Problem 45 - Chemistry: The Central Science

Question

How many hydrogen atoms are in each of the following: (a) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), (b) \(\mathrm{Ca}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\) (c) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} ?\)

Step-by-Step Solution

Answer

(a) There are 6 hydrogen atoms in \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\). (b) There are 6 hydrogen atoms in \(\mathrm{Ca}\left(\mathrm{CH}_{3}\mathrm{COO}\right)_{2}\). (c) There are 12 hydrogen atoms in \(\left(\mathrm{NH}_{4}\right)_{3}\mathrm{PO}_{4}\).

Step 1: Break down the chemical formula

First, we'll identify the hydrogen-containing groups in the formula: \(\mathrm{H}_{5}\) and \(\mathrm{OH}\). Then, we'll count the hydrogen atoms in each group.

Step 2: Count hydrogen atoms

There are 5 hydrogen atoms in \(\mathrm{H}_{5}\) and 1 hydrogen atom in \(\mathrm{OH}\). Adding both counts, we have 5 + 1 = 6 hydrogen atoms in the molecule.

Step 3: Display the answer

There are 6 hydrogen atoms in \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\). (b) \(\mathrm{Ca}\left(\mathrm{CH}_{3}\mathrm{COO}\right)_{2}\)

Step 1: Break down the chemical formula

First, we'll identify the hydrogen-containing groups in the formula: \(\mathrm{CH}_{3}\) and \(\mathrm{COO}\) (as a polyatomic ion). Then, we'll count the hydrogen atoms in each group and remember that there is a subscript 2.

Step 2: Count hydrogen atoms

There are 3 hydrogen atoms in \(\mathrm{CH}_{3}\) and none in \(\mathrm{COO}\). Since the entire polyatomic ion has a subscript 2, we multiply the hydrogen count by 2: 3 * 2 = 6 hydrogen atoms.

Step 3: Display the answer

There are 6 hydrogen atoms in \(\mathrm{Ca}\left(\mathrm{CH}_{3}\mathrm{COO}\right)_{2}\). (c) \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}\)

Step 1: Break down the chemical formula

First, we'll identify the hydrogen-containing groups in the formula: \(\mathrm{NH}_{4}\) and \(\mathrm{PO}_{4}\). Then, we'll count the hydrogen atoms in each group and remember to account for the subscript 3 for \(\mathrm{NH}_{4}\).

Step 2: Count hydrogen atoms

There are 4 hydrogen atoms in \(\mathrm{NH}_{4}\) and none in \(\mathrm{PO}_{4}\). Since the \(\mathrm{NH}_{4}\) group has a subscript 3, we multiply the hydrogen count by 3: 4 * 3 = 12 hydrogen atoms.

Step 3: Display the answer

There are 12 hydrogen atoms in \(\left(\mathrm{NH}_{4}\right)_{3}\mathrm{PO}_{4}\).

Key Concepts

Counting Atoms

Counting atoms in a chemical formula is an essential skill in understanding chemical compositions. Each chemical formula consists of symbols that represent elements, with subscript numbers indicating the quantity of each atom in the compound. The general approach to counting atoms involves:

• Identifying each element or group of elements in the formula
• Locating the subscript numbers which inform you how many atoms of each element are present
• Considering any coefficients or multiplication factors from brackets and parentheses

For instance, in the formula \( \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \), the molecule comprises carbon (\( \mathrm{C} \)), hydrogen (\( \mathrm{H} \)), and oxygen (\( \mathrm{O} \)). Here, the subscripts after \( \mathrm{C} \) and \( \mathrm{H} \) indicate the number of carbon and hydrogen atoms respectively. If a chemical formula seems complex due to polyatomic groups (like \( \mathrm{COO} \) in \( \mathrm{Ca(CH}_3 \mathrm{COO})_2 \)), handle them as units and apply any subscripts to each atom in the group.

Hydrogen Atoms

Hydrogen atoms in chemical formulas often appear as part of larger groups or molecules. Identifying the number of hydrogen atoms means carefully examining these groups. Consider the following examples:

• In \( \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \), hydrogen is part of both \( \mathrm{H}_5 \) and the hydroxyl group, \( \mathrm{OH} \). The total number of hydrogen atoms is calculated by adding 5 (from \( \mathrm{H}_5 \)) and 1 (from \( \mathrm{OH} \)), resulting in 6 hydrogen atoms.
• For \( \mathrm{Ca(CH}_3 \mathrm{COO})_2 \), hydrogen atoms are only in the \( \mathrm{CH}_3 \) group. Since this group is within a bracketed polyatomic ion repeated twice, you multiply the original count (3 hydrogen atoms) by 2, giving you 6 hydrogen atoms.

It's important to realize that every bracketed group or polyatomic ion that's part of a larger compound must be evaluated fully and sometimes multiplied by any number indicating its repeated occurrence in the formula.

Polyatomic Ions

Polyatomic ions are groups of atoms bonded together that carry an electric charge. Like other elements, they appear within chemical formulas as cohesive units and are integral for determining the complete composition. Recognizing polyatomic ions helps in both identifying atoms and calculating their quantities. Consider these details:

• Polyatomic ions can be as simple as hydroxide (\( \mathrm{OH}^- \)) or as complex as phosphate (\( \mathrm{PO}_4^{3-} \)). They are treated as single entities during chemical calculations.
• In \( \mathrm{Ca(CH}_3 \mathrm{COO})_2 \), \( \mathrm{COO}^- \) represents the acetate ion. When multiply repeating units, such as \( \mathrm{NH}_4^+ \) in \( (\mathrm{NH}_4)_3 \mathrm{PO}_4 \), it's crucial to apply the subscript across the entire ion, leading to multiplication of each component inside.

Understanding polyatomic ions is essential not only for balancing reactions but also for determining various properties of compounds. Learning these nuances enhances your ability to accurately interpret and manipulate chemical formulas.