Chemistry - Mole Notes



Chemistry - Mole Notes

I. The Mole

A. When substances react they combine or break apart on the atom or molecule level. To describe what will happen in a reaction, a unit that can compare the number of atoms of each substance present is needed.

B. It was determined in the lab that the atomic mass in grams of carbon - 12 (12 grams) contained 6.02 x 1023 atoms.

C. 6.02 x 1023 is now known as Avagadro’s Number. The gram equivalent mass of any atom or the molecular mass (the mass of all the atoms in a molecule) will always contain Avagadro’s Number of particles.

D. A mole of any substance contains Avagadro’s Number of particles of that substance.

E. Example - How many grams of iron are in one mole?

55.9 grams (always round to the nearest .1 gram).

Example - How many grams are in one mole of sodium chloride?

23.0 g. + 35.5 g. = 58.5 g.

* Read in Prentice Hall Chemistry pages 287 thru 296.

II. Mole Conversions

A. As long as you have a periodic table, the conversion between moles, grams, and number of particles is not difficult.

B. Grams to moles.

Example - How many moles are in 14 grams of sodium chloride?

Na - 23.0 g. Cl - 35.5 g. so NaCl - 58.5 g.

14 g. x 1 mole/58.5 g. = .24 moles

C. Moles to grams.

Example - How many grams are in .576 moles of zinc nitrate?

Zn - 65.4 g. N - 14.0 g. O - 16.0 g. so Zn(NO3)2 - 189.4 g.

.576 moles x 189.4 g./1 mole = 109 grams

D. Moles to particles.

Example - How many atoms are in .111 moles of calcium?

Avagadro’s Number 6.02 x 1023 / 1mole

.111 mole x 6.02 * 1023 atoms/ 1 mole = .668 * 1022 atoms

E. Grams to particles.

Example - How many molecules are in 4.4 g. of water?

H - 1.0 g. O - 16.0 g. so H2O = 18.0 g.

4.4 g. x 1mole/ 18.0 g.=.24 moles x 6.02 * 1023 molec./mole = 1.4 x 1023

F. Particles to moles.

Example - How many moles are in 5.22 x 1023 molecules of silver bromide?

5.22 * 1023 molec. x 1 mole/6.02 x 1023 molec. = .876 moles

* Read in Prentice Hall Chemistry pages 297 thru 299.

III. Percentage Composition

A. The percentage of an elements mass compared to the total mass of the compound.

B. Can be used for identification of unknown substances.

C. Calculating percentage composition.

1. Mass of element / mass of compound X 100

2. Example - What is the percentage of sulfur in lithium sulfate?

Li - 6.9 g. S - 32.1 g. O - 16.0 g. Li2SO4

Total mass of Li2SO4 - 109.9 g.

32.1 g. / 109.9 g. X 100 = 29.2 %

3. Example - What is the percentage composition of magnesium iodide?

Mg - 24.3 g. I - 126.9 g. MgI2 - 278.1 g.

Mg = 24.3 g. / 278.1g. X 100 = 8.7 %

I = 253.8 g. / 278.1 g. X 100 = 91.3 %

IV. Empirical Formula

A. The simplest whole number ratio between elements in a compound.

B. For inorganic chemistry (what we are currently studying) the empirical formula is the same as the molecular formula.

C. To calculate the empirical formula find the number of moles of each element in a compound then determine the ratio between the molar amounts.

D. Example - What is the empirical formula for a compound that contains 7.30 g. of sodium, 5.08 g. of sulfur and 7.62 g. of oxygen.

Na = 7.30 g. / 23.0 g. = .317 g. moles 2

S = 5.08 g. / 32.1 g. = .158 moles 1

O = 7.62 g. / 16.0 g. = .476 moles 3

Na - 2 S - 1 O - 3 Na2SO3

E. Example - What is the empirical formula for a compound that contains 40.0% carbon, 6.67 % hydrogen, and 53.3 % oxygen.

Assume a 100 g. sample so you can directly change the % to grams.

C = 40.0 g. / 12.0 g. = 3.3 moles 1

H = 6.67 g. / 1.0 g. = 6.6 moles 2

O = 53.3 g. / 16.0 g. = 3.33 moles 1

CH2O

* Read in Prentice Hall Chemistry pages 305 thru 312.

V. Hydrates

A. A substance that has water trapped in its crystal.

B. Usually formed when a compound is formed in a solution and the water is evaporated away to isolate the substance.

C. Written as MgCl2 * H2O

D. To separate the water the compounds can be strongly heated.

E. Efflorescent hydrates give off water when the vapor pressure is reduced.

F. Hygroscopic hydrates have a low vapor pressure and remove water from the air increasing their water content. They can be used to keep other substances dry.

G. To calculate hydrated formulas you need to determine the moles of compound compared to the moles of water. The mass lost due to heating is the mass of the water. The mass left after heating is the mass of the compound.

H. Example - What is the formula for a hydrated copper II carbonate compound that masses 1.78 g. before heating and 1.24 g. after heating?

1.78 g. - 1.24 g. = .54 g is the mass of the water.

CuCO3 = 1.24g. / 123.5 g. = .010 mole 1

H2O = .54 g. / 18.0 g. = .030 mole 3

CuCO3 * 3 H2O

* Read in Prentice Hall Chemistry pages 454 thru 456.

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