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Composition of Unknown CompoundsHow do we determine the chemical formula of new & unknown compounds?

http://www.youtube.com/watch?v=6ZJqN8KKrBU

Step 1:Percentage Composition (%Comp)

%Comp is simply the % of each element by mass in the compound %Comp = element molar mass

compound molar mass

Experimentally, %Comp is determined by a process called “combustion analysis” unknown substance is heated to very high

temperatures; chemical behaviour is observed each element has unique behaviour at high

temperatures ∴ we can predict what percent of the sample belongs

to each element

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%Comp by Combustion Analysis

Example 1: If we have a 50g sample of jelly

beans, with 20g of red jelly beans and 30g of green jelly beans, what is the percent composition of each colour?

Red: 20/50 x 100 = 40% Green: 30/50 x 100 = 60%

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Example 2: Iron(ii) Oxide; FeO, mmtotal= 71.84 g/mol

% Fe = 55.85/71.84 x100 = 77.7%

% O = 16.00/71.84 x100 = 22.3 %

Iron(iii) Oxide; Fe2O3,

mmtotal= 159.69 g/mol % Fe =

(2x55.85)/159.69 x100 = 69.9%

% O = (3x16.00)/159.69 x100

= 30.1 %

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Step 2:Empirical Formula (Molar Ratio)

Molecular formulae of many compounds is known, but how?

Knowing %Comp of unknown sample, we can calculate the molar ratio of elements

when identifying a new compound chemists first develop an empirical formula before determining the true molecular formula & structural arrangements

Empirical formula is simply a molar ratio of atoms within in the same molecule

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Empirical Formula (Molar Ratio)

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Molecular Formula

Empirical Formula Ratio

Water

HydrogenPeroxide

Empirical Formula: The Calculation!

A) Given % composition assume a mass of 100g of the unknown compound

B) Convert this to a MOLAR amount of each elements using element’s molar mass (n=m/mm)

C) Create a ratio by dividing each molar value by the lowest total amount of moles (n)

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Example Calculation:If we have an unknown sample with 5.88% H and 94.12%

O, what is the Empirical Formula?

Given % composition assume a mass of 100g of the unknown compound

Therefore; mH = 5.88g mO=94.12g

B) Convert this to a MOLAR amount of each elements using element’s molar mass (n=m/mm)

nH = 5.88g/1.01g/mol = 5.82 mol nO=94.12g/16.00g/mol = 5.88mol

C) Create a ratio by dividing each molar value by the lowest total amount of moles (n)

Hydrogren = 5.82/5.82 = 1 & Oxygen = 5.88/5.82 = 1

Therefore; the empirical formula is 1H: 1O or HO

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Step 3: True Molecular Formula

to find the MOLECULAR MASS of an unknown molecule, chemists use a device called a MASS SPECTROMETER (see pg#166)

1. https://www.youtube.com/watch?v=tOGM2gOHKPc

2. https://www.youtube.com/watch?v=J-wao0O0_qM

the molecular formula is then determined by analyzing the factor difference between the molecule’s empirical molecular mass and the true molecular mass and the true molecular mass

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Example: If the Empirical Formula of our unknown is HO,

and the Mass Spectrometer tells us the molar mass is 34.02g/mol, what is the true molar mass?

Factor Difference = True Mass/Empirical Mass = 34.02/17.01 = 2

THEREFORE,True Molecular Formula is2 x HO = H2O2

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