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Norlethia Harris Danielle Ledbetter Keanon Nelson Kierra Ross Michael Thompson

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

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  1. Chapter 6: Chemical Composition Norlethia HarrisDanielle LedbetterKeanon NelsonKierra RossMichael Thompson Ms.Tam Magnet Chemistry 4th block October 28, 2010

  2. Counting By Weighing

  3. 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

  4. 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

  5. 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.

  6. 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

  7. 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

  8. THE MOLE

  9. 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

  10. 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.

  11. 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

  12. 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.

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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

  20. 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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