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Wize High School Grade 11 Chemistry Textbook > Quantities in Chemistry

Empirical Formulas

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Percent Composition

  • The percent composition shows the amount that each element in a compound contributes to the overall mass of that compound.
  • The law of definite proportions states that the elements in a chemical compound are always present in the same proportions by mass. Example: A pinch of salt will have the same percent composition as a cup of salt.
  • To determine the percent composition of a compound, divide the mass of a particular element by the total mass of the compound and multiply by 100 to get a percentage
% composition by mass=mass contribution of elementtotal mass of compound×100%\boxed{\displaystyle \% \text{ composition by mass} =\frac{\text{mass contribution of element}}{\text{total mass of compound}}\times100\%}


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Example: Percent Composition using Chemical Formula
What is the percent composition by mass of oxygen in sodium hydroxide, NaOH?
40%

% composition by mass of O=mass controbution of oxygentotal mass of NaOH×100%\displaystyle \% \text{ composition by mass of O} =\frac{\text{mass controbution of oxygen}}{\text{total mass of NaOH}}\times100\%
1) Mass contribution of oxygen:
mO=1×15.999 g/mol=15.999 g/molm_O = 1 \times 15.999\ g/mol = 15.999\ g/mol

2) Total mass of NaOH:

mNaOH=(1×MNa)+(1×MO)+(1×MH)=22.990 g/mol+15.999 g/mol+1.008 g/mol=39.997 g/molm_{NaOH}=(1\times M_{Na})+(1\times M_O)+(1\times M_H)=22.990\ g/mol+15.999\ g/mol+1.008\ g/mol=39.997\ g/mol

3) Find percent composition by mass of oxygen

% composition by mass of O=15.999 g/mol39.997 g/mol×100=40.00%\displaystyle \% \text{ composition by mass of O}=\dfrac{15.999\ g/mol}{39.997\ g/mol}\times 100=40.00\%
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Empirical Vs Molecular Formulas

  • Molecular formulas tell us exactly how many atoms make up a molecule. Example: C6H6 tells us that for each molecule of C6H6, there are
    6
    C atoms and
    6
    H atoms.
  • Empirical formulas are the smallest possible "unit" of the molecular formula. Example: The empirical formula of C6H6 would be:
    CH
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  • When a molecular formula cannot be reduced , the molecular formula and empirical formula of the compound are the same Example: NO2
  • Many molecules can have the same empirical formula Example: C2H2 and C6H6 have the same empirical formula of CH

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Example: Percent Composition using Experimental Mass

A 27.0 g sample of a compound contains 7.20 g of carbon, 2.20 g of hydrogen and 17.6 g of oxygen. Calculate the percent composition of the compound.

%composition C=7.20g C27.0g compound×100%=26.7%\% \text{composition}\ C=\dfrac{7.20g\ C}{27.0g\ \text{compound}}\times100\%=26.7\%
%composition H=2.20g H27.0g compound×100%=8.1%\% \text{composition}\ H=\dfrac{2.20g\ H}{27.0g\ \text{compound}}\times100\%=8.1\%
%composition O=17.6g O27.0g compound×100%=65.2%\% \text{composition}\ O=\dfrac{17.6g\ O}{27.0g\ \text{compound}}\times100\%=65.2\%

Practice: Percent Composition

What percent of iron (III) hydroxide, Fe(OH)3, is oxygen? Round your answer to the nearest whole integer; do not include any symbols.

Practice: Finding Empirical Formula

A compound has the following mass composition: C = 86.59%, H = 8.36% and N = 5.05%. What is the empirical formula of this compound?

Practice: Finding Molecular Formula

A sample of a compound contains 1.52g of N atoms and 3.47g of O atoms. The molar mass of the compound is 92.02g/mol. Determine the molecular formula.