Why doesn't the density of water add up? [closed]








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If the density of hydrogen is 0.09 g/ml and the density of oxygen is 1.429 g/ml, how is it that the density of water is 1 g/ml? 




The densities of hydrogen and oxygen actually do approximately add up to the density of water if you use a state (temperature and pressure) where all three compounds may be considered to be an **ideal gas** (whereas at normal temperature and pressure, hydrogen and oxygen are gases, but water is liquid).


For example, at a temperature of $T=200\ \mathrm{^\circ C}$ and a pressure of $p=1\ \mathrm{bar}$, all three compounds are gases with the following densities:


$$\begin{align}
\rho(\ce{H2})&=0.05122\ \mathrm{kg\ m^{-3}}\\ 
\rho(\ce{O2})&=0.81329\ \mathrm{kg\ m^{-3}}\\ 
\rho(\ce{H2O})&=0.46031\ \mathrm{kg\ m^{-3}}
\end{align}$$


You also have to take into account that hydrogen molecules ($\ce{H2}$) consist of two hydrogen atoms, oxygen molecules ($\ce{O2}$) consist of two oxygen atoms, and water molecules ($\ce{H2O}$) consist of two hydrogen atoms and one water atom:


$$\ce{H2 + 1/2O2 -> H2O}$$


Thus, the density of water can be estimated as


$$\begin{align}
\rho(\ce{H2O})&\approx\rho(\ce{H2})+\frac12\rho(\ce{O2})\\[6pt]
&=0.05122\ \mathrm{kg\ m^{-3}}+\frac12\times0.81329\ \mathrm{kg\ m^{-3}}\\[6pt]
&=0.45787\ \mathrm{kg\ m^{-3}}
\end{align}$$


which is quite close to the real value of $\rho(\ce{H2O})=0.46031\ \mathrm{kg\ m^{-3}}$.




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**Explanation**


Density $\rho$ is defined as
$$\rho=\frac{m}{V}$$
where
$m$ is mass, and
$V$ is volume.


The mass of a pure sample is given as
$$m=n\cdot M$$
where
$n$ is amount of substance, and
$M$ is molar mass.


The volume of a gas can be estimated using the ideal gas law as follows.
$$\begin{align}
p\cdot V&=n\cdot R\cdot T\\[6pt]
V&=\frac{n\cdot R\cdot T}{p}
\end{align}$$
where  

$T$ is temperature,  

$p$ is pressure, and  

$R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$[[source](http://physics.nist.gov/cgi-bin/cuu/Value?r)] is the molar gas constant.


Therefore, the density of an ideal gas at a given temperature and pressure only depends on the molar mass of the gas.
$$\begin{align}
\rho&=\frac{m}{V}\\[6pt]
&=\frac{n\cdot M}{\frac{n\cdot R\cdot T}{p}}\\[6pt]
&=\frac{M\cdot p}{R\cdot T}
\end{align}$$


And the molar mass of water can be calculated from the molar mass of hydrogen and oxygen as
$$M(\ce{H2O})=M(\ce{H2})+\frac12M(\ce{O2})$$


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