Molecular Mass and Formula Unit Mass — Long Answer Questions
Medium Level (Application & Explanation)
Q1. Explain the difference between molecular mass and formula unit mass, and give one clear example of each.
Answer:
Molecular mass is the sum of the atomic masses of all atoms in a molecule. It applies to covalent substances where discrete molecules exist. For example, water (H₂O) has a molecular mass = 2 × 1 (H) + 16 (O) = 18 u.
Formula unit mass is the sum of the atomic masses of all ions in the simplest ratio of an ionic compound. It applies to ionic substances that do not exist as discrete molecules but in a lattice. For example, sodium chloride (NaCl) has a formula unit mass = 23 (Na) + 35.5 (Cl) = 58.5 u.
The key difference: molecular mass refers to one actual molecule; formula unit mass refers to the smallest repeating ionic ratio in an ionic solid. Both use the same idea of adding atomic masses but apply to different kinds of chemical bonding and structure.
Q2. Show step-by-step how to calculate the molecular mass of methanol (CH₃OH) and explain why we treat its atoms the way we do.
Answer:
Identify atoms: CH₃OH has 1 carbon (C), 4 hydrogen (H) (three in CH₃ and one in OH), and 1 oxygen (O).
Explanation: We count every atom present in the molecular formula because molecular mass is the sum of the atomic masses of all constituent atoms. Even though hydrogens are grouped (CH₃ and OH), each hydrogen contributes its atomic mass. The formula CH₃OH represents the actual arrangement of atoms in a single molecule of methanol; therefore the molecular mass reflects the total mass of that one molecule.
Q3. Chlorine has an average atomic mass of 35.5 u. Explain why this is so and how it affects calculations like Cl₂ and NaCl.
Answer:
The value 35.5 u is a weighted average of chlorine’s naturally occurring isotopes, mainly ³⁵Cl and ³⁷Cl. Because these isotopes occur in different proportions on Earth, the average atomic mass is not a whole number.
For a molecule like Cl₂, we calculate molecular mass by adding two average atomic masses: Cl₂ = 2 × 35.5 = 71 u.
For an ionic compound like NaCl, the formula unit mass = 23 (Na) + 35.5 (Cl) = 58.5 u.
Effect on calculations: using the average atomic mass gives an accurate average mass for real samples where isotopes are present in natural proportions. It means molecular or formula masses can be fractional (like 58.5 u), which is normal and useful for laboratory and industrial calculations.
Q4. Sugar (sucrose) has formula C₁₂H₂₂O₁₁. Calculate its molecular mass and explain one practical reason why knowing this mass is useful in everyday life or in the kitchen.
Knowing the molecular mass helps scientists and food technologists understand how many molecules correspond to a given mass of sugar, which is important in chemical reactions like fermentation where yeast acts on sugar molecules.
In the kitchen, although home cooks rarely use molecular mass directly, the concept helps explain why smaller amounts of sugar (by weight) still contain a huge number of sugar molecules—this underpins sweetness, energy content, and how sugar participates in browning reactions during baking.
Thus, molecular mass links the microscopic world of molecules to the macroscopic amounts we measure in real life.
Q5. Calculate the formula unit mass of K₂CO₃ and explain each step and why carbonate contributes three oxygens in the calculation.
Answer:
Identify atoms in the formula unit: K₂CO₃ has 2 potassium (K), 1 carbon (C), and 3 oxygen (O).
Use atomic masses: K = 39 u, C = 12 u, O = 16 u.
Calculate: Formula unit mass = 2 × 39 (K) + 1 × 12 (C) + 3 × 16 (O) = 78 + 12 + 48 = 138 u.
Explanation about carbonate: The carbonate ion is CO₃²⁻, which contains one carbon atom bonded to three oxygen atoms. In the formula K₂CO₃, the carbonate ion is present once, so we count all three oxygens. The 2 potassium ions balance the 2− charge of the carbonate. Counting every atom in the simplest ionic ratio gives the correct formula unit mass for the ionic compound.
High Complexity (Analytical & Scenario-Based)
Q6. Both O₂ and CH₃OH have the same molecular mass as some other species (for example, CH₃OH = 32 u and O₂ = 32 u). Analyze why two substances with the same molecular mass can behave differently chemically and physically.
Answer:
Molecular mass alone gives information about mass but not structure, bonding, or polarity, which strongly determine chemical and physical behavior.
O₂ (32 u) is a diatomic nonpolar molecule made of two oxygen atoms with a double bond; it is a gaseous element at room temperature and supports combustion.
