Problem 18 - General Chemistry: Principles and Modern Applications

Question

For the compound \(\operatorname{Ge}\left[\mathrm{S}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{3}\right]_{4},\) determine (a) the total number of atoms in one formula unit (b) the ratio, by number, of C atoms to H atoms (c) the ratio, by mass, of Ge to \(S\) (d) the number of \(g\) S in 1 mol of the compound (e) the number of \(C\) atoms in 33.10 g of the compound

Step-by-Step Solution

Answer

The total number of atoms in one formula unit is 57. The ratio of C atoms to H atoms is 5:8. And the ratio, by mass, of Ge to S is 0.566:1. Meanwhile, one mole of this compound contains roughly \(2.408*10^{24}\) atoms of S. Lastly, 33.10 g of compound contains approximately \(1.018*10^{24}\) atoms of C.

Step 1: Determine the total number of atoms in one formula unit

To find the total number of atoms, simply list out all the atoms that present in the formula and count them. The compound contains 1 Ge atom, 4 S atoms, 20 C atoms, and 32 H atoms. In total, there are 57 atoms in one formula unit.

Step 2: Calculate the ratio by number of C atoms to H atoms

This involves simply counting the number of each type of atom in the formula. There are 20 C atoms and 32 H atoms. Thus, the ratio of C to H is 20:32, which simplifies to 5:8.

Step 3: Determine the ratio by mass of Ge to S

First, calculate the atomic mass of Ge and S using the periodic table. The atomic mass of Ge and S is about 72.63 and 32.06 g/mol respectively. Multiply the atomic mass by the number of each type of atom in the formula. The compound contains 1 Ge atom and 4 S atoms. Therefore, the mass of Ge and S in the compound is 72.63 g/mol and \(4*32.06\) g/mol = 128.24 g/mol, respectively. The ratio by mass of Ge to S is therefore 72.63:128.24, or approximately 0.566:1.

Step 4: Determine the number of S atoms in 1 mol of the compound

In one mole of this compound, there are four moles of S atoms because the subscript after S in the compound formula is 4. So, there are \(4*6.022*10^{23}\) S atoms in 1 mol of compound, based on Avogadro's number, which is approximately \(2.408*10^{24}\) S atoms.

Step 5: Calculate the number of C atoms in 33.10 g of the compound

Calculate the molar mass of the entire compound first. It is approximately 391.296 g/mol. Then, calculate the number of moles of compound in 33.10 grams, which is \(33.10/391.296\) mol = 0.0846 mol. One mole of this compound contains 20 moles of C, so 0.0846 mol of compound contains \(0.0846*20 = 1.692\) mol of C. Thus, the number of C atoms in 33.10 g of compound is \(1.692*6.022*10^{23} = 1.018*10^{24}\) atoms of C.

Key Concepts

Atomic Composition: How to Count and Understand

In understanding the atomic composition of a compound, our primary goal is to tally up all the various types of atoms that exist in one unit of the compound. This begins with analyzing the chemical formula. The given compound, \(\operatorname{Ge}[\mathrm{S}(\mathrm{CH}_{2})_{4}\mathrm{CH}_{3}]_{4}\), is a great example. This indicates each molecule contains:

• 1 Germanium (Ge) atom
• 4 Sulfur (S) atoms
• 20 Carbon (C) atoms
• 32 Hydrogen (H) atoms

Counting these all together provides a total of 57 atoms per formula unit.
Understanding atomic composition is crucial as it forms the base for advanced topics like chemical reactions and molecular interactions.

Mass Ratios: Comparing Masses within a Compound

The concept of mass ratios involves comparing the mass of each type of atom in a compound. To do this, we first need the atomic masses of the elements, obtained from the periodic table.
For instance, the mass of Germanium (Ge) is about 72.63 g/mol. Sulfur (S), on the other hand, has an atomic mass of about 32.06 g/mol.
In our compound, while there is one Ge atom, there are four S atoms. This means that the mass of S is \(4 \times 32.06 = 128.24\) g/mol for the compound. Hence, the mass ratio of Ge to S becomes \(72.63:128.24\), or approximately 0.566:1.
Using mass ratios helps us understand the relative weight contribution of each element in the compound.

Stoichiometry: The Heart of Chemical Calculations

Stoichiometry is central to performing chemical calculations, often linking to relationships of reactants and products in a chemical reaction. However, in this context, it helps us to understand the proportion of elements within a compound.
For \(\operatorname{Ge}[\mathrm{S}(\mathrm{CH}_{2})_{4}\mathrm{CH}_{3}]_{4}\), stoichiometry signifies there are four moles of S atoms in every mole of the compound.
This means one mole of the compound holds exactly four times Avogadro's number in sulfur atoms, approximately \(2.408 \times 10^{24}\) S atoms.
Mastering stoichiometry is crucial as it provides insight into the quantities of substances involved in reactions, which is fundamental for developing balanced equations and predicting products.

Molar Calculations: From Moles to Atoms

Molar calculations form a backbone for translating between mass and number of atoms or molecules in chemistry.
For our compound's complete molar mass of approximately 391.296 g/mol, things start to get interesting when we examine a specific example, like determining how many carbon atoms are present in a 33.10 g sample of this compound.
To find this, first calculate the number of moles of the compound: \(33.10 \text{ g} / 391.296 \text{ g/mol} = 0.0846 \text{ mol}\).
Each mole of compound contains 20 moles of carbon atoms, therefore 0.0846 moles of the compound contains \(0.0846 \times 20 = 1.692\) moles of carbon.
With Avogadro's number (\(6.022 \times 10^{23}\)), this equates to approximately \(1.018 \times 10^{24}\) carbon atoms.
Molar calculations enable us to accurately connect the mass of a chemical to its inherent atomic make-up, invaluable in both laboratory calculations and real-world applications.