QuestionMay 3, 2025

A sample of a nitrogen oxide compound is found to contain 30.4% nitrogen and the rest of the compound is oxygen. What is its empirical formula? N_(30)O_(70) N_(3)O_(7) NO_(2) N_(2)O_(4)

A sample of a nitrogen oxide compound is found to contain 30.4% nitrogen and the rest of the compound is oxygen. What is its empirical formula? N_(30)O_(70) N_(3)O_(7) NO_(2) N_(2)O_(4)
A sample of a nitrogen oxide
compound is found to contain 30.4% 
nitrogen and the rest of the compound
is oxygen.
What is its empirical formula?
N_(30)O_(70)
N_(3)O_(7)
NO_(2)
N_(2)O_(4)

Solution
4.3(307 votes)

Answer

NO_{2} Explanation 1. Convert percentages to grams Assume a 100 g sample. Nitrogen = 30.4 g, Oxygen = 100 - 30.4 = 69.6 g. 2. Convert grams to moles Use molar masses: N = 14.01 \, \text{g/mol}, O = 16.00 \, \text{g/mol}. Moles of N = \frac{30.4}{14.01} \approx 2.17, Moles of O = \frac{69.6}{16.00} \approx 4.35. 3. Find the simplest ratio Divide by the smallest value: \frac{2.17}{2.17} = 1, \frac{4.35}{2.17} \approx 2. 4. Write the empirical formula The ratio is N_1O_2.

Explanation

1. Convert percentages to grams<br /> Assume a 100 g sample. Nitrogen = $30.4$ g, Oxygen = $100 - 30.4 = 69.6$ g.<br /><br />2. Convert grams to moles<br /> Use molar masses: $N = 14.01 \, \text{g/mol}$, $O = 16.00 \, \text{g/mol}$. <br />Moles of $N = \frac{30.4}{14.01} \approx 2.17$, <br />Moles of $O = \frac{69.6}{16.00} \approx 4.35$.<br /><br />3. Find the simplest ratio<br /> Divide by the smallest value: <br />$\frac{2.17}{2.17} = 1$, $\frac{4.35}{2.17} \approx 2$.<br /><br />4. Write the empirical formula<br /> The ratio is $N_1O_2$.
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