Question

question 4 (2 points)
what is the empirical formula for a compound containing 30.4% nitrogen and the remainder oxygen?
no
no3
n2o4
no2

Explanation:

Step1: Assume 100g of the compound

Since it's a percentage - based problem, assuming 100g of the compound means we have 30.4g of nitrogen (N) and \(100 - 30.4=69.6\)g of oxygen (O).

Step2: Calculate the moles of each element

The molar mass of N is \(M_N = 14\space g/mol\), and the number of moles of N, \(n_N=\frac{30.4\space g}{14\space g/mol}\approx2.17\space mol\). The molar mass of O is \(M_O = 16\space g/mol\), and the number of moles of O, \(n_O=\frac{69.6\space g}{16\space g/mol}=4.35\space mol\).

Step3: Find the mole - ratio

Divide each number of moles by the smaller number of moles. Divide \(n_N\) and \(n_O\) by \(n_N\) (since \(n_N = 2.17\space mol\) is smaller). \(\frac{n_N}{n_N}=1\) and \(\frac{n_O}{n_N}=\frac{4.35}{2.17}\approx2\).

Answer:

D. \(NO_2\)