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Ch.2 - Atoms, Molecules, and Ions
All textbooksBrown 15th EditionCh.2 - Atoms, Molecules, and IonsProblem 11c
Chapter 2, Problem 11c

A 1.0-g sample of carbon dioxide (CO2) is fully decomposed into its elements, yielding 0.273 g of carbon and 0.727 g of oxygen. If a sample of a different compound decomposes into 0.429 g of carbon and 0.571 g of oxygen, what is its ratio of the mass of O to C? (c) According to Dalton's atomic theory, what is the empirical formula of the second compound?

Verified step by step guidance
1
Calculate the mass ratio of oxygen to carbon for the second compound by dividing the mass of oxygen by the mass of carbon: \( \frac{0.571 \, \text{g O}}{0.429 \, \text{g C}} \).
Simplify the ratio obtained in step 1 to find the simplest whole number ratio. This can be done by dividing both the numerator and the denominator by the smaller of the two values.
Use the simplified ratio to determine the empirical formula. The ratio represents the relative number of moles of oxygen and carbon atoms in the compound.
Convert the mole ratio into subscripts in a chemical formula. If the ratio from step 2 is, for example, 1.33, this would suggest a subscript of 1 for carbon and approximately 1.33 for oxygen in the empirical formula.
Round the subscripts to the nearest whole number if necessary, to get the empirical formula of the compound. This rounding is based on the assumption that the empirical formula represents the simplest whole-number ratio of atoms in the compound.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Mass Ratio Calculation

To determine the ratio of the mass of oxygen to carbon in a compound, divide the mass of oxygen by the mass of carbon. This ratio provides insight into the relative proportions of elements in a compound, which is essential for understanding its composition and can help in deriving the empirical formula.
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Dalton's Atomic Theory

Dalton's atomic theory posits that matter is composed of indivisible atoms, and compounds are formed when atoms of different elements combine in fixed ratios. This theory underpins the concept of empirical formulas, which represent the simplest whole-number ratio of the elements in a compound, reflecting the relative amounts of each element present.
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Empirical Formula

The empirical formula of a compound is the simplest integer ratio of the elements present in that compound. To derive it, one must first calculate the moles of each element from their masses, then simplify the ratio of these moles to the smallest whole numbers, which provides a clear representation of the compound's composition.
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Related Practice
Textbook Question

In the Millikan oil-drop experiment (see Figure 2.5), the tiny oil drops are observed through the viewing lens as rising, stationary, or falling, as shown here. The arrows indicate the rate of motion. a. What causes their rate of fall to vary from their rate in the absence of an electric field?

Textbook Question

A 1.0-g sample of carbon dioxide (CO2) is fully decomposed into its elements, yielding 0.273 g of carbon and 0.727 g of oxygen. (a) What is the ratio of the mass of O to C?

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Textbook Question

A 1.0-g sample of carbon dioxide (CO2) is fully decomposed into its elements, yielding 0.273 g of carbon and 0.727 g of oxygen. (b) If a sample of a different compound decomposes into 0.429 g of carbon and 0.571 g of oxygen, what is its ratio of the mass of O to C?

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Textbook Question

Hydrogen sulfide is composed of two elements: hydrogen and sulfur. In an experiment, 6.500 g of hydrogen sulfide is fully decomposed into its elements. b. What fundamental law does this experiment demonstrate?

Textbook Question

A chemist finds that 30.82 g of nitrogen will react with 17.60, 35.20, 70.40, or 88.00 g of oxygen to form four different compounds. (a) Calculate the mass of oxygen per gram of nitrogen in each compound. 30.82 g N and 17.60 g O

Textbook Question

A chemist finds that 30.82 g of nitrogen will react with 17.60, 35.20, 70.40, or 88.00 g of oxygen to form four different compounds. (a) Calculate the mass of oxygen per gram of nitrogen in each compound. 30.82 g N and 88.00 g O

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