1. Do Now:
If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is sugar?

2. Hydrate Lab – for tomorrow
Pre-lab
Title
Purpose
Materials
Procedure
Data table
Hydrate Anhydrous

3. Hydrates Hydrates, like zinc acetate dihydrate, Zn(C2H3O2)2 * 2H2O
Water molecules are incorporated into the crystalline structure.
Methane hydrate
Hydrates, like zinc acetate dihydrate, Zn(C2H3O2)2 * 2H2O
are commonly found in skin care products such as moisturizer, shampoo and lip balm.

4. What is the % H2O in nickel chloride dihydrate, NiCl2 * 2H2O?
Element #
g/mol
(molar mass)
TOTAL Ni 1
58.69 g/mol
58.69g Cl 2
35.45 g/mol
70.90g
H2O
18.02 g/mol
36.04g
MOLAR MASS=
165.63g/mol NiCl2 * 2H2O
%H2O =
= % H2O

5. Percentage Composition
(by mass...not atoms)
% = x 100
part
whole Mg
magnesium
24.305 12 Cl
chlorine
35.453 17
% Mg = x 100
24.31 g
95.21 g
25.52% Mg
Law of definite proportions states that a chemical compound always contains the same proportion of elements by mass
Percent composition — the percentage of each element present in a pure substance—is constant
Calculation of mass percentage
1. Use atomic masses to calculate the molar mass of the compound
2. Divide the mass of each element by the molar mass of the compound and then multiply by 100% to obtain percentages
3. To find the mass of an element contained in a given mass of the compound, multiply the mass of the compound by the mass percentage of that element expressed as a decimal
Mg2+ Cl1-
74.48% Cl
MgCl2
It is not 33% Mg and 66% Cl
g= g
2 Cl @ g= g
95.21 g

6. percentage composition: the mass % of each
element in a compound
% of element =
g element
molar mass of compound
x 100
Find % composition.
(see calcs above)‏
PbO2
207.2 g Pb :
239.2 g
= 86.6% Pb
32.0 g O :
239.2 g
= 13.4% O
(NH4)3PO4
42.0 g N :
149.0 g
= 28.2% N
12.0 g H :
149.0 g
= 8.1% H
31.2 g P :
149.0 g
= 20.8% P
64.0 g O :
149.0 g
= 43.0% O

7. Empirical and Molecular Formulas
A pure compound always consists of the same elements combined in the same proportions by weight.
Therefore, we can express molecular composition as PERCENT BY WEIGHT.
Empirical formula gives only the relative numbers of atoms in a substance in the smallest possible ratio
Molecular formula gives the actual number of atoms of each kind present per molecule
Ethanol, C2H6O
52.13% C
13.15% H
34.72% O

8. Different Types of Formulas
Molecular Formula – shows the real # of atoms in one molecule or formula unit
Empirical Formula – shows smallest whole number mole ratio
**Sometimes the empirical &molecular formula can be the same
Structural Formula- molecular formula info PLUS bonding electron and atomic arrangement
C6H6 CH

9. Calculating Empirical formula
Percent to mass (assume 100g)
Mass to mole (molar mass)
Divide by the smallest (ratio)
Multiply til’ whole*

10. Empirical Formula
Quantitative analysis shows that a compound contains 32.38% sodium,
22.65% sulfur, and 44.99% oxygen.
Find the empirical formula of this compound.
sodium sulfate
32.38% Na
22.65% S
44.99% O
32.38 g Na
22.65 g S
44.99 g O
= mol Na
/ mol
= 2 Na
Na2SO4
Na2SO4
= mol S
= 1 S
= mol O
= 4 O
Step 1) %  g
Step 2) g  mol
Step 3) mol
mol

11. Empirical Formula
Quantitative analysis shows that a compound contains 66.75% copper, 10.84% phosphorus
and 22.41% oxygen.
Find the empirical formula of this compound.
copper (I) phosphate
66.75% Cu
10.84 % P
22.41 % O
66.75g Cu
10.84g P
22.41g O
= mol Cu
/ mol
=3 Cu
Cu3PO4
Cu3PO4
= mol P
= 1 P
= mol O
= 4 O
Step 1) %  g
Step 2) g  mol
Step 3) mol
mol