ThePlanMan gives the basics in his answer. In addition: the densest material known to man is Osmium, at $\rho$ = 22 kg/dm$^3$. This compares to water at 1 kg/dm$^3$ and rock at 2-8 kg/dm$^3$. So the densest possible comet is only 22 times as dense as one that consists of water ice. That's not enough to give a gravity of anywhere near 1 G.
Comet 67P/Churyumov-Gerasimenko has a mass of $10^{13}$ kg. To get 1 G, you need 10$^{24}$ kg, or the mass of Earth.
A planet so heavy you can barely move, is either much larger than Earth or contains exotic matter. A neutron star, for instance, but then you're far beyond "can barely move" - a person would be instantly squashed flat.
Could a smaller planet give a 1 G surface gravity?
We can calculate the mass of a planet:
$$ m_1 = 4/3 *{\pi r^3} * \rho $$
When I replace $m_1$ with this formula we get the gravitational force for a person on the surface of a planet:
$$F = \frac{G * 4/3 *{\pi r^3} * \rho * m_2}{r^2}$$
Which can be reduced to:
$$F = {G * 4/3 *{\pi r} * \rho * m_2}$$
The gravitational force scales linearly with radius and density. The smallest possible body (solid osmium) that has 1 G gravity is on the order of 1/20 the mass of Earth, i.e. it has a radius of 0.36 times Earth's radius.