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Percentage Composition

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. It is not 33% Mg and 66% Cl. 1 Mg @ 24.305 amu = 24.305 amu

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Percentage Composition

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  1. 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

  2. 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

  3. / 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

  4. / 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

  5. / 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

  6. 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

  7. 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

  8. / 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

  9. / 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

  10. 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)

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