Chapter 11: Problem 77

A gaseous compound of carbon and hydrogen contains \(80.0 \%\) by mass carbon. A \(2.00\) -L sample has a mass of \(2.678 \mathrm{~g}\) at \(\mathrm{STP}\). (a) Calculate the molar mass of the compound. (b) What is the empirical formula of the compound? (c) What is the molecular formula of the compound?

Short Answer

Expert verified

The molar mass of the gaseous compound is 29.5 g/mol, the empirical formula is CH₃, and the molecular formula is C₂H₆ (ethane).

Step by step solution

Calculate mass of carbon and hydrogen in the sample

Given that the compound is 80.0% carbon by mass, we can calculate the mass of carbon and hydrogen in the sample. Mass of carbon = (80.0% of 2.678 g) = 0.800 * 2.678 g = 2.142 g Mass of hydrogen = 2.678 g - 2.142 g = 0.536 g

Calculate moles of carbon and hydrogen

To determine the molar mass of the compound, let's first calculate the moles of carbon and hydrogen, using their respective molar masses (12.01 g/mol for carbon and 1.008 g/mol for hydrogen). Moles of carbon = 2.142 g / 12.01 g/mol = 0.178 mol Moles of hydrogen = 0.536 g / 1.008 g/mol = 0.532 mol

Use the ideal gas law to calculate molar mass

We can use the ideal gas law (PV = nRT) to find the moles of the compound in the 2-L sample at STP. At STP, the pressure (P) is 1 atm, the temperature (T) is 273 K, and the gas constant (R) is 0.0821 L atm/mol K. 1 atm * 2.00 L = n * 0.0821 L atm/mol K * 273 K n = (1 atm * 2.00 L) / (0.0821 L atm/mol K * 273 K) = 0.0909 moles of compound Molar mass of the compound = mass of the sample / moles of compound = 2.678 g / 0.0909 mol = 29.5 g/mol

Divide moles of carbon and hydrogen by the smallest value to find the empirical formula

Divide the moles of both carbon and hydrogen by the smallest value (0.178 mol for carbon) to determine the ratio for the empirical formula. Carbon ratio = 0.178 mol / 0.178 mol = 1 Hydrogen ratio = 0.532 mol / 0.178 mol = 2.99 ≈ 3 Empirical formula: CH₃

Determine the molecular formula of the compound

To find the molecular formula, divide the molar mass of the compound (29.5 g/mol) by the molar mass of the empirical formula (CH₃). Molar mass of CH₃ = 12.01 g/mol + (3 * 1.008 g/mol) = 15.03 g/mol Molecular formula multiple = 29.5 g/mol / 15.03 g/mol = 1.96 ≈ 2 Molecular formula: 2 (CH₃) = C₂H₆ The molecular formula of the compound is C₂H₆ (ethane).

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.

Start your free trial

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

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Short Answer

Step by step solution

Calculate mass of carbon and hydrogen in the sample

Calculate moles of carbon and hydrogen

Use the ideal gas law to calculate molar mass

Divide moles of carbon and hydrogen by the smallest value to find the empirical formula

Determine the molecular formula of the compound

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Chemical Analysis

Physical Chemistry

Inorganic Chemistry

Chemistry Branches

Chemical Reactions

Study anywhere. Anytime. Across all devices.