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Chapter 5: Problem 115

Cyclopropane, a gas that when mixed with oxygen is used as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. If the density of cyclopropane is \(1.88 \mathrm{~g} / \mathrm{L}\) at \(\mathrm{STP}\), what is the molecular formula of cyclopropane?

Short Answer

Expert verified
The molecular formula of cyclopropane is \(\mathrm{C_3H_6}\).

Step by step solution

01

Find moles of C and H

Assume 100 g of cyclopropane which contains 85.7 g C and 14.3 g H. To find the moles of C and H, we need to divide the mass by each element's molar mass. Moles of C: \(\frac{85.7 \mathrm{~g}}{12.01\mathrm{~g/mol}} = 7.14 \mathrm{mol}\) Moles of H: \(\frac{14.3 \mathrm{~g}}{1.01\mathrm{~g/mol}} = 14.16 \mathrm{mol}\)
02

Obtain the empirical formula

Divide moles of C and H by the smallest mole value, which will give the ratio of atoms in the empirical formula. C: \(\frac{7.14}{7.14} = 1\) H: \(\frac{14.16}{7.14} = 1.98 \approx 2\) The empirical formula is \(\mathrm{CH_2}\).
03

Calculate the molar mass of the empirical formula

Molar mass of the empirical formula is the sum of the molar masses of the individual elements multiplied by the number of atoms in the empirical formula. Molar mass of \(\mathrm{CH_2}\): \(12.01\mathrm{~g/mol} + 2(1.01\mathrm{~g/mol}) = 14.03\mathrm{~g/mol}\)
04

Determine the molar mass of the compound

We know the density of cyclopropane at STP is \(1.88\mathrm{~g/L}\). Recall that STP conditions are 0°C and 1 atm, and the molar volume of any gas at STP is approximately 22.4 L/mol. Use the density to determine the molar mass of the compound. Molar mass of cyclopropane: \(1.88\mathrm{~g/L} \cdot 22.4\mathrm{~L/mol} = 42.11\mathrm{~g/mol}\)
05

Find the molecular formula

To determine the molecular formula of cyclopropane, divide the molar mass of the compound by the molar mass of the empirical formula. \(\frac{42.11\mathrm{~g/mol}}{14.03\mathrm{~g/mol}} = 3\) As the ratio obtained is close to 3, it means the molecular formula is three times the empirical formula, so after multiplying: Cyclopropane molecular formula: \(\mathrm{C_3H_6}\)

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Most popular questions from this chapter

Consider the following chemical equation. $$ 2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ If \(25.0 \mathrm{~mL} \mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g)\), ammonia, \(\mathrm{NH}_{3}\left(\mathrm{~g}\right.\) ), and oxygen, \(\mathrm{O}_{2}(\mathrm{~g})\), at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of \(\mathrm{HCN}(\mathrm{g})\) can be obtained from the reaction of \(20.0 \mathrm{~L} \mathrm{CH}_{4}(g), 20.0 \mathrm{~L} \mathrm{NH}_{3}(g)\), and \(20.0 \mathrm{~L} \mathrm{O}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.

A spherical glass container of unknown volume contains helium gas at \(25^{\circ} \mathrm{C}\) and \(1.960 \mathrm{~atm}\). When a portion of the helium is withdrawn and adjusted to \(1.00\) atm at \(25^{\circ} \mathrm{C}\), it is found to have a volume of \(1.75 \mathrm{~cm}^{3}\). The gas remaining in the first container shows a pressure of \(1.710\) atm. Calculate the volume of the spherical container.

A balloon is filled to a volume of \(7.00 \times 10^{2} \mathrm{~mL}\) at a temperature of \(20.0^{\circ} \mathrm{C}\). The balloon is then cooled at constant pressure to a temperature of \(1.00 \times 10^{2} \mathrm{~K}\). What is the final volume of the balloon?

Metallic molybdenum can be produced from the mineral molybdenite, \(\mathrm{MoS}_{2}\). The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are $$ \begin{aligned} \mathrm{MoS}_{2}(s)+\frac{2}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{MoO}_{3}(s)+2 \mathrm{SO}_{2}(g) \\ \mathrm{MoO}_{3}(s)+3 \mathrm{H}_{2}(g) & \longrightarrow \mathrm{Mo}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Calculate the volumes of air and hydrogen gas at \(17^{\circ} \mathrm{C}\) and \(1.00\) atm that are necessary to produce \(1.00 \times 10^{3} \mathrm{~kg}\) pure molybdenum from \(\mathrm{MoS}_{2}\). Assume air contains \(21 \%\) oxygen by volume and assume \(100 \%\) yield for each reaction.
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