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Chapter 3: Problem 149

Because of its detrimental effect on the environment, the lead compound described in Problem 3.148 has been replaced by methyl tert-butyl ether (a compound of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) ) to enhance the performance of gasoline. (This compound is also being phased out because of its contamination of drinking water.) When \(12.1 \mathrm{~g}\) of the compound are burned in an apparatus like the one shown in Figure \(3.6,30.2 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(14.8 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are formed. What is the empirical formula of the compound?

Short Answer

Expert verified
The empirical formula of the compound is \(C_{3}H_{7}O\).

Step by step solution

01

Determine Mass of Each Element

Using the given masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\), determine the mass of carbon and hydrogen. From \(\mathrm{CO}_{2}\), each carbon atom contributes a mass of 12.01 g/mol. Therefore, given that the mass of \(\mathrm{CO}_{2}\) is 30.2 g, the mass of Carbon is \(\frac{12.01}{44.01} \times 30.2 = 8.21 \mathrm{~g}\), and the mass of Oxygen is \(12.1 - 8.21 = 3.89 \mathrm{~g}\). From \(\mathrm{H}_{2} \mathrm{O}\), each hydrogen atom contributes a mass of 1.01 g/mol. Therefore, given that the mass of \(\mathrm{H}_{2} \mathrm{O}\) is 14.8 g, the mass of Hydrogen is \(\frac{(2 \times 1.01)}{18.02} \times 14.8 = 1.657 \mathrm{~g}\).
02

Calculate Number of Moles

We calculate the number of moles for each of Carbon, Hydrogen and Oxygen. Using the atomic masses, we get: \(\text{Moles of C} = \frac{8.21}{12.01} = 0.683 \mathrm{~mol}\), \(\text{Moles of H} = \frac{1.657}{1.01} = 1.64 \mathrm{~mol}\), \(\text{Moles of O} = \frac{3.89}{16.00} = 0.243 \mathrm{~mol}\).
03

Simplify Ratio of Atoms to Smallest Integer Values

The empirical formula is given by the ratio of moles \(C: H: O = 0.683: 1.64: 0.243\). The smaller ratio is 0.243 for \(O\), so divide each mole value by this: \(C: H: O = \frac{0.683}{0.243} : \frac{1.64}{0.243} : \frac{0.243}{0.243} = 2.81 : 6.75 : 1\). It seems like the ratio is not an integer. This might happen due to experimental errors. An approximate ratio is \(C: H: O = 3 : 7 : 1\), so the empirical formula is \(C_{3}H_{7}O\).

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Most popular questions from this chapter

The aluminum sulfate hydrate \(\left[\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot x \mathrm{H}_{2} \mathrm{O}\right]\)contains 8.10 percent \(\mathrm{Al}\) by mass. Calculate \(x-\) that is, the number of water molecules associated with each \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) unit.

Disulfide dichloride \(\left(\mathrm{S}_{2} \mathrm{Cl}_{2}\right)\) is used in the vulcanization of rubber, a process that prevents the slippage of rubber molecules past one another when stretched. It is prepared by heating sulfur in an atmosphere of chlorine:$$\mathrm{S}_{8}(l)+4 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{~S}_{2} \mathrm{Cl}_{2}(l)$$What is the theoretical yield of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) in grams when \(4.06 \mathrm{~g}\) of \(\mathrm{S}_{8}\) are heated with \(6.24 \mathrm{~g}\) of \(\mathrm{Cl}_{2} ?\) If the actual yield of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) is \(6.55 \mathrm{~g}\), what is the percent yield?

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating \(\mathrm{Si}\) in chlorine gas:$$\mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l)$$In one reaction, 0.507 mole of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?

The following reaction is stoichiometric as written$$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}+\mathrm{NaOC}_{2}\mathrm{H}_{5} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{8}+\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{OH}+\mathrm{NaCl}$$ but it is often carried out with an excess of \(\mathrm{NaOC}_{2} \mathrm{H}_{5}\) to react with any water present in the reaction mixture that might reduce the yield. If the reaction shown was carried out with \(6.83 \mathrm{~g}\) of \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\), how many grams of \(\mathrm{NaOC}_{2} \mathrm{H}_{5}\) would be needed to have a 50 percent molar excess of that reactant?

In the formation of carbon monoxide, CO, it is found that \(2.445 \mathrm{~g}\) of carbon combine with \(3.257 \mathrm{~g}\) of oxygen. What is the atomic mass of oxygen if the atomic mass of carbon is 12.01 amu?
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