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Chapter 8: Problem 137

A compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\). It is \(58.51 \%\) C and \(7.37 \%\) H by mass. Helium effuses through a porous frit \(3.20\) times as fast as the compound does. Determine the empirical and molecular formulas of this compound.

Short Answer

Expert verified
The empirical formula of the compound is \(\mathrm{C_2H_3N}\), and the molecular formula is also \(\mathrm{C_2H_3N}\) as the molecular formula ratio is approximately 1.

Step by step solution

01

Calculate the mole ratio of C, H, and N in the compound

First, we need to find the mole ratio of C, H, and N in the compound. We will assume that we have 100 g of the compound, so the mass percentages can be used as masses in grams. Let's calculate the number of moles of each element using their atomic masses: Moles of C = (58.51 g C) / (12.01 g/mol) = 4.87 mol C Moles of H = (7.37 g H) / (1.01 g/mol) = 7.30 mol H Since the compound contains only C, H, and N (and we know the mass percentage of C and H), we can calculate the mass percentage of N and then determine the moles of N: Mass % of N = 100% - (58.51% + 7.37%) = 34.12 % Moles of N = (34.12 g N) / (14.01 g/mol) = 2.44 mol N
02

Convert the mole ratios to the simplest whole numbers

Now, let's convert the mole ratios calculated above to the simplest whole numbers. We will divide each mole value by the smallest mole value: Moles of C = 4.87 mol / 2.44 ≈ 2 Moles of H = 7.30 mol / 2.44 ≈ 3 Moles of N = 2.44 mol / 2.44 ≈ 1 The simplest whole number ratio for this compound is 2:3:1. So, the empirical formula is \(\mathrm{C_2H_3N}\).
03

Use Graham's Law to find the molar mass of the compound

Graham's Law of Effusion states that the rate of effusion of two different gases at the same temperature and pressure is inversely proportional to the square root of their molar masses: \( \frac{ Rate_{1} } { Rate_{2} } = \sqrt{ \frac{MM_{2}}{MM_{1}} }\) In this problem, helium (He) effuses 3.20 times as fast as the compound: \( \frac{Rate_{He}}{Rate_{Compound}} = \sqrt{ \frac{MM_{Compound}}{MM_{He}} }\) We know the molar mass of helium (He) is 4.00 g/mol: \( 3.20 = \sqrt{ \frac{MM_{Compound}}{4.00} }\) Now, let's solve for the molar mass of the compound: \(MM_{Compound} = 4.00 \times 3.20^{2} \) \(MM_{Compound} ≈ 40.96\, g/mol \)
04

Calculate the molecular formula of the compound

To find the molecular formula, we need to determine the ratio of the molar mass of the compound to the molar mass of the empirical formula: Molecular Formula Ratio = \( \frac{Molar\,Mass\,of\,Compound}{Molar\,Mass\,of\,Empirical\,Formula} \) Molar mass of empirical formula (\(\mathrm{C_2H_3N}\)) = (2 × 12.01) + (3 × 1.01) + (1 × 14.01) = 30.05 g/mol Molecular Formula Ratio = \( \frac{40.96}{30.05} ≈ 1.364 \) Since the molecular formula must consist of whole-number multiples of the empirical formula, we can round this ratio to the nearest whole number: Molecular Formula Ratio ≈ 1 Therefore, the molecular formula of the compound is the same as the empirical formula, which is \(\mathrm{C_2H_3N}\).

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