Considering your answer to Exercise 73 , which type of formula, empirical or molecular, can be obtained from elemental analysis that gives percent composition?

Short Answer

Expert verified
In conclusion, from elemental analysis that gives percent composition, the empirical formula can be determined as it represents the simplest whole number ratio of the elements in a compound.

Step by step solution

01

Understand the information provided by elemental analysis.

Elemental analysis reveals the percent composition of each element in a compound. This means it tells us the percentage of each element's mass in the compound's total mass. This information helps us understand the ratio of elements present in a compound.
02

Relate the percent composition to empirical and molecular formulas.

Percent composition data can help us in finding the empirical formula of a compound, as it reflects the simplest whole number ratio of the elements. However, it does not provide information about the actual number of atoms of each element in a molecule, which is necessary to find the molecular formula. To find the molecular formula, we need additional information: the molar mass of the compound.
03

Determine the type of formula that can be obtained from elemental analysis.

Since elemental analysis provides the percent composition of each element in a compound and allows us to determine the simplest whole number ratio of the elements, we can conclude that the empirical formula can be obtained from elemental analysis. In conclusion, an empirical formula can be obtained from elemental analysis that gives the percent composition.

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Key Concepts

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

Empirical Formula
Elemental analysis in chemistry is instrumental in providing us with the percent composition, which is the cornerstone for determining an empirical formula. This formula represents the simplest ratio of the elements present in a compound using whole numbers. For instance, if we have a compound composed of 40% carbon and 60% oxygen by mass, we can deduce that for every 40 grams of carbon, there is 60 grams of oxygen.

Through a series of calculations, these percentages can be converted into a whole number ratio. It's akin to reducing fractions to their simplest form in mathematics. As a result, the empirical formula might not reflect the actual number of atoms within a single molecule but rather an average scaled down to the most basic representation.

Importance of the Empirical Formula

The empirical formula is essential in chemistry as it conveys a clear and simplified insight into the chemical composition of substances. It is particularly useful for ionic compounds, where the actual number of atoms cannot be easily determined due to the nature of ionic lattices. By knowing the empirical formula, chemists can communicate and compare substances effectively without the complexities of molecular intricacies.
Percent Composition
The percent composition of a chemical compound exposes the relative masses of the constituent elements. This essentially tells us how much of each element exists within a 100-gram sample of the compound. To calculate the percent composition, we take the mass of each element and divide it by the total mass of the compound, multiplying by 100 to obtain a percentage.

Calculating the percent composition is not just a rote mathematical exercise; it provides a deep insight into the substance's identity. For instance, knowing that water (H2O) has a high percentage of oxygen tells us about the significant role oxygen plays in the properties of water.

Applying Percent Composition

Understanding the percent composition is vital for applications such as nutritional labelling, where it's crucial to know the amount of each nutrient in a food item. In environmental science, it helps to assess pollution levels in the air or soil by indicating the presence and quantity of hazardous substances.
Molecular Formula
While the empirical formula is crucial for identifying the basic proportionate composition of a compound, the molecular formula tells the entire story by revealing the actual number of atoms of each element in a molecule of that compound. It's quite like knowing the full cast of a theater play instead of just the lead roles.

The molecular formula always bears a whole number multiple relation to the empirical formula. To determine the molecular formula, one would need the compound's molecular (or molar) mass and the empirical formula's mass. By dividing the molecular mass by the mass of the empirical formula, we get a multiplier that when applied to the empirical formula, gives us the molecular formula.

Relevance of Molecular Formula

Knowing the molecular formula is beneficial when predicting the behavior of molecules in chemical reactions, as it allows chemists to gauge the amount of reactants and products involved. In the pharmaceutical industry, this knowledge is paramount when synthesizing medications to ensure proper dosages and chemical stability.

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Most popular questions from this chapter

Many cereals are made with high moisture content so that the cereal can be formed into various shapes before it is dried. A cereal product containing \(58 \% \mathrm{H}_{2} \mathrm{O}\) by mass is produced at the rate of \(1000 . \mathrm{kg} / \mathrm{h} .\) What mass of water must be evaporated per hour if the final product contains only \(20 . \%\) water?

Dimethylnitrosamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2} \mathrm{O}\), is a carcinogenic (cancercausing) substance that may be formed in foods, beverages, or gastric juices from the reaction of nitrite ion (used as a food preservative) with other substances. a. What is the molar mass of dimethylnitrosamine? b. How many moles of \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}_{2} \mathrm{O}\) molecules are present in \(250 \mathrm{mg}\) dimethylnitrosamine? c. What is the mass of \(0.050\) mol dimethylnitrosamine? d. How many atoms of hydrogen are in \(1.0 \mathrm{~mol}\) dimethylnitrosamine? e. What is the mass of \(1.0 \times 10^{6}\) molecules of dimethylnitrosamine? f. What is the mass in grams of one molecule of dimethylnitrosamine?

Consider the following balanced chemical equation: $$ A+5 B \longrightarrow 3 C+4 D $$ a. Equal masses of \(\mathrm{A}\) and \(\mathrm{B}\) are reacted. Complete each of the following with either "A is the limiting reactant because \(" ;{ }^{\prime *} \mathrm{~B}\) is the limiting reactant because \("\) or "we cannot determine the limiting reactant because i. If the molar mass of \(\mathrm{A}\) is greater than the molar mass of B, then ii. If the molar mass of \(\mathrm{B}\) is greater than the molar mass of A, then b. The products of the reaction are carbon dioxide (C) and water (D). Compound \(\mathrm{A}\) has the same molar mass as carbon dioxide. Compound \(\mathrm{B}\) is a diatomic molecule. Identify \(\mathrm{com}-\) pound \(\mathrm{B}\) and support your answer. c. Compound \(\mathrm{A}\) is a hydrocarbon that is \(81.71 \%\) carbon by mass. Detemine its empirical and molecular formulas.

Determine the molecular formulas to which the following empirical formulas and molar masses pertain. a. SNH \((188.35 \mathrm{~g} / \mathrm{mol})\) b. \(\mathrm{NPCl}_{2}(347.64 \mathrm{~g} / \mathrm{mol})\) c. \(\mathrm{CoC}_{4} \mathrm{O}_{4}(341.94 \mathrm{~g} / \mathrm{mol})\) d. SN \((184.32 \mathrm{~g} / \mathrm{mol})\)

Calculate the molar mass of the following substances. a. \(\mathrm{H}\) b. Q N \(\mathrm{N}\) c. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)

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