Chemical Calculations: The Mole Concept and Chemical Formulas. AW Atomic weight (mass of the atom of an element) was determined by relative weights.

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Introduction to Chemistry

Chemical Calculations: The Mole Concept and Chemical Formulas Atomic Weights (Definitions)

• AW Atomic weight (mass of the atom of an element) was determined by relative weights.

• AMU (Atomic Mass Unit) was derived by taking 1/12 of the carbon-12 atom.

A carbon-12 atom has a mass of 1.99272 x 10−23 g.

1/12 (1.99272 x 10−23 g) = 1.6606 x 10−24g 1 AMU = 1.6606 x 10−24g

Since it was known that hydrogen was 1/12 the mass of carbon, the mass of a hydrogen atom was designated as 1 amu

Carbon is 12 times greater in mass than the hydrogen, therefore, it was given the mass of 12 amu.

The masses of all the elements, in amu, were determined by their relative masses to hydrogen or carbon.

mass of an p+ = 1.0073 amu mass of an no = 1.0073 amu mass of an e = 0.00054858 amu

Formula Masses:

• Calculate the masses of each kind of atom in the formula unit.

• Add the masses of all atoms in the formula unit.

e.g. Calculate the formula mass (amu) of Calcium sulfate (CaSO4) 1 amu = 1.66 x 10–24 g

1 Ca = 1 x 40.08 amu = 40.08 amu 1 S = 1 x 32.07 amu = 32.07 amu 4 O = 4 x 16.00 amu = 64.00 amu 136.15 amu

Calculate the grams of one formula unit of calcium sulfate.

136.15 amu CaSO4 1.66 x 10–24 g CaSO4

= 2.26 x 10–22 g CaSO4

1.0 amu CaSO4

The Mole (Avogadro’s number)

A number of atoms, ions, or molecules that is large enough to see and handle.

A mole = number of things

„ Just like a dozen = 12 things

„ One mole = 6.022 x 1023 things Avogadro’s number = 6.022 x 1023

„ Symbol for Avogadro’s number is NA.

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How do we know when we have a mole?

„ count it out

„ weigh it out

Molar mass - mass in grams numerically equal to the atomic weight of the element in grams.

H has an atomic weight of 1.00794 g

„ 1.00794 g of H atoms = 6.022 x 1023 H atoms Mg has an atomic weight of 24.3050 g

„ 24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms Molecules, atoms, protons and electrons

1 molecule of H2O = 2 hydrogen atoms and 1 oxygen atom

12 molecules of H2O = 2 (12) hydrogen atoms and 1 (12) oxygen atoms doz molecules of H2O = 2 (doz) hydrogen atoms and 1 (12) oxygen atoms

6.02 x 1023 molecules of H2O = 2 (6.02 x 1023) hydrogen atoms and 1 (6.02 x 1023) oxygen atoms mol molecules of H2O = 2 (mol) hydrogen atoms and 1 (mol) oxygen atoms

A mole of hydrogen is equal to its amu in grams.

One atom of hydrogen is equal to 1.0 amu

6.02 x 1023 atoms of hydrogen are equal to 1.0 gram

One mole of hydrogen is equal to 1.0 g Mol of H2O = 3 mol atoms

3 mol of H2O = 3 x 3 mol atoms = 9 total mol atoms

One Mole of Contains

„ Cl2 or 70.90g 6.022 x 1023 Cl2 molecules

2(6.022 x 1023 ) Cl atoms

„ C3H8 or 44.11 g 6.022 x 1023 C3H8 molecules

3 (6.022 x 1023 ) C atoms

8 (6.022 x 1023 ) H atoms

• A mole of atoms of one element is equal to mass of one atom in amu in grams.

e.g. 1 carbon atom = 12.01 amu (1.994 x 10–23 g) 6.02 x 1023 carbon atoms = 12.01 grams ( 1 mole of carbon atoms = 12.01 grams)

Molar Mass: A mole of any substance has a mass equal to the summation of the masses of the elements in the compound.

A mole of calcium sulfate: (a salt)

6.02 x 1023 formula units of CaSO4 = 136.2 g

1 mole of formula units of CaSO4 = 136.2 g (molar mass)

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1. If 1 mole of CaSO4 has a mass of 136.2 g, what is the mass of 0.5 moles of CaSO4 ?

