Question

1. vinegar, a common household item, is used in cleaning, cooking, baking and meat preservation. vinegar contains acetic acid, ch₃cooh, which gives vinegar its sour taste and pungent smell. what is the percent composition of carbon in acetic acid?

options: 40.0%, 41.4%, 33.3%, 20.0%

Explanation:

Step1: Calculate molar mass of \(CH_3COOH\)

Molar mass of \(C\): \(12.01\ g/mol\), \(H\): \(1.008\ g/mol\), \(O\): \(16.00\ g/mol\).
In \(CH_3COOH\) (or \(C_2H_4O_2\)), there are \(2\) \(C\), \(4\) \(H\), \(2\) \(O\).
Molar mass \(= 2\times12.01 + 4\times1.008 + 2\times16.00\)
\(= 24.02 + 4.032 + 32.00 = 60.052\ g/mol\).

Step2: Calculate mass of \(C\) in one mole

Mass of \(C = 2\times12.01 = 24.02\ g/mol\).

Step3: Calculate percent composition of \(C\)

Percent composition \(= \frac{\text{Mass of } C}{\text{Molar mass of } CH_3COOH} \times 100\%\)
\(= \frac{24.02}{60.052} \times 100\% \approx 40.0\%\)? Wait, no—wait, recalculate:
Wait, \(2\times12.01 = 24.02\), molar mass \(60.05\). \(24.02/60.05 \approx 0.400\), but wait, maybe I miscalculated. Wait, no: \(CH_3COOH\) is \(C_2H_4O_2\). Wait, \(2C\): \(24.02\), \(4H\): \(4.032\), \(2O\): \(32.00\). Total: \(24.02 + 4.032 = 28.052 + 32 = 60.052\). Then \(24.02 / 60.052 \approx 0.400\), but the options have 40.0%? Wait, no, maybe I made a mistake. Wait, no—wait, acetic acid is \(C_2H_4O_2\), so molar mass is \(2(12.01) + 4(1.008) + 2(16.00) = 24.02 + 4.032 + 32.00 = 60.052\ g/mol\). Mass of C: \(2\times12.01 = 24.02\ g/mol\). Percent: \((24.02 / 60.052) \times 100 \approx 40.0\%\)? But wait, the option 40.0% is there. Wait, but let me check again. Wait, maybe the molar mass is calculated as \(12.01\times2 + 1.008\times4 + 16.00\times2 = 24.02 + 4.032 + 32 = 60.052\). Then \(24.02 / 60.052 = 0.400\), so 40.0%? But wait, the option 40.0% is an option. Wait, but maybe I messed up the formula. Wait, \(CH_3COOH\) is \(C_2H_4O_2\), correct. So yes, percent composition of C is \(\frac{2\times12.01}{60.05} \times 100 \approx 40.0\%\). Wait, but the option 40.0% is present. Wait, but let me check the calculation again. Wait, \(2\times12.01 = 24.02\), \(60.05\) is molar mass. \(24.02 / 60.05 = 0.4\), so 40.0%. So the answer is 40.0%.

Wait, but wait, maybe I made a mistake. Wait, no—acetic acid's molar mass is 60.05 g/mol, and carbon is 24.02 g/mol. So 24.02/60.05 = 0.4, so 40.0%. So the correct option is 40.0%.

Answer:

The percent composition of carbon in acetic acid is calculated as follows:

1. Molar mass of \(CH_3COOH\) (\(C_2H_4O_2\)):

\(2 \times 12.01\ (C) + 4 \times 1.008\ (H) + 2 \times 16.00\ (O) = 60.052\ \text{g/mol}\).

1. Mass of carbon in 1 mole:

\(2 \times 12.01 = 24.02\ \text{g/mol}\).

1. Percent composition of carbon:

\(\frac{24.02}{60.052} \times 100\% \approx 40.0\%\).

The correct option is: \(\boldsymbol{40.0\%}\) (the first option with "40.0%").