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EMPIRICAL FORMULA

EMPIRICAL FORMULA. Is the simplest whole-number ratio of atoms of each element present in a compound For example, the empirical formula of C 3 H 6 is CH 2 . If the mass of each element in a sample is known, then the empirical formula can be calculated. Steps to consider.

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EMPIRICAL FORMULA

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  1. EMPIRICAL FORMULA Is the simplest whole-number ratio of atoms of each element present in a compound For example, the empirical formula of C3H6 is CH2. If the mass of each element in a sample is known, then the empirical formula can be calculated.

  2. Steps to consider • Calculatethe#ofmoles(mole =mass ¸ molar mass) • Find the mole ratio of atoms by dividing through by the smallest number • Write the simplest formula from the mole ratio • Multiply through by a whole number if the empirical formula is not obtained at step 3.

  3. Simplest formula calculations Q- a compound is found to contain the following % by mass: 69.58% Ba, 6.090% C, 24.32% O. What is the simplest (i.e. empirical) formula? Step 1:calculatethe#ofmoles(mol=g¸g/mol) Step 2: express moles as the simplest ratio by dividing through by the lowest number. Step 3: write the simplest formula from mol ratios.

  4. Simplest formula: sample problem Example, A compound contains 69.58% Ba, 6.090% C, and 24.32% O. What is the empirical (a.k.a. simplest) formula? 1: Ba: 69.58 g ¸ 137.33 g/mol = 0.50666 mol Ba C: 6.090 g ¸ 12.01 g/mol = 0.50708 mol C O: 24.32 g ¸ 16.00 g/mol = 1.520 mol O 2: Divide through by the lowest number 0.50666/ 0.50666 = 1 Ba 0.50708/ 0.50666 = 1.001 C 1.520/ 0.50666 = 3.000 O 3: the simplest formula is BaCO3

  5. A compound consists of 29.1%Na, 40.5% S, and30.4%O. Determinethesimplestformula. • A compound is composed of 7.20 g carbon, 1.20 g hydrogen, and 9.60 g oxygen. Find the empirical formula for this compound • 3. Try question 1 on page 13.

  6. Question 1 1: Assume 100 g: 29.1 g Na, 40.5 g S, 30.4 g O 2: Na: 29.1 g ¸ 22.99 g/mol = 1.266 mol S: 40.5 g ¸ 32.06 g/mol = 1.263 mol O: 30.4 g ¸ 16.00 g/mol = 1.90 mol 3: Na: 1.266¸1.263 = 1.00 S : 1.263 ¸1.263 = 1 O : 1.90 ¸1.263 = 1.50 4: the simplest formula is Na2S2O3

  7. Question 2 1: 7.20 g C, 1.20 g H, 9.60 g O 2: C: 7.20 g ¸ 12.01 g/mol = 0.5995 mol C H: 1.20 g ¸ 1.01 g/mol = 1.188 mol H O: 9.6 g ¸ 16.00 g/mol = 0.60 mol O 3: C: 0.5995/ 0.5995 = 1 H: 1.188/ 0.5995 = 1.98 O: 0.60/ 0.5995 = 1.0 4: the simplest formula is CH2O

  8. Molecular formula Is the formula that gives the actual number of atoms of each element in a molecule. 1. First, determine empirical formula mass of the simplest formula. 2. Divide the molecular mass of the compound by the empirical mass to get a factor 3. Multiply each subscript in the simplest formula by this factor to obtain the molecular formula.

  9. Molecular formula calculations • E.g. Determinethemolecular formula of a compound with empirical formula CH2O if its molar mass is 150 g/mol. Solution • First, determine molar mass of the simplest formula. For CH2O it is 30 g/mol (12+2+16). • Divide the molar mass of the compound by this to get a factor: 150 g/mol ¸ 30 g/mol = 5 • Multiply each subscript in the formula by this factor: C5H10O5 is the molecular formula.

  10. 1. Combustion analysis gives the following: 26.7% C, 2.2% hydrogen, 71.1% oxygen. If the molecular mass of the compound is 90 g/mol, determine its molecular formula. 2. What information must be known to determine a) the empirical formula of a substance? b) the molecular formula of a substance? 3. A compound’s empirical formula is CH, and it weighs 104g/mol. Give the molecular formula. 4. A substance is decomposed and found to consist of 53.2% C, 11.2% H, and 35.6% O by mass. Calculate the molecular formula of the unknown if its molar mass is 90 g/mol.

