Question

Zinc acetate is sometimes prescribed by physicians for the treatment of Wilson's disease, which is a genetically caused condition wherein copper accumulates to toxic levels in the body. If you were to analyze a sample of zinc acetate and find that it contains \(3.33 \times 10^{23}\) acetate ions, how many grams of zinc acetate must be present in the sample?

Short answer

Approximately 50.76 grams of zinc acetate are present.

Step-by-step solution

Determine the Moles of Acetate Ions

Given that there are \(3.33 \times 10^{23}\) acetate ions, use Avogadro's number \(6.022 \times 10^{23}\) to find the moles of acetate ions: \[ \text{Moles of acetate ions} = \frac{3.33 \times 10^{23} \text{ ions}}{6.022 \times 10^{23} \text{ ions/mol}} \approx 0.553 \text{ mol} \]

Determine the Moles of Zinc Acetate

Zinc acetate has the formula \( \text{Zn(CH}_3\text{COO)}_2 \). Each formula unit provides two acetate ions. Therefore, divide the moles of acetate ions by 2 to find the moles of zinc acetate: \[ \text{Moles of zinc acetate} = \frac{0.553 \text{ mol}}{2} \approx 0.2765 \text{ mol} \]

Calculate the Molar Mass of Zinc Acetate

Calculate the molar mass of \( \text{Zn(CH}_3\text{COO)}_2 \):- Zinc (Zn): 65.38 g/mol- Carbon (C): 12.01 g/mol (4 carbons = 48.04 g/mol)- Hydrogen (H): 1.01 g/mol (6 hydrogens = 6.06 g/mol)- Oxygen (O): 16.00 g/mol (4 oxygens = 64.00 g/mol)Total molar mass = 65.38 + 48.04 + 6.06 + 64.00 = 183.48 g/mol.

Find the Mass of Zinc Acetate

Use the moles of zinc acetate from Step 2 and the molar mass from Step 3 to find the mass: \[ \text{Mass of zinc acetate} = 0.2765 \text{ mol} \times 183.48 \text{ g/mol} \approx 50.76 \text{ g} \]

Key concepts

Moles calculation

Calculating moles is a fundamental concept in chemistry because it allows us to count particles by weighing them. In this exercise, we start with a given number of acetate ions, specifically \(3.33 \times 10^{23}\).
To convert from particles to moles, we use Avogadro's number, which is \(6.022 \times 10^{23}\) particles per mole. This conversion is essential because the mole is the bridge between the atomic scale and the one we observe.
By dividing the number of particles (acetate ions) by Avogadro's number, we can find the number of moles of acetate ions:
\[ \text{Moles of acetate ions} = \frac{3.33 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.553 \text{ mol} \]
Understanding this step is crucial, as it shows how to interpret large numbers of atoms or molecules in terms of moles, making further calculations achievable and relatable to "real-world" quantities.

Molar mass

Molar mass is the mass of one mole of a substance and is a critical factor in converting between moles and grams.
For zinc acetate, which has the chemical formula \( \text{Zn(CH}_3\text{COO)}_2 \), we need to calculate its molar mass by summing up the individual atomic masses:

• Zinc (Zn) contributes 65.38 g/mol.
• The acetate group \((\text{CH}_3\text{COO})\) consists of carbon (C), hydrogen (H), and oxygen (O).
  Each carbon atom adds 12.01 g/mol.
  There are four in total, resulting in 48.04 g/mol.
• Each hydrogen atom adds 1.01 g/mol, with six hydrogens totaling 6.06 g/mol.
• Each oxygen atom adds 16.00 g/mol, and with four oxygens, it's 64.00 g/mol.

Summing these up gives a total molar mass of 183.48 g/mol for zinc acetate.
Grasping molar mass helps in translating between moles, a counting measure, and grams, a measure of mass.

Zinc acetate calculation

Zinc acetate plays a crucial role in this stoichiometry exercise. To find out how much zinc acetate is involved, we use the moles of acetate ions calculated previously. Each zinc acetate molecule contains two acetate ions.
This means that the moles of zinc acetate are half the moles of acetate ions:
\[ \text{Moles of zinc acetate} = \frac{0.553 \, \text{mol}}{2} \approx 0.2765 \, \text{mol} \]
Knowing the number of moles of zinc acetate allows us to determine its mass by using the molar mass previously calculated. Multiplying the moles with the molar mass:
\[ \text{Mass of zinc acetate} = 0.2765 \, \text{mol} \times 183.48 \, \text{g/mol} \approx 50.76 \, \text{g} \]
This calculation provides a clear path from counting ions to measuring the weight of a compound.

Avogadro's number

Avogadro's number is a cornerstone of chemistry, defining the relationship between macroscopic and microscopic scales. It's \(6.022 \times 10^{23}\), the number of atoms or molecules in one mole of a substance.
This number allows chemists to convert between the number of particles and the amount of substance measured in moles.

• If you have the number of particles, dividing by Avogadro's number gives you moles.
• If you have moles, multiplying by Avogadro's number gives you the number of particles.

This duality makes Avogadro's number immensely useful in calculations where actual counting at the atomic level is impractical.
In our exercise, it enabled us to determine how many moles were present in \(3.33 \times 10^{23}\) acetate ions, a step important in linking particles to measurable quantities like mass.