Q. \( \textbf{Molar mass}\ \mathrm{CH_3COOH} \).

Q: Molar mass of \mathrm{CH_3COOH} ?

\mathrm{CH_3COOH} = \mathrm{C_2H_4O_2} . Molar mass =2(12.01)+4(1.008)+2(16.00)\approx 60.05\ \mathrm{g/mol}.

Q: Why is \mathrm{CH_3COOH} written as \mathrm{C_2H_4O_2} ?

Expand groups: \mathrm{CH_3} gives \mathrm{C_1H_3} , and \mathrm{COOH} gives \mathrm{C_1H_1O_2} . Total \mathrm{C_2H_4O_2} .

Q: What’s the molar mass of \mathrm{CH_3COOH} if rounding atomic masses to integers?

Use \mathrm{C}=12, \mathrm{H}=1, \mathrm{O}=16: 2(12)+4(1)+2(16)=24+4+32=60\ \mathrm{g/mol}.

Q: How do you compute molar mass from the formula systematically?

Count atoms in \mathrm{C_2H_4O_2} , multiply each count by its atomic mass, then sum: \sum n_i M_i . Here: 2M_C+4M_H+2M_O.

Q: Is the result in grams per mole or daltons?

Molar mass is typically in \mathrm{g/mol} . Numerically, \mathrm{1\ molar\ mass\ unit\ (amu)} equals one dalton, so values match (about 60.05\ \mathrm{Da} per molecule).

Answer

\(\text{CH}_3\text{COOH}\) has atoms: \(\text{C}_2\text{H}_4\text{O}_2\).

Using atomic masses \(\text{C}=12.01\), \(\text{H}=1.008\), \(\text{O}=16.00\) g/mol:

\[
M = 2(12.01) + 4(1.008) + 2(16.00)
\]

\[
M = 24.02 + 4.032 + 32.00 = 60.052 \approx 60.05 \text{ g/mol}
\]

Detailed Explanation

Problem: Find the molar mass of CH₃COOH (acetic acid).

Step 1: Identify the atoms in the formula.

The chemical formula is:

\[
\mathrm{CH_3COOH}
\]

Count each type of atom:

\[
\mathrm{C: 2,\ \ H: 4,\ \ O: 2}
\]

Step 2: Write the molar mass calculation setup.

Use atomic molar masses (typical periodic table values):

\[
\mathrm{C \approx 12.01\ g/mol,\ \ H \approx 1.008\ g/mol,\ \ O \approx 16.00\ g/mol}
\]

Multiply each atomic molar mass by the number of atoms, then add:

\[
M = 2(12.01) + 4(1.008) + 2(16.00)
\]

Step 3: Compute each contribution.

Carbon contribution:

\[
2(12.01) = 24.02\ g/mol
\]

Hydrogen contribution:

\[
4(1.008) = 4.032\ g/mol
\]

Oxygen contribution:

\[
2(16.00) = 32.00\ g/mol
\]

Step 4: Add them to get the molar mass.

\[
M = 24.02 + 4.032 + 32.00 = 60.052\ g/mol
\]

Final Answer:

\[
\boxed{M(\mathrm{CH_3COOH}) \approx 60.05\ g/mol}
\]

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General Chemistry FAQs

Molar mass of \( \mathrm{CH_3COOH} \)?

\( \mathrm{CH_3COOH} = \mathrm{C_2H_4O_2} \). Molar mass \(=2(12.01)+4(1.008)+2(16.00)\approx 60.05\ \mathrm{g/mol}\).

Why is \( \mathrm{CH_3COOH} \) written as \( \mathrm{C_2H_4O_2} \)?

Expand groups: \( \mathrm{CH_3} \) gives \( \mathrm{C_1H_3} \), and \( \mathrm{COOH} \) gives \( \mathrm{C_1H_1O_2} \). Total \( \mathrm{C_2H_4O_2} \).

What is the molar mass using atomic masses \( \mathrm{C}=12.01\), \( \mathrm{H}=1.008\), \( \mathrm{O}=16.00\)?

\(2(12.01)+4(1.008)+2(16.00)=24.02+4.032+32.00=60.052\approx 60.05\ \mathrm{g/mol}\).

What’s the molar mass of \( \mathrm{CH_3COOH} \) if rounding atomic masses to integers?

Use \( \mathrm{C}=12\), \( \mathrm{H}=1\), \( \mathrm{O}=16\): \(2(12)+4(1)+2(16)=24+4+32=60\ \mathrm{g/mol}\).

How do you compute molar mass from the formula systematically?

Count atoms in \( \mathrm{C_2H_4O_2} \), multiply each count by its atomic mass, then sum: \( \sum n_i M_i \). Here: \(2M_C+4M_H+2M_O\).

Is the result in grams per mole or daltons?

Molar mass is typically in \( \mathrm{g/mol} \). Numerically, \( \mathrm{1\ molar\ mass\ unit\ (amu)} \) equals one dalton, so values match (about \(60.05\ \mathrm{Da}\) per molecule).

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