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Chapter 6: Problem 39

A gaseous hydrocarbon weighing \(0.231 \mathrm{g}\) occupies a volume of \(102 \mathrm{mL}\) at \(23^{\circ} \mathrm{C}\) and \(749 \mathrm{mmHg} .\) What is the molar mass of this compound? What conclusion can you draw about its molecular formula?

Short Answer

Expert verified
The molar mass of the given gaseous hydrocarbon is approximately 57.3 g/mol. The molecular formula could be C4H10, C3H8, C2H6 or CH4. However, final judgement cannot be made from molar mass solely.

Step by step solution

01

Convert all quantities to appropriate units

First, convert the volume from mL to L by dividing it by 1000. So, \(102 mL = 0.102 L\). Temperature must be converted from Celsius to Kelvin by adding 273.15. So, \(23^{\circ}C = 296.15 K\). Lastly, the pressure is given in mmHg and we need to change it to atm by dividing the given pressure by 760 (since 1 atm = 760 mmHg), so \(749 mmHg = 0.9868 atm\).
02

Calculate the amount of moles

The equation \(PV = nRT\) is used, which is solved for n (amount of gas in moles), \(n = PV / RT\). Plugging in the values \(n = (0.9868 atm)(0.102 L) / ((0.0821 L atm)/(mol K))(296.15 K)\) we get \(n = 0.00403 mol\).
03

Use the number of moles to find molar mass

The molar mass of an element is determined by the formula \(Molar mass = mass / moles\). Plugging in the values, we have \(Molar mass = 0.231 g / 0.00403 mol = 57.3 g/mol\).
04

Draw a conclusion about the molecular formula

Given molar mass 57.3 g/mol, it leads to the conclusion that the molecular formula could be C4H10 (butane) or C3H8 (propane) or C2H6 (ethane) or CH4 (methane) as these have molar masses close to 57.3 g/mol.

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Most popular questions from this chapter

Chlorine dioxide, \(\mathrm{ClO}_{2}\), is sometimes used as a chlorinating agent for water treatment. It can be prepared from the reaction below: \(\mathrm{Cl}_{2}(\mathrm{g})+4 \mathrm{NaClO}(\mathrm{aq}) \longrightarrow 4 \mathrm{NaCl}(\mathrm{aq})+2 \mathrm{ClO}_{2}(\mathrm{g})\) In an experiment, \(1.0 \mathrm{L} \mathrm{Cl}_{2}(\mathrm{g}),\) measured at \(10.0^{\circ} \mathrm{C}\) and 4.66 atm, is dissolved in 0.750 L of 2.00 M \(\mathrm{NaClO}(\mathrm{aq}) .\) If \(25.9 \mathrm{g}\) of pure \(\mathrm{ClO}_{2}\) is obtained, then what is the percent vield for this experiment?

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