Norlethia harris danielle ledbetter keanon nelson kierra ross michael thompson
This presentation is the property of its rightful owner.
Sponsored Links
1 / 20

Norlethia Harris Danielle Ledbetter Keanon Nelson Kierra Ross Michael Thompson PowerPoint PPT Presentation


  • 96 Views
  • Uploaded on
  • Presentation posted in: General

Chapter 6: Chemical Composition. Norlethia Harris Danielle Ledbetter Keanon Nelson Kierra Ross Michael Thompson. Ms.Tam Magnet Chemistry 4 th block October 28, 2010. Counting By Weighing. Averaging the Mass of Similar Objects. Avg mass = total mass

Download Presentation

Norlethia Harris Danielle Ledbetter Keanon Nelson Kierra Ross Michael Thompson

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Chapter 6:

Chemical Composition

Norlethia HarrisDanielle LedbetterKeanon NelsonKierra RossMichael Thompson

Ms.Tam

Magnet Chemistry

4th block

October 28, 2010


Counting

By

Weighing


Averaging the Mass of Similar Objects

  • Avg mass = total mass

  • Objects do not need to have identical masses to be counted by weighing.

  • We simply need to know the avg. mass of the objecs.

# of object


Averaging the Mass of Different Objects

  • 2 samples containing different types of components, A & B, both contain the same # of components if the ratio of the sample masses is the same as the ratio of the masses of the individual components


EXAMPLE

You have M&M’s and Jolly Ranchers

  • M&M’s avg. mass of 5g

  • Jolly Ranchers avg. mass of 15g

    You scoop the M&M’s on to the scale and it reads 500g. What mass of Jolly Ranchers do you need to give the same number of Jolly Ranchers as there are M&M’s in 500g of M&M’s

    Avg. mass of Jolly Ranchers = 15g = 3

    Avg. mass of M&M’s 5g

    3 x 500g = 1500g

    This means you must weigh out an amount of Jolly Ranchers that is 3x the mass of M&M’s.


Atomic Masses: Counting Atoms by Weighing

  • To count atoms by weighing, we need to know the mass of individual atoms

  • Instead of measuring each isotope we use the average atomic mass

Example – Imagine you have a pile of 1000 natural carbon atom

Mass of 1000 C.Atoms = (1000 atoms)12.01 amu

1 c.atom

12,010 amu = 1000 x 12.01 amu

1.201 x 104amu


Using Mass to Count Atoms

  • Example

    Measured mass of carbons – 3.00 x 1020

    Avg. atomic mass – 12.01 amu

    Conversion factor – 1 carbon

    12.01 amu

    3.00 x 1020 amu x 1 carbon = 2.50 x 1019

    12.01 amu


THE

MOLE


Masses with the Same Number of Atoms

  • If you take 2 different elements with the same number of atoms you will have 2 different masses due to the fact that atomic masses of atoms are different in different elements

    Ex.-- 26.98 g of aluminum has the same number of atoms as 63.55 g copper


Defining the Mole

  • A mole is the number equal to the number of carbon atoms in 12.01 grams of carbon.

  • One mole is equal to 6.022 x 1023 units.

  • That amount is also known as Avagrado's number.


Using the Mole

  • The mole is used to measure how much mass is in 6.022 x 10^23 atoms of a certain element.

    Ex. 1 mol of aluminum is 26.98 g

    -The mole can be used to determine the amount of atoms of a set amount of mass.

    The conversion factor is 6.022 x 10^23

    1 mol __ atoms

    Ex. to determine the amount of atoms in 0.496 mol H atoms we use

    0.496 mol H atoms x 6.022 x 10^23 H atoms = 2.99 x 10^23

    1 mol H atoms


Molar Mass

  • Chemical Compound- a collection of atoms

  • Molar Mass-The mass in grams of 1 mole of substance.

  • The molar mass is obtained by finding the sum of the mass of the component atoms.


Molar mass (SO2)→? g/mol

Atomic Masses

S= 32.07 g/mol

O= 16.00 g/mol

Molar Mass (CaCO3) → ? g/mol

Atomic Masses

Ca= 40.08 g/mol

C= 12.01 g/mol

O= 16.00 g/mol

Example 1 Problem 1


To find the mass fraction fro a given element, you use this equation:

Mass Fraction for a given element

Mass of the element present in 1 mol of compound

=

Mass of 1 mol of compound

e.g. Calculate the mass of each element in C2H5OH and the molar mass.

Mass of C= 2 mol x 12.01 g/mol = 24.02 g

Mass of H= 6 mol x 1.008 g/mol = 6.048 g

Mass of O= 1 mol x 16.00 g/mol = 16.00 g

Mass of 1 mol of C2H5OH = 46.07g


  • Mass Percent (aka weight percent): the percent by mass of a component of a mixture or of a given element in a compound.

  • e.g. Find the mass percent of C

    Mass percent of C= Mass of C in 1 mol C2H5OH

X 100%

Mass of 1 mol C2H5OH

= 24.02 g

X 100% = 52.14%

46.07


Formulas of Compounds

  • To change masses to number of mol, use this formula (that I had to make).

  • Mass of Element in sample (1 mol E atoms)= X mol E Atoms

    Atomic #

  • Ex: .2015g Sample with .0806 g of Carbon, .01353g of Hydrogen, and .1074g of Oxygen.

  • .0806( 1 )= .00671 mol C atoms

    12.01


Cont.

  • As you can see, the variable E stands for any element on the periodic table. The answer X is the number of mol atoms.

  • To Convert Mol Atoms  Atoms

  • (X mol E atoms)(Avogadro’s Number)= E atoms

  • Ex: (.00671)(6.022 x 1023)= 4.04 x 1021 C atoms


Empirical & Molecular Formulas

  • Empirical- The actual formula of a compound that expresses the smallest whole number ratio present.

  • Molecular- The actual formula of a compound, the one that gives the composition of the molecules present.

  • Ex: C4H8O4 = Empirical Formula of CH20

  • Ex: Glucose = Molecular Formula of C6H12O6


Calculation of Empirical Formulas

  • It is important to find the chemical formula of a substance. By using a sample of a pure element found in the compound, you can find the ratio/chemical formula.

  • Subtract the mass of the pure substance from the compound to find the mass of the other element present in the substance. Multiply these respective mass by 1/Atomic number and .2636Ni x 1/58.69=.004491=4491/449=more or less


  • Therefore in the compound NiO you have one Ni for each O.

  • The emperical formula of Nickel Oxide is NiO.

  • For a binary compound, We do the same thin

    Jeopardy


  • Login