0.5 mole CaSO4 136.2 g CaSO4

= 68.1 g CaSO4

1.0 mole CaSO4

2. How many formula units are there in 0.5 moles of CaSO4 ?

1.0 mole CaSO4 = 6.02 x 1023 formula units of CaSO4

0.5 mole CaSO4 6.02 x 1023 formula units of CaSO4

= 3.01 x 1023 formula units of CaSO4

1.0 mole CaSO4

3. How many moles of CaSO4 are there in 88.55 g of CaSO4? 1.0 mole CaSO4 = 136.2 g CaSO4

88.55 g CaSO4 1.0 mole CaSO4

= 0.6501 moles CaSO4 136.2 g CaSO4

4. How many formula units of CaSO4 are there in 56.25 g of CaSO4? 1.0 mole CaSO4 = 136.2 g CaSO4

1.0 mole CaSO4 = 6.02 x 1023 formula units of CaSO4

56.25 g CaSO4 1.0 mole CaSO4 6.02 x 1023 formula units CaSO4

= 2.486 x 1023 formula units CaSO4

136.2 g CaSO4 1.0 mole CaSO4

5. How many oxygen atoms are in 2.10 grams of CaSO4?

1.0 mole CaSO4 = 136.2 g CaSO4

1.0 mole CaSO4 = 6.0 mole O atoms 1.0 mole O = 6.02 x 1023atoms of O

2.10 g CaSO4 1.0 mol CaSO4 6.0 mol O atoms 6.02 x 1023 atoms

= 5.57 x 1022 atoms of oxygen 136.2 g CaSO4 1.0 mol CaSO4 1.0 mol O

A mole of H2O has a molar mass of 18.0 g 2 H = 2 (1.0 g) = 2.0 g

1 O = 1(16.0) g = 16.0g 18.0 g

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If you have 5.00 x 1025 molecules of water, what is the mass of the water?

If you have 52.5 g of H2O, how many hydrogen atoms do you have?

Percent Composition and Formulas of Compounds

The Law of Definite Proportions (Constant Composition) tells us that a compound has the same kinds of atoms in the same ratio. Therefore, we can determine the percent composition of any substance:

% part x 100

whole

=

For example: What is the percent composition of all elements in CH3COOH?

1. Need molar mass of molecule.

2 C = 2 (12.0 g) = 24.0 g 4 H = 4 (1.0 g) = 4.0 g 2 O = 2 (16.0 g) = 32.0 g 60.0 g 2. Take the percentage of each element.

% 24.0 100

60.0 40.0 %

C g x

= g

=

% 4.0 100

60.0 6.67 %

H g x

= g

=

% 32.0 100

60.0 53.3 %

O g x

= g

= 40.0 % + 6.67 % + 53.3 % = 99.97 % = 100. %

What is the percent composition of oxygen in calcium sulfate?

part

% = whole x 100%

% Ca = mass of O x 100 mass of compound

= 64.00 amu x 100 136.15 amu = 47.01 %

Empirical and Molecular Formulas:

Molecular Empirical formula: Simplest formula

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C2H4

C4H8 CH2

C6H12

Experimental Determination of Empirical and Molecular Formulas:

Example 1

Suppose we know that a compound contains only the elements carbon, hydrogen, and oxygen, and we weigh out a 0.2015-g sample for analysis. We find that this 0.2015-g sample of compound contains 0.0806 g of carbon, 0.01353 g of hydrogen, and 0.1074 g of oxygen.

We begin by converting to moles.

We now know that 0.2015 g of the compound contains 0.00671 mol of C atoms, 0.01342 mol of H atoms, and 0.006713 mol of O atoms.

What is the ratio of the atoms in the empirical formula?

• Divide by the smallest number of moles 0.00671 = 1 mol C

0.00671

0.01342 = 2 mol H CH2O empirical formula 0.00671

0.00671 = 1 mol O 0.006713

Suppose we know that the molecular formula mass (molar mass) of this compound is 180.2 gram/mol.

Empirical mass = 30.02 g

180.2 g = 6 30.03 g 6 (CH2O) = C6H12O6

Example 2

When a 0.3546 g sample of vanadium metal is heated in air, it reacts with oxygen to achieve a final mass of 0.6330 g.

Calculate the empirical formula of this vanadium oxide.

0.6330 g vanadium + oxygen 0.3546 g V 1.0 mol V

– 0.3546 g vanadium = 0.006961 mol V atoms 0. 2784 g oxygen 50.94 g V

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0.2784 g O 1.0 mol O

= 0.01740 mol O atoms 16.00 g O

0.006961 mol = 1 mol V x 2 = 2 mol V 0.006961 mol

V2O5

0.01740 mol = 2.5 mol O x 2 = 5 mol O 0.006961 mol

Empirical From % composition:

The most common form of nylon (Nylon-6) is 63.68 % carbon, 12.38% nitrogen, 9.80% hydrogen, and 14.14% oxygen.

Calculate the empirical formula for Nylon-6.

In 100 grams, you would have 63.68 g C, 12.38 g N, 9.80 g H, and 14.14 g O 63.68 g C 1.0 mol C

= 5.302 mol C = 6.000 mol C 12.01 g C 0.8837 mol

12.38 g N 1.0 mol N

= 0.8837 mol N = 1.000 mol N 14.01 g N 0.8837 mol

9.80 g H 1.0 mol H

= 9.72 mol H = 11.0 mol H 1.008 g H 0.8837 mol

14.14 g O 1.0 mol O

= 0.8838 mol O = 1.000 mol O 16.00 g O 0.8837 mol

C6NH11O

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