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Chapter 9: Q54E (page 510)

How could you show experimentally that the molecular formula of propene is \({{\rm{C}}_3}{{\rm{H}}_6}\), not \({\rm{C}}{{\rm{H}}_2}\)?

Short Answer

Expert verified

We can determine propene's molar mass by calculating its mass, volume, and pressure. The projected molar mass would be three times that of \({\rm{C}}{{\rm{H}}_2}\)this leads to the conclusion that propene's molecular formula is \({{\rm{C}}_3}{{\rm{H}}_6}\), not \({\rm{C}}{{\rm{H}}_2}\).

Step by step solution

01

Definition of volume

A substance's volume is the amount of space it occupies

02

Explanation

Let us solve the given problem.

\(\frac{m}{v} = \frac{{P\mathcal{M}}}{{RT}}\)

\(\mathcal{M}\)- Molar mass of the compound.

\(R,T\)are constants.

From the equation, we can see that the molar mass of the compound depends on the mass, volume and pressure of the compound. So if we calculate mass, volume and pressure of propene, we can determine propene's molar mass. The molar mass calculated would be thrice the molar mass of \({\rm{C}}{{\rm{H}}_2}\).

Therefore, molecular formula of propene is\({{\rm{C}}_3}{{\rm{H}}_6}\), not \({\rm{C}}{{\rm{H}}_2}\).

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Most popular questions from this chapter

Ethanol, C2H5OH,is produced industrially from ethylene, C2H4, by the following sequence of reactions:

\begin{aligned}{\rm{3}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{+2}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\to{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{HS}}{{\rm{O}}_{\rm{4}}}{\rm{+(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{{\rm{)}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\\{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{HS}}{{\rm{O}}_{\rm{4}}}{\rm{ +(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{{\rm{)}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{+3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\to{\rm{3}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH + 2}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \end{aligned}

What volume of ethylene at STP is required to produce 1000 metric tons (1000kg) of ethanol if the overall yield of ethanol is 90.1%?

In addition to the data found in Figure \({\rm{9}}{\rm{.13}}\), what other information do we need to find the mass of the sample of air used to determine the graph?

Question: Calculate the relative rate of diffusion of \(^{\rm{1}}{{\rm{H}}_{\rm{2}}}\) (molar mass \({\rm{2}}{\rm{.0 g/mol}}\)) compared to that of \(^{\rm{2}}{{\rm{H}}_{\rm{2}}}\) (molar mass \({\rm{4}}{\rm{.0 g/mol}}\)) and the relative rate of diffusion of \({{\rm{O}}_{\rm{2}}}\) (molar mass \({\rm{32 g/mol}}\)) compared to that of \({{\rm{O}}_{\rm{3}}}\) (molar mass \({\rm{48 g/mol}}\)).

The volume of an automobile air bag was 66.8 Lwhen inflated at \({{\rm{2}}^{\rm{^\circ }}}{\rm{C}}\)with 77.8 gof nitrogen gas. What was the pressure in the bag in kPa?

One method of analyzing amino acids is the van Slyke method. The characteristic amino groups(-NH2) in protein material are allowed to react with nitrous acid, HNO2, to form N2 gas. From the volume of the gas, the amount of amino acid can be determined. A 0.0604-g sample of a biological sample containing glycine, CH2(NH2)CO2H, was analyzed by the van Slyke method and yielded 3.70mL of N2 collected over water at a pressure of 735 torrs and 29 ̊C. What was the percentage of glycine in the sample?

\[{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{(N}}{{\rm{H}}_{\rm{2}}}{\rm{)C}}{{\rm{O}}_{\rm{2}}}{\rm{H+HN}}{{\rm{O}}_{\rm{2}}}\to{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{(OH)C}}{{\rm{O}}_{\rm{2}}}{\rm{H+}}{{\rm{H}}_{\rm{2}}}{\rm{O+}}{{\rm{N}}_{\rm{2}}}\]
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