Calculate the empirical formulas of the compounds containing: nitrogen 47% and oxygen 53%.

My guess was to get the molar mass of N and O and divide that by 47 and 53 and continue from there because that was my teachers example but it doesn’t seem right


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    $\begingroup$ Given is amounts of $\ce{N}$ and $\ce{O}$ in $\pu{g}$. To find amounts of $\ce{N}$ and $\ce{O}$ in $\pu{mol}$, divide $\pu{g}$ amounts by relevant molar masses. Rest is pure mathematics. $\endgroup$ – Mathew Mahindaratne Feb 27 at 2:55

Empirical formula reflects lowest ratio of the amounts of the elements expressed in natural numbers $\mathbb{N}$. Amount $n_i$ of $i$th element is the ratio between mass $m_i$ and molar mass $M_i$; mass $m_i$ can be found via mass fraction $ω_i$ and total mass $m$:

$$n_i = \frac{m_i}{M_i} = \frac{ω_im}{M_i}$$

In general, finding empirical formula $\ce{A_aB_bC_c}$ is just a matter of finding the ratio

$$a:b:c = n(\ce{A}) : n(\ce{B}): n(\ce{C}) = \frac{ω(\ce{A})}{M(\ce{A})} : \frac{ω(\ce{B})}{M(\ce{B})} : \frac{ω(\ce{C})}{M(\ce{C})}$$

where $a,b,c\in\mathbb{N}$ ($m$ is cancelled out as it's a constant).

For $\ce{N_xO_y}$

$$x : y = \frac{ω(\ce{N})}{M(\ce{N})} : \frac{ω(\ce{O})}{M(\ce{O})} = \frac{0.47}{14.01} : \frac{0.53}{16.00} = \pu{3.35e-2} : \pu{3.31e-2} \approx 1:1$$

resulting in empirical formula $\ce{NO}$.


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