You start by determining the empirical formula for the compound

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You start by determining the empirical formula for the compound.

1)Determine the mass in grams of each element in the sample. If you are given , you can directly convert the percentage of each element to grams.

For example, a molecule has a molecular weight of 180.18 g/mol. It is found to contain 40.00% carbon, 6.72% hydrogen and 53.28% oxygen.

Convert the percentages to grams.

40.00 grams of carbon 6.72 grams of hydrogen 53.28 grams of oxygen

1. Determine the number of moles of each element by dividing its mass in grams by its molar mass (atomic weight in g/mol).

moles C = 40.00 g x 1 mol C/12.01 g/mol C = 3.33 moles C

moles H = 6.72 g x 1 mol H/1.01 g/mol H = 6.65 moles H

moles O = 53.28 g x 1 mol O/16.00 g/mol O = 3.33 moles O

1. Calculate the by dividing the number of moles of each element by the lowest number of moles to get the lowest whole number ratio.

C = 3.33/3.33 = 1.00 H = 6.65/3.33 = 2.00 O = 3.33/3.33 = 1.00

1. Use the whole number ratio of moles to determine the empirical formula. The whole number ratio for each element becomes its subscript.

Empirical formula is ##"C""H"_2"O"##

1. Determine the empirical formula weight.

Empirical formula weight = (1 x 12.01g/mol) + (2 x 1.01g/mol) + (1 x 16.00g/mol) = 30.02g/mol.

1. Divide the molar mass for the molecular formula by the empirical formula mass. The result determines how many times to multiply the subscripts in the empirical formula to get the molecular formula.

The molecular weight of this compound is 180.18g/mol. 180.18/30.02 = 6.002

Multiply each subscript in the empirical formula to get the molecular formula. The molecular formula is ##"C"_6"H"_12"O"_6"##.

1. Double check by calculating your molecular molar mass. C = (6 x 12.01g/mol) = 72.06 H = (12 x 1.01g/mol) = 12.12 O = (16.00 x 6g/mol) = 96 Total: 180.18g/mol