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Which of the following statements about \(57.1 \mathrm{g}\) of octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) is (are) not true? (a) \(57.1 \mathrm{g}\) is 0.500 mol of octane. (b) The compound is \(84.1 \%\) C by weight. (c) The empirical formula of the compound is \(\mathrm{C}_{4} \mathrm{H}_{3}\) (d) \(57.1 \mathrm{g}\) of octane contains \(28.0 \mathrm{g}\) of hydrogen atoms.

Short Answer

Expert verified
Statements (c) and (d) are not true.

Step by step solution

01

Determine Molar Mass of Octane

First, we need to calculate the molar mass of octane, \ \(\text{C}_8\text{H}_{18}\). Using the atomic masses, we have 8 carbon atoms and 18 hydrogen atoms: \(8 \times 12.01 \, \text{g/mol} + 18 \times 1.008 \, \text{g/mol} = 96.08 \, \text{g/mol} + 18.144 \, \text{g/mol} = 114.224 \, \text{g/mol}\).
02

Verify Statement (a)

To determine the number of moles in 57.1 g of octane, divide the mass by the molar mass: \[\frac{57.1 \, \text{g}}{114.224 \, \text{g/mol}} \approx 0.500 \, \text{mol}.\] Statement (a) is true.
03

Verify Statement (b)

Calculate the percentage of carbon in octane. The molar mass of carbon in octane is \(8 \times 12.01 = 96.08 \, \text{g/mol}\). The percentage of carbon by weight is \[\frac{96.08}{114.224} \times 100\% \approx 84.17\%.\] Thus, statement (b) is true.
04

Verify Statement (c)

To find the empirical formula, express the subscripts in the lowest terms. The molecular formula is \(\text{C}_8\text{H}_{18}\), which simplifies to \(\text{C}_4\text{H}_9\) (not \(\text{C}_4\text{H}_3\)). Therefore, statement (c) is not true.
05

Verify Statement (d)

Calculate the mass of hydrogen in 57.1 g of octane. Since the molar mass of hydrogen in octane is \(18 \times 1.008 = 18.144 \, \text{g/mol}\), the mass of hydrogen can be calculated by its proportion: \[\frac{18.144}{114.224} \times 57.1 \, \text{g} \approx 9.072 \, \text{g}.\] Thus, statement (d) is not true.

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Key Concepts

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

Empirical Formula
The empirical formula of a compound provides the simplest whole-number ratio of the elements within it. This means that it reflects the lowest terms of molecules present in a compound. To determine the empirical formula, divide the numbers of moles of each element by the smallest number of moles calculated. This may involve dividing by a common factor if applicable.
For octane, the molecular formula is \(\text{C}_8\text{H}_{18}\). To find the empirical formula, we need to express these subscripts in their simplest terms. By dividing 8 by 2 and 18 by 2, the empirical formula simplifies to \(\text{C}_4\text{H}_9\). This shows that the statement given in the original exercise, \(\text{C}_4\text{H}_3\), was incorrect.
The information from empirical formulas is useful as it gives insight into the relative proportions of each element, helping chemists understand compositional makeup and predict how a compound might react.
Percentage Composition
Percentage composition informs us of the proportion, by mass, of each element within a compound. It is calculated by dividing the total mass of each element in a formula unit by the molar mass of the whole compound, then multiplying by 100 to convert it into a percentage.
For calculating the percentage of carbon in octane \((\text{C}_8\text{H}_{18})\), you take the molar mass of all the carbon atoms in the molecule (8 carbons each with a molar mass of 12.01 g/mol for a total of 96.08 g/mol). Divide this by the molar mass of octane (114.224 g/mol) and then multiply by 100% to get \(\approx 84.17\%\).
  • This approach can help chemists determine purity and analyze mixture compositions.
Thus, knowing the percentage composition can validate the composition of a sample or compare it against a theoretical model to check for deviations.
Molecular Formula
The molecular formula specifies the exact number of each type of atom present in a molecule, providing a complete representation of its chemical structure. To determine a molecular formula, one often starts by identifying the empirical formula, then uses the compound's molar mass to determine how many empirical formula units are present in one molecule.
For a compound like octane, we directly use the molecular formula, \(\text{C}_8\text{H}_{18}\), given that the molar mass of octane is approximately 114.224 g/mol. Typically, chemists verify molecular formulas by dividing the compound's molar mass by the empirical formula mass.
  • The molecular formula provides a more specific description of a compound compared to the empirical formula, making it crucial for understanding chemical reactions and properties.
  • Given the right data, any empirical formula can be expanded to a molecular formula; knowing both helps in identifying compounds and predicting their reactivity.
Understanding a substance's molecular formula is essential in fields such as synthetic chemistry and pharmacology, where precise molecular manipulation is key.

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Most popular questions from this chapter

Uranium is used as a fuel, primarily in the form of uranium(IV) oxide, in nuclear power plants. This question considers some uranium chemistry. (a) A small sample of uranium metal (0.169 g) is heated to between 800 and \(900^{\circ} \mathrm{C}\) in air to give 0.199 g of a dark green oxide, Ux \(\mathrm{O}_{y}\) How many moles of uranium metal were used? What is the empirical formula of the oxide, \(\mathrm{U}_{x} \mathrm{O}_{y}\) ? What is the name of the oxide? How many moles of \(\mathrm{U}_{x} \mathrm{O}_{y}\) must have been obtained? (b) The naturally occurring isotopes of uranium are \(^{234} \mathrm{U},^{235} \mathrm{U},\) and \(^{238}\) U. Knowing that uranium's atomic weight is \(238.02 \mathrm{g} / \mathrm{mol},\) which isotope must be the most abundant? (c) If the hydrated compound \(\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2} \cdot z \mathrm{H}_{2} \mathrm{O}\) is heated gently, the water of hydration is lost. If you have \(0.865 \mathrm{g}\) of the hydrated compound and obtain \(0.679 \mathrm{g}\) of \(\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) on heating, how many waters of hydration are in each formula unit of the original compound? (The oxide \(\mathrm{U}_{x} \mathrm{O}_{y}\) is obtained if the hydrate is heated to temperatures over \(\left.800^{\circ} \mathrm{C} \text { in the air. }\right)\)

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