Molar Mass & Molarity
Molar Mass
• Mass in grams of one mole of an element
or compound.
• Numerically equal to the atomic weight of
the element or the sum of all the atomic
weights in the formula.
• Round all atomic masses to keep 2
decimals.
Molar Mass Examples
• NaCl
= 22.99 + 35.45
= 58.44g/mole
• CuSO4 . 5H2O (this is a hydrate; it means there are 5 water
molecules added for each copper sulfate molecule)
= 63.55 + 32.07 + 4(16.00) + 5(18.02)
= 159.62 + 90.10
= 249.72 g/mole
Practice
1) KCl
2) Li2SO4
.
3) (NH4)2C2O4 H2O
Answers
1) KCl= 39.10 + 35.45
= 74.55g/mole
2) Li2SO4= 2(6.94) + 32.07 + 4(16.00)
= 109.95g/mole
3) (NH4)2C2O4 . H2O
= 2(14.01)+8(1.01)+2(12.01)+4(16.00)+18.02
=142.14g/mole
Practice Again
4)potassium hydroxide
5) copper (II) bromide
6) magnesium phosphate
Answers
4) KOH = 39.10 + 16.00 + 1.01
= 56.11g/mole
5) CuBr2 = 63.55 + 2(79.90)
= 383.15g/mole
6) Mg3(PO4)2= 3(24.31)+ 2(30.97) + 8(16.00)
= 262.87g/mole
Molarity, M
• Moles of solute (compound) per liter of
solution.
• Moles divided by volume (in liters).
– You will have to convert mL to L.
1.5M HCl
• This means there are 1.5 moles of HCl in
each liter of HCl
• How many grams of HCl are in 0.45 liters
of 1.5M HCl?
– You must first calculate the number of moles
in that volume.
– Then multiply the moles by the molar mass of
HCl
0.45L of 1.5 M HCl
• 1.5moles/L x 0.45L
=0.675moles; round for sig. figs.=0.68moles
(notice the L cancels out)
• Next, multiply the moles by the molar
mass of HCl
0.68moles x 36.46g/mole= 24.7928g
=24g
(notice the moles cancel out)
0.500M NaOH
• How many grams of NaOH are in 25.0mL
of this solution?
• First, convert the volume to L
25.0mL /1000= 0.0250mL
(keep all sig. figs. from original number)
• Next multiply the molarity by the volume
0.500mol/L x 0.0250L= 0.0125moles
• Finally, multiply the moles by the molar mass
0.0125moles x 40.00g/mol= 0.500g