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Ch.10 - Gases
Chapter 10, Problem 110a

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of 85.7% C and 14.3% H by mass. a. If 1.56 g of cyclopropane has a volume of 1.00 L at 0.984 atm and 50.0°C, what is the molecular formula of cyclopropane?

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1
Calculate the molar mass of cyclopropane using the ideal gas law: PV = nRT. Rearrange to find n (moles of gas): n = PV / RT.
Use the given conditions: P = 0.984 atm, V = 1.00 L, R = 0.0821 L·atm/mol·K, and T = 50.0°C (convert to Kelvin by adding 273.15).
Calculate the molar mass (M) of cyclopropane using the formula: M = mass / n, where mass = 1.56 g.
Determine the empirical formula by converting the percentage composition to moles: 85.7% C and 14.3% H. Assume 100 g of cyclopropane, which gives 85.7 g C and 14.3 g H.
Convert grams to moles: moles of C = 85.7 g / 12.01 g/mol, moles of H = 14.3 g / 1.008 g/mol. Find the simplest whole number ratio of moles to determine the empirical formula. Compare the empirical formula mass to the molar mass to find the molecular formula.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Empirical and Molecular Formulas

The empirical formula represents the simplest whole-number ratio of elements in a compound, while the molecular formula indicates the actual number of atoms of each element in a molecule. To determine the molecular formula, one often starts with the empirical formula, which can be derived from the mass percentages of the elements present.
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Ideal Gas Law

The Ideal Gas Law, expressed as PV = nRT, relates the pressure (P), volume (V), number of moles (n), gas constant (R), and temperature (T) of an ideal gas. This law is essential for calculating the number of moles of a gas from its volume, pressure, and temperature, which is necessary for determining the molecular formula of cyclopropane in this scenario.
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Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on the conservation of mass. In this context, it helps in converting the mass of cyclopropane to moles, which is crucial for finding the molecular formula after determining the empirical formula from the mass composition.
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