/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 Do the empirical and molecular f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Do the empirical and molecular formulas of a compound always have the same percent composition values? Explain your answer.

Short Answer

Expert verified
Explain your answer. Answer: Yes, empirical and molecular formulas have the same percent composition values for a given compound. This is because both formulas represent the same ratio of elements in the compound, and the percent composition is calculated based on this ratio, independent of the actual number of atoms.

Step by step solution

01

What is an Empirical Formula?

An empirical formula represents the simplest whole number ratio of elements present in a compound. This formula shows the minimal information about the composition of the compound but does not provide any information on the actual number of atoms in the molecule.
02

What is a Molecular Formula?

A molecular formula shows the actual number of atoms of each element present in a compound. Molecular formulas can be multiples of empirical formulas, but they retain the same ratio of elements.
03

The difference between Empirical and Molecular Formulas

Empirical formulas give the simplest whole number ratio of elements in a compound, while molecular formulas show the actual number of atoms of each element. A molecular formula can be a whole number multiple of the empirical formula. For example, the empirical formula of ethane is CH3 while its molecular formula is C2H6.
04

Relationship between Percent Composition in Empirical and Molecular Formulas

The percent composition of an element in a compound is calculated as the mass of the element divided by the total mass of the compound, multiplied by 100. Despite the difference in the number of atoms in empirical and molecular formulas, the ratio of elements remains the same. Thus, the percent composition of elements in both empirical and molecular formulas will be the same.
05

Conclusion

In conclusion, the empirical and molecular formulas of a compound always have the same percent composition values. This is because both formulas represent the same ratio of elements in a compound, and the percent composition is calculated based on this ratio, independent of the actual number of atoms.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Percent Composition
Understanding percent composition is essential for mastering chemistry, particularly when investigating the detailed makeup of compounds. Percent composition refers to the percentage by mass of each element in a chemical compound. It is a reflection of how much of a certain element is present compared to the total mass of the compound. To calculate it, one divides the mass of each element by the total mass of the compound and then multiplies the result by 100.

For instance, calculating the percent composition of water (H2O) involves determining the mass of hydrogen and oxygen in a given sample. As water's molecular weight is approximately 18 grams per mole—composed of 2 grams of hydrogen and 16 grams of oxygen—hydrogen makes up about 11% (2/18*100) and oxygen about 89% of water by mass. This calculation is critical not only for understanding the compound's composition but also for various applications in stoichiometry, such as determining reactants' quantities for chemical reactions.
Chemical Compound Representation
Representing a chemical compound accurately is key to conveying its structure and composition efficiently. Chemical compounds are represented in several ways, and two of the primary methods are empirical and molecular formulas. The empirical formula provides the simplest, most reduced ratio of elements in a compound, indicating the relative number of each type of atom. For example, the empirical formula for glucose is CH2O, illustrating that for every carbon atom, there are two hydrogen atoms and one oxygen atom.

On the other hand, the molecular formula details the exact number of each type of atom in a molecule. Using glucose again as an example, its molecular formula is C6H12O6, showing that a glucose molecule contains 6 carbon, 12 hydrogen, and 6 oxygen atoms. While empirical formulas are useful for understanding the elemental proportions, molecular formulas provide a precise idea of the molecule's complexity and are crucial for constructing molecular models and understanding the compound's behavior in chemical reactions.
Mole Ratio
The concept of the mole ratio is a cornerstone in the study of chemistry, encompassing the relationship between the amounts of substances involved in a chemical reaction. It stems from the stoichiometry of a balanced chemical equation, where the coefficients represent the ratio of moles that react with each other. For example, the equation 2H2 + O2 -> 2H2O indicates that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water.

Therefore, when considering a reaction, the mole ratio provides the proportions of reactants needed or products formed. It allows chemists to predict the mass of substances required or produced, making it an invaluable tool for laboratory work and industrial processes. By applying the mole ratio, we can also ensure that chemical reactions occur efficiently, with minimal waste of reactants and energy. Mole ratios are not only theoretical constructs; they are the practical guide for actual chemical synthesis and analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose 25 metric tons of coal that is \(3.0 \%\) sulfur by mass is burned at an electric power plant ( 1 metric ton \(=10^{3} \mathrm{kg}\) ). During combustion, the sulfur is converted into sulfur dioxide. How many tons of sulfur dioxide are produced?

Large quantities of fertilizer are washed into the Mississippi River from agricultural land in the Midwest. The excess nutrients collect in the Gulf of Mexico, promoting the growth of algae and endangering other aquatic life.a. One commonly used fertilizer is ammonium nitrate. What is the chemical formula of ammonium nitrate? b. Corn farmers typically use \(5.0 \times 10^{3} \mathrm{kg}\) of ammonium nitrate per square kilometer of cornfield per year. Ammonium nitrate can be prepared by the following reaction: $$\mathrm{NH}_{3}(a q)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(a q)$$ How much nitric acid would be required to make the fertilizer needed for \(1 \mathrm{km}^{2}\) of cornfield per year? c. The ammonium ions can be converted into \(\mathrm{NO}_{3}^{-}\) by bacterial action. \(\mathrm{NH}_{4}^{+}(a q)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{NO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{H}^{+}(a q)\) If \(10 \%\) of the ammonium component of \(5.0 \times 10^{2} \mathrm{kg}\) of fertilizer ends up as nitrate, how much oxygen would be consumed?

Why is the quantity of \(\mathrm{CO}_{2}\) obtained in a combustion analysis not a direct measure of the oxygen content of the starting compound?

A 1 -liter sample of seawater contains \(19.4 \mathrm{g}\) of \(\mathrm{Cl}^{-}, 10.8 \mathrm{g}\) of \(\mathrm{Na}^{+},\) and \(1.29 \mathrm{g}\) of \(\mathrm{Mg}^{2+}\) a. How many moles of each ion are present? b. If we evaporated the seawater, would there be enough Cl "present to form the chloride salts of all the sodium and magnesium present?

In Brazil automobiles use ethanol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O},\) as fuel, whereas in the United States we rely on gasoline. Using \(\mathrm{C}_{8} \mathrm{H}_{18}\) (octane) to represent gasoline, write balanced chemical equations for the complete combustion of ethanol and octane. Which fuel produces more \(\mathrm{CO}_{2}\) per gram of fuel?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.