Question

Cyclopropane, a gas that when mixed with oxygen is used as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. If the density of cyclopropane is \(1.88 \mathrm{~g} / \mathrm{L}\) at \(\mathrm{STP}\), what is the molecular formula of cyclopropane?

Short answer

The molecular formula of cyclopropane is \(\mathrm{C_3H_6}\).

Step-by-step solution

Find moles of C and H

Assume 100 g of cyclopropane which contains 85.7 g C and 14.3 g H. To find the moles of C and H, we need to divide the mass by each element's molar mass. Moles of C: \(\frac{85.7 \mathrm{~g}}{12.01\mathrm{~g/mol}} = 7.14 \mathrm{mol}\) Moles of H: \(\frac{14.3 \mathrm{~g}}{1.01\mathrm{~g/mol}} = 14.16 \mathrm{mol}\)

Obtain the empirical formula

Divide moles of C and H by the smallest mole value, which will give the ratio of atoms in the empirical formula. C: \(\frac{7.14}{7.14} = 1\) H: \(\frac{14.16}{7.14} = 1.98 \approx 2\) The empirical formula is \(\mathrm{CH_2}\).

Calculate the molar mass of the empirical formula

Molar mass of the empirical formula is the sum of the molar masses of the individual elements multiplied by the number of atoms in the empirical formula. Molar mass of \(\mathrm{CH_2}\): \(12.01\mathrm{~g/mol} + 2(1.01\mathrm{~g/mol}) = 14.03\mathrm{~g/mol}\)

Determine the molar mass of the compound

We know the density of cyclopropane at STP is \(1.88\mathrm{~g/L}\). Recall that STP conditions are 0°C and 1 atm, and the molar volume of any gas at STP is approximately 22.4 L/mol. Use the density to determine the molar mass of the compound. Molar mass of cyclopropane: \(1.88\mathrm{~g/L} \cdot 22.4\mathrm{~L/mol} = 42.11\mathrm{~g/mol}\)

Find the molecular formula

To determine the molecular formula of cyclopropane, divide the molar mass of the compound by the molar mass of the empirical formula. \(\frac{42.11\mathrm{~g/mol}}{14.03\mathrm{~g/mol}} = 3\) As the ratio obtained is close to 3, it means the molecular formula is three times the empirical formula, so after multiplying: Cyclopropane molecular formula: \(\mathrm{C_3H_6}\)

Key concepts

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of atoms of each element present in the compound. It differs from the molecular formula, which indicates the actual number of each type of atom in a molecule. Let's consider cyclopropane as an example. This compound is composed of 85.7% carbon and 14.3% hydrogen by mass.
To determine the empirical formula, assume a 100 g sample. This gives 85.7 g of carbon and 14.3 g of hydrogen. By dividing these masses by the respective atomic masses (12.01 g/mol for carbon and 1.01 g/mol for hydrogen), we calculate the number of moles of each element.
Carbon yields about 7.14 moles, and hydrogen gives 14.16 moles. To find the simplest ratio, divide each molar amount by the smallest value obtained, in this case, 7.14. This yields a ratio of 1 carbon atom to approximately 2 hydrogen atoms. Hence, the empirical formula for cyclopropane is \(\mathrm{CH_2}\). It suggests a basic structure but not the real composition of a molecule in that form.

Molar Mass

The molar mass is the mass of one mole of a substance, typically expressed in g/mol. This value is key to determining the molecular formula of a compound like cyclopropane.
For cyclopropane, after deducing its empirical formula \(\mathrm{CH_2}\), we calculate its molar mass by adding up the molar masses of the constituent atoms. Each carbon atom contributes 12.01 g/mol, and each hydrogen atom contributes 1.01 g/mol. Thus, the molar mass of the empirical formula \(\mathrm{CH_2}\) is calculated as:
- Carbon: 1 atom \(\times\) 12.01 g/mol = 12.01 g/mol- Hydrogen: 2 atoms \(\times\) 1.01 g/mol = 2.02 g/mol
The total molar mass of \(\mathrm{CH_2}\) becomes 14.03 g/mol. To find the molecular formula, this empirical molar mass is then compared to the actual molar mass of the gas determined through experiments, such as density measurements at standard conditions.

Density at STP

Density at STP (Standard Temperature and Pressure) is a vital parameter that assists in calculating the molar mass of gaseous compounds. STP is defined at 0°C (273.15 K) and 1 atm pressure. Under these conditions, one mole of any gas occupies approximately 22.4 liters.
For cyclopropane, with a given density of 1.88 g/L at STP, this density can be used to determine the molar mass of the gas. The calculation involves multiplying the gas density by the molar volume at STP:
- Density of cyclopropane: 1.88 g/L- Molar volume at STP: 22.4 L/mol
When these values are multiplied, the molar mass of cyclopropane is obtained: \(1.88 \times 22.4 = 42.11\,\mathrm{g/mol}\). This result is crucial for comparing against the empirical formula's molar mass to deduce the true molecular formula of cyclopropane.

Cyclopropane

Cyclopropane is an interesting compound, notable for its use as an anesthetic in medical settings when mixed with oxygen. Its chemical composition is predominantly carbon and hydrogen. Understanding its complete makeup involves deducing the molecular formula from empirical data.
Starting with the empirical formula \(\mathrm{CH_2}\), calculated from percentage masses of its elements, the real breakthrough comes from aligning molar masses. The empirical formula's molar mass (14.03 g/mol) falls short compared to the experimental value (42.11 g/mol) derived from density and molar volume analysis.
Dividing the compound’s molar mass by the empirical molar mass gives the multiplier for finding the true molecular formula. For cyclopropane: - Compound Molar Mass: 42.11 g/mol- Empirical Molar Mass: 14.03 g/mol- Multiplier: \(\frac{42.11}{14.03} \approx 3\)
Hence, the molecular formula of cyclopropane is \(\mathrm{C_3H_6}\), which details three carbons and six hydrogens per molecule, reflecting its complete and actual molecular structure.