Chem 109 - General Chemistry - Spring 2011
Lecture Notes 7: 2 February
Stoichiometry: Moles, Atoms, & Formulae
Most samples of matter consist of combinations of atoms or ions to give compounds characterized by molecules or formulae. We are thus interested in molecular masses, formula masses etc.
Examples:
- What is the weight of 0.25 mole of:
0.25 mole x (39.10 g/mole + 79.90 g/mole) = 29.75 g = 3.0 x 101g
so weight = 0.25 mole x (6 [12.01]g/mole + 12 x [1.008 g/mole] + 6 [16.00] g/mole) = 45.04 g = 45 g
- How many moles are there in 23.5 g of:
so MW = (12 [12.01]g/mole + 22 x [1.008 g/mole] + 11 [16.00 ] g/mole) = 342.30 g/mole
and moles = (23.5 g)/(342.30 g/mole) = 6.865 x 10-2mole = 6.87 x 10-2mole.
- How many moles are there in 23.5 g of NaCl?
- How many moles are there in 23.5 g of propane (C3H8)?
- How many atoms are there in 1.00 g of
First need FW = (39.10 g/mole + 79.90 g/mole) = 119.00 g/mole
then find moles = (1.00 g)/(119.00 g/mole) = 8.403 x10-3mole
then multiply by Avogadro's Number: (8.403 x10-3mole) x (6.022 x 1023 atoms/mole) = 5.061 x1021 KBr "units"
But - there are two atoms/formula, So 1.01 x1022 atoms
(Of course there are no atoms actually present, just ions. However, all the parts for the atoms are there, so we still ask the question in terms of atoms.)
- How many atoms are there in 1.00 g of glucose? (C6H12O6)
- How many atoms are there in 1.00 g of ethanol? (C2H5OH)
Formula Stoichiometry
Determination of the Percentage Composition of Compounds: There are two common ways to describe the composition of compounds: by the ratios of elements by atom or mole in them, and by the ratio of elements by percentage. Let's look at the percentage elemental analysis of a compound.
- What is the percentage composition of sodium superoxide (NaO2)?
First we need to determine the formula weight = 22.99 g/mol Na + 2 (16.00 g/mol O) = 54.99 g/mol NaO2
% Na = {(23.00 g/mol) / (54.99 g/mol)} x 100% = 41.81%
% O = {(32.00 g/mol) / (55.99 g/mol)} x 100% = 58.19%
Check: 41.81 % + 58.19% = 100.00% (Note that frequently the addition will not quite work out due to rounding errors.)
Determination of Empirical (Simplest) Formulae: Want to determine the ratios, in moles, of elements in an analysis.
- A compound of mercury, oxygen, and nitrogen gave an analysis of 61.80% mercury and 8.63% nitrogen by weight. What is the empirical formula for this compound? (Note that oxygen is commonly not given in these analyses because it is used to react with other substances in the analysis process, and so is difficult to measure itself.)
First need to find the amount of oxygen: 100% - 61.80% - 8.63% = 29.57%
Next we need to find the number of moles. Easiest to assume 100 g total, and find moles of each, as we will see next time.
© R A Paselk
Last modified 2 February 2011
