# Exercise 1 | Chemical Calculations

How many molecules of CO2 do you have in 28.0 g of CO2?

nCO2 = $\frac{{\mathrm{m}}_{\mathrm{CO}2}}{{\mathrm{M}}_{\mathrm{CO}2}}$ and nCO2 = $\frac{{\mathrm{N}}_{\mathrm{CO}2}}{{\mathrm{N}}_{\mathrm{A}}}$

So, $\frac{{\mathrm{N}}_{\mathrm{CO}2}}{{\mathrm{N}}_{\mathrm{A}}}$ =  $\frac{{\mathrm{m}}_{\mathrm{CO}2}}{{\mathrm{M}}_{\mathrm{CO}2}}$

NCO2 = $\frac{{\mathrm{m}}_{\mathrm{CO}2}}{{\mathrm{M}}_{\mathrm{CO}2}}$ x NA =  x NA = $\frac{28.0}{44.0}$ x 6.022 x 1023 = 3.83 x 1023 atoms