Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write the empirical formula corresponding to each of the following molecular formulas: (a) \(\mathrm{Al}_{2} \mathrm{Br}_{6}\), (b) \(\mathrm{C}_{8} \mathrm{H}_{10}\), (c) \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\), (d) \(\mathrm{P}_{4} \mathrm{O}_{10}\), (e) \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\), (f) \(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6}\).

Short Answer

Expert verified
The empirical formulas corresponding to each of the given molecular formulas are: (a) \(\mathrm{AlBr}_{3}\) (b) \(\mathrm{C}_{4} \mathrm{H}_{5}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) (d) \(\mathrm{P}_{2} \mathrm{O}_{5}\) (e) \(\mathrm{C}_{3} \mathrm{H}_{2} \mathrm{Cl}\) (f) \(\mathrm{BNH}_{2}\)

Step by step solution

01

Find the GCD of subscripts

Calculate the greatest common divisor of the subscripts for each of the given molecular formulas. (a) \(\mathrm{Al}_{2} \mathrm{Br}_{6}\) GCD(2, 6) = 2 (b) \(\mathrm{C}_{8} \mathrm{H}_{10}\) GCD(8, 10) = 2 (c) \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\) GCD(4, 8, 2) = 2 (d) \(\mathrm{P}_{4} \mathrm{O}_{10}\) GCD(4, 10) = 2 (e) \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\) GCD(6, 4, 2) = 2 (f) \(\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6}\) GCD(3, 3, 6) = 3
02

Divide the subscripts by their GCD

Divide the subscripts in each molecular formula by their greatest common divisor. (a) \(\mathrm{Al}_{\frac{2}{2}} \mathrm{Br}_{\frac{6}{2}}\), which simplifies to \(\mathrm{Al}_{1} \mathrm{Br}_{3}\) (b) \(\mathrm{C}_{\frac{8}{2}} \mathrm{H}_{\frac{10}{2}}\), which simplifies to \(\mathrm{C}_{4} \mathrm{H}_{5}\) (c) \(\mathrm{C}_{\frac{4}{2}} \mathrm{H}_{\frac{8}{2}} \mathrm{O}_{\frac{2}{2}}\), which simplifies to \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{1}\) (d) \(\mathrm{P}_{\frac{4}{2}} \mathrm{O}_{\frac{10}{2}}\), which simplifies to \(\mathrm{P}_{2} \mathrm{O}_{5}\) (e) \(\mathrm{C}_{\frac{6}{2}} \mathrm{H}_{\frac{4}{2}} \mathrm{Cl}_{\frac{2}{2}}\), which simplifies to \(\mathrm{C}_{3} \mathrm{H}_{2} \mathrm{Cl}_{1}\) (f) \(\mathrm{B}_{\frac{3}{3}} \mathrm{N}_{\frac{3}{3}} \mathrm{H}_{\frac{6}{3}}\), which simplifies to \(\mathrm{B}_{1} \mathrm{N}_{1} \mathrm{H}_{2}\)
03

Write the empirical formulas

Write down the empirical formula for each molecular formula. (a) \(\mathrm{AlBr}_{3}\) (b) \(\mathrm{C}_{4} \mathrm{H}_{5}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) (d) \(\mathrm{P}_{2} \mathrm{O}_{5}\) (e) \(\mathrm{C}_{3} \mathrm{H}_{2} \mathrm{Cl}\) (f) \(\mathrm{BNH}_{2}\)

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 Vaia!

Key Concepts

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

Molecular Formula
A molecular formula provides a significant amount of information about a compound, specifically its exact number of each type of atom. For example, glucose has a molecular formula of \( C_6H_{12}O_6 \), which indicates that one molecule of glucose contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. This explicit representation is crucial in chemistry as it helps understand the composition and the structure of the compound.

When delving into molecular formulas, it is essential to differentiate them from empirical formulas. The molecular formula represents the actual numbers of atoms in a molecule while the empirical formula is the simplest whole-number ratio of those atoms. For example, the empirical formula for glucose is \( CH_2O \) because the ratio of carbon to hydrogen to oxygen is 1:2:1, which brings us to the importance of simplifying molecular formulas to their empirical counterparts.
Greatest Common Divisor (GCD)
The greatest common divisor (GCD), sometimes known as the greatest common factor (GCF), is the largest positive integer that evenly divides each number in a set without leaving a remainder. In the context of chemistry, GCD is used to simplify chemical formulas by finding the simplest ratio of elements within a compound.

