Reacting Masses: Calculations Involving Masses in Reactions
The Foundation: Laws and The Mole Concept
All calculations involving reacting masses are built upon two key scientific laws. The first is the Law of Conservation of Mass[1], which states that mass is neither created nor destroyed in a chemical reaction. This means the total mass of the reactants equals the total mass of the products. The second is the Law of Constant (or Definite) Proportions[2], which tells us that a pure chemical compound always contains the same elements in the same proportion by mass.
The molar mass of a substance (in grams per mole, g/mol) is numerically equal to its relative atomic mass (Ar) or relative formula/molecular mass (Mr)[3]. For example, the Ar of carbon is 12.0, so one mole of carbon atoms has a mass of 12.0 g. For water (H2O), the Mr is (2 x 1.0) + 16.0 = 18.0, so one mole of water has a mass of 18.0 g.
Interpreting the Balanced Chemical Equation
A balanced chemical equation provides the recipe for the reaction. The numbers in front of the chemical formulas are called stoichiometric coefficients. They tell us the ratio of moles of each substance involved.
Consider the combustion of methane:
This equation reads: 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O. This mole ratio (1:2:1:2) is the heart of all reacting masses calculations. We can scale these ratios up or down using the mole concept to find actual masses.
The Step-by-Step Calculation Method
The general approach to solving reacting mass problems involves a clear sequence of steps. Following this method prevents errors and builds a strong understanding.
| Step | Action | Description |
|---|---|---|
| 1 | Write the Balanced Equation | This is non-negotiable. It gives you the correct mole ratios. |
| 2 | Calculate Molar Masses | Find the molar mass (M) for all relevant substances using the periodic table. |
| 3 | Convert Given Mass to Moles | Use $n = m / M$ for the substance whose mass is known. |
| 4 | Use Mole Ratio | Use the coefficients from the balanced equation to find the moles of the substance you are asked about. |
| 5 | Convert Moles to Mass | Use $m = n \times M$ to find the final answer in grams. |
From Simple to Advanced: Worked Examples
Example 1: Basic Calculation (Elementary/Middle School Level)
What mass of magnesium oxide (MgO) is produced when 24 g of magnesium (Mg) burns in air? The reaction is: $2Mg + O_2 \rightarrow 2MgO$.
Step 1: Balanced equation is given.
Step 2: Molar masses. M(Mg) = 24.3 g/mol, M(MgO) = 24.3 + 16.0 = 40.3 g/mol.
Step 3: Moles of Mg. $n(Mg) = 24 / 24.3 = 0.988$ mol.
Step 4: Mole ratio. From equation, $2Mg : 2MgO$ is a 1:1 ratio. So, $n(MgO) = 0.988$ mol.
Step 5: Mass of MgO. $m(MgO) = 0.988 \times 40.3 = 39.8 g$.
Example 2: Involving a Gaseous Product (High School Level)
Calculate the mass of aluminum required to produce 1.50 L of hydrogen gas (H2) at STP[4] from reaction with hydrochloric acid. The reaction is: $2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2$. At STP, 1 mole of any gas occupies 22.4 L.
Step 1: Balanced equation is given.
Step 2: Molar masses. M(Al) = 27.0 g/mol. We also need the molar volume: 22.4 L/mol for H2.
Step 3: Moles of H2. $n(H_2) = 1.50 / 22.4 = 0.06696$ mol.
Step 4: Mole ratio. From $2Al : 3H_2$, the ratio is $2/3$ or $Al:H_2 = 2:3$. So, $n(Al) = (2/3) \times n(H_2) = (2/3) \times 0.06696 = 0.04464$ mol.
Step 5: Mass of Al. $m(Al) = 0.04464 \times 27.0 = 1.21 g$.
Limiting Reactants and Percentage Yield
In real-life chemistry, we often don't have the exact perfect amounts of reactants. The limiting reactant (or limiting reagent) is the substance that is completely used up first in a reaction, thus determining how much product can be formed. The other reactants are in excess.
How to identify it: Calculate the moles of each reactant, then use the mole ratio from the equation to see which one would produce the least amount of product. That reactant is limiting.
Furthermore, the actual mass of product obtained in an experiment (actual yield) is often less than the maximum possible calculated mass (theoretical yield). The efficiency is expressed as percentage yield:
A yield below 100% can be due to incomplete reactions, side reactions, or loss during transfer.
Practical Application: Baking Soda and Vinegar Volcano
A classic and safe experiment is the reaction of baking soda (sodium bicarbonate, NaHCO3) with vinegar (acetic acid, CH3COOH) to produce carbon dioxide gas for a ‘volcano’ effect: $NaHCO_3 + CH_3COOH \rightarrow CH_3COONa + CO_2 + H_2O$.
Suppose you have 8.4 g of baking soda and excess vinegar. How much carbon dioxide gas (at room conditions) could you theoretically produce? First, find moles of NaHCO3: M(NaHCO3) = 23.0+1.0+12.0+(3x16.0)=84.0 g/mol, so $n = 8.4 / 84.0 = 0.10$ mol. The equation shows a 1:1 mole ratio between NaHCO3 and CO2. Therefore, 0.10 mol of CO2 is produced. The mass would be $m = 0.10 \times 44.0 = 4.4 g$ of CO2. This application shows how reacting mass calculations allow you to predict the amount of gas for your volcano's ‘eruption’.
Important Questions
A: An unbalanced equation gives incorrect ratios of atoms and molecules. The coefficients in a balanced equation directly provide the correct mole ratios between all reactants and products, which is essential for accurate calculations. Using an unbalanced equation would be like using a wrong recipe—your amounts would be completely off.
A: Relative atomic mass (Ar) is a dimensionless number that compares the average mass of an atom of an element to 1/12th the mass of a carbon-12 atom. Molar mass (M) is the mass of one mole of a substance and is expressed in grams per mole (g/mol). Numerically, they are the same. For example, Ar(Carbon)=12.0, and M(Carbon)=12.0 g/mol.
A: Convert the given masses of all reactants to moles. Then, using the balanced equation's mole ratios, calculate how many moles of the same product each reactant could produce independently. The reactant that produces the smallest number of moles of that product is the limiting reactant. It’s like figuring out which ingredient will run out first when making sandwiches.
Footnote
[1] Law of Conservation of Mass: A fundamental principle stating that in a closed system, the total mass of substances present before a chemical reaction (reactants) is equal to the total mass of substances after the reaction (products).
[2] Law of Constant (Definite) Proportions: A law stating that a given chemical compound always contains its component elements in fixed and definite proportions by mass.
[3] Mr (Relative Molecular/Formula Mass): The sum of the relative atomic masses (Ar) of all the atoms in a molecule or formula unit of a compound. It is a dimensionless number.
[4] STP (Standard Temperature and Pressure): A standard set of conditions for measuring gases. Commonly, STP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atmosphere (101.3 kPa). Under these conditions, one mole of any ideal gas occupies 22.4 liters.