Question

Calculate the number of atoms of each element present in each of the following samples. a. \(4.21 \mathrm{~g}\) of water b. \(6.81 \mathrm{~g}\) of carbon dioxide c. \(0.000221 \mathrm{~g}\) of benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\) d. 2.26 moles of \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)

Short answer

a. Water: \(2.813\times10^{23}\) H atoms, \(1.407\times10^{23}\) O atoms b. Carbon dioxide: \(9.315\times10^{22}\) C atoms, \(1.862\times10^{23}\) O atoms c. Benzene: \(1.02\times10^{19}\) C atoms, \(1.02\times10^{19}\) H atoms d. Sucrose: \(1.632\times10^{25}\) C atoms, \(2.992\times10^{25}\) H atoms, \(1.496\times10^{25}\) O atoms

Step-by-step solution

Find the molar mass of each compound

We first need to determine the molar mass of each compound: - Water (H₂O) = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol - Carbon dioxide (CO₂) = 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol - Benzene (C₆H₆) = 6(12.01 g/mol) + 6(1.01 g/mol) = 78.12 g/mol - Sucrose (C₁₂H₂₂O₁₁) = 12(12.01 g/mol) + 22(1.01 g/mol) + 11(16.00 g/mol) = 342.32 g/mol

Convert mass to moles for given samples

Using the given mass of each compound, contact the mass into moles using their molar mass: a. Water: \(\frac{4.21 \mathrm{~g}}{18.02\mathrm{~g/mol}}=0.2336 \mathrm{~moles}\) b. Carbon dioxide: \(\frac{6.81 \mathrm{~g}}{44.01\mathrm{~g/mol}}=0.1547 \mathrm{~moles}\) c. Benzene: \(\frac{0.000221 \mathrm{~g}}{78.12\mathrm{~g/mol}}=2.828\times10^{-6} \mathrm{~moles}\)

Determine the number of atoms of each element per mole of the compound

For each compound, find the number of atoms of each element per mole of the compound by analyzing their molecular formula: - Water (H₂O) = 2 H atoms and 1 O atom - Carbon dioxide (CO₂) = 1 C atom and 2 O atoms - Benzene (C₆H₆) = 6 C atoms and 6 H atoms - Sucrose (C₁₂H₂₂O₁₁) = 12 C atoms, 22 H atoms, and 11 O atoms

Calculating the number of atoms in each sample

Finally, multiply the number of moles by the number of atoms of each element per mole to find the total number of atoms in the given samples: a. Water: \(0.2336 \mathrm{~moles}\times2\mathrm{~atoms/mole} =0.4672 \mathrm{~moles}~\text{of H}\), \(0.2336 \mathrm{~moles} \times1\mathrm{~atom/mole}=0.2336 \mathrm{~moles}~\text{of O}\) b. Carbon dioxide: \(0.1547 \mathrm{~moles}\times1\mathrm{~atom/mole}=0.1547 \mathrm{~moles}~\text{of C}\), a_calc_carbon_dioxide: \(0.1547 \mathrm{~moles}\times2\mathrm{~atoms/mole}=0.3094 \mathrm{~moles}~\text{of O}\) c. Benzene: \(2.828\times10^{-6} \mathrm{~moles}\times6\mathrm{~atoms/mole} = 1.697 \times10^{-5}\mathrm{~moles}~\text{of C}\), \(2.828\times10^{-6} \mathrm{~moles}\times6\mathrm{~atoms/mole} =1.697\times10^{-5}\mathrm{~moles}~\text{of H}\) d. Sucrose: \(2.26 \mathrm{~moles}\times12\mathrm{~atoms/mole} =27.12 \mathrm{~moles}~\text{of C}\), \(2.26 \mathrm{~moles}\times22\mathrm{~atoms/mole} =49.72 \mathrm{~moles}~\text{of H}\), \(2.26 \mathrm{~moles}\times11\mathrm{~atoms/mole} =24.86 \mathrm{~moles}~\text{of O}\) Now, convert moles of atoms to atoms: a. Water: H atoms: \(0.4672 \mathrm{~moles}\times6.022\times10^{23}\mathrm{~atoms/mole}=2.813\times10^{23}\mathrm{~atoms}\), O atoms: \(0.2336 \mathrm{~moles}\times6.022\times10^{23}\mathrm{~atoms/mole}=1.407\times10^{23}\mathrm{~atoms}\) b. Carbon dioxide: C atoms: \(0.1547 \mathrm{~moles}\times6.022\times10^{23}\mathrm{~atoms/mole}=9.315\times10^{22}\mathrm{~atoms}\), O atoms: \(0.3094 \mathrm{~moles}\times6.022\times10^{23}\mathrm{~atoms/mole}=1.862\times10^{23}\mathrm{~atoms}\) c. Benzene: C atoms: \(1.697\times10^{-5} \mathrm{~moles}\times6.022\times10^{23}\mathrm{~atoms/mole}=1.02\times10^{19}\mathrm{~atoms}\), H atoms: \(1.697\times10^{-5} \mathrm{~moles}\times6.022\times10^{23}\mathrm{~atoms/mole}=1.02\times10^{19}\mathrm{~atoms}\) d. Sucrose: C atoms: \(27.12 \mathrm{~moles}\times6.022\times10^{23}\mathrm{~atoms/mole}=1.632\times10^{25}\mathrm{~atoms}\), H atoms: \(49.72 \mathrm{~moles}\times6.022\times10^{23}\mathrm{~atoms/mole}=2.992\times10^{25}\mathrm{~atoms}\), O atoms: \(24.86 \mathrm{~moles}\times6.022\times10^{23}\mathrm{~atoms/mole}=1.496\times10^{25}\mathrm{~atoms}\)

