How to find the density of silver if given the density of gold?

So gold's density is 19.3 g/#cm^3# and the placement of silver and gold's atoms in the crystallstructure are the same and also both have the atomradius of 144 pm.
What i don't understand is how to calculate the density of silver from all of this. What formula should i use to find this?

Thank you!

1 Answer
Jan 15, 2017

Answer:

Here's what I got.

Explanation:

Your tool of choice here will be this equation

#color(blue)(ul(color(black)(rho = n/(V * N_A) * M_"M")))#

Here

  • #rho# is the density of the element
  • #n# is the number of atoms per unit cell
  • #M_"M"# is the molar mass of the element
  • #V# is the volume of the unit cell
  • #N_A# is Avogadro's constant, equal to #6.022 * 10^(23)"mol"^(-1)#

Now, the problem tells you that the placement of the silver atoms in the crystal structure is identical to that of the gold atoms. Moreover, both have the same radius.

So, if you have the same number of atoms in a unit cell, you will have the same value for #n# for both elements. Moreover, if these atoms are arranged the same and have the same radius, you will also have the same value for #V# for both elements.

You don't need to know what those values are, but you need to realize that they are the same.

Since Avogadro's constant is also the same for both elements, you can rearrange the above equation as

#rho = overbrace(n/(V * N_A))^(color(blue)("the same for both elements")) * M_"M"#

For gold, you will have

#rho_"Au" = n/(V * N_A) * M_"M Au"#

For silver, you will have

#rho_"Ag" = n/(V * N_A) * M_"M Ag"#

Divide these two equations to get

#rho_"Au"/rho_"Ag" = color(red)(cancel(color(black)(n/(V * N_A)))) * 1/color(red)(cancel(color(black)(n/(V * N_A)))) * M_"M Au"/M_"M Ag"#

This is equivalent to

#rho_"Ag" = M_"M Ag"/M_"Au" * rho_"Au"#

This equation tells you that the difference between the density of gold and that of silver can only be attributed to the difference in the mass of the atoms that make up the two elements

A quick look in the periodic table will show that

#M_"M Au" = "196.9666 g mol"^(-1)" "# and #" "M_"M Ag" = "107.8682 g mol"^(-1)#

The density of silver will thus be equal to

#rho_"Ag" = (107.8682 color(red)(cancel(color(black)("g mol"^(-1)))))/(196.9666color(red)(cancel(color(black)("g mol"^(-1))))) * "19.3 g cm"^(-3)#

#color(darkgreen)(ul(color(black)(rho_"Ag" = "10.6 g cm"^(-3)#

The answer is rounded to three sig figs.