Question

The observed heat capacity per gram of a compound containing rubidium and oxygen is 0.64 J·K–1·g–1. Use Dulong and Petit\'s rule to determine the empirical formula of the compound.

Answer 1

if i understand dulong-petit correctly, the heat capacity of a solid is 3R per mol of atoms.

let's say the formula is RbxOy and we have one gram of it. then the number of atoms (counting Rb and O separately) is approx:

(1 gram / (85x + 16y grams per mole) ) ( x + y )

then the heat capacity of one gram of RbxOy is equal to:

3R * (1 gram / (85x + 16y grams per mole) ) ( x + y ) = 0.64 J·K–1·g–1

when i set x=1 and y=2, the LHS is approx. 0.6396

so my guess is RbO2, or rubidium superoxide. the oxygen carries a -1/2 oxidation state in each case. it's a little different from expected, but it's not unheard of. it's analogous to potassium superoxide KO2 (see chemistry: the central science 8th ed., p.245).

Answer 2

So the rule is: C_p=N*3R
R=8.3145 so 3R=approx. 25 J/K*mol
heat capacity =0.64 J/g*K
N=x+y
C_p= molar heat capacity

Step 1: (We are looking for the subscripts of Rb and O so I will use the variables Rb_x and O_y. )
We are given the heat capacity, but we want the molar heat capacity so we need to multiply it by the molar mass. Since we don't know the molar mass because x and y are unknown we can write out the formula: (85.5x+16y)g/mol
so C_p= 0.64(8.5x+16y)

Step 2: we know that N=x+y and that 3R=25 so plug it all into the equation:

0.64(85.5x+16y)=(x+y)(25)

Step 3: plug 1 in for x and solve for y, then use y to solve for x

54.72+10.24y=25+25y
29.72=14.76y
y=2.01 so the subscript is 2

54.72x+20.48=25x+50
29.72x=29.52
x=approx 1

Answer= RbO_2