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Stoichiometry and the MOLE!!

By: Claire Chrisman and Allison Wells. Stoichiometry and the MOLE!!. The Mole!. A mole is equal to 6.02 x 10 23 Discovered by Avogadro Chosen because 6.02 x 10 23 atoms of an element have a mass that is equal to the element's atomic mass.

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Stoichiometry and the MOLE!!

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  1. By: Claire Chrisman and Allison Wells Stoichiometry and the MOLE!!

  2. The Mole! • A mole is equal to 6.02 x 1023 • Discovered by Avogadro • Chosen because 6.02 x 1023 atoms of an element have a mass that is equal to the element's atomic mass. • the # of moles = (the # of atoms / 6.02 x 1023 ) • Mols=mass/molar mass • Units= g/mol

  3. Percent Composition • Percent Composition is the mass percent of each element in a compound. % Element A = (Mass of element A / total mass of sample) x 100 • Another formula that is useful if you are given the formula of the compound is: % Element A= (mass of A / molar mass of compound) x 100

  4. Stoichiometry! • 1. Balance the equation. • 2. Convert to moles. • 3. Then multiply by (wanted / given) • 4. Ta-Da! It's really that easy.

  5. Stoichiometry example problem • _Na2CO3 (s) _Na2O(s) + _CO2(g) • a) Balance the equation. • b) If you start with 47.2g of sodium carbonate, how many moles of sodium oxide can you make? What mass is this?

  6. Atomic mass is the average mass of all an element's atoms. An isotope occurs when the same element has a different number of neutrons. To find atomic mass: (% abundance of 1st isotope)(mass of 1st isotope)(% abundance of 2nd isotope)(mass of 2nd isotope)...... Now for an example: Atomic Mass

  7. If the abundance of 6Li (6.015121 amu) is 7.500% and the abundance of 7Li (7.016003 amu) is 92.500%, what is the average atomic mass? (a) 6.0750 amu (b) 6.0902 amu (c) 6.9250 amu (d) 6.9409 amu

  8. The number of protons and neutrons in an element. Mass Number Atomic Number • Abbreviated, Z. • The number of protons in an element.

  9. Limiting Reactants • The limiting reactant is the reactant that is used up first in the reaction causing the reaction to stop (sometimes.)

  10. Simple, you simply assume that each reactant is limiting and do some calculations to see how much product each one will make. For example: 16 Ag(s) + S8(s)→ 8 Ag2S(s) What mass of Ag2S can be produced from 2.0g Ag and 2.0g S8? Now we do stoich to find the mass of Ag2S that is made from 2.0g Ag. Then we do stoich again to find the mass of Ag2s that is made from 2.0g of S8. Whichever mass is smaller will be the limiting reactant. Cool....now how do I find a limiting reactant?

  11. Multiple Choice Carbon disulfide and oxygen react to yield carbon dioxide and sulfur dioxide. If 1 mole of CS2 is combined with 1 mole of oxygen, what is the limiting reactant? A. carbon disulfide ? B. Oxygen ? C. sulfur dioxide ? D. carbon dioxide

  12. Percent Yield • The measure of how efficient you and the reaction were at producing your product. % Yield=(Actual Yield / Theoretical Yield) x 100 • Actual=mass measured in lab • Theoretical=mass calculated from stoich

  13. Molecular Formula- the formula that shows what the compound actually is. ex. C6H2O6 Empirical and Molecular Formulas • Empirical Formula- the most reduced form of the formula. ex. CH2O

  14. 1. Convert to moles. 2. Divide all amounts bythe smallest amount. 3. The numbers you get will be the subscripts. Finding Empirical from empirical data. Finding empirical formulas from molecular. • Just reduce the subscripts. Ex. If a given sample contains 87.5% N and 12.5% H by mass, find the empirical formula.

  15. Finding molecular formulas from empirical formulas. • Pretty simple, just divide the molar mass given to you by the molar mass of the empirical formula. ex. Styrene has an empirical formula of CH. If styrene has a molar mass of 104g/mol, what is its molecular formula?

  16. Hydrates • A hydrate is a compound that has whole water molecules bonded to its structure. • Can be heated to remove the water, which leaves the anhydrous compound behind. • Can solve them in a way similar to empirical formulas ex. 2.681g hydrated NiSO4 *xH2O is heated. Anhydrous NiSO4 has a mass of 1.549g. Determine the hydration. 1. Subtract 2.681-1.549 to find the mass of H2O. Then convert both 1.549 and the 1.132 to moles. 2. Next divide the moles of NiSO4 by moles H2O.

  17. The End!

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