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Lecture 3

CHEM 06100 Lecture Notes - Lecture 3: Molar Mass, Unified Atomic Mass Unit, Stoichiometry


Department
Chemical Engineering Biochemistry
Course Code
CHEM 06100
Professor
Grinias
Lecture
3

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Stoichiometry and Chemical Reactions
Stoichiometry- to determine how much mass, moles, and the making of product.
The mole (mol) if we take 12 grams of carbon 12 we have 1 mol.
One Mole (1mol) * 6.22x1023
Calculating Molar Mass
An Atom: Na= 1 x M of Na = 22.99g/ mol (amu)
A molecule molar mass:
O2= 2 x M of O = 2 x 16.00 g/mol = 32.00g/mol (amu)
H2O= 2 M of H= 2.00+ 1 MO= 16.00=18.00g/mol
An Ion (particle)= CI= 1 x < =35.45 g/mol
A formula Unit NaCl= 1 x M of Na = 22.99+1 x M of CI = 35.45 g/mol = 58.44g/mol
Need a transfer electron and electronegativity to make Ion.
Mass Percent from the chemical formula
Mass % of elements x =
[ atoms of x in formula] x [ atomic mass of x (amu)]
Molecular (or formula) mass of compound (amu)
Mass of elements x=
[moles of x in formula]x [ molar mass of x(/mol 0]
Mass (g) of 1 mol of compound
Mass Fraction and the Mass of an Element
Mass of any element in sample
Mass of compound x mass of element in 1 mol of compound
Mass of 1 mol of compound
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