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Mass fraction for a given element =

Percent Composition. Mass fraction for a given element =. Mass of the element present in 1 mol of compound Mass of 1mole of compound. Sometimes called the mass percent or weight percent. Percent Composition from Masses. Example:

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Mass fraction for a given element =

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  1. Percent Composition Mass fraction for a given element = Mass of the element present in 1 mol of compound Mass of 1mole of compound Sometimes called the mass percent or weight percent

  2. Percent Composition from Masses Example: Calculate the percent by mass of the element copper in the compound CuBr2. The molar mass of Copper is 63.55 g/mol The molar mass of Bromine is 79.90 g/mol The molar mass of Copper (II) bromide is 63.55g + 79.90g + 79.90g = 223.35 g/mol

  3. Percent Composition from Masses Example: Calculate the percent by mass of the element copper in the compound CuBr2. Mass percent = mass element x 100 mass compound Mass percent CuBr2 = 63.55 g x 100 223.35 g = 28.45%

  4. Calculation of Empirical Formulas Empirical Formula: Chemical formula determined from lab data. Lowest possible ratio of atoms.

  5. Calculation of Empirical Formulas Example: CH2 is the lowest ratio (and empirical formula) of the molecule C3H6 Subscripts in a chemical formula show the ratio of atoms (or ions) in a molecule. In this case 1 : 2

  6. Calculation of Empirical Formulas A sample of CaCl2 has ~ 1 calcium ion : 2 chlorine ions Or a ratio of 1 : 2

  7. Calculation of Empirical Formulas We can use the unit “mole” to count things. If the subscripts give the ratio of atoms, then they also give the ratio of moles of atoms.

  8. Calculation of Empirical Formulas A sample of CaCl2 has ~ 1 mole of calcium ions : 2 moles of chlorine ions Or a ratio of 1 : 2 # atoms A : # atoms B = # moles A : # moles B

  9. Calculation of Empirical Formulas Therefore, if the ratio of moles of each atom is found… Then the subscripts of the chemnical formula are known. Example: 1 mole C 2 mole H CH2

  10. Calculation of Empirical Formulas Calculate the molar mass of: CH2 = 14.027 g/mol C2H4 = 28.054 g/mol = 42.081 g/mol C3H6 C4H8 = 56.108 g/mol Ratio stays the same 1 : 2

  11. Finding Empirical Formulas (%) If given percents, use those percents as grams (as if you assume you have a 100 gram sample) – Remember percents add up to 100! Change grams to moles for each atom Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles) Use the ratio as subscripts for writing the chemical formula

  12. Example: Find the empirical formula if a sample contains Ca and Cl and is 36.1% Ca If given percents, use those percents as grams (as if you assume you have a 100 gram sample). – Remember percents add up to 100! 1 mol Ca = 40.08 g 1 mol Cl = 35.45 g

  13. Example: Find the empirical formula if a sample contains Ca and Cl and is 36.1% Ca Change grams to moles for each atom 1 mol Ca = 40.08 g 1 mol Cl = 35.45 g 36.1 g Ca 1 mol Ca = 0.901 mol Ca 40.08 g Ca 63.9 g Cl 1 mol Cl = 1.80 mol Cl 35.45 g Ca

  14. Example: Find the empirical formula if a sample contains Ca and Cl and is 36.1% Ca Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles) 0.901 mol Ca = 1 mol Ca 0.901 1.80 mol Cl = 2 mol Cl 0.901

  15. Example: Find the empirical formula if a sample contains Ca and Cl and is 36.1% Ca Use the ratio as subscripts for writing the chemical formula 1 mol Ca 2 mol Cl = CaCl2

  16. Finding Empirical Formulas (g) If given grams, change grams to moles for each atom Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles) Use the ratio as subscripts for writing the chemical formula If one or more of these numbers are not integers, multiply each by the smallest integer that will make all of them whole numbers

  17. Example: An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula for this compound If given grams, change grams to moles for each atom 1 mol Al = 26.98 g 1 mol O = 16.00 g

  18. Example: An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g or oxygen. Calculate the empirical formula for this compound If given grams, change grams to moles for each atom 1 mol Al = 26.98 g 1 mol O = 16.00 g 4.151 g Al 1 mol Al = 0.1589 mol Al 26.98 g Al 3.692 g O 1 mol O = 0.2308 mol O 16.00 g O

  19. Example: An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g or oxygen. Calculate the empirical formula for this compound Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles) 0.1589 mol Al = 1.000 mol Al 0.1589 0.2308 mol O = 1.500 mol O 0.1589

  20. Example: An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g or oxygen. Calculate the empirical formula for this compound Use the ratio as subscripts for writing the chemical formula 1.000 mol Al 1.500 mol O = Al1O1.5 !! Can’t do this

  21. Example: An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g or oxygen. Calculate the empirical formula for this compound If on or more of these numbers are not integers, multiply each by the smallest integer that will make all of them whole numbers Al1 x 2O1.5 x 2= Al2O3

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