Problem 11
Question
A 1.0 -g sample of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is fully decomposed into its elements, yielding \(0.273 \mathrm{~g}\) of carbon and \(0.727 \mathrm{~g}\) of oxygen. (a) What is the ratio of the mass of \(\mathrm{O}\) to \(\mathrm{C} ?(\mathbf{b})\) If a sample of a different compound decomposes into \(0.429 \mathrm{~g}\) of carbon and \(0.571 \mathrm{~g}\) of oxygen, what is its ratio of the mass of \(\mathrm{O}\) to \(\mathrm{C} ?(\mathbf{c})\) According to Dalton's atomic theory, what is the empirical formula of the second compound?
Step-by-Step Solution
Verified Answer
(a) 2.66, (b) 1.33, (c) C₃O₄.
1Step 1: Calculate Mass Ratio for CO₂
To find the mass ratio of oxygen to carbon in carbon dioxide, we divide the mass of oxygen by the mass of carbon: \(\text{Ratio for CO}_2 = \frac{0.727}{0.273} \approx 2.66\). This means for every gram of carbon, there are approximately 2.66 grams of oxygen.
2Step 2: Calculate Mass Ratio for Second Compound
Calculate the mass ratio of oxygen to carbon for the other compound using the masses: \(\text{Ratio for second compound} = \frac{0.571}{0.429} \approx 1.33\). This implies for every gram of carbon, there are approximately 1.33 grams of oxygen.
3Step 3: Determine Empirical Formula of Second Compound
The empirical formula is determined by the simplest ratio of elements in the compound. The ratio of O to C is 1.33, which is close to the ratio of 4 to 3. Hence, the empirical formula of the second compound is likely \(\text{C}_3\text{O}_4\) since 1.33 simplifies close to the ratio 4:3 when considering whole numbers.
Key Concepts
Mass RatioEmpirical FormulaDalton's Atomic Theory
Mass Ratio
Understanding the concept of mass ratio is crucial for analyzing chemical compositions in compounds.
Mass ratio refers to the ratio of the mass of one element to the mass of another in a compound.
In our exercise with carbon dioxide (\(\text{CO}_2\)), we find the mass of oxygen and carbon when it decomposes. Oxygen weighs \(0.727 \text{ g}\) and carbon weighs \(0.273 \text{ g}\).
This concept is not exclusive to \(\text{CO}_2\). It applies to all compounds and is a fundamental principle in chemical analysis and stoichiometry. Calculating mass ratios helps us understand the relative proportions of elements within any chemical compound.
This information is essential when diving deeper into empirical formulas and Dalton's atomic theory.
Mass ratio refers to the ratio of the mass of one element to the mass of another in a compound.
In our exercise with carbon dioxide (\(\text{CO}_2\)), we find the mass of oxygen and carbon when it decomposes. Oxygen weighs \(0.727 \text{ g}\) and carbon weighs \(0.273 \text{ g}\).
- To calculate the mass ratio of oxygen to carbon, you divide the mass of oxygen by the mass of carbon:
- This ratio means that for each gram of carbon in the compound, there are approximately 2.66 grams of oxygen.
This concept is not exclusive to \(\text{CO}_2\). It applies to all compounds and is a fundamental principle in chemical analysis and stoichiometry. Calculating mass ratios helps us understand the relative proportions of elements within any chemical compound.
This information is essential when diving deeper into empirical formulas and Dalton's atomic theory.
Empirical Formula
The empirical formula represents the simplest integer ratio of the elements present in a compound.
To find it, calculate the ratio of the masses of each element to one another.
In our example with the second compound, which decomposed into \(0.429 \text{ g}\) of carbon and \(0.571 \text{ g}\) of oxygen, we calculated the mass ratio:
Hence, the empirical formula for the second compound becomes \(\text{C}_3\text{O}_4\).
Understanding empirical formulas is valuable because they indicate the essential makeup of a compound, providing a clear picture of its composition.
They provide the basic structure from which molecular formulas derive by simply multiplying the empirical formula by an integer.
It is important to use the simplest ratio of elements to form the empirical formula, as it represents the foundation of chemical identification.
To find it, calculate the ratio of the masses of each element to one another.
In our example with the second compound, which decomposed into \(0.429 \text{ g}\) of carbon and \(0.571 \text{ g}\) of oxygen, we calculated the mass ratio:
- \[\text{Ratio for second compound} = \frac{0.571}{0.429} \approx 1.33\]
- This is approximately a 4:3 ratio when expressed in simplest whole numbers.
Hence, the empirical formula for the second compound becomes \(\text{C}_3\text{O}_4\).
Understanding empirical formulas is valuable because they indicate the essential makeup of a compound, providing a clear picture of its composition.
They provide the basic structure from which molecular formulas derive by simply multiplying the empirical formula by an integer.
It is important to use the simplest ratio of elements to form the empirical formula, as it represents the foundation of chemical identification.
Dalton's Atomic Theory
Dalton's atomic theory laid the foundation for modern chemistry.
John Dalton proposed that matter is made up of very small, indivisible particles called atoms.
This theory can explain why chemical reactions occur in fixed proportions, which relates to the mass ratios and empirical formulas discussed.
In our exercise, Dalton's theory helps interpret the data from decomposition, indicating that the second compound, with a mass ratio close to 4:3, led to the empirical formula \(\text{C}_3\text{O}_4\).
Overall, Dalton's atomic theory not only provided an insight into the nature of matter but also underpins the methodology for determining empirical formulas and understanding chemical decomposition.
Without these fundamental concepts, the analysis of chemical compounds would be much less precise and comprehensive.
John Dalton proposed that matter is made up of very small, indivisible particles called atoms.
This theory can explain why chemical reactions occur in fixed proportions, which relates to the mass ratios and empirical formulas discussed.
- Dalton asserted that atoms of the same element are identical in mass and properties.
- He also stated that compounds form when atoms of different elements combine in fixed ratios.
In our exercise, Dalton's theory helps interpret the data from decomposition, indicating that the second compound, with a mass ratio close to 4:3, led to the empirical formula \(\text{C}_3\text{O}_4\).
- This aligns with the idea that atoms combine in simple whole number ratios to form compounds.
- It showcases how empirical formulas represent these small whole number ratios, which is consistent with Dalton's idea of combining proportions.
Overall, Dalton's atomic theory not only provided an insight into the nature of matter but also underpins the methodology for determining empirical formulas and understanding chemical decomposition.
Without these fundamental concepts, the analysis of chemical compounds would be much less precise and comprehensive.
Other exercises in this chapter
Problem 13
A chemist finds that \(30.82 \mathrm{~g}\) of nitrogen will react with 17.60 , 35.20,70.40 , or \(88.00 \mathrm{~g}\) of oxygen to form four different compounds
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In a series of experiments, a chemist prepared three different compounds that contain only iodine and fluorine and determined the mass of each element in each c
View solution Problem 17
What fraction of the \(\alpha\) particles in Rutherford's gold foil experiment are scattered at large angles? Assume the gold foil is two layers thick, as shown
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