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Chapter 11: Problem 66

When analyzing the results of combustion analysis, we use the mass percentage of carbon in the \(\mathrm{CO}_{2}(g)\) produced to find the mass of carbon from our original sample and the mass percentage of hydrogen in the \(\mathrm{H}_{2} \mathrm{O}(g)\) produced to find the mass of hydrogen from our original sample. Why can't the same procedure be used to determine the mass of oxygen present in the original sample?

Short Answer

Expert verified
The oxygen's source ambiguity (sample vs. air) prevents direct mass determination via combustion product analysis.

Step by step solution

01

Understanding Combustion Analysis

In combustion analysis, a compound containing carbon and hydrogen (and sometimes oxygen) is burned, typically producing carbon dioxide (\(\mathrm{CO}_2(g)\)) and water (\(\mathrm{H}_2O(g)\)). The mass percentages of carbon and hydrogen in these products are used to calculate the amounts of these elements in the original sample.
02

Calculation of Carbon and Hydrogen Masses

The mass of carbon in the original sample is determined from the mass of \(\mathrm{CO}_2(g)\) produced, using the fact that each mole of carbon dioxide contains one mole of carbon. Similarly, the mass of hydrogen is determined from the mass of \(\mathrm{H}_2O(g)\) produced, knowing that each mole of water contains two moles of hydrogen.
03

Challenge with Determining Oxygen Mass

Oxygen in the original sample can potentially come from two sources: the organic compound itself and the oxygen in the air used for combustion. The \(\mathrm{CO}_2(g)\) and \(\mathrm{H}_2O(g)\) products do not distinguish between oxygen from these different sources.
04

Limitation of Direct Oxygen Measurement

Since we cannot differentiate between the oxygen from the original compound and the oxygen from the air used in combustion, it is impossible to directly calculate the mass of oxygen initially present using just the combustion products' mass percentages.

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Key Concepts

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

Mass Percentage
Mass percentage is a critical concept in combustion analysis used to determine the composition of an original sample. In simple terms, it tells us how much of a particular element is present in a compound relative to the total mass of the compound. This is expressed as a percentage.
  • For carbon, we measure the mass of carbon dioxide (\( \mathrm{CO}_2 \)) formed during combustion.
  • For hydrogen, we measure the mass of water (\( \mathrm{H}_2\mathrm{O} \)) produced.
By measuring these products and knowing the mass percentages, we can calculate the masses of carbon and hydrogen in the original sample. This method is straightforward for carbon and hydrogen because their combustion products directly reveal their presence in the original sample.
Carbon Dioxide
Carbon dioxide is a key product in combustion analysis, as it helps us determine the mass of carbon in an original sample. When the sample undergoes combustion, the carbon atoms combine with oxygen from the air to form \( \mathrm{CO}_2 \).
  • Each molecule of carbon dioxide contains exactly one atom of carbon.
  • The mass of \( \mathrm{CO}_2 \) produced is used to calculate the amount of carbon in the original sample.
To find the mass of carbon, use the equation:\[\text{Mass of carbon} = \text{Mass of } \mathrm{CO}_2 \times \left( \frac{12}{44} \right)\]Here, 12 represents the molar mass of carbon, and 44 is the molar mass of carbon dioxide. Calculating it this way ensures we understand how much carbon was present in the original sample.
Water Production
Water production is another vital indicator in combustion analysis, specifically for determining the mass of hydrogen in the original sample. When the sample is burned, hydrogen atoms form water ( \( \mathrm{H}_2\mathrm{O} \)) by combining with oxygen.
  • Each water molecule consists of two hydrogen atoms.
  • From the mass of \( \mathrm{H}_2\mathrm{O} \) produced, we can calculate the amount of hydrogen in the original sample.
The calculation is straightforward, using the equation:\[\text{Mass of hydrogen} = \text{Mass of } \mathrm{H}_2\mathrm{O} \times \left( \frac{2}{18} \right)\]Here, 2 is the molar mass of hydrogen atoms in one water molecule, and 18 is the total molar mass of water. This method offers a precise way to determine hydrogen content based on water production.
Original Sample
The original sample in combustion analysis is the sample being studied, which typically contains carbon, hydrogen, and sometimes oxygen. Understanding the composition of the original sample is the main goal of the analysis.
  • We can directly measure the amounts of carbon and hydrogen using \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \) produced.
  • However, determining the oxygen content is challenging.
Oxygen in the original sample might come from the sample itself or from air used in combustion. This makes it difficult to calculate exactly how much oxygen originated from the sample.Unfortunately, this limitation means that the mass percentages of combustion products cannot effectively reveal the oxygen content. Instead, alternative methods or additional information are required to fully understand the original sample's oxygen content.

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Most popular questions from this chapter

11-76. Glucose is used as an energy source by the human body. The overall reaction in the body is described by the equation $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \rightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Calculate the number of grams of oxygen required to convert \(28.0\) grams of glucose to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Also compute the number of grams of \(\mathrm{CO}_{2}(g)\) produced.

II-53. Cryolite, \(\mathrm{Na}_{3} \mathrm{AlF}_{6}(s)\), an ore used in the production of aluminum, can be synthesized by the reaction described by the equation \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{NaOH}(l)+\mathrm{HF}(g) \stackrel{\text { High } T}{\longrightarrow}\) \(\mathrm{Na}_{3} \mathrm{AlF}_{6}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad(\) unbalanced \()\) (a) Balance this equation. (b) If \(10.0\) kilograms of \(\mathrm{Al}_{2} \mathrm{O}_{3}(s), 50.00\) kilograms of \(\mathrm{NaOH}(l)\), and \(50.0\) kilograms of \(\mathrm{HF}(g)\) react completely, how many kilograms of cryolite will be produced? (c) Which reactants will be in excess and how many kilograms of each of these reactants will remain?

The arsenic in an ore sample was converted into water-soluble sodium arsenate, \(\mathrm{Na}_{9} \mathrm{AsO}_{4}(a q)\). From the solution of \(\mathrm{Na}_{9} \mathrm{AsO}_{4}(a q)\), insoluble silver arsenate, \(\mathrm{Ag}_{9} \mathrm{AsO}_{4}(s)\), was precipitated and weighed. \(\mathrm{A}\) \(5.00\) -gram ore sample produced \(3.09\) grams of silver arsenate. Calculate the mass percentage of arsenic in the ore sample.

A 2.46-gram sample of copper metal is reacted completely with chlorine gas to produce \(5.22\) grams of copper chloride. Determine the empirical formula of this chloride.

A 9.87-gram sample of an alloy of aluminum and magnesium is completely reacted with hydrochloric acid and yields \(0.998\) grams of hydrogen gas. Calculate the percentage by mass of each metal in the alloy.
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