1 / 11

PERCENTAGE COMPOSITION and EMPIRICAL & MOLECULAR FORMULA

PERCENTAGE COMPOSITION and EMPIRICAL & MOLECULAR FORMULA. part whole. 24 g 95 g. % = x 100. % Mg = x 100. 24.305. 35.453. Mg. Cl. 12. 17. magnesium. chlorine. Percentage Composition. (by mass...not atoms). 25.52% Mg. Mg 2+ Cl 1-. 74.48% Cl. MgCl 2.

ltodd
Download Presentation

PERCENTAGE COMPOSITION and EMPIRICAL & MOLECULAR 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. PERCENTAGE COMPOSITION and EMPIRICAL & MOLECULAR FORMULA

  2. part whole 24 g 95 g % = x 100 % Mg = x 100 24.305 35.453 Mg Cl 12 17 magnesium chlorine Percentage Composition (by mass...not atoms) 25.52% Mg Mg2+ Cl1- 74.48% Cl MgCl2 It is not 33% Mg and 66%Cl 1 Mg @ 24.305 amu = 24.305 amu 2 Cl @ 35.453 amu = 70.906 amu 95.211 amu

  3. Empirical and Molecular Formulas A pure compound always consists of the same elements combined in the same proportions by weight. Therefore, we can express molecular composition as PERCENT BY WEIGHT. Ethanol, C2H6O 52.13% C 13.15% H 34.72% O

  4. / 0.708 mol 32.38 g Na 22.65 g S 44.99 g O / 0.708 mol / 0.708 mol Empirical Formula Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound. sodium sulfate = 2 Na 32.38% Na 22.65% S 44.99% O = 1.408 mol Na Na2SO4 Na2SO4 = 0.708 mol S = 1 S = 2.812 mol O = 4 O Step 1) %  g Step 2) g  mol Step 3) mol mol

  5. / 1.19 mol = 1 Na / 1.19 mol = 1 H NaHCO3 / 1.19 mol = 1 C / 1.19 mol = 3 O Empirical Formula A sample weighing 250.0 g is analyzed and found to contain the following: 27.38% sodium 1.19% hydrogen 14.29% carbon 57.14% oxygen 27.38 g Na 1.19 g H 14.29 g C 57.14 g O Assume sample is 100 g. Determine the empirical formula of this compound. Step 1) convert %  gram Step 2) gram  moles Step 3) mol / mol

  6. / 6.917 mol = 1 C CH2.5 / 6.917 mol = 2.5 H (2.4577 H) Empirical & Molecular Formula (contains only hydrogen + carbon) (~17% hydrogen) A 175 g hydrocarbon sample is analyzed and found to contain ~83% carbon. The molar mass of the sample is determined to be 58 g/mol. Determine the empirical and molecular formula for this sample. Determine the empirical formula of this compound. 2 C @ 12 g = 24 g 5 H @ 1 g = 5 g 29 g Step 1) convert %  gram Step 2) gram  moles Step 3) mol / mol Assume sample is 100 g. Then, 83 g carbon and 17 g hydrogen. MMempirical = 29 g/mol C2H5 MMmolecular = 58 g/mol 58/29 = 2 Therefore 2(C2H5) = C4H10 butane

  7. 17.8 17.8 0.555 mol 0.555 mol Common Mistakes when Calculating Empirical Formula Given: Compound consists of 36.3 g Zn and 17.8 g S. Find: empirical formula 36.3 g Zn = 2 Zn Zn2S Chemical formula indicates MOLE ratio, not GRAM ratio 17.8 g S = 1 S 1 mol Zn 1 36.3 g Zn Zn = 0.555 mol Zn 65.4 g Zn ZnS 17.8 g S 1 mol S 1 S zinc sulfide = 0.555 mol S 32.1 g S

  8. Empirical Formula of a Hydrocarbon 1 mol CO2 44.01 g x 2 mol C 1 mol CO2 x burn in O2 g CO2 mol CO2 mol C mol H Empirical formula CxHy g H2O mol H2O 2 mol H 1 mol H2O x 1 mol H2O 18.02 g x Kotz & Treichel, Chemistry & Chemical Reactivity, 3rd Edition , 1996, page 224

  9. / 183.4 g x 100% = 35.6 % Zn / 183.4 g x 100% = 26.2 % C / 183.4 g x 100% = 3.3 % H / 183.4 g x 100% = 34.9 % O Find the molar mass and percentage composition of zinc acetate CH3COO1- Zn2+ acetate = CH3COO1- Zn(CH3COO)2 1 Zn @ 65.4 g/mol = 65.4 g 4 C @ 12 g/mol = 48 g 6 H @ 1 g/mol = 6 g 4 O @ 16 g/mol = 64 g 183.4 g Zn(CH3COO)2

  10. / 0.5118 mol / 0.5118 mol A compound is found to be 45.5% Y and 54.5% Cl. Its molar mass (molecular mass) is 590 g. Assume a 100 g sample size a) Find its empirical formula 1 mol Y 45.5 g Y = 0.5118 mol Y = 1 Y 88.9 g Y YCl3 1 mol Cl 54.5 g Cl = 1.535 mol Cl = 3 Cl 35.5 g Cl 1 Y @ 88.9 g/mol = 88.9g b) Find its molecular formula 3 Cl @ 35.5 g/mol = 106.5 g = 3 590 / 195.4 195.4 g YCl3 3 (YCl3) Y3Cl9

  11. part whole % = x 100 % 6.02x1023 Molar Mass vs. Atomic Mass 2 g H2 = _____ H2 = _______ 2 amu 18 g H2O = _____ H2O = ________ 18 amu 120 g MgSO4 = _____ MgSO4 = ________ 120 amu 149 g (NH4)3PO4 = _____ (NH4)3PO4 = ________ 149 amu (by mass) Percentage Composition Empirical Formula • %  g • g  mol • mol • mol Empirical vs. Molecular Formula (lowest ratio)

More Related