Simple Solution #4
0.20 x 75 g/mol
0.30 x 100 g/mol
0.50 x 106 g/mol
This is called the weighted average mass of the three isotopes of Zeelandium.
The atomic masses on the periodic table are weigh
Your Turn Again:
Calculate the percentage composition for aluminum carbonate. (This means
find the percentage by mass of each element in the compound.)
% Composition for aluminum carbonate
aluminum carbonate is Al2(CO3)3
atomic mass / molar mass
An empirical formula is the mathematically reduced form of a chemical
formula. Several examples.
Empirical Formulas from Percentages
C4.1b Calculate the empirical formula of a compound based on the weight of
each element in a compound.
Molar Mass of Hydrates
The mass of the water is ADDED to the mass of the salt.
manganese II bromide tetrahydrate
2 x 79.9
4 H2O + 4 x
Remember to ADD the mass of water.
Nomenclature and Formula Mass
The Percentage Composition Calculation allows us to calculate the
percentage of mass for each element in a compound.
Sodium hydroxide is NaOH.
Na 23.0 g/mol
+ 1.0 g/mol
Find The Percentage of Each Element
% Na = 23 x 100 = 57.5 % N
Most Common Isotope
C4.10d Predict which isotope will have the greatest abundance given the
possible isotopes for an element and the average atomic mass in the periodic
For the most precise data, this is an item one would normally look up.
30 grams hydrogen sulfate (answer)
Hydrogen sulfate is H2SO4
The molar mass is 98.1 g/mol
30 grams x 1mol
= 0.31 mol H2SO4
30 grams x 1mol
x 6.022 x 1023 formula units
= 1.84 x 1023 formula units hydrogen sulfate
Warm-Downsnot in packet
What is the mass of 6.78x1024 formula units of barium hydroxide?
Formula units moles grams
171.3 g Ba(OH)2
Warm Downs-not in packet
How many molecules in 1 teaspoon of water?
Simple Sample #2
If 33.3 % of our class weighs 100 lbs, 33.3 % weighs 120 lbs and 33.3 %
weighs 140 lbs what is the average weight?
Simple Sample #3
If 20 % of the textbooks in our school weigh 500 grams each and 80 % weigh
600 grams each, what is the ave
Empirical Formula Example 2
Find the empirical formula of a sample having 37.49 % C, 12.58 % H, 49.93 %
37.49 g C
12.58 g H
49.93 g O
= 3.121 mol C
12.481 mol H
3.121 mol O