# Exercise 1 | The periodic table and Chemical Periodicity

What is the molar mass M of chromium(III) bromide?

Calculate number of moles n in 26.0 g of chromium(III) bromide?

How many molecules N do you have in 26.0 g of chromium(III) bromide?

How many atoms of bromine NBr do you have in 26.0 g of chromium(III) bromide?

chromium(III) bromide = CrBr3

M = MCrBr3 = MCr + 3 x MBr = 51.9961 + 3 x 79.904 = 291.71 g.mol-1

n = nCrBr3 = $\frac{{\mathrm{m}}_{\mathrm{CrBr}3}}{{\mathrm{M}}_{\mathrm{CrBr}3}}$ = $\frac{26.0}{291.71}$ = 8.91 x 10-2 mol

nCrBr3 $\frac{\mathbf{N}}{{\mathrm{N}}_{\mathrm{A}}}$  ⇒ N = nCrBr3 x NA = 5.37 x 1022 molecules of CrBr3

3 atoms of Br in CrBr3:

NBr = 3 x N = 1.61 x 1023 atoms of Br