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

Expert verified
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

01

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.
02

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: \(12.01\, \mathrm{g/mol} + 2\cdot 1.01\, \mathrm{g/mol} = 14.03\, \mathrm{g/mol}\)
03

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free