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A \(0.513-\mathrm{g}\) sample containing only carbon, hydrogen, and nitrogen burns in excess \(\mathrm{O}_{2}\) to produce \(1.04 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.704 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) Calculate the mass of each element in the sample and the mass percentage of each element in the compound.

Short Answer

Expert verified
C: 55.2%, H: 15.4%, N: 29.4%

Step by step solution

01

Calculate Moles of CO2 and H2O Produced

First, calculate the moles of carbon dioxide (CO2) and water (H2O) produced using their respective molar masses. The molar mass of CO2 is approximately 44.01 g/mol, and the molar mass of H2O is approximately 18.02 g/mol. \[\text{Moles of } \mathrm{CO}_{2} = \frac{1.04 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.0236 \, \text{mol} \] \[\text{Moles of } \mathrm{H}_{2} \mathrm{O} = \frac{0.704 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.0391 \, \text{mol} \]
02

Determine Mass of Carbon and Hydrogen

Calculate the mass of carbon and hydrogen in the products. Each mole of CO2 contains one mole of carbon, and each mole of H2O contains two moles of hydrogen.\[ \text{Mass of } \text{C} = 0.0236 \, \text{mol} \times 12.01 \, \text{g/mol} \approx 0.283 \, \text{g} \] \[ \text{Mass of } \text{H} = 0.0391 \, \text{mol} \times 2 \, \times 1.008 \, \text{g/mol} \approx 0.079 \, \text{g} \]
03

Determine Mass of Nitrogen

Since the sample only contains carbon, hydrogen, and nitrogen, the mass of nitrogen is found by subtracting the masses of carbon and hydrogen from the total mass of the sample.\[ \text{Mass of } \text{N} = 0.513 \, \text{g} - (0.283 \, \text{g} + 0.079 \, \text{g}) = 0.151 \, \text{g} \]
04

Calculate Mass Percentages

Calculate the mass percentage of each element by dividing the mass of each element by the total mass of the sample and multiplying by 100.\[ \% \text{C} = \left(\frac{0.283 \, \text{g}}{0.513 \, \text{g}}\right) \times 100 \approx 55.2\% \] \[ \% \text{H} = \left(\frac{0.079 \, \text{g}}{0.513 \, \text{g}}\right) \times 100 \approx 15.4\% \] \[ \% \text{N} = \left(\frac{0.151 \, \text{g}}{0.513 \, \text{g}}\right) \times 100 \approx 29.4\% \]

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Key Concepts

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

Mass Percentage in Elemental Analysis
In chemistry, determining the mass percentage of elements in a compound provides valuable insights into its composition. The mass percentage is calculated by taking the mass of a specific element and dividing it by the total mass of the sample, then multiplying by 100 to express it as a percentage. This helps us understand the proportion of each element within a compound or sample.

For example, if a compound contains carbon, hydrogen, and nitrogen, we calculate the mass of each from experimental data, as shown in our exercise. The mass percentage for carbon (\(\%\text{C}\)) is:
  • \[ \%\text{C} = \frac{\text{Mass of C}}{\text{Total Mass of Sample}} \times 100 \]
This percentage tells us the fraction of the compound's mass attributable to carbon. By understanding mass percentages, we can compare different compounds and understand their fundamental make-up.

This is especially useful in fields like material science and pharmacology.
Combustion Analysis for Determining Composition
Combustion analysis is a powerful method for determining the elemental composition of organic compounds. During combustion, the sample is burned in oxygen, converting carbon to carbon dioxide (CO2) and hydrogen to water (H2O). By measuring these products, we can deduce the amounts of carbon and hydrogen originally present in the sample. This process is central to understanding how much of each element is in a compound, particularly for unknown or newly synthesized materials.

In the exercise, we used the products of combustion to find the moles of CO2 and H2O:
  • Moles of CO2: \(\frac{1.04\, \text{g}}{44.01\, \text{g/mol}} \approx 0.0236\, \text{mol}\)
  • Moles of H2O: \(\frac{0.704\, \text{g}}{18.02\, \text{g/mol}} \approx 0.0391\, \text{mol}\)
These calculations are essential steps in combustion analysis, allowing us to calculate the mass of carbon and hydrogen released. It provides a detailed picture of the sample composition and is essential for evaluating unknown or complex organic compounds.
Molar Mass Calculation and its Relevance
Calculating molar mass is a fundamental step in analyzing chemical substances. The molar mass of a compound is the mass of one mole of its entities (atoms, molecules, etc.). It is crucial for converting between the mass of a substance and the amount in moles, which underpins much of chemistry.
  • Molar masses of products guide calculations, such as determining moles of CO2 and H2O.
  • For CO2, each mole is 44.01 g, while H2O is 18.02 g.
Converting these masses into moles allows chemists to quantify the proportions of elements in the sample. For instance, knowing the molar mass of CO2, we estimated the carbon originating from it. Such calculations enable a deeper understanding of chemical compositions, necessary for reactions and stoichiometry. This ability to move between mass and moles aids in precise lab work, pharmaceutical formulations, and even in too many industry applications to list here.

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Most popular questions from this chapter

A backup system on the space shuttle that removes carbon dioxide is canisters of \(\mathrm{LiOH}\). This compound reacts with \(\mathrm{CO}_{2}\) to produce \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) and water. How many grams of \(\mathrm{CO}_{2}\) can be removed from the atmosphere by a canister that contains \(83 \mathrm{~g} \mathrm{LiOH}\) ?

Predict the formula of an ionic compound formed from calcium and nitrogen. Calculate the mass percentage composition of the elements in this compound.

What is the empirical formula of a substance that contains \(0.80 \mathrm{~g}\) carbon and \(0.20 \mathrm{~g}\) hydrogen?

One of the ways to remove nitrogen monoxide gas, a serious source of air pollution, from smokestack emissions is by reaction with ammonia gas, \(\mathrm{NH}_{3}\). The products of the reaction, \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\), are not toxic. Write the balanced equation for this reaction. Assign an oxidation number to each element in the reactants and products, and indicate which element is oxidized and which is reduced.

\(\mathrm{In}_{2} \mathrm{~S}_{3}\) can be converted into metallic indium by a twostep process. First, it is converted into \(\mathrm{In}_{2} \mathrm{O}_{3}\) by reaction with oxygen. The other product of the reaction is \(\mathrm{SO}_{2}\). Indium metal is obtained by reaction of \(\mathrm{In}_{2} \mathrm{O}_{3}\) with carbon. Assume that the other product of the second reaction is carbon dioxide. (a) Write the two equations for this process. (b) Calculate the mass, in kilograms, of indium produced from \(35.7 \mathrm{~kg} \operatorname{In}_{2} \mathrm{~S}_{3}\), assuming excesses of the other reactants.

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