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Chapter 8: Problem 75

A quantity of \(1.4 \mathrm{~g}\) of a hydrocarbon gives \(1.8 \mathrm{~g}\) water on complete combustion. The empirical formula of hydrocarbon is (a) \(\mathrm{CH}\) (b) \(\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{3}\) (d) \(\overline{\mathrm{CH}}\),

Short Answer

Expert verified
Answer: (b) CH₂

Step by step solution

01

Write the general combustion reaction for a hydrocarbon

A hydrocarbon contains only carbon (C) and hydrogen (H). The general combustion reaction for any hydrocarbon can be represented as: \(\mathrm{C_x H_y} + \frac{(\mathrm{2x+y})}{2} \mathrm{O_2} \rightarrow \mathrm{x CO_2} + \frac{y}{2} \mathrm{H_2 O}\)
02

Determine the number of moles of water produced

We are given that \(1.8 \mathrm{~g}\) of water is produced. We need to convert this mass into moles to use the stoichiometry for further calculations. The molar mass of water (H₂O) is 18 g/mol. Therefore: Number of moles of water produced = \(\frac{1.8 \mathrm{~g}}{18 \mathrm{~g/mol}} = 0.1 \mathrm{~mol}\)
03

Find the number of moles of hydrogen in the hydrocarbon

From the balanced combustion equation, the ratio of moles of water to moles of hydrogen in hydrocarbon is: \(\frac{1}{2}\mathrm{H_2 O} : \frac{1}{2} \mathrm{H_y}\) Since there are 0.1 moles of water produced, the number of moles of hydrogen (H) in the hydrocarbon can be calculated using the ratio as follows: Number of moles of hydrogen = \(2 \times 0.1 = 0.2 \mathrm{~mol}\)
04

Calculate the mass of hydrogen in the hydrocarbon

Using the number of moles of hydrogen and the molar mass of hydrogen (1 g/mol), we can find the mass of hydrogen in the hydrocarbon: Mass of hydrogen = \(0.2 \mathrm{~mol} \times 1.0 \mathrm{~g/mol} = 0.2 \mathrm{~g}\)
05

Calculate the mass of carbon in the hydrocarbon

Since the total hydrocarbon mass is given as 1.4 g, we can find the mass of carbon by subtracting the mass of hydrogen from the total mass: Mass of carbon = \(1.4 \mathrm{~g} - 0.2 \mathrm{~g} = 1.2 \mathrm{~g}\)
06

Calculate the number of moles of carbon in the hydrocarbon

We can now calculate the number of moles of carbon in the hydrocarbon using the mass of carbon and the molar mass of carbon (12 g/mol): Number of moles of carbon = \(\frac{1.2 \mathrm{~g}}{12 \mathrm{~g/mol}} = 0.1 \mathrm{~mol}\)
07

Determine the empirical formula

The empirical formula is the simplest whole number ratio of moles of elements in a compound. In this case, it is the ratio of carbon (C) and hydrogen (H) moles: Carbon to hydrogen ratio = \(\frac{0.1}{0.1} : \frac{0.2}{0.1} = 1:2\) So the empirical formula of the hydrocarbon is \(\mathrm{CH_2}\). Therefore, the correct answer is (b) \(\mathrm{CH_2}\).

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Most popular questions from this chapter

Large quantities of ammonia are burned in the presence of a platinum catalyst to give nitric oxide, as the first step in the preparation of nitric acid. \(\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \stackrel{\mathrm{Pt}}{\longrightarrow} \mathrm{NO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (Unbalanced) Suppose a vessel contains \(0.12\) moles \(\mathrm{NH}_{3}\) and \(0.14\) moles \(\mathrm{O}_{2}\). How many moles of NO may be obtained?

A volume of \(500 \mathrm{ml}\) of a \(0.1 \mathrm{M}\) solution of \(\mathrm{AgNO}_{3}\) added to \(500 \mathrm{ml}\) of \(0.1 \mathrm{M}\) solution of \(\mathrm{KCl}\). The concentration of nitrate ion in the resulting solution is (a) \(0.05 \mathrm{M}\) (b) \(0.1 \mathrm{M}\) (c) \(0.2 \mathrm{M}\) (d) Reduced to zero

How much \(\mathrm{BaCl}_{2}\) would be needed to make \(250 \mathrm{ml}\) of a solution having the same concentration of \(\mathrm{Cl}^{-}\) as one containing \(\begin{array}{llll} & 3.78 & \mathrm{~g} & \mathrm{NaCl} \text { per } & 100 \mathrm{ml} ?\end{array}\) \((\mathrm{Ba}=137)\) (a) \(16.8 \mathrm{~g}\) (b) \(67.2 \mathrm{~g}\) (c) \(33.6 \mathrm{~g}\) (d) \(22.4 \mathrm{~g}\)

A mixture of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) is caused to react in a closed container to form \(\mathrm{NH}_{3} .\) The reaction ceases before either reactant has been totally consumed. At this stage, \(2.0 \mathrm{moles}\) each of \(\mathrm{N}_{2}, \mathrm{H}_{2}\) and \(\mathrm{NH}_{3}\) are present. The moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) present originally were, respectively, (a) 4 and 4 moles (b) 3 and 5 moles (c) 3 and 4 moles (d) 4 and 5 moles

A sample of ozone gas is found to be \(40 \%\) dissociated into oxygen. The average molecular mass of sample should be
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