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Chapter 2: Problem 81

Write the empirical formula for the ionic compound made from each of the following pairs of elements. (a) calcium and chlorine (b) rubidium and sulfur (c) lithium and nitrogen (d) yttrium and selenium

Short Answer

Expert verified
(a) CaCl鈧�, (b) Rb鈧係, (c) Li鈧僋, (d) Y鈧係e鈧�

Step by step solution

01

Determine the Ionic Charges

For each pair of elements, determine the common ionic charge for each. - Calcium (Ca) typically forms a 2+ ion (Ca^{2+}). - Chlorine (Cl) typically forms a 1- ion (Cl^-). - Rubidium (Rb) typically forms a 1+ ion (Rb^+). - Sulfur (S) typically forms a 2- ion (S^{2-}). - Lithium (Li) typically forms a 1+ ion (Li^+). - Nitrogen (N) typically forms a 3- ion (N^{3-}). - Yttrium (Y) typically forms a 3+ ion (Y^{3+}). - Selenium (Se) typically forms a 2- ion (Se^{2-}).
02

Determine the Ratio of Ions

Use the charges to find the ratio of ions required to balance the charges to neutral. - For **CaCl_2**: Combine Ca^{2+} and two Cl^- to balance the charge. - For **Rb_2S**: Combine two Rb^+ ions with one S^{2-} to balance the charge. - For **Li_3N**: Combine three Li^+ ions with one N^{3-} to balance the charge. - For **Y_2Se_3**: Combine two Y^{3+} ions with three Se^{2-} ions to balance the charge.
03

Write the Empirical Formula

For each pair of elements, write the chemical formula using the simplest whole number ratio from Step 2. - For calcium and chlorine: **CaCl_2** - For rubidium and sulfur: **Rb_2S** - For lithium and nitrogen: **Li_3N** - For yttrium and selenium: **Y_2Se_3**

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Key Concepts

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

Empirical Formula
The empirical formula of an ionic compound provides the simplest whole-number ratio of ions within the compound. It is essential for indicating the composition of the compound, using the smallest integers that can represent the relative number of atoms in the formula. This is especially useful for substances composed of ions, where the overall charge must be balanced to ensure neutrality.

To determine the empirical formula, follow these steps:
  • Identify the charges on each of the ions involved.
  • Use these charges to find the smallest ratio that results in a completely neutral compound, where positive and negative charges cancel each other out.
  • Write down the chemical symbols of the elements, adding subscripts to indicate the number of atoms needed from each element to achieve this balance.
Solving for the empirical formula ensures that the final chemical representation is correct and consistent with the law of electroneutrality, where the sum of positive charges equals the sum of negative charges.
Ionic Charges
Ionic charges are crucial for forming stable compounds. Each ion exhibits a specific charge, which depends on how many electrons it tends to lose or gain to achieve a stable electronic configuration similar to the nearest noble gas.

For example:
  • Calcium (Ca) loses two electrons to form a Ca虏鈦� ion.
  • Chlorine (Cl) gains one electron, becoming Cl鈦�.
  • Rubidium (Rb) forms Rb鈦� by losing one electron.
  • Sulfur (S) gains two electrons to form S虏鈦�.
  • Lithium (Li) loses one electron, resulting in Li鈦�.
  • Nitrogen (N) gains three electrons and becomes N鲁鈦�.
  • Yttrium (Y) forms a Y鲁鈦� ion by losing three electrons.
  • Selenium (Se) gains two electrons, resulting in Se虏鈦�.
Understanding these charges helps in predicting and balancing the compounds formed between different pairs of elements. Correctly balancing these charges is key to writing correct empirical formulas for ionic compounds.
Element Pairing
Element pairing in ionic compounds involves combining different ions together to form a neutral compound. Each ion's charge plays an important role in determining the ratio in which different ions combine. The goal is to balance the overall charge to make a stable, neutral compound.

Here's how pairing works with some examples:
  • Calcium (Ca虏鈦�) pairs with two Chlorine (Cl鈦�) ions to make CaCl鈧�, balancing two positive charges with two negative charges.
  • Rubidium (Rb鈦�) needs to pair two of its ions with one Sulfur (S虏鈦�) to form Rb鈧係, with two +1 charges from rubidium ions matching the -2 charge of sulfur.
  • Three Lithium (Li鈦�) ions combine with one Nitrogen (N鲁鈦�) ion to form Li鈧僋.
  • Yttrium (Y鲁鈦�) pairs with three Selenium (Se虏鈦�) ions to yield Y鈧係e鈧�, balancing six positive charges from yttrium ions with six negative charges from selenium ions.
These examples illustrate that successfully pairing elements requires finding the lowest common multiple of the charges, ensuring that the compound remains electrically neutral.

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Most popular questions from this chapter

Give a name and symbol for an element that is in the same group with (a) argon. (b) \(\mathrm{N}\). (c) Os. (d) tungsten.

Write the name of each of the following ionic compounds. (a) \(\mathrm{NH}_{4} \mathrm{Br}\) (b) \(\mathrm{BaCl}_{2}\) (c) \(\mathrm{K}_{2} \mathrm{O}\) (d) \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\)

Name each of the following compounds, and indicate whether each is ionic or molecular. (a) \(\mathrm{NO}\) (b) \(\mathrm{Y}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (c) \(\mathrm{Na}_{2} \mathrm{O}\) (d) \(\mathrm{NBr}_{3}\)

Name and give the symbols for two elements that (a) are metals. (b) are nonmetals. (c) are metalloids. (d) consist of diatomic molecules.

Which two elements would you expect to exhibit the greatest similarity in physical and chemical properties: Na. Kr. P. Ra, Sr. Te? Explain vour choice.
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