For instance, if you have a compound with the molecular formula \( C_8H_{10} \) and you wish to find the empirical formula, the GCD of the subscripts (in this case, 8 and 10) is 2. This GCD is utilized to divide each subscript, effectively simplifying the molecular formula to its lowest terms, yielding the empirical formula \( C_4H_5 \). Recognizing and calculating the GCD is a critical step in chemical formula simplification, which streamlines many aspects of chemical analysis and synthesis.
Chemical Formula Simplification
Chemical formula simplification is about reducing a molecular formula to its most basic form, the empirical formula, by using the GCD of the subscripts. This process is like simplifying fractions in mathematics. Taking the molecular formula \( C_6H_{12}O_6 \) again, we find that the GCD of 6, 12, and 6 is 6. Dividing each subscript by 6, the empirical formula is determined to be \( CH_2O \).

Keeping It Simple

Often, students encounter complex molecular formulas especially in organic compounds. Simplification removes complexity and helps them understand the key components of a compound. As seen in the exercise with \( C_8H_{10} \) and \( C_6H_{12}O_6 \) A simplified formula is invaluable in stoichiometry, where chemists calculate reactants and products in chemical reactions. Additionally, simplifying a chemical formula can uncover symmetries and patterns in chemical behavior, easing the prediction of reaction outcomes or properties of the substance. The journey from a molecular to an empirical formula is not simply a mathematical exercise, but a path to deeper insight into the nature of the substance being studied.

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

Elements in the same group of the periodic table often form oxyanions with the same general formula. The anions are also named in a similar fashion. Based on these observations, suggest a chemical formula or name, as appropriate, for each of the following ions: (a) \(\mathrm{BrO}_{4}^{-}\), (b) \(\mathrm{SeO}_{3}{ }^{2-}\), (c) arsenate ion, (d) hydrogen tellurate ion.

Which of the following pairs of atoms are isotopes of one another? \((\) a \(){ }^{11} \mathrm{~B},{ }^{11} \mathrm{C}\); (b) \({ }^{55} \mathrm{Mn},{ }^{54} \mathrm{Mn} ;(\mathbf{c}){ }_{50}^{118} \mathrm{Sn},{ }_{50}^{120} \mathrm{Sn}\)

The diameter of a rubidium atom is \(4.95 \AA\). We will consider two different ways of placing the atoms on a surface. In arrangement A, all the atoms are lined up with one another to form a square grid. Arrangement B is called a close-packed arrangement because the atoms sit in the "depressions" formed by the previous row of atoms: (a) Using arrangement \(A\), how many Rb atoms could be placed on a square surface that is \(1.0 \mathrm{~cm}\) on a side? (b) How many Rb atoms could be placed on a square surface that is \(1.0 \mathrm{~cm}\) on a side, using arrangement B? (c) By what factor has the number of atoms on the surface increased in going to arrangement \(B\) from arrangement A? If extended to three dimensions, which arrangement would lead to a greater density for Rb metal?

In a series of experiments, a chemist prepared three different compounds that contain only iodine and fluorine and determined the mass of each element in each compound: (a) Calculate the mass of fluorine per gram of iodine in each compound. (b) How do the numbers in part (a) support the atomic theory?

Write the chemical formula for each substance mentioned in the following word descriptions (use the front inside cover to find the symbols for the elements you do not know). (a) Zinc carbonate can be heated to form zinc oxide and carbon dioxide. (b) On treatment with hydrofluoric acid, silicon dioxide forms silicon tetrafluoride and water. (c) Sulfur dioxide reacts with water to form sulfurous acid. (d) The substance phosphorus trihydride, commonly called phosphine, is a toxic gas. (e) Perchloric acid reacts with cadmium to form cadmium(II) perchlorate. (f) Vanadium(III) bromide is a colored solid.

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.

Sign-up for free