Deriving Empirical and Molecular Formulas

We can use our understanding of molar masses to derive the chemical formula for an unknown

compound. To do this follow the following sequence of measurement and calculations:

data from:

combustion analysis apparatus data from:

mass spectrometry

percent composition empirical formula . molecular formula .

Percent Composition

A hydrocarbon sample can be burned in a combustion analysis apparatus such as the one shown below.

The resulting change in mass in the water scrubber and carbon dioxide scrubber can be used to precisely

detect the mass of carbon and of hydrogen in the sample. Similar apparatus can be used to detect the

presence and mass of other elements too.

The resulting masses can be used to calculate the percentage of the sample that are composed of each element.

%𝐶 = 𝑚𝑐𝑎𝑟𝑏𝑜𝑛

𝑚𝑠𝑎𝑚𝑝𝑙𝑒× 100% %𝐻 =

𝑚ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛

𝑚𝑠𝑎𝑚𝑝𝑙𝑒× 100% %𝑒𝑙𝑒𝑚𝑒𝑛𝑡

𝑜𝑡ℎ𝑒𝑟 = 𝑚𝑜𝑡ℎ𝑒𝑟 𝑒𝑙𝑒𝑚𝑒𝑛𝑡

𝑚𝑠𝑎𝑚𝑝𝑙𝑒× 100%

Eg. A 500.00 mg tablet of Aspirin, C9H8O4, contains 300.00 mg carbon and 8.08 mg hydrogen. The remaining mass is oxygen. Determine the percentage composition of Aspirin.

%𝐶 = 𝑚𝑐𝑎𝑟𝑏𝑜𝑛

𝑚𝑠𝑎𝑚𝑝𝑙𝑒

× 100%

%𝐶 = 300.00𝑚𝑔

500.00𝑚𝑔× 100%

%𝐶 = 60.00%

%𝐻 = 𝑚ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛

𝑚𝑠𝑎𝑚𝑝𝑙𝑒

× 100%

%𝐶 = 8.08𝑚𝑔

500.00𝑚𝑔× 100%

%𝐶 = 1.62%

%𝑂 = 𝑚𝑜𝑥𝑦𝑔𝑒𝑛

𝑚𝑠𝑎𝑚𝑝𝑙𝑒

× 100%

%𝐶 = (500.00𝑚𝑔 − 300.00𝑚𝑔 − 8.08𝑚𝑔)

500.00𝑚𝑔× 100%

%𝐶 = 38.38%

Calculating the Empirical Formula Knowing the % composition of a compound by mass means that we can now derive the simplest molar ratio of each element in the compound. A chemical formula showing the simplest whole number ratio of each element is called the empirical formula. Determining an Empirical Formula: 1. Assume you have 100g sample of the compound. This makes it easy to convert % to grams. 2. Convert the mass of each element to moles using the molar mass of the element. 3. Divide by the moles of the element in the smallest proportion 4. If this does not result in whole numbers, multiply all by the simplest factor to produce a whole

number molar ratio between each element Eg. 1 – Determine the empirical formula for a compound that is 52.14% carbon, 13.13% hydrogen, and 34.73% oxygen Ratio %C = 52.14% mC = 52.14 g MC = 12.01 g/mol nC = 4.35 mol /2.17 = 2

%H = 13.13% mH = 13.13 g MH = 1.01 g/mol nH = 13.00 mol /2.17 = 6

%O = 34.73 % mO = 34.73 g MO = 16.00 g/mol nO = 2.17 mol /2.17 = 1 The empirical formula is C2H6O Eg. 1 – Determine the empirical formula for a compound that is 60.0% carbon, 4.5% hydrogen, and 35.5% oxygen Ratio X4 %C = 60.0% mC = 60.0 g MC = 12.01 g/mol nC = 4.99 mol /2.22 = 2.25 = 9

%H = 4.5% mH = 4.5 g MH = 1.01 g/mol nH = 4.45 mol /2.22 = 2 = 8

%O = 35.5 % mO = 35.5 g MO = 16.00 g/mol nO = 2.22 mol /2.22 = 1 = 4 The empirical formula is C9H8O4

Calculating the Molecular Formula

The simplest ratio of elements in a molecule is not necessarily the true ratio of elements in that compound.

The chemical formula reflecting the true number of atoms of each element in a compound is called the molecular formula.

Mass spectrometry will provide a molar mass for your unknown compound. If the experimentally determined molar mass matches the molar mass of your empirical formula, that means that the subscripts in your empirical formula are a true reflection of the number of atoms of each element per molecule of the sample. If not, then the compound must contain some multiple of the subscripts of each element in the empirical formula. Likewise, the experimentally determined molar mass must be a whole number multiple of the molar mass of the empirical formula. Determining a Molecular Formula: 1. Calculate the molar mass of the empirical formula. 2. Divide the experimentally determined molar mass derived from mass spectrometry by the molar

mass of the empirical formula. This will give you a multiplication factor, x. 3. If x = 1, the empirical formula is also the molecular formula 4. If x > 1, multiply the subscripts of the empirical formula by x to derive the molecular formula. Eg. If a sample of a compound with the empirical formula C2H4O has a molar mass of 88.12 g/mol, determine the molecular formula. Msample of C2H4O = 88.12g/mol MC2H4O = 44.06g/mol

𝑥 = 𝑀𝑠𝑎𝑚𝑝𝑙𝑒

𝑀𝐶2𝐻4𝑂=

88.12 𝑔/𝑚𝑜𝑙

44.06 𝑔/𝑚𝑜𝑙= 2

C2H4O x2 C4H8O2

The molecular formula is twice the empirical formula. The molecular formula is C4H8O2

Summary The general sequence of measurements and calculations to derive a molecular formula:

Analysis of a sample Calculation Calculation Analysis of a sample Calculation

Place a sample into a combustion analysis apparatus. This provides the mass per element in your sample

Use the mass values to calculate the percent composition of the sample

Assuming a 100g sample, convert % comp to mass. Calculate moles of each element Find the simplest whole number mole ratio of elements in the compound

Place a sample into a mass spectrometer. This provides the molar mass of your sample

Compare the experimentally determined molar mass to the molar mass of the empirical formula.

𝑥 =𝑀𝑠𝑎𝑚𝑝𝑙𝑒

𝑀𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎

Adjust all subscripts by the multiple x

% Composition Empirical Formula Molecular Formula