<< Chapter < Page Chapter >> Page >
  1. Problems where you will be given the formula of the substance and asked to calculate the percentage by mass of each element in the substance.
  2. Problems where you will be given the percentage composition and asked to calculate the formula.
  3. Problems where you will be given the products of a chemical reaction and asked to calculate the formula of one of the reactants. These are often referred to as combustion analysis problems.
  4. Problems where you will be asked to find number of moles of waters of crystallisation.

Calculate the percentage that each element contributes to the overall mass of sulphuric acid ( H 2 SO 4 ).

  1. Hydrogen = 1,008 × 2 = 2,016 u

    Sulphur = 32,07 u

    Oxygen = 4 × 16 = 64 u

  2. Use the calculations in the previous step to calculate the molecular mass of sulphuric acid.

    Mass = 2 , 016 + 32 , 07 + 64 = 98 , 09 u
  3. Use the equation:

    Percentage by mass = atomic mass molecular mass of H 2 SO 4 × 100 %

    Hydrogen

    2 , 016 98 , 09 × 100 % = 2 , 06 %

    Sulphur

    32 , 07 98 , 09 × 100 % = 32 , 69 %

    Oxygen

    64 98 , 09 × 100 % = 65 , 25 %

    (You should check at the end that these percentages add up to 100%!)

    In other words, in one molecule of sulphuric acid, hydrogen makes up 2,06% of the mass of the compound, sulphur makes up 32,69% and oxygen makes up 65,25%.

A compound contains 52.2% carbon ( C ), 13.0% hydrogen ( H ) and 34.8% oxygen ( O ). Determine its empirical formula.

  1. Carbon = 52,2 g , hydrogen = 13,0 g and oxygen = 34,8 g

  2. n = m M

    Therefore,

    n ( Carbon ) = 52 , 2 12 , 01 = 4 , 35 mol
    n ( Hydrogen ) = 13 , 0 1 , 008 = 12 , 90 mol
    n ( Oxygen ) = 34 , 8 16 = 2 , 18 mol
  3. In this case, the smallest number of moles is 2.18. Therefore...

    Carbon

    4 , 35 2 , 18 = 2

    Hydrogen

    12 , 90 2 , 18 = 6

    Oxygen

    2 , 18 2 , 18 = 1

    Therefore the empirical formula of this substance is: C 2 H 6 O . Do you recognise this compound?

207 g of lead combines with oxygen to form 239 g of a lead oxide. Use this information to work out the formula of the lead oxide (Relative atomic masses: Pb = 207 u and O = 16 u ).

  1. 239 - 207 = 32 g
  2. n = m M

    Lead

    207 207 = 1 mol

    Oxygen

    32 16 = 2 mol
  3. The mole ratio of Pb : O in the product is 1:2, which means that for every atom of lead, there will be two atoms of oxygen. The formula of the compound is PbO 2 .

Vinegar, which is used in our homes, is a dilute form of acetic acid. A sample of acetic acid has the following percentage composition: 39,9% carbon, 6,7% hyrogen and 53,4% oxygen.

  1. Determine the empirical formula of acetic acid.
  2. Determine the molecular formula of acetic acid if the molar mass of acetic acid is 60 g · mol - 1 .
  1. In 100 g of acetic acid, there is 39,9 g C , 6,7 g H and 53,4 g O

  2. n = m M

    n C = 39 , 9 12 = 3 , 33 mol n H = 6 , 7 1 = 6 , 7 mol n O = 53 , 4 16 = 3 , 34 mol
  3. Empirical formula is CH 2 O

  4. The molar mass of acetic acid using the empirical formula is 30 g · mol - 1 . Therefore the actual number of moles of each element must be double what it is in the empirical formula.

    The molecular formula is therefore C 2 H 4 O 2 or CH 3 COOH

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Chemistry grade 10 [caps]. OpenStax CNX. Jun 13, 2011 Download for free at http://cnx.org/content/col11303/1.4
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Chemistry grade 10 [caps]' conversation and receive update notifications?

Ask