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 STP, what is the molecular formula of cyclopropane?

Short answer

The molecular formula of cyclopropane is C3H6, which was determined by first finding the empirical formula (CH2) using the given percentage masses of C and H, and then using the known density at STP to calculate the molecular weight of cyclopropane (41.85 g/mol). The molecular formula is 3 times the empirical formula, resulting in C3H6.

Step-by-step solution

Calculate the empirical formula

Based on the information provided, we know that cyclopropane is composed of 85.7% carbon (C) and 14.3% hydrogen (H) by mass. Assume we have 100 g of cyclopropane; then we would have 85.7 g of carbon and 14.3 g of hydrogen. To find the empirical formula, we can convert these masses to moles: Moles of carbon: \[\frac{85.7\, \mathrm{g}}{12.01\, \mathrm{g/mol}} = 7.14\, \mathrm{mol}\] Moles of hydrogen: \[\frac{14.3\, \mathrm{g}}{1.01\, \mathrm{g/mol}} = 14.16\, \mathrm{mol}\] Next, divide each of the mole values by the smallest mole value: \[\frac{7.14\, \mathrm{mol}}{7.14\, \mathrm{mol}} = 1.00\] \[\frac{14.16\, \mathrm{mol}}{7.14\, \mathrm{mol}} = 1.98 \approx 2\] The empirical formula is CH2.

Find the molecular weight of the empirical formula

Now that we have the empirical formula (CH2), we can calculate its molecular weight: Molecular weight of CH2: \(12.01\, \mathrm{g/mol} + 2\cdot 1.01\, \mathrm{g/mol} = 14.03\, \mathrm{g/mol}\)

Determine the molecular weight of cyclopropane and the molecular formula

With the molecular weight of the empirical formula and the density of cyclopropane at STP, we can now calculate the molecular weight of cyclopropane. Since we know the density at STP, we can use the equation \(d = \frac{PM}{RT}\), where d is the density, P is the pressure (1 atm), R is the gas constant (\(0.0821\, \mathrm{L\cdot atm/mol\cdot K}\)), and T is the temperature (273 K). Rearrange this equation for M (molecular weight): \[M = \frac{dRT}{P} = \frac{(1.88\, \mathrm{g/L})(0.0821\, \mathrm{L\cdot atm/mol\cdot K})(273\, \mathrm{K})}{1\, \mathrm{atm}} = 41.85\, \mathrm{g/mol}\] Now, to find the molecular formula, divide the molecular weight of cyclopropane by the molecular weight of the empirical formula: \[\frac{41.85\, \mathrm{g/mol}}{14.03\, \mathrm{g/mol}} = 2.98 \approx 3\] The molecular formula of cyclopropane is therefore 3 times the empirical formula CH2, which is C3H6.

Key concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of elements in a compound. For cyclopropane, we are given its composition by mass: 85.7% carbon (C) and 14.3% hydrogen (H). To determine the empirical formula, we assume a 100 g sample of cyclopropane, making calculations straightforward. This way, we would have 85.7 g of carbon and 14.3 g of hydrogen.

Converting these masses to moles involves dividing by the respective atomic masses:

• Moles of carbon: \(\frac{85.7 \, \mathrm{g}}{12.01 \, \mathrm{g/mol}} = 7.14 \, \mathrm{mol}\)
• Moles of hydrogen: \(\frac{14.3 \, \mathrm{g}}{1.01 \, \mathrm{g/mol}} = 14.16 \, \mathrm{mol}\)

To find the simplest ratio, divide each value by the smaller mole value, 7.14 in this case:

• Carbon: \(\frac{7.14}{7.14} = 1.00\)
• Hydrogen: \(\frac{14.16}{7.14} = 1.98 \, \approx 2\)

Thus, the empirical formula for cyclopropane is \( \text{CH}_2 \). This formula gives a basic sense of the composition but not the actual structure.

Molecular Weight Calculation

Calculating molecular weight using an empirical formula involves summing the atomic weights of each element in the empirical formula. For cyclopropane's empirical formula, \( \text{CH}_2 \), you add the atomic weights of carbon and hydrogen:

• Carbon: \(12.01 \, \mathrm{g/mol}\)
• Hydrogen (2 moles): \(2 \times 1.01 \, \mathrm{g/mol} = 2.02 \, \mathrm{g/mol}\)

The total molecular weight of \( \text{CH}_2 \) is:\[12.01 \, \mathrm{g/mol} + 2.02 \, \mathrm{g/mol} = 14.03 \, \mathrm{g/mol}\]This calculation helps understand the relative scale of the empirical formula. However, it does not reflect the full molecular weight of cyclopropane.

Density at STP

The concept of density at Standard Temperature and Pressure (STP), which is 0°C or 273 K and 1 atm, is crucial for determining the molecular formula. Density, a measure of mass per unit volume, combined with the gas laws, allows us to calculate the molecular weight of a gas. For cyclopropane, the density is given as 1.88 g/L.

We can use the formula\[d = \frac{PM}{RT}\]to find the molecular weight \(M\). Here, \(d\) is the density, \(P\) is the pressure, \(R\) is the gas constant \(0.0821 \, \mathrm{L\cdot atm/mol\cdot K}\), and \(T\) is the temperature in kelvin:\[M = \frac{(1.88 \, \mathrm{g/L}) (0.0821 \, \mathrm{L\cdot atm/mol\cdot K}) (273 \, \mathrm{K})}{1 \, \mathrm{atm}} = 41.85 \, \mathrm{g/mol}\]This gives us the actual molecular weight of cyclopropane, indicating how many empirical units fit into the actual molecule.

Chemical Composition Analysis

Determining the molecular formula involves analyzing the chemical composition and combining it with calculations of empirical and molecular weights. Starting with the empirical formula \( \text{CH}_2 \), we find the actual molecular weight to be approximately \(41.85 \, \mathrm{g/mol}\) through density at STP.

The next step is to determine how the empirical formula relates to the molecular formula. This is done by dividing the molecular weight of cyclopropane by the molecular weight of the empirical formula:\[\frac{41.85 \, \mathrm{g/mol}}{14.03 \, \mathrm{g/mol}} = 2.98 \, \approx 3\]This means the molecular formula is three times the empirical formula, giving us \( \text{C}_3\text{H}_6 \).

In essence, chemical composition analysis bridges empirical observations and molecular reality, providing a clear path from initial mass percentages to a complete molecular formula.