams to moles
Use the molar mass of the given substance.
Apply the mole ratio
Use coefficients to convert from moles of given to moles of wanted.
Convert moles to grams
Use the molar mass of the product or requested substance.
When more than one reactant mass is given, calculate product from each reactant. The smaller product mass comes from the limiting reactant.
Summary
What to remember
Mass is not the mole ratio
Coefficients compare moles, not grams.
Molar mass is the converter
Use molar mass to move between grams and moles.
The balanced equation is the map
Use coefficients to connect moles of reactant to moles of product.
Theoretical yield is predicted mass
It is the maximum product mass calculated from stoichiometry.