Question

calculate the percent composition by mass of hydrogen in the fictional compound below. include the units, but do not write the chemical formula. round the answer to 3 significant figures. \\(\ce{nh_6(c_5h_5o_3)_3}\\) hydrogen =

Explanation:

Step1: Calculate moles of H

In \( \text{NH}_6(\text{C}_5\text{H}_5\text{O}_3)_3 \), moles of H: from \( \text{NH}_6 \): 6, from \( (\text{C}_5\text{H}_5\text{O}_3)_3 \): \( 5\times3 = 15 \). Total \( n_{\text{H}} = 6 + 15 = 21 \).

Step2: Calculate molar mass of H

Molar mass of H (\( M_{\text{H}} \)) = \( 1.008 \, \text{g/mol} \). Mass of H: \( m_{\text{H}} = 21 \times 1.008 = 21.168 \, \text{g/mol} \).

Step3: Calculate molar mass of compound

Molar mass of N: \( 14.01 \, \text{g/mol} \), H: \( 21.168 \, \text{g/mol} \), C: \( 5\times3\times12.01 = 180.15 \, \text{g/mol} \), O: \( 3\times3\times16.00 = 144.00 \, \text{g/mol} \). Total molar mass (\( M_{\text{total}} \)) = \( 14.01 + 21.168 + 180.15 + 144.00 = 359.328 \, \text{g/mol} \).

Step4: Calculate percent composition of H

Percent composition (\( \% \text{H} \)) = \( \frac{m_{\text{H}}}{M_{\text{total}}} \times 100 = \frac{21.168}{359.328} \times 100 \approx 5.89\% \).

Answer:

\( 5.89\% \)