CH₃OH (32 u) is methanol, a polar molecule with an –OH group, capable of hydrogen bonding; it is a liquid at room temperature and dissolves many polar substances.
Properties such as melting/boiling points, solubility, reactivity, and state of matter depend on intermolecular forces and molecular geometry, not just mass. For example, hydrogen bonding in CH₃OH raises its boiling point relative to simple nonpolar molecules of similar mass.
Therefore, substances with equal molecular masses can behave very differently because mass is only one factor among structure, bonding, and intermolecular forces.
Q7. You are given a sample labeled “unknown ionic salt” with formula unit mass close to 111 u. Propose a method to test whether the sample could be CaCl₂, and explain how the formula unit mass helps narrow down possibilities.
Answer:
Calculation: CaCl₂ formula unit mass = 40 (Ca) + 2 × 35.5 (Cl) = 111 u. So the number matches.
How formula unit mass helps: it narrows candidates by matching the observed mass to sums of atomic masses; 111 u specifically suggests a metal plus two chlorines or equivalent composition.
Simple tests to confirm CaCl₂:
Solubility test: CaCl₂ is soluble in water. If the sample dissolves readily, it supports an ionic salt.
Flame test: A small dissolved sample on a clean wire heated in a flame gives a brick-red (orange-red) color characteristic of Ca²⁺.
Precipitation test: Add dilute sodium carbonate (Na₂CO₃) solution; formation of a white precipitate of calcium carbonate (CaCO₃) indicates Ca²⁺. The precipitate dissolves on adding acid confirming carbonate.
Combining formula unit mass with these qualitative tests provides strong evidence whether the unknown is CaCl₂.
Q8. Heavy water (D₂O) uses deuterium (D), an isotope of hydrogen with mass 2 u. Calculate the molecular mass of D₂O and compare it with H₂O. Discuss one practical implication of this difference.
Answer:
Calculation: Deuterium mass ≈ 2 u, oxygen = 16 u. Molecular mass of D₂O = 2 × 2 + 16 = 20 u.
For normal water H₂O, molecular mass = 2 × 1 + 16 = 18 u.
Comparison: D₂O is approximately 11% heavier than H₂O (20 u vs 18 u).
Practical implication:
The greater mass affects physical properties: heavy water has slightly higher boiling and melting points, greater density, and different vibrational frequencies. These differences affect reaction rates where hydrogen transfer is involved because bonds with deuterium are stronger and vibrate more slowly; this is called the kinetic isotope effect.
In nuclear reactors, D₂O is used as a moderator because deuterium slows neutrons without capturing them as readily as protium (¹H). This property is unrelated to chemical reactivity but demonstrates a real-world application that arises from the mass difference.
Q9. Suppose a chemist has two samples: 32 u of O₂ (one molecule) and 32 u consisting of a single molecule of CH₃OH. Analyze why these single-molecule samples would not be directly comparable in experiments, and what would be a meaningful way to compare amounts of substances.
Answer:
Single molecules have identical mass numbers here but are entirely different chemical species. Handling or experimenting with a single molecule of anything is not realistic in a laboratory because chemical measurements require enormous numbers of molecules.
Even if both have mass 32 u per molecule, their properties (state at room temperature, reactivity, polarity) differ, so experimental outcomes would differ widely.
A meaningful comparison uses amounts corresponding to many molecules, e.g., by mass or by counting units like the mole (Avogadro’s number). The mole links macroscopic mass to the number of particles and allows comparison on equal particle basis.
For practical experiments, chemists weigh macroscopic amounts (grams) and use molecular/formula masses to calculate how many molecules or formula units are present. This ensures reproducible and meaningful comparisons.
Q10. You find a gas sample labeled “molecular mass ≈ 44 u.” Suggest at least three possible molecular formulas that have molecular masses near 44 u, and explain how you would distinguish among them using simple laboratory observations.
Answer:
Three possible molecules near 44 u:
CO₂: 12 (C) + 2 × 16 (O) = 44 u.
C₃H₈ (propane) is 3×12 + 8×1 = 44 u.
NO₂ is 14 + 2×16 = 46 u (slightly higher), but a mixture or measurement error could confuse it; another possibility close to 44 is C₂H₄O (acetaldehyde) = 44 u.
How to distinguish:
CO₂ test: Bubble the gas through limewater (Ca(OH)₂); CO₂ turns limewater milky due to CaCO₃ formation. CO₂ is also non-flammable.
Flame test/combustibility: Propane (C₃H₈) is flammable and burns with a yellowish flame; CO₂ will not burn and extinguishes flame.
