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By now, we know that atoms and molecules are incredibly small. This makes the idea of counting them seem impossible. In fact it would make counting a barrel full of pennies seem easy in comparison!

Elements on the periodic table have two numbers associated with them. The atomic number gives information regarding the number of protons and electrons. The average atomic mass describes how heavy the elements relative to the lightest element, hydrogen. Below is a list of three elements, their actual masses in grams and their masses relative to the lightest element, hydrogen.

Element

Mass per atom (g/atom)

Relative Mass

H

1.67 x 10^-24

1

Li

1.15 x 10^-23

6.941

C

2.01 x 10^-23

12.011

If we massed out 1 g of H, 6.941 g of Li and 12.011 g of C, we would have the same number of atoms of each element. This number corresponds to Avogadro’s number, 6.02 x 10²³. We refer to this amount as 1 mole.

The mole is a term we use to indicate the quantity of an element or compound. In the lab we mass out various substances, but equations refer to the number of compounds or elements that react or are formed. Moles help us bridge the gap between the mass of a substance and the total number of atoms or molecules there are in that amount. Without going into great detail we can say that:

The mass of an element or compound in grams that is equal to the numerical value of the atomic or molecular mass is 1 mole of that substance.

For example:

1 g H = 1 mole H = 6.02 x 10²³ H atoms

12 .001 g C = 1 mole C = 6.02 x 10²³ C atoms

58.5 g NaCl = 1 mole NaCl = 6.02 x 10²³ NaCl formula units

For an element, the molar mass is simply the atomic mass in grams. For compounds, the molar mass can be calculated by taking the sum of the molar masses of each atom in the compound.

For example:

Glucose   C₆H₁₂O₆          

6(C) + 12(H) + 6(O) = 6(12.0) + 12(1.0) + 6(16) = 180

Calcium Phosphate  Ca₃(PO₄)₂       

3(Ca) + 2(P) + 8(O) = ???

Calculations With Moles

The molar mass has units of grams / mole and can be thought of as a conversion factor. If we know the mass of an element, we can calculate how many moles of that element.

Mass to moles

240 g C | 1 mole C    = 20 moles C
               |   12 g C

Likewise, if we know how many moles of an element there are, we can calculate what the mass is.

Moles to mass

3 moles Ca | 40 g Ca   =  120 g Ca
                      | 1 mole Ca

Because we know that 1 mole = 6.02 x 10²³ of anything, we can use this conversion factor to calculate how many atoms are present in a sample of a particular element.

Moles to how many atoms

2 moles Li | 6.02 x 10²³ Li   = 1.204 x 10²⁴ Li
                    | 1 mole Li

Knowing how many atoms there are, we can calculate how many moles

Number to moles

3.01 x 10²³ Mg | 1 mole Mg            = 0.5 moles Mg
                           |   6.02 x 10²³ Mg

It may be easier using the mole "bubble" diagram.

Percent Composition

Determining the percent composition of elements in a compound requires that you understand percent and molar mass.

Percent = Part / Whole x 100

Example: Calculate the percent Ca and Cl in CaCl₂.

Step 1: Determine the molar mass

CaCl₂ = 1(Ca) + 2(Cl) = 1(40) + 2(35.5) = 111

Step 2: Determine the percent of each

[1(Ca) / CaCl₂ ] x 100 = (40 / 111) x 100 = 36 %

Empirical Formula

The empirical formula of a compound is the simplest whole number ratio of the elements in a compound. In the lab you made MnCl_x. You massed the Mn, the MnCl_x and calculated the mass of Cl that the Mn combined with. Knowing the mass of Mn and Cl, you determined the moles of each and then found the ratio in which they combined.

Example: A similar experiment was done with Ce and I. It was determined that 1.67 g of Ce reacted with 4.54 g of I. Determine the empirical formula.

Step 1: Determine the moles of each

1.67 g Ce | 1 mole Ce  = 0.012 moles Ce
                  |  140.1 g Ce

4.54 g I | 1 mole I   = 0.036 moles I
              | 127 g I

Step 2: Determine the simplest whole number ratio

0.036 moles I / 0.012 moles Ce =

3 moles I / 1mole Ce

Step 3: Write the formula with the metal first.

Empirical formulas can be determined by percentages of each element. To convert the percentages to masses, assume you have 100 g of your compounds. Now that all the percentages are now masses, you can calculate the moles of each, then determine the simplest whole number ratio.

Beware of ratios that come out 1.5 / 1 or 2.5 / 2. DO NOT be tempted to round up or down. To make these ratios into simple whole number ratios multiply the numerator and denominator by a number (2 in this case) that gets rid of the decimal.
1.5 / 1 x 2/2 = 3 / 2    ,    2.5 / 1 x 2/2 = 5 / 2

Calculating Molecular Formulas From Empirical Formulas

While one laboratory method may only yield an empirical formula, another experiment may give information on the molecular mass. These two "formulas" should help.

Empirical Formula x n = Molecular Formula

Empirical Mass x n = Molecular Mass

n = a whole number multiple (the same for formula and mass equations)

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