Norlethia harris danielle ledbetter keanon nelson kierra ross michael thompson
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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

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Norlethia harris danielle ledbetter keanon nelson kierra ross michael thompson

Chapter 6:

Chemical Composition

Norlethia HarrisDanielle LedbetterKeanon NelsonKierra RossMichael Thompson

Ms.Tam

Magnet Chemistry

4th block

October 28, 2010


Norlethia harris danielle ledbetter keanon nelson kierra ross michael thompson

Counting

By

Weighing


Averaging the mass of similar objects

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

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

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

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

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


Norlethia harris danielle ledbetter keanon nelson kierra ross michael thompson

THE

MOLE


Masses with the same number of atoms

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

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

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

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.


Example 1 problem 1

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


Norlethia harris danielle ledbetter keanon nelson kierra ross michael thompson

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


Norlethia harris danielle ledbetter keanon nelson kierra ross michael thompson

  • 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

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


Norlethia harris danielle ledbetter keanon nelson kierra ross michael thompson

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 & 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

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


Norlethia harris danielle ledbetter keanon nelson kierra ross michael thompson

  • 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


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