Which weighs more in atmosphere, $1\,{\rm kg}$ of steel or $1\,{\rm kg}$ of feathers? I'm having a discussion at the moment regarding the mass of $1\,{\rm kg}$ of feathers and $1\,{\rm kg}$ of steel.
The person I'm arguing with states that $1\,{\rm kg}$ of feathers will be lighter when weighed, compared to the $1\,{\rm kg}$ of steel, because the feathers are more buoyant.
She has done her calculations for the density of feathers and works out that $1\,{\rm kg}$ of feathers will displace $400\,{\rm L}$ of water. And because the feathers are more buoyant than the steel they would actually weigh less.
We are talking about the kilogram as a unit of mass, not weight.
The argument goes that if a scale is balanced perfectly in vacuum, it is not in air. The feathers, being more buoyant in air, would cause the scale to tip toward the steel.
I’m sure she is wrong, and even though buoyancy may be a factor, she calculates that the feathers would only have about half the weight in air that they do in vacuum.
So the question is, if equal masses of feathers and steel were balanced in a vacuum, would they still balance in $1\,{\rm atm}$? If not, what would be the difference in weight?
 A: Feathers are made from keratin, with a density of about $1.3\ \mathrm{g/cm^3}$. The net volume displaced by a kilogram of feathers is then $751\ \mathrm{cm^3}$. Steel has a density of $7.86\ \mathrm{g/cm^3}$ and a kilogram of it displaces $127\ \mathrm{cm^3}$.
Sea level air has a density of $0.0012\ \mathrm{g/cm^3}$, so the buoyant force on $751\ \mathrm{cm^3}$ of keratin is then $0.92\ \mathrm {gf}$ and the buoyant force on $127\ \mathrm{cm^3}$ of steel is $0.155\ \mathrm {gf}$.
This means that if we weigh the keratin in a vacuum, it will weigh $1\ \mathrm{kgf}$ but in air it will weigh $999.08\ \mathrm {gf}$. If we weigh the steel in a vacuum, it will weigh $1\ \mathrm{kgf}$ but in air it will weigh $999.85\ \mathrm {gf}$.
If we place the two bodies – keratin and steel, one kilogram of each – on a pivoting balance equidistant from the pivot in a vacuum, they will be in balance, but in air the keratin will be lighter by $(999.85-999.08)\ \mathrm {gf}$ or $0.77\ \mathrm {gf}$.
A: This depends on how you define "weight". If "weight" is defined as the gravitational force from the environment gravitator (here the Earth), then both will, of course, weight the same. If "weight", however, is defined by a scale reading and the scale and objects weighed are immersed in an atmosphere, then you're right: the feathers will "weigh" less as buoyancy helps them to float a little.
How much? Well, we need some figures for the density of all three materials involved. Archimedes' principle tells you that the buoyant force equals the weight (in the gravitational sense) of the displaced air, which, in turn, is determined by the volume of the object. If we denote that by $V_\mathrm{obj}$, we have, then,
$$F_B = \rho_\mathrm{air} V_\mathrm{obj} g$$
In terms of the mass $m_\mathrm{obj}$ and density $\rho_\mathrm{obj}$ of the object, we of course have
$$V_\mathrm{obj} = \frac{m_\mathrm{obj}}{\rho_\mathrm{obj}}$$
hence
$$F_B = \frac{\rho_\mathrm{air}}{\rho_\mathrm{obj}} m_\mathrm{obj}g$$
If we denote the gravitational weight $w$, equal to
$$w = m_\mathrm{obj}g$$
and the scale or effective weight $w_\mathrm{eff}$, we have the very neat expression
$$w_\mathrm{eff} = w - F_B = \left(1 - \frac{\rho_\mathrm{air}}{\rho_\mathrm{obj}}\right) w$$
in other words, all you need is simply the ratio of densities between the object and the immersing medium, plus the actual gravitational weight - for an exact one kilogram, and standard Earth gravity, this is $w = 9.806\ 65\ \mathrm{N}$.
Now iron has a density of 7800 g/L ("steel" will vary depending on the amount of added carbon, but this should be close), equiv. $\mathrm{kg/m^3}$, while air has about 1.2 g/L, hence the iron piece will have
$$\left(1 - \frac{\rho_\mathrm{air}}{\rho_\mathrm{obj}}\right) = \left(1 - \frac{1.2}{7800}\right) \approx 0.99984$$
hence its $w_\mathrm{eff}$ is reduced from its original weight by about 0.016%, or to 9.8051 N.
For the feathers, this:
https://wat.lewiscollard.com/archive/www.newton.dep.anl.gov/askasci/gen06/gen06451.htm
suggests their density is actually surprisingly higher - at about 1000 g/L, though considerably variable. In this case, the feathers should retain about 99.88% of their "true" weight, meaning a weight reduction of 0.12%, or a $w_\mathrm{eff}$ of 9.7948 N.
Neither of these are very noticeable, hence you would likely not feel anything interesting holding either in your hands.
(As to why the density reported there is more than you might think, it's because this is the density of the actual material feathers are made from. This is, also, the relevant density to use in this calculation because the air interacts, as a gas with no surface tension, directly at the atomic scale of the fibers in the feathers, thus the material, while the "apparent" low density is by considering density to be the sparse appearance of the material, which is how we would hold feathers in our hand, but not how air "holds" them.)
A: Due to buoyancy one kilogram of hot air balloon indeed weighs less than one kilogram of steel! 
A: You can argue that her feathers contain a lot of air (never perfectly compressed) so that the volume is not true.  You could also take your kilo of steel and make it into a hollow sphere (empty) of the same volume as the feathers (whatever that is)  and then it would actually weigh less because you have no air inside.  But you could be nice and let her make the point about a bigger volume being more bouyant, she is correct about this but ..... even  volume is tricky.
