THE MOLE CONCEPT

A sensitive balance can weight to the nearest 0.0001 g, but a typical atom has a mass of only 0.000 000 000 000 000 000 000 01 g

Is it possible to keep track of atoms by counting them? Yes, but not directly. We count them in groups. That is we count atoms in the same way we count eggs by the dozen (12), bottle rockets by the gross (144), beer by the case (24), and sheets of paper by the ream (500).



 

The relative mass of an atom is compared to a reference atom using an instrument called a mass spectrometer. Carbon-12 has been chosen as a reference and it is assigned a mass of exactly 12 amu. The masses of other atoms are compared relative to carbon-12.



 

Experiments have shown that 12.01 g of carbon contains 6.02 x 1023 atoms of carbon.

In fact, the gram atomic mass of each element contains 6.02x1023 atoms of that element.

 



AVOGADRO’S #:

Number of atoms in 12.01 grams of carbon

6.02x1023 atoms

 



MOLE:

Amount of a substance containing 6.02x1023 particles

 

 



Why do we need this tool (the mole)?

Chemists and many others, need to know relative amounts of substances.




FORMULA MASS:

The sum of the atomic masses, for each element, found in the compound.

Exp: Calculate the formula mass of H2O

2 hydrogen atoms @ 1.008 amu

1 oxygen atom @ 15.999 amu

2.016+15.9999= 18.015 amu

One molecule of water has a mass of 18.015 amu


 

 

MOLAR MASS:

Mass in grams of one mole of any substance, element, or compound.

Has the same numerical value as formula mass.



Q: What is the mass of 6.02x1023 H2O molecules?

A: 18.015 g

 



RELATE MOLES TO PARTICLES AND PARTICLES TO MOLES

How many H2 molecules are there in 0.25 moles of H2?

7.9 billion carbon atoms represents how many moles of carbon?

 



Know what molar mass is and how to calculate molar mass.

The molar mass of an element is equal to its atomic mass expressed in grams.

Carbon has a molar mass of 12.01 g/mol

The molar mass of a compound is equal to the sum of the atomic masses, for each element, found in the compound.

CuSO4 has a molar mass of 159.5g/mol

63.5 + 32 + 4(16) = 159.5



 

What is the mass of one H2O molecule in grams?

 

Know the molar volume for a gas at STP

1 mol of gas = 6.02x1023 molecules = 22.4L

 

 


Percent Composition

The percent composition is a list of the mass percent of each element in a compound.

Calculate the percent composition of proply chloride (C3H7Cl)

Frist, calculate the formula mass for the compound by adding up the masses of all the elements:

3 carbons at 12.01amu each = 36.03 amu

7 hydrogen at 1.01amu each= 7.07 amu

1 chlorine at 35.45 amu

Summing these values yield:

36.03+7.07+35.45= 78.55 amu = formula mass of C3H7Cl

2. For each element found in the compound set up ratio of its mass over the total mass times 100 to convert to percent.

Carbon (36.03 / 78.55)x100 = 45.91% Carbon

Hydrogen (7.07 / 78.55)x100 = 9.00% Hydrogen

Chlorine (35.45 / 78.55)x100 = 45.13% Chlorine

These values add up to a hundred percent (100.04%).

 



 

Empirical formula Simplest formula for a compound; it is the smallest whole-number ratio of the atoms present.

Molecular formula Expression of the formula for a compound; it shows the actual number of atoms of each element present in one molecule of the compound.



 

Calculate the empirical formula for a compound composed of 26.6% potassium, 35.4% chromium, and 38.1% oxygen.

Percent means a fraction of 100. Thus, if we assume a100 grams of sample we can then use the percent values as grams of each element. From grams we find moles of each element in the formula.

Find moles potassium.

Next find moles of chromium.

Then find moles oxygen.

Next, set up a mole ratio that relates the moles of each element to the moles of the element that is least present.

Because a chemical formula must have only whole numbers we multiply by 2 to yield: K2Cr2O7 as the empirical formula.



 

An oxide of nitrogen gave the following analysis: 3.04g of nitrogen combined with 6.95g of oxygen. Experiments show that the molecular mass of this compound is 91.0 amu. Determine its molecular formula