Question

How many atoms of each type are represented in each of the following? \(\begin{array}{ll}{\text { a. } 3 \mathrm{N}_{2}} & {\text { f. } 5 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}} \\ {\text { b. } 2 \mathrm{H}_{2} \mathrm{O}} & {\text { g. } 4 \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}} \\ {\text { c. } 4 \mathrm{HNO}_{3}} & {\text { h. } 2\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}}\end{array}\) \(\begin{array}{ll}{\text { d. } 2 \mathrm{Ca}(\mathrm{OH})_{2}} & {\text { i. } 6 \mathrm{Al}_{2}\left(\mathrm{SeO}_{4}\right)_{3}} \\ {\text { e. } 3 \mathrm{Ba}\left(\mathrm{ClO}_{3}\right)_{2}} & {\text { j. } 4 \mathrm{C}_{3} \mathrm{H}_{8}}\end{array}\)

Short answer

a. 6 Nb. 4 H, 2 Oc. 4 H, 4 N, 12 Od. 2 Ca, 4 O, 4 He. 3 Ba, 6 Cl, 18 Of. 5 Fe, 10 N, 30 Og. 12 Mg, 8 P, 32 Oh. 4 N, 16 H, 2 S, 8 Oi. 12 Al, 18 Se, 72 Oj. 12 C, 32 H

Step-by-step solution

Analyze each molecule separately

Examine the chemical formula of each molecule to identify the type and number of atoms in a single unit of the molecule.

Multiply by the given quantity

Multiply the number of each type of atom in the single molecule's formula by the coefficient in front of the molecule to find the total number of each atom.

Sum the atoms for each molecule

Add up the totals for each type of atom in the full quantity specified in the problem.

Solution for Part a: 3 \( \mathrm{N}_{2} \)

Each \( \mathrm{N}_{2} \) contains 2 Nitrogen (N) atoms.Hence, 3 \( \mathrm{N}_{2} \) will contain \( 3 \times 2 = 6 \) Nitrogen (N) atoms.

Solution for Part b: 2 \( \mathrm{H}_{2}\mathrm{O} \)

Each \( \mathrm{H}_{2}\mathrm{O} \) contains 2 Hydrogen (H) atoms and 1 Oxygen (O) atom.Hence, 2 \( \mathrm{H}_{2}\mathrm{O} \) will contain \( 2 \times 2 = 4 \) Hydrogen (H) and \( 2 \times 1 = 2 \) Oxygen (O) atoms.

Solution for Part c: 4 \( \mathrm{HNO}_{3} \)

Each \( \mathrm{HNO}_{3} \) contains 1 Hydrogen (H), 1 Nitrogen (N), and 3 Oxygen (O) atoms.Hence, 4 \( \mathrm{HNO}_{3} \) will contain \( 4 \times 1 = 4 \) Hydrogen (H), \( 4 \times 1 = 4 \) Nitrogen (N), and \( 4 \times 3 = 12 \) Oxygen (O) atoms.

Solution for Part d: 2 \( \mathrm{Ca} (\mathrm{OH})_{2} \)

Each \( \mathrm{Ca} (\mathrm{OH})_{2} \) contains 1 Calcium (Ca), 2 Oxygen (O), and 2 Hydrogen (H) atoms.Hence, 2 \( \mathrm{Ca} (\mathrm{OH})_{2} \) will contain \( 2 \times 1 = 2 \) Calcium (Ca), \( 2 \times 2 = 4 \) Oxygen (O), and \( 2 \times 2 = 4 \) Hydrogen (H) atoms.

Solution for Part e: 3 \( \mathrm{Ba} (\mathrm{ClO}_{3})_{2} \)

Each \( \mathrm{Ba} (\mathrm{ClO}_{3})_{2} \) contains 1 Barium (Ba), 2 Chlorine (Cl), and 6 Oxygen (O) atoms.Hence, 3 \( \mathrm{Ba} (\mathrm{ClO}_{3})_{2} \) will contain \( 3 \times 1 = 3 \) Barium (Ba), \( 3 \times 2 = 6 \) Chlorine (Cl), and \( 3 \times 6 = 18 \) Oxygen (O) atoms.

Solution for Part f: 5 \( \mathrm{Fe} (\mathrm{NO}_{3})_{2} \)

Each \( \mathrm{Fe} (\mathrm{NO}_{3})_{2} \) contains 1 Iron (Fe), 2 Nitrogen (N), and 6 Oxygen (O) atoms.Hence, 5 \( \mathrm{Fe} (\mathrm{NO}_{3})_{2} \) will contain \( 5 \times 1 = 5 \) Iron (Fe), \( 5 \times 2 = 10 \) Nitrogen (N), and \( 5 \times 6 = 30 \) Oxygen (O) atoms.

