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\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
& 1.008 & 2 & & & & & & & & & & \\
\hline
\begin{tabular}{c}
3 \\
Li \\
6.94
\end{tabular} &
\begin{tabular}{c}
4 \\
Be \\
9.01
\end{tabular} & & & & & & & & & & & \\
\hline
& & & & & & & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\begin{tabular}{c}
22.99 \\
Na \\
\end{tabular} &
\begin{tabular}{c}
24.31 \\
Mg \\
\end{tabular} &
\begin{tabular}{c}
26.98 \\
Al \\
\end{tabular} &
\begin{tabular}{c}
28.09 \\
Si \\
\end{tabular} &
\begin{tabular}{c}
30.97 \\
P \\
\end{tabular} &
\begin{tabular}{c}
32.07 \\
S \\
\end{tabular} &
\begin{tabular}{c}
35.45 \\
Cl \\
\end{tabular} &
\begin{tabular}{c}
39.95 \\
Ar \\
\end{tabular} &
\begin{tabular}{c}
39.10 \\
K \\
\end{tabular} &
\begin{tabular}{c}
40.08 \\
Ca \\
\end{tabular} &
\begin{tabular}{c}
44.96 \\
Sc \\
\end{tabular} &
\begin{tabular}{c}
47.87 \\
Ti \\
\end{tabular} &
\begin{tabular}{c}
50.94 \\
V \\
\end{tabular} \\
\hline
\begin{tabular}{c}
52.00 \\
Cr \\
\end{tabular} &
\begin{tabular}{c}
54.94 \\
Mn \\
\end{tabular} &
\begin{tabular}{c}
55.85 \\
Fe \\
\end{tabular} &
\begin{tabular}{c}
58.93 \\
Co \\
\end{tabular} &
\begin{tabular}{c}
58.69 \\
Ni \\
\end{tabular} &
\begin{tabular}{c}
63.55 \\
Cu \\
\end{tabular} &
\begin{tabular}{c}
65.38 \\
Zn \\
\end{tabular} & & & & & & \\
\hline
\begin{tabular}{c}
85.47 \\
Rb \\
\end{tabular} &
\begin{tabular}{c}
87.62 \\
Sr \\
\end{tabular} &
\begin{tabular}{c}
88.91 \\
Y \\
\end{tabular} &
\begin{tabular}{c}
91.22 \\
Zr \\
\end{tabular} &
\begin{tabular}{c}
92.91 \\
Nb \\
\end{tabular} &
\begin{tabular}{c}
95.95 \\
Mo \\
\end{tabular} &
\begin{tabular}{c}
98 \\
Tc \\
\end{tabular} &
\begin{tabular}{c}
101.07 \\
Ru \\
\end{tabular} &
\begin{tabular}{c}
102.91 \\
Rh \\
\end{tabular} &
\begin{tabular}{c}
106.42 \\
Pd \\
\end{tabular} &
\begin{tabular}{c}
107.87 \\
Ag \\
\end{tabular} &
\begin{tabular}{c}
112.41 \\
Cd \\
\end{tabular} & \\
\hline
\end{tabular}

Select the correct answer.

A compound has a molar mass of 92.02 grams/mole, and its percent composition is [tex]$30.4\%$[/tex] nitrogen (N) and [tex]$69.6\%$[/tex] oxygen (O). What is its molecular formula?

A. [tex][tex]$N_2O_4$[/tex][/tex]
B. [tex]$NO_2$[/tex]
C. [tex]$N_2O$[/tex]
D. [tex][tex]$N_4O_2$[/tex][/tex]


Sagot :

To determine the molecular formula of a compound given its molar mass and percent composition, we follow these steps:

1. Calculate the Mass of Each Element in the Compound:
- The molar mass of the compound is 92.02 grams/mole.
- The percent composition is 30.4% nitrogen (N) and 69.6% oxygen (O).

Therefore, the mass of each element in 1 mole of the compound:
[tex]\[ \text{Mass of N} = \left( \frac{30.4}{100} \right) \times 92.02 \text{ g} = 27.97 \text{ g} \][/tex]
[tex]\[ \text{Mass of O} = \left( \frac{69.6}{100} \right) \times 92.02 \text{ g} = 64.05 \text{ g} \][/tex]

2. Convert Masses to Moles:
- The molar mass of nitrogen (N) is 14.01 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.

Therefore, the number of moles of each element:
[tex]\[ \text{Moles of N} = \frac{27.97 \text{ g}}{14.01 \text{ g/mol}} = 1.997 \approx 2 \text{ moles} \][/tex]
[tex]\[ \text{Moles of O} = \frac{64.05 \text{ g}}{16.00 \text{ g/mol}} = 4.003 \approx 4 \text{ moles} \][/tex]

3. Determine the Simplest Whole Number Ratio:
- The ratio of moles of nitrogen to moles of oxygen is approximately:
[tex]\[ N : O = 2 : 4 \][/tex]

Simplifying the ratio by dividing both numbers by the smallest value (2):
[tex]\[ N : O = 1 : 2 \][/tex]

4. Establish the Empirical and Molecular Formula:
- The simplest whole number ratio gives us the empirical formula [tex]\( NO_2 \)[/tex].
- To see if this is the molecular formula, we must check if the molar mass of [tex]\( NO_2 \)[/tex] matches the given molar mass of the compound.

The molar mass of [tex]\( NO_2 \)[/tex]:
[tex]\[ \text{Molar mass of } NO_2 = 14.01 \text{ (N)} + 2 \times 16.00 \text{ (O)} = 14.01 + 32.00 = 46.01 \text{ g/mol} \][/tex]

Since the molar mass of the compound (92.02 g/mol) is twice the molar mass of [tex]\( NO_2 \)[/tex], the molecular formula must be:
[tex]\[ \text{Molecular formula} = 2 \times NO_2 = N_2O_4 \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{N_2O_4} \][/tex]