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Chapter 6: Problem 46

A particular gaseous hydrocarbon that is \(82.7 \%\) C and \(17.3 \%\) H by mass has a density of \(2.33 \mathrm{g} / \mathrm{L}\) at \(23^{\circ} \mathrm{C}\) and \(746 \mathrm{mm} \mathrm{Hg} .\) What is the molecular formula of this hydrocarbon?

Short Answer

Expert verified
The molecular formula of the given hydrocarbon is \( C_2H_5 \).

Step by step solution

01

Calculate molar mass and ratio of elements

First, convert the percentages to grams (since the total molar mass is 100g). 82.7% C is \( 82.7g \) of C and 17.3% H is \( 17.3g \) of H. The number of moles of C and H can then be calculated using atomic masses (C is \(12.01 \, g/mol \) and H is \( 1.008 \, g/mol \)) which yields: \( n_C = \frac{82.7g}{12.01 \, g/mol}= 6.88 mol \) and \ \( n_H=\frac{17.3g}{1.008 \, g/mol}= 17.16 mol \).
02

Determine the empirical formula

The ratio of number of moles calculated in step 1 gives \(\frac{17.16 mol}{6.88 mol}= 2.49 \) which rounds approximately to \( 2.5 \). Given that molecular formulas must be in whole numbers, this suggests the empirical formula is \( C_2 H_5 \) .
03

Apply the Ideal Gas Law

To calculate the molar mass of the gas, the Ideal Gas Law: \( PV=nRT \) can be rearranged to: \( \frac{PM}{RT} = d \), where the molar mass M can be expressed as: \( M=\frac{dRT}{P} \). With temperatures in Kelvin, pressure in atm, R as \(\frac{0.0821 \, L \cdot atm}{mol \cdot K} \) and converting the given condition of pressure and temperature, you can find the molar mass. Replacing values, you get: \( M=\frac{2.33g/L \cdot 0.0821 \, L \cdot atm / (mol \cdot K) \cdot (23+273) \, K}{0.9803 \, atm}=30.07 \, g/mol \)
04

Determine the Molecular Formula

Next, determine how many empirical formula units are in the molecular formula. The molecular formula is found by simply multiplying the subscripts of the empirical formula by the factor. By: \( n=\frac{M_{molar}}{M_{empirical}}=\frac{30.07 \, g/mol}{29.07 \, g/mol}=1.03 \) which rounded off gives \(1\) . Hence, the molecular formula is \( C_2H_5 \) which is the same as the empirical formula because the factor is close to \(1\) .

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Most popular questions from this chapter

A nitrogen molecule ( \(\mathrm{N}_{2}\) ) having the average kinetic energy at \(300 \mathrm{K}\) is released from Earth's surface to travel upward. If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give your answer in kilometers. [Hint: When the molecule comes to rest, the potential energy of the molecule will be \(m g h\) where \(m\) is the molecular mass in kilograms, \(g=9.81 \mathrm{m} \mathrm{s}^{-2}\) is the acceleration due to gravity, and \(h\) is the height, in meters, above Earth's surface.]

Calculate the total kinetic energy, in joules, of \(155 \mathrm{g} \mathrm{N}_{2}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and 1.00 atm. \([\text {Hint}:\) First calculate the average kinetic energy, \(\bar{e}_{k}\).

Calculate \(u_{\mathrm{rms}},\) in meters per second, for \(\mathrm{Cl}_{2}(\mathrm{g})\) molecules at \(30^{\circ} \mathrm{C}\)

What is the molar volume of an ideal gas at (a) \(25^{\circ} \mathrm{C}\) and 1.00 atm; \((b) 100^{\circ} \mathrm{C}\) and 748 Torr?

The Haber process is the principal method for fixing nitrogen (converting \(\mathrm{N}_{2}\) to nitrogen compounds). $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})$$ Assume that the reactant gases are completely converted to \(\mathrm{NH}_{3}(\mathrm{g})\) and that the gases behave ideally. (a) What volume of \(\mathrm{NH}_{3}(\mathrm{g})\) can be produced from 152 \(\mathrm{L} \mathrm{N}_{2}(\mathrm{g})\) and \(313 \mathrm{L}\) of \(\mathrm{H}_{2}(\mathrm{g})\) if the gases are measured at \(315^{\circ} \mathrm{C}\) and 5.25 atm? (b) What volume of \(\mathrm{NH}_{3}(\mathrm{g}),\) measured at \(25^{\circ} \mathrm{C}\) and\(727 \mathrm{mmHg},\) can be produced from \(152 \mathrm{L} \mathrm{N}_{2}(\mathrm{g})\) and \(313 \mathrm{L} \mathrm{H}_{2}(\mathrm{g}),\) measured at \(315^{\circ} \mathrm{C}\) and \(5.25 \mathrm{atm} ?\)
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