Mass–mass calculations Knowledge-Based Study

Study notes

1. What Are Mass–Mass Calculations?

   In chemical reactions, reactants combine and products form in exact ratios governed by the law of conservation of mass. While laboratory balances measure quantities in grams, chemical equations describe reactions in terms of moles. Therefore, calculating the mass of a substance produced or consumed requires converting between mass and amount of substance.

   A mass–mass calculation determines the mass of one substance involved in a chemical reaction from a known mass of another substance using a balanced chemical equation.

   Key Principles

   • Conservation of Mass: Matter is neither created nor destroyed; atoms are simply rearranged.
   • Balanced Equation: Coefficients represent mole ratios, not direct mass ratios.
   • Molar Mass ($M$): The conversion factor between grams and moles for any pure substance ($\text{g/mol}$).

   The Core Pathway: The Mole Bridge

   You cannot convert directly from grams of reactant to grams of product because different substances have different molar masses. Instead, you must always cross the mole bridge.

   The General Roadmap

   $$\text{Mass of Given (g)} \xrightarrow{\div M_{\text{given}}} \text{Moles of Given (mol)} \xrightarrow{\times \text{Mole Ratio}} \text{Moles of Unknown (mol)} \xrightarrow{\times M_{\text{unknown}}} \text{Mass of Unknown (g)}$$

   Step-by-Step Procedure

   1. Write and balance the chemical equation: Ensure all stoichiometric coefficients correctly reflect atom conservation.
   2. Convert given mass to moles: Divide the given mass by the molar mass of the initial substance.
      $$n_{\text{given}} = \frac{m_{\text{given}}}{M_{\text{given}}}$$
   3. Apply the mole ratio: Multiply by the ratio of stoichiometric coefficients from the balanced equation.
      $$n_{\text{unknown}} = n_{\text{given}} \times \left(\frac{\text{coefficient of unknown}}{\text{coefficient of given}}\right)$$
   4. Convert moles of unknown to mass: Multiply the moles of the target substance by its molar mass.
      $$m_{\text{unknown}} = n_{\text{unknown}} \times M_{\text{unknown}}$$

   Sample Problem

   Calculate the mass of ammonia ($\text{NH}_3$) produced when $28.0\text{ g}$ of nitrogen gas ($\text{N}_2$) reacts completely with excess hydrogen gas ($\text{H}_2$).

   Molar masses: $M(\text{N}_2) = 28.02\text{ g/mol}$, $M(\text{NH}_3) = 17.03\text{ g/mol}$

   Solution

   Step 1: Balanced Equation

   $$\text{N}_2(g) + 3\text{H}_2(g) ightarrow 2\text{NH}_3(g)$$

   Step 2: Convert Mass of $\text{N}_2$ to Moles

   $$n(\text{N}_2) = \frac{28.0\text{ g}}{28.02\text{ g/mol}} \approx 0.9993\text{ mol of N}_2$$

   Step 3: Determine Moles of $\text{NH}_3$ via Mole Ratio

   According to the balanced equation, $1\text{ mol of N}_2$ yields $2\text{ mol of NH}_3$:

   $$n(\text{NH}_3) = 0.9993\text{ mol of N}_2 \times \left(\frac{2\text{ mol NH}_3}{1\text{ mol N}_2}\right) \approx 1.999\text{ mol of NH}_3$$

   Step 4: Convert Moles of $\text{NH}_3$ to Grams

   $$m(\text{NH}_3) = 1.999\text{ mol} \times 17.03\text{ g/mol} \approx 34.0\text{ g}$$

   Answer: Approximately $34.0\text{ g}$ of $\text{NH}_3$ is produced.

   Pitfalls to Avoid

   • Directly using mass ratios: Never use the coefficients as gram multipliers. In the equation $\text{N}_2 + 3\text{H}_2 ightarrow 2\text{NH}_3$, $1\text{ g}$ of $\text{N}_2$ does not react with $3\text{ g}$ of $\text{H}_2$.
   • Diatomic elements: Remember that common elemental gases like nitrogen, oxygen, and hydrogen exist as diatomic molecules ($\text{N}_2$, $\text{O}_2$, $\text{H}_2$), so their molar masses are double their atomic masses.
   • Unbalanced equations: Always check atom counts on both sides of the reaction before beginning any calculation.
   • Inverting the mole ratio: Ensure the unit you want to cancel is in the denominator and the unit you seek is in the numerator.

   Quick Verification Checklist

   1. Is the chemical equation properly balanced?
   2. Did you use the correct molar masses including subscripts?
   3. Did all intermediate units cancel out cleanly?