Considering your answer to Exercise 79, which type of formula, empirical or molecular, can be obtained from elemental analysis that gives percent composition?

Short Answer

Expert verified
From elemental analysis that provides percent composition, only the empirical formula can be obtained directly. The empirical formula represents the simplest integer ratio of elements in a compound. To find the molecular formula, additional information such as the molar mass of the compound is required.

Step by step solution

01

Definition of Empirical Formula and Molecular Formula

An empirical formula represents the simplest integer ratio of elements in a compound, while the molecular formula represents the actual number of atoms of each element in a molecule of the compound. For example, the molecular formula of glucose is C₆H₁₂O₆, while its empirical formula is CH₂O, which is a simplified ratio of the elements present in glucose.
02

Elemental Analysis and Percent Composition

Elemental analysis is a technique used to determine the elemental composition of a substance. It provides the percent composition of each element in the compound. Percent composition can be used to derive the simplest ratio of the elements present in a compound, which is the empirical formula.
03

Empirical Formula from Elemental Analysis

Using the percent composition obtained from elemental analysis, we can find the empirical formula of a compound by following these steps: 1. Convert the percentages to grams (assuming a 100g sample, the percentages would be equivalent to the gram amount for each element). 2. Convert grams to moles by dividing the mass of each element by its molar mass. 3. Determine the simplest whole number ratio by dividing each of the moles obtained by the smallest moles value among the elements. Regardless of the size of the sample, the steps would still yield the simplest ratio of elements in the compound, which is the empirical formula.
04

Molecular Formula from Elemental Analysis

By itself, elemental analysis cannot provide us with the molecular formula of a compound, as it only gives the percent composition. Furthermore, the molecular formula requires additional information, such as the molar mass of the compound. However, once we know the empirical formula and molar mass, we can determine the molecular formula by finding the ratio between the actual molar mass and the empirical formula molar mass. In conclusion, from elemental analysis that provides percent composition, only the empirical formula can be obtained directly. To find the molecular formula, we would need additional information such as the molar mass of the compound.

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Key Concepts

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

Elemental Analysis
Elemental analysis is a crucial scientific technique used to establish the qualitative and quantitative composition of chemical substances. This process determines what elements are present in a compound and in what quantities. For example, by performing elemental analysis on a water sample, we would discover that it is composed of hydrogen and oxygen, and a deeper analysis would show the ratio of these elements.

When conducting elemental analysis for a compound, chemists often use methods like combustion analysis, mass spectrometry, or X-ray fluorescence. These techniques provide detailed information about the percentages of each element within the substance. Such precise data is vital for identifying the compound and understanding its properties and potential uses. It's like having a detailed recipe that shows exactly how much of each ingredient is in your favorite cake!

In a classroom or laboratory exercise, a problem may present you with a compound's percent composition. Your goal as a student is to use this information to deduce the simplest form of its empirical formula. Taking elemental analysis one step further, you could also determine environmental impacts, the purity of a substance, or even trace elements in medicinal and forensic applications. It's like being a detective, where elemental analysis gives you the clues to solve the chemical mystery.
Percent Composition
Understanding percent composition is like breaking down a work of art to see how much of each color was used. It tells us the relative amounts of each element in a compound. Specifically, percent composition reveals the percentage, by mass, of each element in the compound, providing a deeper insight into the substance's chemical makeup.

For example, if you have a compound with 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass, you know that for every 100 grams of this compound, you have 40 grams of carbon, 6.7 grams of hydrogen, and 53.3 grams of oxygen.
  • These percentages help us to establish the empirical formula by showing the mass proportion of each element present.
It's essential to remember that the empirical formula derived from percent composition will not tell us how many atoms of each element are in a molecule; it only provides the simplest whole-number ratio. This is where the relation between percent composition and molecular formula becomes more complex because obtaining the molecular formula requires additional information such as the molar mass.

Percent composition is an invaluable tool in fields such as nutrition, pharmacology, and materials science. For example, knowing the percent composition of nutrients in food can help dieticians create balanced diets, and in pharmacology, it's fundamental for understanding drug formulations.
Molar Mass
Discussing molar mass is like talking about the weight of a cloud; it's not about the weight of one tiny droplet but the overall mass of a defined volume of water vapor. In chemistry, the molar mass is the mass in grams of one mole of a substance, which is Avogadro's number (\(6.022 \times 10^{23}\)) of molecules or atoms.
  • It serves as a bridge between the microscopic world of atoms and molecules and the macroscopic world we measure in the laboratory.


Molar mass is necessary for converting between the mass of a substance and the amount in moles, allowing us to use the ideal gas laws, calculate concentrations, and reactants or products in chemical reactions. For instance, knowing the molar mass of carbon dioxide helps environmental scientists quantify the amount released in the atmosphere.

To find the molar mass of a compound, we add together the atomic masses of each element in the compound as listed on the periodic table, multiplied by the number of atoms of each element in the molecular formula. It's like a shopping list where you add up the total cost of multiple items to find the total expenditure.
  • This figure is instrumental when working out chemical equations and stoichiometry, which involves the quantitative relationships between reactants and products in a reaction.

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Most popular questions from this chapter

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of \(0.157 \mathrm{~g}\) of the compound produced \(0.213 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.0310 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that \(0.103 \mathrm{~g}\) of the compound produces \(0.0230 \mathrm{~g} \mathrm{NH}_{3} .\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

A compound contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\). Combustion of \(35.0 \mathrm{mg}\) of the compound produces \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg}\) \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula of the compound?

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(g)\) or \(\mathrm{NO}_{2}(g) \mathrm{ac}-\) cording to these unbalanced equations: $$ \begin{array}{l} \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ In a certain experiment \(2.00\) moles of \(\mathrm{NH}_{3}(g)\) and \(10.00 \mathrm{moles}\) of \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, \(6.75\) moles of \(\mathrm{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

Calculate the percent composition by mass of the following compounds that are important starting materials for synthetic polymers: a. \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2}\) (acrylic acid, from which acrylic plastics are made) b. \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{2}\) (methyl acrylate, from which Plexiglas is made) c. \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}\) (acrylonitrile, from which Orlon is made)

A \(2.077-\mathrm{g}\) sample of an element, which has an atomic mass between 40 and 55, reacts with oxygen to form \(3.708 \mathrm{~g}\) of an oxide. Determine the formula of the oxide (and identify the element).

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