To determine the mass percent of hydrogen (H) in the sample, we need to use the information provided about the combustion products and work through the following steps:

Step 1: Determine the moles of $\text{CO}_2$ and $\text{H}_2\text{O}$

• Molar mass of $\text{CO}_2$ = 12.01 (C) + 2 × 16.00 (O) = 44.01 g/mol
• Molar mass of $\text{H}_2\text{O}$ = 2 × 1.008 (H) + 16.00 (O) = 18.016 g/mol

Now calculate the moles of $\text{CO}_2$ and $\text{H}_2\text{O}$:

$\text{Moles of } \text{CO}_2 = \frac{4.06 \, \text{g}}{44.01 \, \text{g/mol}} = 0.0923 \, \text{mol}$ $\text{Moles of } \text{H}_2\text{O} = \frac{1.66 \, \text{g}}{18.016 \, \text{g/mol}} = 0.0921 \, \text{mol}$

Step 2: Determine the mass of hydrogen

Since each mole of water ($\text{H}_2\text{O}$) contains 2 moles of hydrogen atoms, we can determine the moles of hydrogen:

$\text{Moles of H} = 2 \times 0.0921 \, \text{mol} = 0.1842 \, \text{mol}$

Now, using the molar mass of hydrogen (1.008 g/mol), we calculate the mass of hydrogen:

$\text{Mass of H} = 0.1842 \, \text{mol} \times 1.008 \, \text{g/mol} = 0.1856 \, \text{g}$

Step 3: Calculate the mass percent of hydrogen

Finally, to find the mass percent of hydrogen in the compound:

$\text{Mass percent of H} = \left( \frac{0.1856 \, \text{g}}{2.77 \, \text{g}} \right) \times 100 = 6.70\%$

Conclusion:

The mass percent of hydrogen in the sample is 6.70%.

Would you like more details on any part of this solution?

Here are 5 related questions to consider:

1. How do we calculate the mass percent of carbon in this sample?
2. What is the empirical formula of this compound based on the given data?
3. How does the combustion analysis technique work in determining elemental composition?
4. What is the significance of using "excess oxygen" in combustion reactions?
5. Can you determine the molecular formula from empirical data?

Tip: Always make sure to carefully balance combustion reactions when working with elemental analyses!