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. Percentage Composition
(by mass...not atoms)
24.31 g
95.21 g
% Mg = x 100 Mg
magnesium
24.305 12 Cl
chlorine
35.453 17
70.90 g
95.21 g
% Cl = x 100
25.53% Mg
Mg2+ Cl1-
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
74.47% Cl
MgCl2
It is not 33% Mg and 66% Cl
g= g
2 Cl @ g= g
95.21 g

3. What’s the Percent composition of sugar? Sucrose Formula C12H22O11

4. Hydrate Lab
CuSO4 ∙ ?H2O
CuSO4
Hydrate Anhydrous
Water molecules are incorporated into the crystalline structure.
Hydrates, like zinc acetate dihydrate, Zn(C2H3O2)2 ∙ 2H2O are commonly found in skin care products (ex: moisturizer, shampoo & lip balm).

5. 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

6. 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

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.
Ethanol, C2H6O
52.13% C
13.15% H
34.72% O
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

8. The Empirical Formula March
Percent to mass
Mass to mole
Divide by smallest
Return to whole

9. Empirical Formula Find the empirical formula of this compound.
Quantitative analysis shows that a compound contains 50.04% carbon, 5.59% hydrogen, and 44.37% oxygen.
Find the empirical formula of this compound.
50.04% C
5.59% H
44.37% O
50.04g C
5.59g H
44.37g O
= 4.17 mol C
/ 2.77 mol
= 1.5 C X x2
= mol H
= 2 H
C3H4O2
= mol O
= 1 O
Step 1) %  g
Step 2) g  mol
Step 3) mol
mol
Step 4) return
to whole

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

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