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Empirical  Formulae
 

Empirical Formulae

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How to find the empirical formula of a compound from given data. Also calculating the molecular formula. ...

How to find the empirical formula of a compound from given data. Also calculating the molecular formula.
**More good stuff available at:
www.wsautter.com
and
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    Empirical  Formulae Empirical Formulae Presentation Transcript

    • EMPIRICAL FORMULAE & PERCENT COMPOSITION Copyright Sautter 2003
    • The next slide is a quick promo for my books after which the presentation will begin Thanks for your patience! Walt S. [email_address] More stuff at: www.wsautter.com
    • Books available at: www. wsautter .com www.smashwords.com www.amazon.com www.bibliotastic.com www.goodreads.com Walt’s Books for Free!
    • EMPIRICAL FORMULAE
      • An empirical formula is the simplest formula for a compound. For example, H 2 O 2 can be reduced to a simpler formula, HO. Similarly, N 2 O 4 can be reduced to NO 2 . Both HO and NO 2 are empirical formulae.
      • Some formulae cannot be reduced and are the empirical formula. An example is H 2 O which cannot be reduced to lower terms.
      • Empirical formulae for compounds can be determined from experimental information such as percent composition data.
    • Empirical Formula and Molecular Formula Empirical Formula = CH Possible Molecular Formulae C H The molecular formula is always a whole number multiple of the empirical formula 2 2 3 3 4 4 n n 2 2 4 3 6 n 2n
    • PERCENT COMPOSITION
      • Percent composition refers to the mass percentage of each element contained in a compound.
      • Percent composition can be determined experimentally or from the formula of a substance. For example, the percentage of carbon and hydrogen contained in methane can be found from its formula CH 4 .
      • One mole of CH 4 contains 1 mole of carbon and 4 moles of hydrogen. One mole of carbon has a mass of 12.0 grams and 4 moles of hydrogen have a mass of 4.0 grams (4 x 1.0 grams per mole). The mass of one mole of CH 4 is 16.0 grams (12.0 + 4.0 = 16.0 g)
      • % C = (12.0 / 16.0) x 100 % = 75.0 % carbon
      • % H = (4.0/ 16.0) x 100 % = 25.0 % hydrogen
      • The total of the percents for each element must = 100 %
      • 75.0 % + 25.0 % = 100 %
    • PERCENT COMPOSITION
      • Problem : A compound is found to contain 0.1417 g of nitrogen and 0.4045 g of oxygen. The sample has a mass of 0.5462 g. What is the percent composition of the compound?
      • Solution : % of X = (grams of X / total mass) x 100 %
      • % N = (0.1417 / 0.5462) x 100 % = 25.6 %
      • % O = (0.4045 / 0.5462) x 100 % = 74.1 %
      • Check : 25.6 % + 74.1 % = 99.7 % ~ 100 %
      • It is often required to find the empirical formula of a compound from data such as that given in this problem. Remember, empirical formula is the lowest whole number ratio of atoms in a compound.
    • Empirical Formula Calculations
      • Problem : A compound is found to contain 0.1417 g of nitrogen and 0.4045 g of oxygen. The sample has a mass of 0.5462 g. What is the empirical formula of the compound?
      • Solution : All formulae are based on moles . First find the moles of each element present.
      • Moles N = 0.1417 g / 14.0 grams per mole = 0.0101
      • Moles O = 0.4045 g / 16.0 grams per mole = 0.0253
      • Now divide the smallest mole value into the other mole values
      • Moles O / moles N = 0.0253moles / 0.0101 = 2.50
      • Moles N / moles N is of course 1.00
      • The formula at this point is then NO 2.5
      • Empirical formula requires the lowest “whole number ratio”
      • Multiplying the formula through by 2 to obtain whole numbers we get N 2 O 5 (dinitrogen pentaoxide)
    • Empirical Formula Calculations Using Percent Composition
      • Problem : What is the empirical formula for a compound consisting of 37.5 % carbon, 12.5 % hydrogen and oxygen?
      • Solution : We will assume a 100 g sample of the compound. It then consists of 37.5 g C, 12.5 g H and 50.0 g O (100 g sample – 37.5g C – 12.5 g H)
      • Moles C = 37.5 g / 12.0 g per mole = 3.125
      • Moles H = 12.5 g / 1.00 g per mole = 12.5
      • Moles O = 50.0 g / 16.0 g per mole = 3.125
      • Divide each by the smallest mole value
      • Moles C / moles O = 3.125 / 3.125 = 1.00
      • Moles H / moles O = 12.5 / 3.125 = 4.00
      • The empirical formula is CH 4 O or CH 3 OH (methyl alcohol)
    • Empirical Formula Calculations
      • Steps for finding empirical formulae:
      • (1) If percent composition data is given assume a 100 gram sample
      • (2) Convert grams to moles for each element
      • (3) Divide the lowest mole value into each of the other mole values
      • (4) If all the ratios are not whole numbers multiply through to obtain whole number values for each ratio
      • (5) Write the empirical formula using the whole number ratios as subscripts
    • Empirical Formula Calculations From Combustion (Burning) Data
      • Problem: A compound containing only C, H and O is burned in O 2 to obtain CO 2 and H 2 O. A 0.5438 g sample gives 1.039 g CO 2 and 0.6369 g H 2 O. Find its empirical formula.
      • Solution : All the carbon is contained in the CO 2. Carbon dioxide is 27.3 % C (12.0 / 44.0) so 0.273 x 1.039 g gives 0.283 grams of C
      • The hydrogen is contained in the water which is 11.1 % H (2 / 18) so .111 x 0.6369 g gives 0.0708 g H
      • The rest of the sample is O (0.5438 – 0.283 – 0.0709) or 0.190 grams of O
      • Moles C = 0.283 / 12 = 0.0236
      • Moles H = 0.0708 / 1.00 = 0.0708
      • Moles O = 0.190 / 16 = 0.0119
      • Moles C / moles O = 0.0236 / 0.0119 = 1.98 ~ 2
      • Moles H / moles O = 0.0708 / 0.0119 = 5.95 ~ 6
      • Empirical Formula equals C 2 H 6 O or C 2 H 5 OH (ethyl alcohol)
    • Finding the Molecular Formula from the Empirical Formula and Molar Mass
      • Problem : The empirical formula for polystyrene is CH. It has a molecule mass of 104. What is its molecular formula?
      • Solution : Since the molecular formula is always some whole number multiple of the empirical formula, the molecular mass is always some whole number multiple of the empirical formula mass.
      • The empirical formula mass is 13 (12 for C + 1 for H)
      • Dividing 104 by 13 we get 8. Polystyrene must contain 8 CH units and have a molecular formula of C 8 H 8
    • The End