Key concepts

Mole Concept

Understanding the mole concept is essential in chemistry to connect the microscopic world of atoms and the macroscopic world of grams and liters we deal with in the laboratory. A mole is a fundamental measuring unit in the realm of chemistry used to describe the amount of a chemical substance. It's like a bridge between

• microscopic entities such as atoms and molecules
• macroscopic measurements often used in lab settings, i.e., grams.

This unit designates an amount of substance that contains as many entities - be it atoms, molecules, or ions - as there are atoms in 12 grams of pure carbon-12. This specific number is known as Avogadro's number, which is approximately \(6.022 \times 10^{23}\). By using this big number, chemists can easily translate between the chemicals’ tiny atomic world and the larger, tangible world. For example, if you have a mole of water, it means you possess \(6.022 \times 10^{23}\) water molecules. This concept also helps in calculating the number of atoms present in a given sample when combined with the understanding of molar mass. Without translating moles, it would be incredibly challenging to measure or discuss atomic-scale interactions.

Stoichiometry

Stoichiometry is the heart and soul of chemical reactions, providing an equation that balances on both sides. It involves using the stoichiometric coefficients in chemical equations to understand the quantitative relationships between reactants and products. When you look at a balanced chemical equation, you observe that the law of conservation of mass is maintained, meaning matter isn't created or destroyed, just transformed.
For any chemical reaction, stoichiometry will tell you how much of each reactant is needed to generate a specific amount of product or how much by-product remains. By using stoichiometric calculations, you can determine the proportionate relationship between different substances involved. Calculations often require:

• Finding molar masses from the periodic table.
• Using these molar masses to convert grams to moles.
• Applying mole ratios derived from the balanced equation.

For instance, balancing water formation from hydrogen and oxygen, \(2H_2 + O_2 \rightarrow 2H_2O\), reveals that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water. This illustrates how the stoichiometric coefficients inform us about the exact proportions of molecules required and formed during the reaction process.

Atomic Structure

Atoms are the basic units of matter that make up the universe. They consist of a nucleus which houses protons (positively charged) and neutrons (neutral), with electrons (negatively charged) considered in orbit-like regions around the nucleus. The atomic structure is essential because it determines how atoms interact and bond with each other. Different elements have different numbers of protons, neutrons, and electrons, which defines the element’s identity and its place on the periodic table.
Understanding an atom's structure helps in explaining:

• The arrangement of elements in the periodic table.
• Atomic number, mass number, and isotopes.
• How atoms form chemical bonds.

For instance, carbon-12, a common isotope of carbon, has 6 protons and 6 neutrons in its nucleus. These subatomic particles have almost the entire mass of the atom. Electrons orbit the nucleus, and their arrangements define the atom's size and chemical properties. The atomic structure helps us understand how elements can join to form compounds, dictates the shape, and determines the strength of a molecule. Moreover, it's closely connected to the mole concept, as knowing how many atoms are present in a sample allows chemists to perform more precise calculations in experiments and industrial applications.