
Atomic Number
› Number of protons found in the nucleus of
the element.

Mass Number
› The number of protons and neutrons in a
given element.

Atomic Mass
› The total mass of the protons, neutrons, and
electrons in the element without being an
ion or isotope.
Mass
Number
Atomic
Number
Atomic Mass

Empirical Formula
› is the reduced version of the formula

A compound was analyzed and found
to contain 13.5 g Ca, 10.8 g O2 and 0.675
g H. What is the empirical formula?
13.5 g Ca /40.08 MM Ca = 0.337 mol Ca
 10.8 g O / 16 MM O = 0.675 mol O
 0.675 g H / 1.01 MM H = 0.668 mol H

› Divide by the smallest
0.337 mol Ca / 0.337 = 1
 0.675 mol O / 0.337 = 2
 0.668 mol H / 0.337 = 2

› Answer: CaO2H2

Molecular Formula
› All the atoms in a formula

The simplest formula for vitamin C is
C3H4O3. If the molecular mass of vitamin
C is 180, what is the molecular formula?

Calculate the atomic mass of the
empirical formula.
› MM = 88.07

Divide the molecular mass from the
molar mass
› 180/88.07 = 2

Times by the empirical formula
› (C3H4O3)(2) = C6H8O6

RULES:
› Digits and zeros between nonzero digits are
significant.
 7.907 has?
 907.08 has?
› Zeros to the left of the first nonzero digit is not
significant
 0.0709 has?
› Zeros at the right of the decimal point are
significant
 12.000 has?
› Zeros at the end of a number greater than 1 are
not significant, unless there is a decimal point
 1,200 has?
 1,200. has?

When multiplying, dividing, adding, or
subtracting the answer should have the
same number of significant figures as the
smallest one in the equation.
› 0.352 x 0.90876 = 0.320
› 26 + 45.88 + 0.09534 = 72

Mole = Avogradro’s # = 6.022 x 1023
› The most important number in chemistry that
you’ll ever need to know.
› Converts between moles and molecules

Moles and Grams
› Moles = grams/molar mass

Moles and Gas
› PV = nRT

Molarity
› Moles = M x liters

When doing stoich, you need wanted over
given.
› You get wanted / given from the balancing of
the equation
__NH3 + __O2  __N2 + __H2o
 If the equation above were balanced with
lowest whole number coefficients, the
coefficient for NH3 would be?
a)
b)
c)
d)
e)
1
2
3
4
5
1)
2)
3)
Convert whatever quantity you are
given into moles
Use the balanced equations exponents
to figure out the wanted/given to find
the wanted moles of the reactant
Times the moles by the molar mass of
your reactant to find the grams
2 HBr + Zn  ZnBr2 + H2
 A piece of solid zinc weighing 98 grams
was added to a solution containing 324
grams of HBr. What is the volume of H2
produced at standard temperature and
pressure if the reaction above run to
completion?
a)
b)
c)
d)
e)
11 liters
22 liters
34 liters
45 liters
67 liters
When you are doing stoich and trying to
find how much of product you can
make, you need to find the grams for
each of your reactants.
 After doing stoich for both of your
reactants, you will find that one of them
is smaller, or in some cases equal.
 The smallest answer is your limiting
reactant and the right answer.

Mg3N2 + 6H2O  3Mg(OH)2 + 2NH3
 If you initially have 58.1 g of Mg3N2 and
6H2O how much NH3 can you make?

(58.1 g Mg3N2) / (100.95 MM of Mg3N2) = 0.575
mol Mg3N2
› (0.575 mol Mg3N2)(2/1) = 1.15 mol NH3
› (1.15 mol NH3)(17.01) = 19.5 g NH3

(20.4 g H2O) / (18 MM of H2O) = 1.13 mol H2O
› (1.13 mol H2O )(2/6) = 0.378 mol NH3
› (0.378 mol NH3) (17.01) = 6.42 g NH3

So H2O is the limiting reactant
Hydrate: a substance that contains water
Anhydrate: a substance that does not
contain water
 A sample of a hydrate BaCl2 with a mass of
61 grams was heated until all the water was
removed. The sample was then weighed
and found to have a mass of 52 grams.
What is the formula of the hydrate?


a)
b)
c)
d)
e)
BaCl2  5 H2O
BaCl2  4 H2O
BaCl2  3 H2O
BaCl2  2 H2O
BaCl2  H2O
61 g / 244 MM = 0.25 mols BaCl2
 Water added to the mass, so 1 mole has
to have 9 grams of H2O

› 9 g / 18 MM = 0.5 mols H2O

So, if 0.25 mole of hydrate contains 0.5
mole of H2O there must be 2 moles of
H2O for every mole of hydrate