Q. \(\text{Molar mass of } \mathrm{H_2CO_3}\)

Q: What is the molar mass of \mathrm{H_2CO_3} ?

\mathrm{H_2CO_3} has 2 H, 1 C, 3 O. Using atomic masses \mathrm{H}=1.01, \mathrm{C}=12.01, \mathrm{O}=16.00: 2(1.01)+12.01+3(16.00)=62.03\ \mathrm{g/mol}.

Q: Does using \mathrm{O}=15.999} or \mathrm{16.00} change the answer?

Slightly. For example, with \mathrm{H}=1.008, \mathrm{C}=12.011, \mathrm{O}=15.999: 2(1.008)+12.011+3(15.999)=62.024\ \mathrm{g/mol}. Rounding gives \approx 62.03\ \mathrm{g/mol}.

Q: What is the molar mass of just the carbon part in \mathrm{H_2CO_3} ?

The carbon contribution is 1 \times A_{\mathrm{C}} \approx 12.01\ \mathrm{g/mol}. It’s one carbon atom per formula unit.

Q: What is the total molar mass contributed by oxygen atoms in \mathrm{H_2CO_3} ?

There are 3 oxygen atoms. Oxygen contribution is 3 \times A_{\mathrm{O}} = 3(16.00)=48.00\ \mathrm{g/mol} (or 3(15.999)=47.997\ \mathrm{g/mol} with more precise values).

Q: What is the molar mass contributed by hydrogen atoms in \mathrm{H_2CO_3} ?

There are 2 hydrogen atoms. Hydrogen contribution is 2 \times A_{\mathrm{H}} = 2(1.01)=2.02\ \mathrm{g/mol} (or 2(1.008)=2.016\ \mathrm{g/mol} with precise values).

Answer

To find the molar mass of hydrogen carbonate, use atomic masses: H = 1.008, C = 12.01, O = 16.00 (g/mol).

\[
M\left(\mathrm{H_2CO_3}\right)=2(1.008)+1(12.01)+3(16.00)
\]

\[
=2.016+12.01+48.00=62.03 \text{ g/mol}
\]

Final result: \(62.03\ \text{g/mol}\).

Detailed Explanation

To find the molar mass of \( \mathrm{H_2CO_3} \), you add up the atomic masses of all atoms in one formula unit.

Step 1: Identify how many atoms of each element are in \( \mathrm{H_2CO_3} \)

\( \mathrm{H_2CO_3} \) contains:

\(2\) atoms of hydrogen, \( \mathrm{H} \)
\(1\) atom of carbon, \( \mathrm{C} \)
\(3\) atoms of oxygen, \( \mathrm{O} \)

Step 2: Write the molar mass expression using atomic masses

Use the standard atomic masses (in \( \mathrm{g\ mol^{-1}} \)):

\( \mathrm{H} \approx 1.01 \)
\( \mathrm{C} \approx 12.01 \)
\( \mathrm{O} \approx 16.00 \)

Now compute:

\[
M(\mathrm{H_2CO_3}) = 2\bigl(1.01\bigr) + 1\bigl(12.01\bigr) + 3\bigl(16.00\bigr)
\]

Step 3: Perform the arithmetic carefully

Calculate each part:

\(2(1.01) = 2.02\)
\(1(12.01) = 12.01\)
\(3(16.00) = 48.00\)

Add them:

\[
M(\mathrm{H_2CO_3}) = 2.02 + 12.01 + 48.00
\]
\[
M(\mathrm{H_2CO_3}) = 62.03\ \mathrm{g\ mol^{-1}}
\]

Final Answer

The molar mass of \( \mathrm{H_2CO_3} \) is \( \boxed{62.03\ \mathrm{g\ mol^{-1}}} \).

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General Chemistry FAQs

What is the molar mass of \( \mathrm{H_2CO_3} \)?

\( \mathrm{H_2CO_3} \) has \(2\) H, \(1\) C, \(3\) O. Using atomic masses \( \mathrm{H}=1.01\), \( \mathrm{C}=12.01\), \( \mathrm{O}=16.00\): \(2(1.01)+12.01+3(16.00)=62.03\ \mathrm{g/mol}\).

How do you compute molar mass from atomic masses?

Multiply each element’s atomic mass by its subscript in the formula, then sum: \(M=\sum n_i A_i\). For \( \mathrm{H_2CO_3} \): \(M=2A_{\mathrm{H}}+1A_{\mathrm{C}}+3A_{\mathrm{O}}\).

Does using \( \mathrm{O}=15.999} \) or \( \mathrm{16.00} \) change the answer?

Slightly. For example, with \( \mathrm{H}=1.008\), \( \mathrm{C}=12.011\), \( \mathrm{O}=15.999\): \(2(1.008)+12.011+3(15.999)=62.024\ \mathrm{g/mol}\). Rounding gives \(\approx 62.03\ \mathrm{g/mol}\).

What is the molar mass of just the carbon part in \( \mathrm{H_2CO_3} \)?

The carbon contribution is \(1 \times A_{\mathrm{C}} \approx 12.01\ \mathrm{g/mol}\). It’s one carbon atom per formula unit.

What is the total molar mass contributed by oxygen atoms in \( \mathrm{H_2CO_3} \)?

There are \(3\) oxygen atoms. Oxygen contribution is \(3 \times A_{\mathrm{O}} = 3(16.00)=48.00\ \mathrm{g/mol}\) (or \(3(15.999)=47.997\ \mathrm{g/mol}\) with more precise values).

What is the molar mass contributed by hydrogen atoms in \( \mathrm{H_2CO_3} \)?

There are \(2\) hydrogen atoms. Hydrogen contribution is \(2 \times A_{\mathrm{H}} = 2(1.01)=2.02\ \mathrm{g/mol}\) (or \(2(1.008)=2.016\ \mathrm{g/mol}\) with precise values).

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