# The vd for a certain gas is 70. If the molecular formula for this gas is ("CO")_x, then what is x?

Jun 6, 2016

Thanks @michael-2 for clarifying what "vd" is.

The vapor density of a compound is:

\mathbf(rho_"vap" = ("mass of n molecules of gas")/("mass of n molecules of H"_2(g)))

Let's take $n = 6.0221413 \times {10}^{23}$, i.e. the number of things in $\text{1 mol}$. Then what we have is:

rho_"vap" = (M_(r,"gas"))/(M_(r,"H"_2))

where ${M}_{r}$ is the relative molar mass of the substance in $\text{g/mol}$.

So, since we know that ${\rho}_{\text{vap}} = 70$ and is unitless, that gives us:

color(green)(M_(r,"gas") = M_(r,"H"_2) * rho_"vap")

$= \left(2 \times \text{1.0079 g/mol}\right) \times 70$

$=$ $\textcolor{g r e e n}{\text{141.106 g/mol}}$

Based on that, and the molar mass of $\text{CO}$ to be $12.011 + 15.999 = \text{28.01 g/mol}$, we have that:

$\textcolor{b l u e}{x} = \left({M}_{r , \text{gas"))/(M_(r,"empirical formula}}\right)$

$= \left(\text{141.106 g/mol")/("28.01 g/mol}\right)$

$= 5.04 \approx \textcolor{b l u e}{5}$