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Two compounds A and B have the formula CH2: finding molecular formulas from molar mass

Two compounds A and B have the formula CH2 as an empirical formula. If the molar mass of A is 42.0 g/mol and the molar mass of B is 84.0 g/mol, what are the molecular formulas of A and B?

Subject: General Chemistry Chapter: Chemical Compounds Topic: Deducing Molecular Formula Answer included
two compounds a and b have the formula empirical formula molecular formula molar mass formula mass simplest whole-number ratio multiple n deducing molecular formula
Accepted answer Answer included

two compounds a and b have the formula \(\mathrm{CH_2}\) is a standard cue that the written formula is an empirical formula, meaning the simplest whole-number ratio of atoms. A molecular formula describes the actual numbers of atoms in one molecule and is an integer multiple of the empirical formula.

Empirical-formula unit mass

Atomic masses \(\mathrm{C} \approx 12.01\ \mathrm{g\ mol^{-1}}\) and \(\mathrm{H} \approx 1.008\ \mathrm{g\ mol^{-1}}\) give an empirical-formula mass for \(\mathrm{CH_2}\):

\[ M(\mathrm{CH_2}) = 12.01 + 2 \cdot 1.008 = 14.026 \approx 14.03\ \mathrm{g\ mol^{-1}} \]

Scaling relationship

A molecular formula has the form \((\mathrm{CH_2})_n = \mathrm{C}_n\mathrm{H}_{2n}\) where \(n\) is a positive integer given by \(n = \dfrac{M_{\text{molecule}}}{M(\mathrm{CH_2})}\).

Molecular formulas of A and B from molar mass

The molar mass of compound A is \(42.0\ \mathrm{g\ mol^{-1}}\) and the molar mass of compound B is \(84.0\ \mathrm{g\ mol^{-1}}\). Each is compared to the empirical-formula mass \(14.03\ \mathrm{g\ mol^{-1}}\).

\[ n_A = \frac{42.0}{14.03} \approx 2.99 \approx 3 \qquad\Rightarrow\qquad \mathrm{A} = (\mathrm{CH_2})_3 = \mathrm{C_3H_6} \]
\[ n_B = \frac{84.0}{14.03} \approx 5.99 \approx 6 \qquad\Rightarrow\qquad \mathrm{B} = (\mathrm{CH_2})_6 = \mathrm{C_6H_{12}} \]

Summary table

Compound Empirical formula Molar mass (g/mol) Multiple \(n\) Molecular formula
A \(\mathrm{CH_2}\) \(42.0\) \(3\) \(\mathrm{C_3H_6}\)
B \(\mathrm{CH_2}\) \(84.0\) \(6\) \(\mathrm{C_6H_{12}}\)

Visualization of empirical-unit stacking

Molecular formula as an integer multiple of the empirical formula CH2 Two rows show CH2 empirical units repeated n times: n equals 3 for compound A (C3H6) and 6 for compound B (C6H12), matching their molar masses. Empirical formula CH₂ → molecular formula (CH₂)ₙ Each colored block represents one empirical unit CH₂ (mass ≈ 14.03 g/mol). Compound A Molar mass 42.0 g/mol → n = 3 → molecular formula C₃H₆ CH₂ CH₂ CH₂ n = 3 Compound B Molar mass 84.0 g/mol → n = 6 → molecular formula C₆H₁₂ CH₂ CH₂ CH₂ CH₂ CH₂ CH₂ n = 6
Empirical units stack in whole-number multiples. A uses three \(\mathrm{CH_2}\) units (\(\mathrm{C_3H_6}\)) and B uses six \(\mathrm{CH_2}\) units (\(\mathrm{C_6H_{12}}\)), matching their molar masses.

Consistency checks

The ratios \(42.0/14.03\) and \(84.0/14.03\) land very close to integers because the given molar masses are rounded while atomic masses are not exact integers. Integer \(n\) values are the chemically meaningful outcome, since molecular formulas require whole-number subscripts.

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