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Chapter 8: Problem 72

A compound has carbon, hydrogen, and oxygen in \(3: 3: 1\) atomic ratio. If the number of moles in \(1 \mathrm{~g}\) of the compound is \(6.06 \times 10^{-3}\), the molecular formula of the compound will be (a) \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{9} \mathrm{H}_{9} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{O}_{4}\)

Short Answer

Expert verified
Answer: The molecular formula of the compound is C9H9O3.

Step by step solution

01

Determine the empirical formula from the given atomic ratio

We are given the atomic ratio of carbon, hydrogen, and oxygen as \(3:3:1\). This indicates that the ratio of the number of moles of each element in the simplest form of the compound (the empirical formula) is: - Carbon: 3 moles - Hydrogen: 3 moles - Oxygen: 1 mole Thus, the empirical formula of the compound is \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}\).
02

Calculate the molar mass of the empirical formula

Now we'll find the molar mass of the empirical formula \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}\). Molar mass = 3 × (mass of \(\mathrm{C}\)) + 3 × (mass of \(\mathrm{H}\)) + (mass of \(\mathrm{O}\)) The molar masses of \(\mathrm{C}\), \(\mathrm{H}\), and \(\mathrm{O}\) are 12 g/mol, 1 g/mol, and 16 g/mol, respectively. Molar mass = 3 × 12 + 3 × 1 + 16 = 36 + 3 + 16 = 55 g/mol The molar mass of the empirical formula \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}\) is 55 g/mol.
03

Find the molecular formula

We are given the number of moles in 1 gram of the compound as \(6.06 \times 10^{-3}\) moles. From this given, we can find the molar mass of the molecular formula: Molar mass of molecular formula = Mass / Number of moles Molar mass of molecular formula = 1 g / (6.06 × 10^{-3}\,\text{moles}) Molar mass of molecular formula = 165 g/mol Now, we'll find the ratio between the molar mass of the molecular formula and the molar mass of the empirical formula: Ratio = (Molar mass of molecular formula) / (Molar mass of empirical formula) Ratio = 165 / 55 = 3 As the above ratio is 3, this indicates that the molecular formula is 3 times the empirical formula. Therefore, the molecular formula of the compound is \(\mathrm{C}_{9} \mathrm{H}_{9} \mathrm{O}_{3}\). Hence, the correct answer is (c) \(\mathrm{C}_{9} \mathrm{H}_{9} \mathrm{O}_{3}\).

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Key Concepts

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

Empirical Formula
Understanding the empirical formula is essential to mastering molecular formula calculations. The empirical formula represents the simplest ratio of atoms of each element present in a compound. It's a kind of chemical shorthand that indicates the relative numbers of atoms of different elements in the substance.

For instance, in our exercise, the given atomic ratio of carbon to hydrogen to oxygen is 3:3:1. This ratio simplifies to the empirical formula of \(\mathrm{C}_3\mathrm{H}_3\mathrm{O}\), where the subscripts denote the number of atoms of each element in the simplest form. It's necessary to note that empirical formulas do not provide information about the actual number of atoms in a molecule, just the simplest ratio.
Mole Concept
The mole concept is a fundamental pillar of chemistry, as it relates the mass of a substance to its chemical equivalent. A mole is defined as the amount of a substance that contains as many entities (such as atoms, molecules, or ions) as there are atoms in 12 grams of pure carbon-12.

The number of entities in a mole is known as Avogadro's number (\(6.022 \times 10^{23}\)). It allows chemists to count particles by weighing, as moles link the macroscopic world we observe to the microscopic world of atoms and molecules. In our problem, the information refers to moles, with 1 gram of the compound containing \(6.06 \times 10^{-3}\) moles, highlighting the critical role of mole concept in calculating the molecular formula.
Atomic Ratio
The atomic ratio is critical in determining the empirical formula of a compound. It gives a clear picture of the proportion of the constituent elements in a molecule. This ratio needs to be converted into moles, serving as the first step toward finding both empirical and molecular formulas.

In the provided exercise, the atomic ratio is given directly, which speeds up the process of determining the empirical formula. Remember that the atomic ratio does not always represent whole numbers of atoms in a molecule, but rather the simplest, whole-number ratio of atoms.
Molar Mass
Molar mass is the mass of one mole of a substance (usually expressed in grams per mole). It's a vital concept when it comes to converting between grams and moles, and is directly related to the molecular weight of a substance from the periodic table. For our exercise, we calculate the molar mass of the empirical formula by adding together the molar masses of each element present, multiplied by their respective quantities in the formula.

The molar mass allows us to compare the relate the amount of substance to its mass and it ties together the empirical formula with the actual molecular formula. Knowing that our empirical formula has a molar mass of 55 g/mol and comparing it to the molar mass of the molecular formula yields the factor necessary to upscale the empirical formula to the molecular formula.

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Most popular questions from this chapter

A mixture of methane and ethylene in the ratio of a:b by volume occupies \(30 \mathrm{ml}\). On complete combustion, the mixture yield \(40 \mathrm{ml}\) of \(\mathrm{CO}_{2} .\) What volume of \(\mathrm{CO}_{2}\) would have been obtained if the ratio would have been b:a? (a) \(50 \mathrm{ml}\) (b) \(30 \mathrm{ml}\) (c) \(40 \mathrm{ml}\) (d) \(60 \mathrm{ml}\)

An amount of \(1.0 \times 10^{-3}\) moles of \(\mathrm{Ag}^{+}\) and \(1.0 \times 10^{-3}\) moles of \(\mathrm{CrO}_{4}^{2-}\) reacts together to form solid \(\mathrm{Ag}_{2} \mathrm{CrO}_{4} \cdot\) What is the amount of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) formed? \((\mathrm{Ag}=108, \mathrm{Cr}=52)\) (a) \(0.332 \mathrm{~g}\) (b) \(0.166 \mathrm{~g}\) (c) \(332 \mathrm{~g}\) (d) \(166 \mathrm{~g}\)

When \(12 \mathrm{~g}\) graphite is burnt in sufficient oxygen, \(\mathrm{CO}\) as well as \(\mathrm{CO}_{2}\) is formed. If the product contains \(40 \%\) CO and \(60 \%\) \(\mathrm{CO}_{2}\) by mass and none of the reactant is left, what is the mass of oxygen gas used in combustion? (a) \(24.0 \mathrm{~g}\) (b) \(21.33 \mathrm{~g}\) (c) \(23.8 \mathrm{~g}\) (d) \(15.6 \mathrm{~g}\)

Iron form two oxides. If for the same mass of iron, mass of oxygen combined in the first oxide is two-third of the mass of oxygen combined in the second oxide, the ratio of valency of iron in first and second oxide is (a) \(1: 1\) (b) \(2: 3\) (c) \(3: 2\) (d) \(2: 5\)

A volume of \(10 \mathrm{~m}\) l chlorine gas combines with \(25 \mathrm{~m}\) l of oxygen gas to form \(10 \mathrm{ml}\) of a gaseous compound. If all the volumes are measured at the same pressure and temperature, what is the molecular formula of compound formed? (a) \(\mathrm{Cl}_{2} \mathrm{O}\) (b) \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) (c) \(\mathrm{ClO}_{2}\) (d) \(\mathrm{Cl}_{2} \mathrm{O}_{5}\)
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