Q. The molar mass of \( \text{Zn}(\text{C}_2\text{H}_3\text{O}_2)_2 \).

Q: What is the molar mass expression for \mathrm{Zn(C_2H_3O_2)_2} ?

\mathrm{Zn(C_2H_3O_2)_2} has 1 Zn, 2 groups of \mathrm{C_2H_3O_2}. So moles = \mathrm{Zn_1C_4H_6O_4}.

Q: How many atoms of each element are in \mathrm{Zn(C_2H_3O_2)_2} ?

\mathrm{Zn_1C_4H_6O_4}. Count: \mathrm{C: 2\times2=4}, \mathrm{H: 3\times2=6}, \mathrm{O: 2\times2=4}, \mathrm{Zn: 1}.

Q: What atomic masses should be used for Zn, C, H, and O?

Common values: \mathrm{Zn=65.38}, \mathrm{C=12.01}, \mathrm{H=1.008}, \mathrm{O=16.00} g/mol.

Q: If someone mistakenly used \mathrm{C_2H_3O_2} once, what error happens?

They would get \mathrm{Zn(C_2H_3O_2)} instead of \mathrm{Zn(C_2H_3O_2)_2}, undercounting atoms by a factor of 2 for \mathrm{C}, \mathrm{H}, and \mathrm{O}.

Answer

Step 1: Interpret the formula. \( \text{Zn(C}_2\text{H}_3\text{O}_2)_2 \) means \( \text{Zn} \), and two acetate groups per formula unit.

Step 2: Count atoms. Each \( \text{C}_2\text{H}_3\text{O}_2 \) has \( \text{C}_2\text{H}_3\text{O}_2 \). With the subscript \(2\):
\( \text{C}_4\text{H}_6\text{O}_4 \) from the anion, plus \( \text{Zn}_1 \).

Step 3: Add molar masses. Using atomic masses \( \text{Zn} = 65.38 \), \( \text{C} = 12.01 \), \( \text{H} = 1.008 \), \( \text{O} = 16.00 \):
\[
M = 1(65.38) + 4(12.01) + 6(1.008) + 4(16.00)
\]
\[
M = 65.38 + 48.04 + 6.048 + 64.00 = 183.468 \text{ g/mol}
\]

Final result: \(\boxed{183.47\ \text{g/mol}}\)

Detailed Explanation

Goal: Find the molar mass of zinc acetate, \( \text{Zn}(\text{C}_2\text{H}_3\text{O}_2)_2 \).

Step 1: Write the formula clearly.

The compound is \( \text{Zn}(\text{C}_2\text{H}_3\text{O}_2)_2 \). The subscript \(2\) applies to every atom inside the parentheses.

So first count atoms:

Zn: \(1\) atom
\(\text{C}_2\text{H}_3\text{O}_2\) times \(2\):
- C: \(2 \times 2 = 4\)
- H: \(3 \times 2 = 6\)
- O: \(2 \times 2 = 4\)

Overall formula: \( \text{Zn}\text{C}_4\text{H}_6\text{O}_4 \).

Step 2: List atomic masses.

Use standard atomic molar masses (in g/mol):

\(\text{Zn} = 65.38\)
\(\text{C} = 12.01\)
\(\text{H} = 1.008\)
\(\text{O} = 16.00\)

Step 3: Multiply each atomic mass by its number of atoms.

Compute each contribution:

Zn contribution: \(1 \times 65.38 = 65.38\)
C contribution: \(4 \times 12.01 = 48.04\)
H contribution: \(6 \times 1.008 = 6.048\)
O contribution: \(4 \times 16.00 = 64.00\)

Step 4: Add all contributions to get the molar mass.

\[
\text{Molar mass} = 65.38 + 48.04 + 6.048 + 64.00
\]
\[
\text{Molar mass} = 183.468 \text{ g/mol}
\]

Final Answer: \( \boxed{183.47 \text{ g/mol}} \)

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General Chemistry FAQs

What is the molar mass expression for \( \mathrm{Zn(C_2H_3O_2)_2} \)?

\(\mathrm{Zn(C_2H_3O_2)_2}\) has \(1\) Zn, \(2\) groups of \( \mathrm{C_2H_3O_2}\). So moles = \( \mathrm{Zn_1C_4H_6O_4}\).

How many atoms of each element are in \( \mathrm{Zn(C_2H_3O_2)_2} \)?

\(\mathrm{Zn_1C_4H_6O_4}\). Count: \(\mathrm{C: 2\times2=4}\), \(\mathrm{H: 3\times2=6}\), \(\mathrm{O: 2\times2=4}\), \(\mathrm{Zn: 1}\).

What atomic masses should be used for Zn, C, H, and O?

Common values: \(\mathrm{Zn=65.38}\), \(\mathrm{C=12.01}\), \(\mathrm{H=1.008}\), \(\mathrm{O=16.00}\) g/mol.

What is the full molar mass calculation setup?

\[\[65.38(1)+12.01(4)+1.008(6)+16.00(4)\]\]

What is the final molar mass of \( \mathrm{Zn(C_2H_3O_2)_2} \) in g/mol?

\(\mathrm{=65.38+48.04+6.048+64.00=183.47\ g/mol}\approx \mathrm{183.47}\).

If someone mistakenly used \( \mathrm{C_2H_3O_2} \) once, what error happens?

They would get \(\mathrm{Zn(C_2H_3O_2)}\) instead of \(\mathrm{Zn(C_2H_3O_2)_2}\), undercounting atoms by a factor of \(2\) for \(\mathrm{C}\), \(\mathrm{H}\), and \(\mathrm{O}\).

Molar mass of Zn(C2H3O2)2.
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