  11. Question 1 1: Assume 100 g total. Thus: 26.7 g C, 2.2 g H, and 71.1 g O 2: C: 26.7 g ¸ 12.01 g/mol = 2.223 mol H: 2.2 g ¸ 1.01 g/mol = 2.18 mol O: 71.1 g ¸ 16.00 g/mol = 4.444 mol 3: C: 2.223 ¸2.18 = 1.02 H: 2.18 ¸ 2.18 = 1 O: 4.444 ¸2.18 = 2.04 4: the simplest formula is CHO2 5: factor = 90/45=2. Molecular formula: C2H2O4

  12. Question 2, 3 • For the empirical formula we need to know the moles of each element in the compound (which can be derived from grams or %). For the molecular formula we need the above information & the molar mass of the compound • Molar mass of CH = 13 g/mol Factor = 104 g/mol ¸ 13 g/mol = 8 Molecular formula is C8H8

  13. Question 4 1: Assume 100 g total. Thus: 53.2 g C, 11.2 g H, and 35.6 g O 2: C: 53.2 g ¸ 12.01 g/mol = 4.430 mol H: 11.2 g ¸ 1.01 g/mol = 11.09 mol O: 35.6 g ¸ 16.00 g/mol = 2.225 mol 3: C: 4.43/2.225 = 1.99 H: 11.09/2.225 = 4.98 O: 2.225/2.225 = 1 4: the simplest formula is C2H5O 5: factor = 90/45=2. Molecular formula: C4H10O2

  14. Assignment • Calculate the percentage composition of each substance: a) SiH4, b) FeSO4 • Calculate the simplest formulas for the compounds whose compositions are listed: a) carbon, 15.8%; sulfur, 84.2% b) silver,70.1%; nitrogen,9.1%; oxygen,20.8% c) K, 26.6%; Cr, 35.4%, O, 38.0% • The simplest formula for glucose is CH2O and its molar mass is 180 g/mol. What is its molecular formula?

  15. Determine the molecular formula for each compound below from the information listed. substancesimplest formulamolar mass(g/mol) a) octane C4H9 114 b) ethanol C2H6O 46 c) naphthalene C5H4 128 d) melamine CH2N2 126 • The percentage composition and approximate molar masses of some compounds are listed below. Calculate the molecular formula of each percentage compositionmolar mass(g/mol) 64.9% C, 13.5% H, 21.6% O 74 39.9% C, 6.7% H, 53.4 % O 60 40.3% B, 52.2% N, 7.5% H 80

  16. C S Mol 1.315 2.626 Mol reduced 1.315/1.315 = 1 2.626/1.315 = 2.00 2 a) Assume 100 g. Thus: 15.8 g C, 84.2 g S. C: 15.8 g ¸ 12.01 g/mol = 1.315 mol C S: 84.2 g ¸ 32.06 g/mol = 2.626 mol S 1 a) Si= 87.43% (28.09/32.13 x 100), H= 12.57% b) Fe= 36.77% (55.85/151.91 x 100), S= 21.10% (32.06/151.91 x 100), O= 42.13% the simplest formula is CS2

  17. 2 b) Ag: 70.1 g ¸ 107.87 g/mol = 0.6499 mol Ag N: 9.1 g ¸ 14.01 g/mol = 0.6495 mol N O: 20.8 g ¸ 16.00 g/mol = 1.30 mol O AgNO2 Ag N O Mol 0.6499 0.6495 1.30 Mol reduced .6499/.6495 = 1.0 .6495/.6495 = 1 1.30/.6495 = 2.00 K2Cr2O7 K Cr O Mol 0.6803 0.6808 2.375 Mol reduced .6803/.6803 = 1 .6808/.6803= 1.00 2.375/.6495 = 3.49 2 c) K: 26.6 g ¸ 39.10 g/mol = 0.6803 mol K Cr: 35.4 g ¸ 52.00 g/mol = 0.6808 mol Cr O: 38.0 g ¸ 16.00 g/mol = 2.375 mol O

  18. C4H10O C H O Mol 5.404 13.37 1.35 Mol reduced 5.404/1.35 = 4.00 13.37/1.35 = 9.90 1.35/1.35 = 1 4 a) C8H18 (C4H9 = 57 g/mol, 114/57 = 2) b) C2H6O (C2H6O = 46 g/mol, 46/46 = 1) c) C10H8 (C5H4 = 64 g/mol, 128/64 = 2) d) C3H6N6 (CH2N2 = 54 g/mol, 126/42 = 3) 3 C6H12O6(CH2O = 30 g/mol, 180/30 = 6) 5 a) C: 64.9 g ¸ 12.01 g/mol = 5.404 mol C H: 13.5 g ¸ 1.01 g/mol = 13.37 mol H O: 21.6 g ¸ 16.00 g/mol = 1.35 mol O C4H10O (C4H10O = 74 g/mol, 74/74 = 1)

  19. 5 b) C: 39.9 g ¸ 12.01 g/mol = 3.322 mol C H: 6.7 g ¸ 1.01 g/mol = 6.63 mol H O: 53.4 g ¸ 16.00 g/mol = 3.338 mol O CH2O C H O Mol 3.322 6.63 3.338 Mol reduced 3.322/3.322 = 1 6.63/3.322 = 2.0 3.338/3.322 = 1.00 B N H Mol 3.728 3.726 7.43 Mol reduced 3.728/3.726 = 1.00 3.726/3.726 = 1 7.43/3.726 = 2.0 C2H4O2 (CH2O = 30 g/mol, 60/30 = 2) 5 c) B3N3H6 (BNH2 = 26.84 g/mol, 80/26.84= 2.98) For more lessons, visit www.chalkbored.com

  20. TRY QUESTIONS 1 – 2 PAGE 13

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