INTRO CHEMISTRY I

CHEM 1305

Calculation of Formula or Molecular Massbookmark.gif (981 bytes)

To determine the molecular or formula mass of a compound (also referred to as the atomic mass unit; amu), multiply the number of atoms of a particular element in a compound by the atomic mass of that element, then add the partial molecular masses together to determine the total molecular mass of the compound (right click here to practice formula mass).

Example:

Calculate the molecular mass of H2O.

Element number of atoms atomic mass partial mass of element
H 2 1.0 2.0
O 1 16.0 16.0
Molecular mass of H2O 18.0 amu

Molecular mass is a term which applies only to compounds that exist as molecules and are held together by covalent bonds. Formula mass is a term which applies to compounds that exist as ions and form ionic bonds. These ionic compounds are referred to as formulas.

Molesbookmark.gif (981 bytes)

The mole is the same number of atoms, formula units, molecules, or ions as there are atoms in exactly 12 g of carbon-12.  One mole of carbon-12 contains 6.02 x 1023 atoms, which has a mass of exactly 12 g (the atomic mass of carbon-12).  The number 6.02 x 1023 is referred to as Avogadro’s number.

One mole of any element contains 6.02 x 1023 atoms of the element and is equal to the atomic mass of the element expressed in grams. One mole of a compound contains 6.02 x 1023 formula units or molecules, and this number of formula units or molecules has a mass equal to the formula or molecular mass expressed in grams.

The atomic mass, formula mass, or molecular mass expressed in grams can also be referred to as the molar mass.  The molar mass is the mass in grams of one mole of any substance, element, or compound (right click here a to practice moles).

Example:   Calculate the number of moles in 80 grams of NaOH.

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Example:   Calculate the number of grams in 0.5 moles of NaOH.

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Example:   Calculate the number of molecules in 0.5 moles of NaOH.

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Example:   Calculate the number of moles in 12.04 x 1023 molecules of NaOH.

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Calculation of Percent Composition of Compoundsbookmark.gif (981 bytes)

First determine the amu or mass of various elements in the compound. Determine the total amu or mass of the compound. Divide the amu or mass of each element by the total mass and multiply by 100 (right click here practice percent composition).

Element number of atoms atomic mass partial mass of element % composition
H 2 1.0 2.0 2/18 x 100 = 11.1
O 1 16.0 16.0 16/18 x 100 = 88.9%
Sum     18.0 amu 100%

Calculation of Empirical and Molecular Formulasbookmark.gif (981 bytes)

The empirical formula of a compound contains the smallest whole-number ratio of the atoms present in a molecule or formula unit of the compound. This empirical formula is found using the percent composition of the compound. The percent composition is determined experimentally in the laboratory. The empirical formula gives only the ratio of the atoms present expressed as the smallest whole numbers.

To determine the empirical formula of a compound containing 80% carbon and 20% hydrogen with a molecular mass of 45 amu, convert the percentages to grams of each element assuming a total mass of 100 grams. This would give 80 grams of carbon and 20 grams of hydrogen.

Next, convert the grams of each element to moles: the moles of carbon would be 6.67 moles and the moles of hydrogen would be 20 moles.

Next, find the smallest whole number ratio of each element by dividing the element with the least number of moles into the moles of all the other elements. This would give a ratio of one carbon for every three hydrogens.

Finally, write the empirical formula using the whole-number ratios of each element: CH3

Note: if one of the ratios is a decimal (i.e. 1.5), multiple ALL the rations by 2 to obtain whole numbers.

% composition grams moles ratio
%C = 80 % 80 g 80/12 = 6.67 6.67/6.67 = 1
%H = 20% 20 g 20/1 = 20 20/6.67 = 3
empirical formula =  CH3

The molecular formula of the compound contains the actual number of atoms of each element present in one molecule of the compound. The molecular formula is a whole-number multiple of the empirical formula.

To determine the molecular formula, the formula mass (amu) of the formula must be known.

First, determine the empirical formula mass of the empirical formula.

Element number of atoms atomic mass partial mass of element
C 1 12.0 12.0
H 3 1.0 3.0
Empirical mass of CH3 = 15.0 amu

Next, divide the molecular mass (45 amu) by the empirical mass (15 amu). This will always yield a whole number: 45 amu / 15 amu = 3.

This whole number is the number of times the molecular formula is bigger than the empirical formula. Simply multiple this whole number by the empirical formula to obtain the molecular formula.

The molecular formula is CH3 x 3 = C3H9

right click here to practice empirical and molecular formulas)