Percent composition
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Percent Composition. Example: MgO Find molar mass of whole compound: MgO = 24.3 + 16.00 = 40.3 grams % Mg (by mass) = mass of Mg = 24.3 g x 100 = 60.3% molar mass of MgO 40.3 g

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Percent composition

Percent Composition

Example: MgO

Find molar mass of whole compound:

MgO = 24.3 + 16.00 = 40.3 grams

% Mg (by mass) = mass of Mg = 24.3 g x 100 = 60.3%

molar mass of MgO 40.3 g

% O (by mass) = mass of O = 16.0 g x 100 = 39.7%

molar mass of MgO 40.3 g

Percent = part / whole


Empirical formulas

Empirical Formulas

  • 4Steps:

    • Change the percent to grams

    • Convert grams to moles

    • Divide each of the moles by the smallest number to find the ratio

    • Round ratio to a whole number. If the ratio is not a whole number (ex. 1.5) multiply each element by 2 to get a whole number

Gives the lowest whole-number ratio of elements and compounds in a formula


Examples
Examples

A compound contains 94.1% Oxygen and 5.9% Hydrogen. What is its empirical formula?

  • 94.1% = 94.1g O

  • 5.9% = 5.9g H

5.9g H

1 mol H

1 mol O

  • 94.1g O

= 5.9 mol H

= 5.88 mol O

16 g O

1 g H

5.9

5.88

  • 5.88

  • 5.88

= 1.003

(can round to 1.0)

= 1.0

4) Ratio = 1:1 Formula = OH


Examples1
Examples

A compound contains 67.6% Mercury and 10.8% Sulfur and 21.6% Oxygen. What is its empirical formula?

  • 67.6% = 67.6g Hg

  • 10.8% = 10.8g S

  • 21.6% = 21.6g O

67.6g Hg = 0.336 mol Hg

10.8g S = 0.338 mol S

21.6g O = 1.35 mol O

  • 0.336

  • 0.336

0.338

0.336

1.35

0.336

= 1.00 Hg

= 1.00 S

= 4.02 O

4)Ratio = 1:1:4 Formula = HgSO4


Examples2
Examples

What is the empirical formula for a compound containing 70.0% Fe and 30.0% O?

  • 70.0% = 70.0g Fe

  • 30.0% = 30.0g O

70.0g Fe = 1.25 mol Fe

30.0g O = 1.875 mol O

1.875

1.25

  • 1.25

  • 1.25

= 1.5 O

= 1.00 Fe

4) Ratio = 1 : 1.5

5) Mult ratio by 2 = 2 : 3

Formula = Fe2O3


Molecular formulas
Molecular Formulas

  • Same as empirical formula, or a simple whole-number multiple of it

  • Steps:

  • Calculate the mass in grams of the empirical formula provided

  • Divide the molar mass by the mass of the empirical formula

  • Multiply this whole number ratio by the empirical formula


Examples3
Examples

Calculate the molecular formula of the compound whose molar mass is 60.0g and empirical formula is CH4N.

1) CH4N = 12 + 4(1) + 14 = 30.0 g

  • 60.0g

  • 30.0g

= 2.0

3) 2.0 (CH4N)

= C2H8N2


Examples4
Examples

What is the molecular formula of ethylene glycol (CH3O), used in antifreeze. It has a molar mass of 62 g/mol.

1) CH3O = 31.0 g

3) 2.0 (CH3O)

= C2H6O2

  • 62.0g

  • 31.0g

= 2.0

Find the molecular formula of C3H2Cl, which is mothballs. Its molar mass is 147 g/mol.

1) C3H2Cl = 73.0 g

3) 2.0 (C3H2Cl)

= C6H4Cl2

  • 147.0g

  • 73.0g

= 2.0


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