If you wanted to convert 800 mmHg to bars, using the horizontal list, you could do it directly:

${\displaystyle bars=800{\mbox{ mmHg}}\times {\frac {1.013{\mbox{ bar}}}{760{\mbox{ mmHg}}}}=1.066{\mbox{ bar}}}$

Using the tables from Wikipedia, you need to convert to an intermediate (the metric unit) and then convert from the intermediate to the desired unit. We would find that

${\displaystyle 1{\mbox{ mmHg}}=133.322{\mbox{ Pa}}}$ and ${\displaystyle 1{\mbox{ bar}}=10^{5}{\mbox{ Pa}}}$

Again, we have to set it up using the same general form, just we have to do it twice:

${\displaystyle bars=800{\mbox{ mmHg}}*{\frac {133.322{\mbox{ Pa}}}{1{\mbox{ mmHg}}}}*{\frac {1{\mbox{ bar}}}{10^{5}{\mbox{ Pa}}}}=1.066{\mbox{ bar}}}$

We could check this equation by plugging in our units:

${\displaystyle P{\dot {=}}M/(L*t^{2}){\mbox{ , }}h{\dot {=}}L{\mbox{ , }}g{\dot {=}}L/t^{2}}$
${\displaystyle g*h{\dot {=}}L^{2}/t^{2}\neq M/(L*t^{2})}$

Since g*h doesn't have the same units as P, the equation must be wrong regardless of the system of units we are using! The correct equation, in fact, is:

${\displaystyle P=\rho *g*h}$
where ${\displaystyle \rho }$ is the density of the fluid. Density has base units of ${\displaystyle M/L^{3}}$ so

${\displaystyle \rho *g*h{\dot {=}}L^{2}/t^{2}*M/L^{3}{\dot {=}}M/(L*t^{2})}$ which are the units of pressure.

This does not tell us the equation is correct but it does tell us that the units are consistent, which is necessary though not sufficient to obtain a correct equation. This is a useful way to detect algebraic mistakes that would otherwise be hard to find. The ability to do this with an algebraic equation is a good argument against plugging in numbers too soon!