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.

Steps to consider

• Calculate the # of moles (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.

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: calculate the # of moles (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.

Simplest formula: sample problemExample, 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 = 1.001C

1.520/ 0.50666 = 3.000 O

3: the simplest formula is BaCO3

1. A compound consists of 29.1 % Na, 40.5 % S, and 30.4 % O. Determine the simplest formula.

2. 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.

Question 11: 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

Question 21: 7.20 g C, 1.20 g H, 9.60 g O2: C: 7.20 g 12.01 g/mol = 0.5995 mol C

H: 1.20 g 1.01 g/mol = 1.188 mol HO: 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

Molecular formulaIs 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.

Molecular formula calculations• E.g. Determine the molecular 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.

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 104 g/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.

Question 11: 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

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/molFactor = 104 g/mol 13 g/mol = 8Molecular formula is C8H8

Question 41: 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 C2H5O5: factor = 90/45=2. Molecular formula: C4H10O2

1. Calculate the percentage composition of each substance: a) SiH4, b) FeSO4

2. 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%

3. The simplest formula for glucose is CH2O and

its molar mass is 180 g/mol. What is its molecular formula?

Assignment

4. Determine the molecular formula for each compound below from the information listed. substance simplest formula molar mass(g/mol)

a) octane C4H9 114 b) ethanol C2H6O 46 c) naphthalene C5H4 128 d) melamine CH2N2 126

5. The percentage composition and approximate molar masses of some compounds are listed below. Calculate the molecular formula of each percentage composition molar mass(g/mol)64.9% C, 13.5% H, 21.6% O 7439.9% C, 6.7% H, 53.4 % O 6040.3% B, 52.2% N, 7.5% H 80

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%

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

the simplest formula is CS2

2.626/1.315

= 2.00

1.315/1.315

= 1

Mol reduced

2.6261.315Mol

SC

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.6499/.6495

= 1.0

0.6499Ag

1.30/.6495= 2.00

.6495/.6495= 1

Mol reduced

1.300.6495MolONAgNO2

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

.6803/.6803= 1

0.6803

K

2.375/.6495= 3.49

.6808/.6803= 1.00

Mol reduced

2.3750.6808Mol

OCrK2Cr2O7

3 C6H12O6 (CH2O = 30 g/mol, 180/30 = 6)

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)

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 HO: 21.6 g 16.00 g/mol = 1.35 mol O

5.404/1.35= 4.00

5.404C

1.35/1.35= 1

13.37/1.35= 9.90

Mol reduced

1.3513.37MolOHC4H10O

C4H10O (C4H10O = 74 g/mol, 74/74 = 1)

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 HO: 53.4 g 16.00 g/mol = 3.338 mol

O

3.322/3.322= 1

3.322C

3.338/3.322= 1.00

6.63/3.322= 2.0

Mol reduced

3.3386.63MolOHCH2O

C2H4O2 (CH2O = 30 g/mol, 60/30 = 2)5 c)

3.728/3.726= 1.00

3.728B

7.43/3.726= 2.0

3.726/3.726= 1

Mol reduced

7.433.726MolHN

B3N3H6 (BNH2 = 26.84 g/mol, 80/26.84= 2.98)For more lessons, visit www.chalkbored.com

TRY QUESTIONS 1 – 2

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