Solution for Part g: 4 \( \mathrm{Mg}_{3} (\mathrm{PO}_{4})_{2} \)

Each \( \mathrm{Mg}_{3} (\mathrm{PO}_{4})_{2} \) contains 3 Magnesium (Mg), 2 Phosphorus (P), and 8 Oxygen (O) atoms.Hence, 4 \( \mathrm{Mg}_{3} (\mathrm{PO}_{4})_{2} \) will contain \( 4 \times 3 = 12 \) Magnesium (Mg), \( 4 \times 2 = 8 \) Phosphorus (P), and \( 4 \times 8 = 32 \) Oxygen (O) atoms.

Solution for Part h: 2 \( (\mathrm{NH}_{4})_{2} \mathrm{SO}_{4} \)

Each \( (\mathrm{NH}_{4})_{2} \mathrm{SO}_{4} \) contains 2 Nitrogen (N), 8 Hydrogen (H), 1 Sulfur (S), and 4 Oxygen (O) atoms.Hence, 2 \( (\mathrm{NH}_{4})_{2} \mathrm{SO}_{4} \) will contain \( 2 \times 2 = 4 \) Nitrogen (N), \( 2 \times 8 = 16 \) Hydrogen (H), \( 2 \times 1 = 2 \) Sulfur (S), and \( 2 \times 4 = 8 \) Oxygen (O) atoms.

Solution for Part i: 6 \( \mathrm{Al}_{2} (\mathrm{SeO}_{4})_{3} \)

Each \( \mathrm{Al}_{2} (\mathrm{SeO}_{4})_{3} \) contains 2 Aluminum (Al), 3 Selenium (Se), and 12 Oxygen (O) atoms.Hence, 6 \( \mathrm{Al}_{2} (\mathrm{SeO}_{4})_{3} \) will contain \( 6 \times 2 = 12 \) Aluminum (Al), \( 6 \times 3 = 18 \) Selenium (Se), and \( 6 \times 12 = 72 \) Oxygen (O) atoms.

Solution for Part j: 4 \( \mathrm{C}_{3} \mathrm{H}_{8} \)

Each \( \mathrm{C}_{3} \mathrm{H}_{8} \) contains 3 Carbon (C) and 8 Hydrogen (H) atoms.Hence, 4 \( \mathrm{C}_{3} \mathrm{H}_{8} \) will contain \( 4 \times 3 = 12 \) Carbon (C) and \( 4 \times 8 = 32 \) Hydrogen (H) atoms.

Key concepts

Chemical Formulas

Chemical formulas are like recipes for molecules. They tell us what elements (like ingredients) are in a molecule and how many atoms of each element are present. For example, water's chemical formula is \( \text{H}_2\text{O} \), which means each water molecule has 2 hydrogen (H) atoms and 1 oxygen (O) atom.

Chemical formulas are fundamental because they allow us to understand what makes each substance unique. By analyzing a chemical formula, we can determine the molecular composition and predict how substances will react with one another. This makes chemical formulas an essential tool in chemistry.

Molecular Composition

Molecular composition refers to the types and numbers of atoms that make up a molecule. Using the chemical formula, we can break down the composition of any molecule. For example, the formula for glucose is \( \text{C}_6\text{H}_{12}\text{O}_6 \). This means that one glucose molecule consists of 6 carbon (C) atoms, 12 hydrogen (H) atoms, and 6 oxygen (O) atoms.

Understanding molecular composition helps us identify the molecule's properties. It guides us in predicting how molecules interact with each other, allowing scientists to create new compounds and materials. By knowing the molecular composition, we can understand a molecule's behavior in different chemical reactions.

Atom Counting

Atom counting is the process of determining the number of each type of atom in a molecule. This is done using the molecule's chemical formula and multiplying by the amount specified in the problem. For instance, if given 3 \( \text{N}_2 \) (nitrogen gas), we know that each \( \text{N}_2 \) molecule contains 2 nitrogen (N) atoms.

We multiply 2 (atoms in one molecule) by 3 (the amount of molecules given): \( 3 \times 2 = 6 \) nitrogen atoms. Similarly, for 2 \( \text{H}_2\text{O} \) (water), since each \( \text{H}_2\text{O} \) molecule has 2 hydrogen atoms and 1 oxygen atom, you simply multiply those numbers accordingly to find the total amount of each atom present in the combined molecules.

Chemical Equations

Chemical equations show us the transformation of reactants into products in a chemical reaction. They use chemical formulas to represent substances, and coefficients to denote the number of molecules involved. For example, the combustion of methane is represented by:

\[ \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \]
Each side of the equation must balance; the number of each type of atom must be the same on both sides. This ensures the law of conservation of mass is followed. Balancing chemical equations is crucial because it provides a quantitative understanding of the chemical reactions, allowing us to predict the amounts of reactants and products involved precisely.