# Stoichiometry

Topics: Stoichiometry, Chemical element, Isotope Pages: 6 (1576 words) Published: April 22, 2013
AP Chemistry Unit 2 Notes
Stoichiometry  You should understand all that is presented in chapter 3 of your text (Zumdahl: Chemistry, 8th edition). Some of the highlights are presented below.

Atomic Masses (Section 3.1)
      Nearly every element is made up of atoms of more than one isotope for that element. A few, like Be, only have one isotope. Others can have a large number of isotopes. Tin (Sn) has ten isotopes. (No pun intended.) Isotopic abundance is determined by the use of a mass spectrometer. Using a magnetic field, this device can sort the isotopes of a pure element into sub-groups based on mass. This way, we can see what percentage of an element is composed of each of the isotopes. The atomic mass values given on the periodic table are average masses, reflecting the weighted average of the isotopes for each element.

average atomic mass =  % of each isotope    atomic mass of each isotope 

For example, Gallium has two isotopes. Ga-69 is 60.108% abundant and has a mass of 68.92558 amu. Ga-71 is 39.892% abundant and has a mass of 70.9247005 amu.

average atomic mass = .60108 x 68.92558amu   .39892 x 70.9247005  average atomic mass = 69.723amu

The Mole (Section 3.2)
    In chemistry, we use the counting unit of a mole (mol). This is defined as the amount of particles that are in exactly 12.00 g of the isotope 12C. This number is 6.022 x 1023. Using this same relationship, it can be reasoned that 6.022 x 10 23 amu = 1.000 g. Therefore, the average mass of one mole of any element in grams is equal to the average atomic weight, also called the average molar mass.

Percent Composition of Compounds and Empirical Formulas (Sections 3.5 and 3.6)   The chemical formula of a compound shows the exact molar ratio of the different elements in the compound. The numbers of each element are recorded using a subscript after each chemical symbol. The subscript of one is understood and never written. The percent composition is determined based on mass percent. It is a ratio of the mass of the number of moles of an element present compared to the total mass of the compound.

EXAMPLE Carvone is responsible for the smell of spearmint oil and has the molecular formula of C 10H14O. Find its molar mass and the mass percentage of each of the three elements. 12.01g = 120.1 g 1 mol 1.008g Mass of H in 1 mol = 14 mol x = 14.11 g 1 mol 16.00g Mass of O in 1 mol = = 1 mol x = 16.00 g 1 mol Mass of C in 1 mol = 10 mol x Molar Mass of 1 mol C H O = 150.2 g 10 14

Mass % of C = Mass % of H = Mass % of C =

120.1g x 100% = 79.96% C 150.2g 14.11g x 100% = 9.394% H 150.2g 16.00g x 100% = 10.65% O 150.2g

AP Chemistry Zumdahl, 8th edition, Ch. 3

Unit 3: Stoichiometry

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AP Chemistry Unit 2 Notes
Stoichiometry 

The empirical formula is the simplest whole number ratio of the atoms of each element present in a compound and can be calculated from the mass percent data. 1. When given the percent of each element present, assume the sample of the compound has a mass of exactly 100 g. 2. The percent of each element present is now the mass of each element present. (10.65% of 100 g is 10.65 g) 3. Divide the mass of each element present by its molar mass to determine the number of moles present. 4. Divide the amount of moles of each element present by the smallest value of these numbers. This will give you a molar ratio. 5. If needed, multiply the ratio by an integer to attain the lowest whole number ratio of moles of elements present in the compound.

EXAMPLE A compound is analyzed to contain 43.64% phosphorus and 56.36% oxygen by mass. What is the empirical formula of the compound?

P
% by mass Mass in 100 g of compound Moles of each element Divide by smallest Smallest whole # mole ratio Empirical formula  43.64g 

O
56.36% 56.36 g
56.36g  1 mol O  3.523 mol O 16.00g

43.64% 43.64 g
1 mol P  1.409 mol P 30.97g

1.409 mol  1 1.409 mol

3.523 mol  2.5...