4.4: Chemical Equations and Quantitative Analysis

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Combustion Analysis

Exercise \(\PageIndex{1}\)

The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon and hydrogen. A 3.000-mg sample of naphthalene burns to give 10.3 mg of CO₂. Determine its empirical and molecular formulas.

Answer

First find mass of Carbon and Hydrogen

\(m_{c}=\left ( \frac{1 g}{1000 mg} \right )\left ( \frac{1 mol CO_{2}}{44.01 g CO_{2}} \right )\left ( \frac{1 mol C}{1 mol CO_{2}} \right )\left ( \frac{12.01 g C}{1 molC} \right )\left ( \frac{1000 mg}{1 g} \right )=2.811 mg C\)

\(m_{H}=\left ( 3.000 mg - 2.811 mg \right )=0.189 mg H\)

Second find the moles of Carbon and Hydrogen.

\(n_{1}=\frac{2.811*10^{-3}g}{12.01g*mol^{-1}}=2.34*10^{-4}mol)

\(n_{2}=\frac{0.189*10^{-3}g}{1.008g*mol^{-1}}=1.875*10^{-4}mol\)

Now solve for Empirical Formula

\(C = \frac{2.34*10^{-4}mol}{1.875*10^{-4}mol}=1.2492\)

\(H = \frac{1.875*10^{-4}mol}{1.875*10^{-4}mol}=1\)

Multiply each by 4 to get the smallest integer. (1.2492 * 4 = 4.9968 or 5). C₅H₄

Lastly solve for the molecular formula using empirical formula

\(M_{1}=5(M_{C})+4(M_{H})\)

\(M_{1}=5(12.01 g*mol^{-1})+4(1.008 g*mol^{-1})=64.085 amu\)

\(n_{f}=\frac{130 amu}{64.085 amu}=2\)

Now multiply the empirical formula by 2 to get the molecular formula. C₁₀H₈.

Exercise \(\PageIndex{2}\)

Mesitylene is a liquid hydrocarbon. Burning 0.115 g of the compound in oxygen gives 0.379 g of CO2 and 0.1035 g of H2O. What is the empirical formula of mesitylene?

Answer: C₃H₄

Exercise \(\PageIndex{3}\)

A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of B₂O₃. What are the empirical and molecular formulas of the compound.

Answer

Empirical Formula: BH₃

Molecular Formula: B₂H₆

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