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During a chemical reaction, equal moles of two diatomic molecules $\ce{X}$ and $\ce{Y}$ are allowed to react to form one gaseous compound $\ce{X_nY_m}$. After the reaction is complete, it was observed that the density of the product formed is $1.50$ greater than the density of the reactant mixture. Specify the formula of the product $\ce{X_nY_m}$.

Here is my work:

$$\ce{X_m + Y_n -> X_nY_m}$$

Density of product: $D_\mathrm{p}$ = $(m/V)_\mathrm{p}$

Density of reactant: $D_\mathrm{r}$ = $(m/V)_\mathrm{r}$

\begin{align} D_\mathrm{p} &= 1.5 \cdot D_\mathrm{r} \\ (m/V)_\mathrm{p} &= 1.5 \cdot (m/V)_\mathrm{r} \\ m_\mathrm{p} &= 1.5 \cdot m_\mathrm{r} \end{align}

I really dont know where to go from there. I would appreciate it if someone could give me step by step how they solved it so I could not just have the answer but also understand how they got to it.

Also, would the formula found be empirical or molecular? How do you know that it is?

andselisk
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Melanie
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    possible duplicate of Density of Gas during after reaction greater than reactant mixture –  Apr 30 '15 at 17:40
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    Ok, First of all, MASS is always conserved. So, the density change should only come from the change in volume. Now, V=nRT/P and you are producing one mole of product in the expense of two moles of reactants. So it should be, vr=1.5*vp
    Can you please clarify your question a little bit more. What do you actually want to find? At first you are saying it's diatomic molecule but according to question statement i think you are asked to find the molecular formula.     – Osman Mamun Apr 30 '15 at 19:43

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