# Empirical Formula  %  g  g  mol  mol / mol. % Mg = x 100 24 g 95 g Percentage Composition Mg magnesium 24.305 12 Cl chlorine 35.453 17 Mg 2+ Cl 1-

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Empirical Formula  %  g  g  mol  mol / mol

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

Empirical and Molecular Formulas A pure compound always consists of the same elements combined in the same proportions by weight. PERCENT BY WEIGHT Therefore, we can express molecular composition as PERCENT BY WEIGHT. Ethanol, C 2 H 6 O 52.13% C 13.15% H 34.72% O

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. = 1.408 mol Na = 0.708 mol S = 2.812 mol O / 0.708 mol = 2 Na = 1 S = 4 O Na 2 SO 4 32.38% Na 22.65% S 44.99% O 32.38 g Na 22.65 g S 44.99 g O sodium sulfate Step 1) %  gStep 2) g  mol Step 3) mol mol Na 2 SO 4

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 Determine the empirical formula of this compound. Step 1) convert %  gram Step 2) gram  moles Step 3) mol / mol Assume sample is 100 g. 27.38 g Na 1.19 g H 14.29 g C 57.14 g O / 1.19 mol = 1 Na / 1.19 mol = 1 H / 1.19 mol = 1 C / 1.19 mol = 3 O NaHCO 3

Empirical & Molecular Formula 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. Step 1) convert %  gram Step 2) gram  moles Step 3) mol / mol Assume sample is 100 g. / 6.917 mol = 1 C / 6.917 mol = 2.5 H (2.4577 H) CH 2.5 Then, 83 g carbon and 17 g hydrogen. C2H5C2H5 MM molecular = 58 g/mol 58/29 = 2 Therefore 2(C 2 H 5 ) = C 4 H 10 MM empirical = 29 g/mol 2 C @ 12 g = 24 g 5 H @ 1 g = 5 g 29 g butane (~17% hydrogen) (contains only hydrogen + carbon)

1 mol S 1 mol Zn 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 65.4 g Zn = 2 Zn 17.8 g S 17.8 = 1 S Zn 2 S 36.3 g Zn 17.8 g S 32.1 g S = 0.555 mol Zn = 0.555 mol S Chemical formula indicates MOLE ratio, not GRAM ratio 0.555 mol 1 1 Zn S ZnS zinc sulfide

Empirical Formula of a Hydrocarbon CxHyCxHy g H 2 O g CO 2 mol H 2 O mol CO 2 mol C mol H burn in O 2 1 mol CO 2 44.01 g x 1 mol H 2 O 18.02 g x 2 mol H 1 mol H 2 O x 2 mol C 1 mol CO 2 x Empirical formula Kotz & Treichel, Chemistry & Chemical Reactivity, 3 rd Edition, 1996, page 224

Empirical and Molecular Formulas Keys Empirical and Molecular Formulas Empirical & Molecular Formula Empirical and Molecular Formulas http://www.unit5.org/chemistry/Nomenclature.html

Errors in Chemical Formulas and Nomenclature Keys Errors in Chemical Formulas and Nomenclature http://www.unit5.org/chemistry/Nomenclature.html Airs n Knomenclayture

Find the molar mass and percentage composition of zinc acetate acetate = CH 3 COO 1- Zn 2+ CH 3 COO 1- Zn(CH 3 COO) 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 Zn(CH 3 COO) 2 183.4 g / 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

1 mol Cl 35.5 g Cl 1 mol Y 88.9 g Y A compound is found to be 45.5% Y and 54.5% Cl. Its molar mass (molecular mass) is 590 g. a) Find its empirical formula b) Find its molecular formula Assume a 100 g sample size 45.5 g Y 54.5 g Cl = 0.5118 mol Y = 1.535 mol Cl / 0.5118 mol = 1 Y = 3 Cl YCl 3 1 Y @ 88.9 g/mol = 88.9g 3 Cl @ 35.5 g/mol = 106.5 g YCl 3 195.4 g 590 / 195.4 = 3 3 (YCl 3 ) Y 3 Cl 9

Molar Mass vs. Atomic Mass 6.02x10 23 H 2 = _______ H 2 = _____ H 2 O = ________ H 2 O = _____ MgSO 4 = ________ MgSO 4 = _____ (NH 4 ) 3 PO 4 = _____ (NH 4 ) 3 PO 4 = ________ Percentage Composition Empirical vs. Molecular Formula  %  g  g  mol  mol mol Empirical Formula 2 amu 18 amu 120 amu 149 amu 2 g 18 g 120 g 149 g % = x 100 % part whole (lowest ratio) (by mass)

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