Q. What is the molar mass of \( \mathrm{C_6H_{12}O_6} \)?

Answer

To find the molar mass of \( \mathrm{C_6H_{12}O_6} \), add the atomic masses of all atoms in one mole of the compound.

\[
\begin{aligned}
M &= 6(12.01) + 12(1.008) + 6(16.00) \\
&= 72.06 + 12.096 + 96.00 \\
&= 180.156 \,\text{g/mol} \approx 180.16 \,\text{g/mol}
\end{aligned}
\]

Final result: \( \mathrm{C_6H_{12}O_6} \) has a molar mass of \( \boxed{180.16 \,\text{g/mol}} \).

Detailed Explanation

To find the molar mass of \( \mathrm{C_6H_{12}O_6} \), you add the atomic masses of all atoms in one formula unit, then convert that total into grams per mole.

Step 1: Identify how many atoms of each element are in the formula

\( \mathrm{C_6H_{12}O_6} \) contains:

\(6\) carbon atoms

\(12\) hydrogen atoms

\(6\) oxygen atoms

Step 2: Write atomic masses you will use

Use standard atomic masses (in \(\mathrm{g/mol}\)):

\(\mathrm{C} \approx 12.01\)

\(\mathrm{H} \approx 1.008\)

\(\mathrm{O} \approx 16.00\)

Step 3: Multiply each atomic mass by the number of atoms

Compute the contribution from each element:

Carbon contribution:

\[ 6 \times 12.01 = 72.06 \]

Hydrogen contribution:

\[ 12 \times 1.008 = 12.096 \]

Oxygen contribution:

\[ 6 \times 16.00 = 96.00 \]

Step 4: Add the contributions to get the total molar mass

\[ 72.06 + 12.096 + 96.00 = 180.156 \]

Step 5: Round appropriately

Rounding to a reasonable number of significant figures gives:

\[ \text{Molar mass of } \mathrm{C_6H_{12}O_6} \approx 180.16 \ \mathrm{g/mol} \]

Final Answer: \( \boxed{180.16 \ \mathrm{g/mol}} \)

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General Chemistry FAQs

How do I find the molar mass of \( \mathrm{C_6H_{12}O_6} \)?

Add atomic masses: \(6\times M(\mathrm{C}) + 12\times M(\mathrm{H}) + 6\times M(\mathrm{O})\). Using \(M(\mathrm{C})=12.01\), \(M(\mathrm{H})=1.008\), \(M(\mathrm{O})=16.00\), total is about \(180.16\ \mathrm{g/mol}\).

What atomic masses should I use for \( \mathrm{C} \), \( \mathrm{H} \), and \( \mathrm{O} \)?

Common choices are \(M(\mathrm{C})=12.01\ \mathrm{g/mol}\), \(M(\mathrm{H})=1.008\ \mathrm{g/mol}\), and \(M(\mathrm{O})=16.00\ \mathrm{g/mol}\) (periodic table values).

Why do I multiply by the subscripts in \( \mathrm{C_6H_{12}O_6} \)?

Subscripts tell how many atoms of each element are in one molecule. Each atom’s mass contributes, so you multiply each element’s molar (atomic) mass by its count.

What is the exact arithmetic for \(6\times 12.01 + 12\times 1.008 + 6\times 16.00\)?

\(6\times 12.01=72.06\), \(12\times 1.008=12.096\), \(6\times 16.00=96.00\). Sum: \(72.06+12.096+96.00=180.156\ \mathrm{g/mol}\approx 180.16\ \mathrm{g/mol}\).

Should I round the molar mass, and to how many decimals?

Typically round to 2 decimal places for school problems: \(180.16\ \mathrm{g/mol}\). If using simplified atomic masses, you might get \(180.18\) or about \(180.0\ \mathrm{g/mol}\).

Is \( \mathrm{C_6H_{12}O_6} \) molar mass the same as glucose molar mass?

Yes. \( \mathrm{C_6H_{12}O_6} \) is the molecular formula of glucose, so its molar mass is the molar mass of glucose, about \(180.16\ \mathrm{g/mol}\).

Find glucose molar mass.
It’s 180.16 g/mol for C6H12O6.

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