Mass fraction for a given element =

Mass of the element present in 1 mol of compoundMass of 1mole of compound

Percent Composition

Sometimes called the mass percent or weight percent

Example:

Calculate the percent by mass of the element copper in the compound CuBr2.

The molar mass of Copper is 63.55 g/mol

The molar mass of Bromine is 79.90 g/mol

The molar mass of Copper (II) bromide is63.55g + 79.90g + 79.90g = 223.35 g/mol

Example:

Calculate the percent by mass of the element copper in the compound CuBr2.

Mass percent = mass element x 100mass compound

Mass percent CuBr2 = 63.55 g x 100223.35 g

= 28.45%

Empirical Formula: Chemical formula determined from lab data. Lowest

possible ratio of atoms.

Example: CH2 is the lowest ratio (and empirical formula) of the molecule C3H6

Subscripts in a chemical formula show the ratio of atoms (or ions) in a molecule. In this case 1 : 2

A sample of CaCl2 has ~

1 calcium ion : 2 chlorine ionsOr a ratio of

1 : 2

We can use the unit “mole” to count things. If the subscripts give the ratio of atoms, then they also give the ratio of moles of atoms.

A sample of CaCl2 has ~

1 mole of calcium ions : 2 moles of chlorine ions

Or a ratio of1 : 2

# atoms A : # atoms B =

# moles A : # moles B

Therefore, if the ratio of moles of each atom is found…

Then the subscripts of the chemnical formula are known.

Example: 1 mole C2 mole H CH2

Calculate the molar mass of:

CH2 = 14.027 g/mol

C2H4 = 28.054 g/mol

C3H6

C4H8

= 42.081 g/mol

= 56.108 g/mol

Ratio stays the same 1 : 2

1. If given percents, use those percents as grams (as if you assume you have a 100 gram sample) – Remember percents add up to 100!

2. Change grams to moles for each atom

3. Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles)

4. Use the ratio as subscripts for writing the chemical formula

1. If given percents, use those percents as grams (as if you assume you have a 100 gram sample). – Remember percents add up to 100!

1 mol Ca = 40.08 g

1 mol Cl = 35.45 g

2. Change grams to moles for each atom1 mol Ca = 40.08 g

1 mol Cl = 35.45 g

36.1 g Ca 1 mol Ca = 0.901 mol Ca

40.08 g Ca

63.9 g Cl 1 mol Cl = 1.80 mol Cl

35.45 g Ca

0.901 mol Ca= 1 mol Ca

0.901

1.80 mol Cl = 2 mol Cl

0.901

3. Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles)

1 mol Ca2 mol Cl

4. Use the ratio as subscripts for writing the chemical formula

= CaCl2

1. If given grams, change grams to moles for each atom

2. Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles)

3. Use the ratio as subscripts for writing the chemical formula

4. If one or more of these numbers are not integers, multiply each by the smallest integer that will make all of them whole numbers

1. If given grams, change grams to moles for each atom

1 mol Al = 26.98 g

1 mol O = 16.00 g

1. If given grams, change grams to moles for each atom

1 mol Al = 26.98 g

1 mol O = 16.00 g

4.151 g Al 1 mol Al = 0.1589 mol Al

26.98 g Al

3.692 g O 1 mol O = 0.2308 mol O

16.00 g O

0.1589 mol Al= 1.000 mol Al

0.15890.2308 mol O

= 1.500 mol O 0.1589

2. Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles)

1.000 mol Al 1.500 mol O

3. Use the ratio as subscripts for writing the chemical formula

= Al1O1.5 !!

Can’t do this

4. If on or more of these numbers are not integers, multiply each by the smallest integer that will make all of them whole numbers

Al1 x 2O1.5 x 2 = Al2O3

Example: An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g or oxygen. Calculate the empirical formula for this compound