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A solution is prepared by mixing \(1.28 \mathrm{mol} \mathrm{C}_{7} \mathrm{H}_{16}\) \(2.92 \mathrm{mol} \mathrm{C}_{8} \mathrm{H}_{18},\) and \(2.64 \mathrm{mol} \mathrm{C}_{9} \mathrm{H}_{20} .\) What is the (a) mole fraction and (b) mole percent of each component of the solution?

Short Answer

Expert verified
The mole fraction of the components are approximately 0.187 for \(C_{7}H_{16}\), 0.427 for \(C_{8}H_{18}\), and 0.386 for \(C_{9}H_{20}\). The mole percent of the components approximates to 18.7% for \(C_{7}H_{16}\), 42.7% for \(C_{8}H_{18}\), and 38.6% for \(C_{9}H_{20}\).

Step by step solution

01

Calculate Total Number of Moles

Firstly, calculate the total number of moles in the mixture by adding the number of moles of each component. This is done as: \(1.28 \, mol \, C_{7}H_{16} + 2.92 \, mol \, C_{8}H_{18} + 2.64 \, mol \, C_{9}H_{20} = 6.84 \, mol\) total.
02

Calculate Mole Fraction

Secondly, calculate the mole fraction of each component by dividing the number of moles of the component by the total number of moles in the mixture. For \(C_{7}H_{16}\): \(\frac{1.28 \, mol}{6.84 \, mol} = 0.187\). For \(C_{8}H_{18}\): \(\frac{2.92 \, mol}{6.84 \, mol} = 0.427\). For \(C_{9}H_{20}\): \(\frac{2.64 \, mol}{6.84 \, mol} = 0.386\)
03

Calculate Mole Percent

Thirdly, calculate the mole percent of each component by multiplying the mole fraction by 100. For \(C_{7}H_{16}\): \(0.187 * 100 = 18.7\% \). For \(C_{8}H_{18}\): \(0.427 * 100 = 42.7\% \). For \(C_{9}H_{20}\): \(0.386 * 100 = 38.6 \% \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Mole Fraction
Imagine you have a pot of soup made with different ingredients. In solution chemistry, the soup represents a mixture, and the ingredients are different chemical components like our molecules C\(_{7}\)H\(_{16}\), C\(_{8}\)H\(_{18}\), and C\(_{9}\)H\(_{20}\). Each component contributes a specific amount of substance to the mixture, and the mole fraction helps tell us about the proportion of each component relative to the whole mixture.

The mole fraction, \( X_i \), of a component in a mixture is defined as the number of moles of that component, \( n_i \), divided by the total number of moles in the mixture, \( n_{\text{total}} \). The formula looks like this:
\[ X_i = \frac{n_i}{n_{\text{total}}} \]

For example, in the exercise, once we summed up the moles of all components and found it to be 6.84 mol, each mole fraction tells us the fraction of the total moles that a particular component contributes:
  • C\(_{7}\)H\(_{16}\): \( 0.187 \)
  • C\(_{8}\)H\(_{18}\): \( 0.427 \)
  • C\(_{9}\)H\(_{20}\): \( 0.386 \)
These values not only show us the proportion of each component but they also sum up to 1, demonstrating that together, they represent the complete mixture.
Exploring Mole Percent
Sometimes, expressing fractions as percentages can be more intuitive, just like saying something is 50% full instead of half full. That's where mole percent comes into play in solution chemistry. It uses the mole fraction and converts it into a percentage to make it simpler to understand and compare.

To calculate the mole percent, you take the mole fraction and multiply it by 100. This doesn't change the information, but it changes how we express it, making it easier to interpret. So, for our components, here's what the mole percent looks like:
  • C\(_{7}\)H\(_{16}\): \( 18.7 \% \)
  • C\(_{8}\)H\(_{18}\): \( 42.7 \% \)
  • C\(_{9}\)H\(_{20}\): \( 38.6 \% \)
This conversion helps in understanding and communicating the composition of a mixture more effectively, especially when comparing different samples or assessing concentration levels.
Defining the Mixture
A mixture is a combination of two or more components in which each retains its own chemical identity and properties. In our exercise, the mixture consists of different hydrocarbons: C\(_{7}\)H\(_{16}\), C\(_{8}\)H\(_{18}\), and C\(_{9}\)H\(_{20}\). Each of these components contributes to the whole by adding to the total amount of substance.

The beauty of mixtures lies in their variability and the ability to optimize their composition for specific uses. Different mixtures can exhibit diverse properties based on the types and ratios of components involved. For instance:
  • Homogeneous Mixtures: The composition is uniform throughout, like the solution in our example, where molecules are evenly distributed at a molecular level.
  • Heterogeneous Mixtures: The components are distinctly separate, and composition varies within the mixture, like a salad.
In chemistry, solutions like the one in the exercise are typically homogeneous because the dissolved substances are spread evenly. Understanding mixtures allows scientists to predict how they will behave in different chemical reactions and processes.

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Most popular questions from this chapter

Two of the substances listed here are highly soluble in water, two are only slightly soluble in water, and two are insoluble in water. Indicate the situation you expect for each one. (a) iodoform, \(\mathrm{CHI}_{3}\) (b) benzoic acid, (c) formic acid, (d) 1-butanol, (e) chlorobenzene, (f) propylene glycol, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}\)

How many grams of iodine, \(I_{2}\), must be dissolved in \(725 \mathrm{mL}\) of carbon disulfide, \(\mathrm{CS}_{2}(d=1.261 \mathrm{g} / \mathrm{mL}),\) to produce a \(0.236 \mathrm{m}\) solution?

A saturated solution prepared at \(70^{\circ} \mathrm{C}\) contains \(32.0 \mathrm{g}\) CuSO \(_{4}\) per 100.0 g solution. A 335 g sample of this solution is then cooled to \(0^{\circ} \mathrm{C}\) and \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) crystallizes out. If the concentration of a saturated solution at \(0^{\circ} \mathrm{C}\) is \(12.5 \mathrm{g} \mathrm{CuSO}_{4} / 100 \mathrm{g}\) soln, what mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) would be obtained? [Hint: Note that the solution composition is stated in terms of \(\mathrm{CuSO}_{4}\) and that the solid that crystallizes is the hydrate \(\left.\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} .\right]\)

An important test for the purity of an organic compound is to measure its melting point. Usually, if the compound is not pure, it begins to melt at a lower temperature than the pure compound.(a) Why is this the case, rather than the melting point being higher in some cases and lower in others?(b) Are there any conditions under which the melting point of the impure compound is higher than that of the pure compound? Explain.

An ideal liquid solution has two volatile components. In the vapor in equilibrium with the solution, the mole fractions of the components are (a) both \(0.50 ;\) (b) equal, but not necessarily \(0.50 ;\) (c) not very likely to be equal; (d) 1.00 for the solvent and 0.00 for the solute.

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