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

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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!

So gold's density is 19.3 g/

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

#### 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 thedensityof the element#n# is thenumber of atoms per unit cell#M_"M"# is themolar massof the element#V# is thevolumeof the unit cell#N_A# isAvogadro'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

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 canonly be attributedto the difference in themass of the atomsthat 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**.