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Chapter 3: Problem 18

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

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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.
04

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.
05

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.

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Most popular questions from this chapter

Assign suitable names to the compounds (a) \(\mathrm{CS}_{2}\) (b) \(\operatorname{SiF}_{4} ;\) (c) \(\operatorname{ClF}_{5} ;\) (d) \(\mathrm{N}_{2} \mathrm{O}_{5} ;\) (e) \(\mathrm{SF}_{6} ;\) (f) \(\mathrm{I}_{2} \mathrm{Cl}_{6}\).

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