1 / 10

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.

Download Presentation

TYPES OF FORMULA

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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 = 25.81 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 the EFM. 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

More Related