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TYPES OF FORMULA

TYPES OF FORMULA. EMPIRICAL FORMULA. - lowest whole number ratio of the elements in a compound STEPS 1. Given the % composition, change the percent to grams. 2. Change the grams to moles using the formula n = GW MM 3. Calculate the lowest whole number ratio of the moles.

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TYPES OF FORMULA

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  1. TYPES OF FORMULA

  2. EMPIRICAL FORMULA - lowest whole number ratio of the elements in a compound STEPS 1. Given the % composition, change the percent to grams. 2. Change the grams to moles using the formula n = GW MM 3. Calculate the lowest whole number ratio of the moles.

  3. Determine the empirical formula of a compound containing 74.19% Na and 25.81% O. Step 1. Change % to grams mass of Na = 74.19 g mass of O = 25.81 g Step 2. Solve for number of moles mole Na = 74.19 g = 3.2256 moles 23 g/mole Na mole O = 25.82 g = 1.613 moles 16 g/ mole O

  4. Step 3. Divide all number of moles by the lowest number. number of atoms of Na 3.2256 moles = number of atoms of O 1.613 moles = 2 atoms 1.613 moles 1 atom 1.613 moles EMPIRICAL FORMULA : Na2O

  5. Find the empirical formula of the compound that contains 33.32% Na, 20.30% N and 46.38% O. • Step 1 – Change % to mass Na = 33.32 g O = 46.38 g N= 20.30 g • Step 2 – Solve for number of moles mole Na = 33.32 g = 1.45 moles 23 g/Na mole N = 20.30 g = 1.45 moles 14 g/N mole O = 46.38 g = 2.90 moles 16 g/O

  6. Step 3 – Divide all mole values by the lowest number of mole to get number of atoms. atom Na = 1.45 moles = 1 atom 1.45 moles atom N = 1.45 moles = 1 atom 1.45 moles atom O = 2.90 moles = 2 atoms 1.45 moles EMPIRICAL FORMULA : NaNO2

  7. MOLECULAR FORMULA Indicates the actual number of atoms of each element present in the molecule. STEPS • Find the empirical formula • Find the empirical formula mass (EFM) • Divide the given molecular mass (MM) by the EFM to get the multiplier of the empirical formula • Multiply the subscript of each atom of the empirical formula by the multiplier to get the molecular formula.

  8. Find the molecular formula of a compound with a molar mass of 32 g/mole and percent composition of 87.5% N and 12.5% H. • Step 1 – Find the empirical formula N = 87.5 g H =12.5 g Solving for number of moles: mole N = 87.5 g = 6.25 moles 14 g/N mole H = 12.5 g = 12.5 moles 1 g/H

  9. Solving for the empirical formula atom of N = 6.25 moles = 1 atom 6.25 moles atom of H = 12.5 moles = 2 atoms 6.25 moles EMPIRICAL FORMULA : NH2 • Step 2. Get theEFM. EFM of NH2= (1 N X 14 g) + (2 H X 1 g) =16 g

  10. Step 3. Divide the MM by the EFM to get the multiplier. multiplier = MM = 32 g = 2 EFM 16 g • Step 4. Multiply the subscript of each atom in the empirical formula to get the molecular formula of the compound. N1X2 H2X2 = N2H4 MOLECULAR FORMULA

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