Exercise 1 | Chemical Calculations

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

nCO2 = mCO2MCO2 and nCO2 = NCO2NA


So, NCO2NA =  mCO2MCO2


NCO2 = mCO2MCO2 x NA = mCO2MC + 2 MO x NA = 28.044.0 x 6.022 x 1023 = 3.83 x 1023 atoms