Chem 107 

Fall 2009 
Lecture Notes: Concentration Examples 





Masses and particles:
Each carbon atom weighs 12.01 amu, thus for 3.01 x 10^{20} atoms have mass = (3.01 x 10^{20} atoms)(12.01 amu/atom) = 3.62 x 10^{21}amu
First we need to find how many moles of carbon are present using Avogadro's Number: (3.01 x 10^{20}atoms)/(6.02 x 10^{23}atom/mole) = 5.00 x 10^{4}mole. Now we multiply by the molecular weight to get the number of grams;
(5.00 x 10^{4}mole)x 12.01g/mole = 6.00 x 10^{3}g
First need to determine the formula in order to know how many iron atoms are in it: Fe_{2}(SO_{4})_{3} So for every mole of iron(III) sulfate there are two moles of iron, so we have 2 x 200.0 mmoles = 400.0 mmoles of Fe Finally, (0.4000 mole Fe) x (55.85 g/mole Fe) = 22.34 g

© R A Paselk
Last modified 23 September 2009