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Chapter 5: Problem 167

A gaseous hydrocarbon (containing C and H atoms) in a container of volume \(20.2 \mathrm{~L}\) at \(350 \mathrm{~K}\) and 6.63 atm reacts with an excess of oxygen to form \(205.1 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(168.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the molecular formula of the hydrocarbon?

Short Answer

Expert verified
The molecular formula of the hydrocarbon is \(C_2H_6\).

Step by step solution

01

Determine the moles of \(CO_2\)

First, the number of moles of carbon dioxide can be determined. This is done by dividing the mass of the carbon dioxide produced by its molar mass. The molar mass of \(CO_2\) is approximately 44 g/mol. Therefore, using the formula \(n=\frac{m}{M}\), where \(n\) is the number of moles, \(m\) is mass and \(M\) is molar mass, we get \(n_{_{CO2}}=\frac{205.1 g}{44 g/mol}=4.66 mol\).
02

Determine number of carbon atoms

Knowing that the number of moles of carbon in \(CO_2\) is 4.66 mol and each molecule of \(CO_2\) contains one atom of carbon, therefore, there are 4.66 moles of carbon in the unknown hydrocarbon.
03

Determine moles of \(H_2O\)

Next, determine the number of moles of hydrogen present in the water produced. Divide the mass of the \(H_2O\) produced by its molar mass to get \(n_{_{H2O}}=\frac{168.0 g}{18 g/mol} = 9.33 mol\). Each \(H_2O\) molecule contains two hydrogen atoms, so the total number of moles of hydrogen in the hydrocarbon is \(2 \times 9.33 mol = 18.66 mol\).
04

Find the volume of the Hydrocarbon

Using the Ideal Gas Law \(PV = nRT\), determine the number of moles of the hydrocarbon, where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant and \(T\) is the temperature. It results in \(n_{_{HC}}=\frac{PV}{RT}=\frac{6.63 atm \times 20.2 L}{0.0821 L atm mol^{-1} K^{-1} \times 350 K}=1.83 mol\).
05

Determine Molecular Formula

Finally, using the number of moles of carbon and hydrogen, determined as 4.66 mol and 18.66 mol respectively, and the total number of moles of the hydrocarbon, calculated as 1.83 mol, we can determine the molecular formula. Since there are approximately 3 times as many moles of hydrogen as carbon, the molecular formula of the hydrocarbon is \(C_{1.83}H_{5.54}\), rounding the subscripts to the nearest whole number gives the empirical formula as \(C_2H_6\). This is also the molecular formula, since the hydrocarbon contains 1.83 moles (approximately equal to 2 moles).

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Most popular questions from this chapter

A mixture of helium and neon gases is collected over water at \(28.0^{\circ} \mathrm{C}\) and \(745 \mathrm{mmHg} .\) If the partial pressure of helium is \(368 \mathrm{mmHg}\), what is the partial pressure of neon? (Vapor pressure of water at \(28^{\circ} \mathrm{C}=\) \(28.3 \mathrm{mmHg} .\) )

A relation known as the barometric formula is useful for estimating the change in atmospheric pressure with altitude. The formula is given by \(P=P_{0} e^{-g .1 h / R T},\) where \(P\) and \(P_{0}\) are the pressures at height \(h\) and sea level, respectively; \(g\) is the acceleration due to gravity \(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right) ; \mathscr{M}\) is the average molar mass of air \((29.0 \mathrm{~g} / \mathrm{mol}) ;\) and \(R\) is the gas constant. Calculate the atmospheric pressure in atm at a height of \(5.0 \mathrm{~km}\), assuming the temperature is constant at \(5^{\circ} \mathrm{C}\) and \(P_{0}=1.0 \mathrm{~atm} .\)

If \(10.00 \mathrm{~g}\) of water are introduced into an evacuated flask of volume \(2.500 \mathrm{~L}\) at \(65^{\circ} \mathrm{C},\) calculate the mass of water vaporized. (Hint: Assume that the volume of the remaining liquid water is negligible; the vapor pressure of water at \(65^{\circ} \mathrm{C}\) is \(187.5 \mathrm{mmHg} .\)

What pressure will be required for neon at \(30^{\circ} \mathrm{C}\) to have the same density as nitrogen at \(20^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm} ?\)

At a certain temperature the speeds of six gaseous molecules in a container are \(2.0 \mathrm{~m} / \mathrm{s}, 2.2 \mathrm{~m} / \mathrm{s}, 2.6 \mathrm{~m} / \mathrm{s}\) \(2.7 \mathrm{~m} / \mathrm{s}, 3.3 \mathrm{~m} / \mathrm{s},\) and \(3.5 \mathrm{~m} / \mathrm{s} .\) Calculate the root- mean-square speed and the average speed of the molecules. These two average values are close to each other, but the root-mean-square value is always the larger of the two. Why?
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